THE
ELEMENTS OF GEOMETRY. BY
GEORGE BRUCE HALSTED, AND EX-FELLOW OF PRINCETON COLLEGE; PH.D. AND EX-FELLOW OF JOHNS HOPKINS university; instructor in post-graduate mathematics, PRINCETON college; PROFESSOR OF PURE AND APPLIED mathematics, UNIVERSITY OF TEXAS.
A.B., A.M.,
NEW YORK
JOHN WILEY 15
:
& SONS,
ASTOR PLACE. 1885.
.
Copyright,
1885,
By JOHN WILEY & SONS.
CAJORI
ELECTROTVPED AND PRINTED
BY RAND, AVERY, AND COMPANY,
Betricatetr
(by permission) TO
J. A.M., CAM.; F.R.S., L.
J.
AND
E.;
ACADEMY OF SCIENCES
SYLVESTER, CORRESPONDING MEMBER INSTITUTE OF FRANCE*. MEMBER
IN BERLIN, GOTTINGEN, NAPLES, MILAN, ST. PETERSBURG,
ETC.; LL.D., UNIV. OF DUBLIN,
FELLOW OF
AND
ST. JOHN'S COL.,
OF GEOMETRY
3En
IN
U.
OF
E.; D.C.L.,
OXFORD; HON.
CAM.; SAVILIAN PROFESSOR
THE UNIVERSITY OF OXFORD;
(Grateful
B^mcmbrance
OF BENEFITS CONFERRED THROUGHOUT TWO
FORMATIVE YEARS.
918182
PREFACE.
In America the geometries most in vogue at present are vitiated by the immediate assumption and misuse of that " and teachers who know something term " direction the Non-Euclidian, or even the modern synthetic geom-
subtile of
;
etries, are
" " directional seeing the evils of this superficial
method.
Moreover the attempt, in these books, to take away by from the familiar word "distance" its abstract
definition
character and connection with length-units, only confuses the
ordinary student.
A
reference to the article Measurement
" in the " Encyclopaedia Britannica will " "
word
show that around the
centers the most abstruse advance in pure
distance
science and philosophy.
An
elementary geometry has no need of the words direction and distance.
The use
in
teaching,
Geometry they
composed with
present work,
in a
may be
yet
way
strives
to
special
present
the
reference
to
Elements of
so absolutely logical and compact, that
ready as
rock-foundation for
more advanced
study.
to
Besides the acquirement of facts, there properly belongs Geometry an educational value beyond any other element-
PREFACE.
VI
In
ary subject.
mind
the
it
first
finds
logic
a practical
instrument of real power.
The method published ment
my Mensuration my words steregon
in
solid angles, with
of
for the treat-
and steradian,
having been adopted by such eminent authorities, may I venture to recommend the use of the word sect suggested in the
same volume
From 1877 Book IX.
I
t
regularly gave
In 1883
my
pupil,
my
classes
H. B. Fine,
at
the method of
my
suggestion,
wrote out a Syllabus of Spherical Geometry on the lines of my teaching, which I have followed in Book IX.
The
figures,
which
I
think give this geometry a special
advantage, owe all their beauty to my colleague. Professor A. V. Lane, who has given them the benefit of his artistic skill
and mastery
of
The whole work friend, Dr. F.
Any
graphics. is
greatly indebted
A. C. Perrine.
We
to
my
pupil
and
have striven after accuracy.
corrections or suggestions relating to the book will be
thankfully received.
GEORGE BRUCE HALSTED. 2004 Matilde Street, Austin, Texas.
CONTENTS. THE ELEMENTS OF GEOMETRY.
BOOK CHAPTER ON
IV.
1.
A
2.
Immediate inference defined
statement defined
i
...
4.
Equivalent statements defined A statement implied
5
Statements reduced to two terms
and a copula 6.
Logically simple sentence
.
.
....
Contranominal defined
4
i
19.
Converse defined
4
i
20.
Inverse defined
4
21.
Obverse defined
2
22.
Two
2
23.
Classes proved identical
the typical form
2
9.
Subject and predicate
2
10.
2
11.
Class terms defined
2
12.
How
13.
14.
defined
....
Contradictory of a class Contradictories include the universe, III.
V.
24.
— Classes. Individual terms defined .... is
theorems prove four
.... ....
5 5
5
2
Definitions.
a class
PAGE
18.
8.
II.
Obverse.
ARTICLE
i
Logically composite sentence , . Conditional statements reduce to
7.
Contranominal, Converse, Inverse,
PAGE
ARTICLE
3.
I.
LCXJIC.
— Statements. Definitions.
I.
I.
25-
Theorems.
....
A
theorem defined Demonstration defined
.
27. 28.
A
29.
3 303
,
.....
Corollary defined Hypothesis and conclusion
26.
3
On
.
geometric theorem defined
Use of type in this work . Inverses of theorem with composite ,
6
hypothesis
The Universe of Discourse.
15.
Universe of discourse defined
16.
The
17.
tradictory Different divisions into contradic-
classes
tories
x and non-x
still
.
.
VI.
3
On Proving Inverses.
con-
3
3132.
3
33-
Inverses often proved geometrically,
6
Rule of identity Rule of inversion
6 7 vii
CONTENTS.
VIU
CHAPTER
II.
THE PRIMARY CONCEPTS OF GEOMETRY. Definitions of Geometric Magnitudes. PAGE ARTICLE 8 Geometry defined 34. I.
35.
A solid defined A surface defined A line defined A point defined A magnitude defined
66.
Right angle defined
67.
A
8
68.
8
69.
8
perpendicular defined Complement defined
13
....
13 13 13
70.
Acute angle defined Obtuse angle defined
8
71.
Perigon defined
13
8
72.
Sum
....
8
73.
perigon Reflex angle defined
44.
Why space is called tri-dimensional,
9
74.
Oblique angles and lines defined
45.
Straight line defined curve defined
9
75.
46.
9
76.
47.
Line
36.
39-
Point,
40-43.
line,
surface, solid,
fined, using
motion
de-
A
henceforth
means
48. 49. 50.
A A
sect defined
plane defined Plane defined, using motion
52.
A figure defined A plane figure defined
53.
Circle
51.
54. 55. 56.
and center defined Radius defined Diameter defined Arc defined
....
...
defined
sum
15
9
80.
Sides defined
15
9
81.
Interior angle defined
15
9
82.
15
9 10
83.
Exterior angle defined Congruent defined
84.
Equivalent defined
15
63.
When
14
triangle defined
15
15
II.
Properties of Distinct Things.
85-93.
"^^^ so-called axioms of arith-
metic III.
16
Some Geometrical Assumptions about Euclid'' s Space.
II
94-99.
The
12
IV.
two adjacent angles equal a 12
straight angle
Supplement defined Supplemental adjacent angles
I5>
10
de-
fined
Straight angle defined
65.
Vertices defined
trace defined
11
62.
64.
79.
11
their
14
14
A A
10
58.
61.
Vertical angles defined Bisector defined
10
Parallel lines defined
Explement Equal angles defined Adjacent angles and
14
14
.
78.
Tj.
10
Angle, vertex, arms, defined
59.
14
10
57.
60.
...
of angles about a point is a
9 9
straight
line
13
100-103.
16
so-called geometric axioms,
The Assumed Constructions.
The
so-called postulates
.
.
17
.
12 .
.
12
Table of symbols used
CHAPTER
III.
PRIMARY RELATIONS OF LINES, ANGLES, AND TRL^I^GLES, II. Angles about Two Points. I. Angles about a Point. PAGE
ARTICLE 104-105.
Right and straight angles
106-107. 1 08-1 10.
Vertical angles
Perigons
.
18, 19
19
20,21
ARTICLE
....
PAGE 22
III.
Transversal defined
112-113.
Corresponding and alternate angles
22
III.
CONTENTS.
IX
ARTICLE
PAGE
Triangles.
PAGE
ARTICLE 114. [15.
16. 17.
Equilateral triangle defined Isosceles triangle defined
Scalene triangle defined Base and vertex defined
...
.... .... ....
24 24
120.
24
121.
Obtuse-angled triangle defined Acute-angled triangle defined
24
122.
Equiangular triangle defined
123.
Homologous defined
Angles of triangle mean interior 24
angles
Right-angled triangle and hypothenuse defined
119.
CHAPTER
.
.
.
Congruent triangles
124-129,
IV.
PROBLEMS. PAGE
ARTICLE
130.
Problem defined
30
132-40.
131.
Solution defined
30
ARTICLE
CHAPTER
PAGE
Deduced constructions
30-35
V.
JNEQUALITIES. PAGE
ARTICLE
Sign of inequality explained 36 Theorems on one triangle . 36-39 .
141.
142-150.
151.
Oblique sect defined
152-164.
ObUques and
CHAPTER
....
39
.
40-47
Application to the triangle .
51,52
triangles
.
VI.
PARALLELS. PAGE
ARTICLE 165-172.
Fundamental theorems
48-51
ARTICLE I
173-175.
CHAPTER
PAGE
VII.
TRIANGLES. PAGE 176-179. 180.
181.
. . Congruent triangles 53-55 Conditions of congruence of .
triangles
55
Locus defined
56
ARTICLE
PAGE
182.
How
183-186.
Loci
187-191.
Intersection of loci
CHAPTER
...
to prove a locus
56
57,58 .
.
.
58-60
VIII.
POLYGONS. I.
192.
A
193.
Its vertices defined
polygon defined
PAGE
ARTICLE
Definitions.
ARTICLE
....
PAGE
194.
Its sides defined
61
195.
Perimeter defined
61
196.
Interior angles of a
.
.
.
polygon
.
61
.
6i
.
62
CONTENTS. PAGE
ARTICLE 197.
Convex defined
198.
Diagonals defined
199.
200.
An An
201.
A
equilateral polygon defined
equiangular polygon defined . . regular polygon defined
Mutually equilateral defined Mutually equiangular defined 203. 204, 205. Polygons may be either .
.
206.
.
.
227-231.
62
232.
. Intercepts Medial defined
.
74
233.
Concurrent defined
•
75
.
234.
Collinear defined
.
•
75
.
235.
Medials concurrent
•
75
.
236.
Centroid, orthocenter, circumcenter, defined
.
,
.
.
.
Trigon, tetragon or quadrilateral,
pentagon,
hexagon,
....
208.
209.
decagon, quindecagon The surface of a polygon defined
.
Parallelogram defined Trapezoid defined
II.
.
.
.
.
73,74
75
IV. Equivaleiice.
heptagon,
octagon, nonagon, decagon, do-
207.
PAGE
62
how proved
237.
Equivalence,
64
238-240.
Base and altitude
64
241-244.
Squares
64
245-255.
Rectangles
64
256,257.
Complements grams
.
76
.
76
77-79 80-86 of
parallelo87
General Properties. V. Problems,
Congruence, angles
210-215.
64-66
.
Parallelograms
258-264. III.
216-220.
Sides and angles
221.
A rectangle defined A rhombus defined A square defined
222. 223.
67-69
.
.
Properties
224, 225. 226.
.
.
to
equivalent
polygons
Parallelograms.
.
Triangle altitudes concurrent
VI. Axial
88-90
and Central Symmetry.
•
70
•
70 70
265. 266.
Corresponding points defined . Axial symmetry defined .
91
.
70,71
267.
Central symmetry defined
92
.
72
91
.
Symmetry
268. 269.
BOOK
in
one figure
.
.
.
.
92,
93
II
RECTANGLES. PAGE
ARTICLE 270. 271.
A A
continuous aggregate defined discrete aggregate defined .
95
.
95
Counting and number . 95, 96 The commutative law for addi-
272-279. 280.
.
.
.
tion
96
282.
The sum of two sects defined The commutative law holds
283.
The
281.
.
.
sura
fined
two rectangles deq8
holds
for
98
rectangles associative law
287.
288.
The commutative law
289.
The commutative law
99
associative law holds for sects,
99
associative law holds for rect-
99
angles for multipli-
100
cation
97 of
The commutative law The The The
285. 286.
97
for
sects
284.
of sects
for rectangle
100
CONTENTS. PAGE
ARTICLE 290.
291-293.
The The
distributive law
.
.
.100
and rectangles
The square on
296-302.
Rectangles and squares
the
100-102
.
sum
294, 295.
of
sects .
303.
The
304.
The
Primary
305-311.
103-106
312-314.
107-111
.
1 11-
Sum
Center, radius, surface of circle, 115 116 Point and circle
369-374-
Unequal
375-378.
Inscribed angles
chord defined
A
segment defined
326.
Explemental arcs defined
327.
Semicircle defined
328.
Major arc and minor arc de-
secant
of
two minor arcs
129, 130
.
....
arcs
.
.
130-132 132, 133
.
116,117 117 .
.
.
.
.
.
.
.117 .117 .117
fined
III.
379-386.
Angles in Segments.
Angles made by chords and secants
134-138
117
333>334-
Major segment and minor segment defined 117 Circle and center 118 Diameter 118
335,336.
Congruent
337-340.
Concentric circles
341-343.
Symmetry
344-348.
Chords
349-353-
Sects
354-356.
Circumscribed
357.
scribed polygon Concyclic defined
358, 359.
Perpendiculars on chords
circles
119 120
in the circle
.
IV. Tangents. 387, 388.
Tangent and point of contact
389-402.
Properties of tangents
defined
V.
120, 121
122-124
and
.
.
.
Circles.
Common
405.
Tangent
circles defined
126
406.
Common
axis of
126
407-411.
Relations between the center-
in-
.
139-143
403, 404.
125,126 circle
Two
138 .
chord
sect, radii,
127
and
144
symmetry
.
144
.
144
relative posi-
tion II.
113
PAGE
366-368.
324.
330-332.
,
ARTICLE
325.
329.
.
III
PAGE
A A
322,323.
Squares on sides Problems
CIRCLE.
Properties.
ARTICLE
320,321.
projection of a sect defined 107
102, 103
THE 315-319.
projection of a point defined 106
two
BOOK I.
PAGE
ARTICLE
distributive law holds for
sects
XI
145-147
Aftgles at the Center.
VI. Problems. 360.
Angles subtended by arcs
361, 362.
Sector
363.
Angle subtended by a
364.
Inscribed angle
365.
Corresponding arcs, sectors, chords
sect
.
.
.128 128
412.
To
.128
413.
Circumscribed polygon and
414.
Escribed
415-417.
Problems
128 angles,
129
bisect a given arc
.
.
.
scribed circle circle defined
.148 in-
148 .
.
.148
149, 150
CONTENTS.
xu
BOOK
IV
REGULAR POLYGONS. ARTICLE
Partition of a Perigon.
I.
PAGE
ARTICLE 418,419.
Successive bisection
420-422.
Trisection
423-426.
Five and
II.
....
To
436.
Radius
fifteen
.
.
inscribe
and
.
equals radius
151
153-154
III.
159
—
Least Perimeter in Equivalent Figures, Greatest Surface in Isoperimetric Fig-
ures. Circles.
circum-
scribe
438.
Isoperimetric defined .
439-443.
Equivalent and isoperimetric 160-162 triangles
444-446.
The
447-450.
Regular polygons
154-159
and
de-
apothem
fined
159
BOOK PAGE
ARTICLE
169,170
451-453.
Multiples
454.
455.
Commensurable defined Incommensurable defined
456.
Commensurability exceptional,
457.
Greatest
.
common
.
.
,
.
submulti-
pie
461, 462.
Incommensurable magnitudes, The side and diagonal of a square are incommensur-
463.
Scales of multiples defined
able
172,173
.
circle
V.
RATIO AND PROPORTION.
458-460.
hexagon
inscribed
152
Regular Polygons and
427-435.
PAGE Side of
437.
.
.159
162-164 .
.
.
165-167
CONTENTS. II.
PAGE
523, 524.
.188
525-528.
Similar figures defined Similar triangles . .
514.
Homologous,
515,516.
.192 Similar polygons similarly
517,518.
placed Center of similitude
....
519-522.
The
.
of
ratio
.
.
.
188-192
Rectangles
and Polygons,
Rectangle of extremes equivalent to rectangle of
193
540-547.
Composition of ratios
.194
548-552.
Theorems and problems,
means
192, 193
.
bisector of an angle, 195, 196 Harmonic division 196
529-539.
.
.
The
III.
simili-
tude, similarly placed
right triangle
PAGE
ARTICLE
Similar Figures.
508-513.
507.
XIU
197-204 .
GEOMETRY OF THREE DIMENSIONS.
BOOK
VII.
OF PLANES AND PAGE
ARTICLK 553.
The plane
554-556.
Assumptions about the plane
557-562.
defined
213
213 Determination of a plane, 214-216
563-565.
Intersection of planes
566.
Principle of duality
567-578.
Perpendicular to plane
.
.
216,217
.
218-224
.
.
.217
ARTICLE
LINES.
204-208 208-212
CONTENTS.
XIV
PAGE
ARTICLE 650-652.
Spherical angles and perpen-
....
dicular arcs
ARTICLE
PAGE
653,654.
Smallest line
252, 253
between two
points
BOOK
254,255
IX.
TWO-DIMENSIONAL SPHERICS. PAGE
ARTICLE 655.
Introduction
656.
Two-dimensional definition of
257
Fundamental Definitions
667-675.
Assumptions with
properties,
The
.
270, 271
708-719.
Spherical triangles
720, 721.
Loci
722, 723.
Symmetrical triangles equiva-
725.
Spherical excess defined
724-728.
Triangles lunes
equivalent
729-731.
Polygons
equivalent
679-684.
Definitions
....
687.
263, 264 Perimeter of polygon 264, 265 Symmetry without congru-
688-691.
Congruent
692-697.
Problems
.
ence
265
.,,,..
.
259
677, 678.
angles
pole
258,259
corolla-
spherical
its
Bi-rectangular triangle
259-261 All spherical lines intersect . 261 A lune and its angle 262
685, 686.
and
Polar triangles
ries
676.
line
705-707.
258
657-662.
PAGE
698-703. 704.
the sphere
663-666.
ARTICLE
.
.... .
.
.
272
274-280 280, 281 281, 282
lent
.
.
282
to 282, 283
lunes
to 283, 284
Equivalent triangles on same base 284
732.
tri-
266-268
IZZiTA'
268-270
735-739.
BOOK
.
272, 273
Tangency Theorems on
285 circles
.
.
286, 287
X.
POLYHEDRONS. ARTICLE 740-744.
PAGE Polyhedron, faces, edges, summits, section
745-747-
....
Pyramid, apex, base, faces
and edges
289, 290
lateral
....
748.
Parallel polygons
749-757.
Prism, altitude, right prism, oblique
758-760. 761.
290
762-764.
Descartes' theorem
765.
Euler's theorem
Parallelopiped, quader, cube 291 All summits joined by one line, 292 .
....
293
294
766-768.
Ratio between quaders
769.
Parallelopiped
290 290, 291
PAGE
ARTICLE
.
295-297 to
equivalent
298
quader 770, 771.
Triangular prism
299
1T^^ 773-
Pyramid
sections
300
774. 775.
. 301 Equivalent tetrahedra Triangular pyramid and prism, 302 .
.
CONTENTS.
XV
BOOK XL MENSURATION, OR METRICAL GEOMETRY. CHAPTER THE METRIC SYSTEM. PAGE
ARTICLE 776-778. 779-782.
783,784. 785-787.
Measurement 303 Meter and metric system, 303, 304 Length 304>305 Area 305
I.
— LENGTH,
AREA. PAGB
ARTICLE
789, 790.
Area Area
791, 792.
Metric units of surface
793.
Given hypothenuse and side
'j^i.
CHAPTER
of rectangle of square
305
307 .
.
.
308
.
308
II.
RATIO OF ANY CIRCLE TO ITS DIAMETER. ARTICLE
To compute
....
PAGB
795.
Length
of circle
312
799-802.
Principle of limits Value of TT
796.
Variable defined
313
803,804.
Circular measure of an
797.
Limit defined
313
794.
perimeters
310
798.
.
.
.313
315,3x6
angle
CHAPTER
.
3^6,317
III.
MEASUREMENT OF SURFACES. ARTICLE 805, 806.
807,808. 809, 810.
PAGE Area of parallelogram . 318 Area of triangle 318,319 Area of regjular polygon and .
.
.
.
.
circle
811-813.
814.
320
Area of sector and sector of annulus 321,322 Lateral area of prism . 322 .
ARTICLE
PAGE
823,
Cylinder Mantel defined
824.
Conical surface
825-832.
Cone and frustum
833.
Area of sphere Zone
815-822.
834, 835.
323-325 325
326 .
.
.
326-329 330 331
.
CHAPTER
IV.
SPACE ANGLES. ARTICLE ^Z^i^yi'
PAGE
Plane and space angles defined
839.
Symmetrical space angles Steregon
840.
Steradian
838.
.
.
.
841.
Space angle made by two planes Spherical pyramid
.
......
PAGE
Space angles proportional . 332 844. To transform any polyhedral angle, 332 845. 846. Area of a lune 332 847. Plane angle as measure of dihedral, 843.
333
842.
ARTICLE
333 ^2>^
.
.
333 334 334
335 Properties of space angles . . .335 849-851. Area of spherical triangle or 848.
polygon
336
CONTENTS.
XVI
CHAPTER
V.
THE MEASUREMENT OF VOLUMES. ARTICLE
ARTICLE 852-854.
855,856. 857.
858, 859.
Volume and metric units .337 869,870. Volume of quader and cube 338 Volume of parallelepiped 338 871, 872. Volume of prism or cyl.
.
.
.
PAGE Tetrahedron and wedge are prismatoids 343 Altitude and cross-section of a
prismatoid
343
860-862.
Volume of pyramid or cone, 339-341
874, 875.
863-865. 866-868.
Prismatoid defined
876.
Volume of any prismatoid Frustum of pyramid or cone Volume of globe
%^T, 878.
Similar solids
inder
Prismoid
Direction Principle of Duality Linkage Cross-ratio
Exercises
338,339 .
.
341, 342
343
873,
.
.
344
.
346 347 347 347 349
350 352
357-366
THE ELEMENTS OF GEOMETRY.
BOOK CHAPTER ON I. 1.
I.
I.
LOGIC.
Definitions.
— Statements.
A statement is
true or false
;
e.g.
any combination of words that " for example "), {exempli gratiuy
is
either
All x^s are fs, 2. To pass from one statement to another, with a consciousness that belief in the first warrants belief in the second, is to
infer. 3.
much
Two
statements are equivalent when one asserts just as more nor less e.g., A equals By and
as the other, neither
B equals
A
;
A.
implied in a previous statement when its truth follows of necessity from the truth of the previous state^ 4.
ment;
statement
is
e.g..
All x's are fs implies some fs are
x*s.
THE ELEMENTS OF GEOMETRY.
'
;5»-
Any
5.
declarative sentence can be reduced to one or
more
simple statements, each consisting of three parts, namely, two terms or classes, and a copula, or relation, connecting
them.
A
x
equals B,
is
examples of this typical form for any word or
/, are simple
Here x and y each stand
of all statements.
group of words that may have the force of a substantive
naming a class e.g., x\?>y may mean all the ^s are y^ is mortal, means, all men are mortals. 6. sentence containing only one such statement
;
;
A
logically simple sentence
;
e.g.,
Man
is
in
man is
a
the only picture-making
animal.
A
sentence that contains more than one such statement 7. a logically composite sentence e.g., A and B are respectively equal to C and Z>, is composite, containing the two statements is
A
;
B equals D. Statements that are expressly conditional, such as, if A By then C is Z>, reduce to the typical form as soon as we see equals C^ and 8.
is
that they
mean {The consequence of
M)
is
N.
In the typical form, bird is biped, x is y, we call x and y terms of the statement; the first being called the subject, the 9.
and the
last the predicate.
II. Definitions.
Terms
10.
terms
;
11.
e.g.,
that
name a
— Classes.
single object are called individual
Newton.
Terms
that
called class terms;
name any one e.g.,
man^
of a
crystal.
group of objects are
The name
stands for
any object that has certain properties, and these properties are possessed in common by the whole group for which the name stands.
ON 12.
LOGIC.
A class is defined by stating enough properties
whether
a thing belongs to
it
to decide
or not.
Thus, "rational animal" was given as a definition of "man." 13. If we denote by x the class possessing any given property, all things not possessing this property form another class, which is called the contradictory of the first, and is denoted by " not x " non-;i;, meaning e.g., the contradictory of animate is ;
inanimate. 14.
Any
one thing belongs either to the class no thing belongs to both.
x
or to the
class non-;tr, but
It follows that;r is just as
So any
much
the contradictory of
non-;ir
y and the class non-j are mututhe contradictories of each other, and both together include ally all things in the universe e.g., unconscious and conscious. as non-jir
is
of x.
class
;
III.
The Universe
of Discourse.
In most investigations, we are not really considering all in the world, but only the collection of all objects which things 15.
are contemplated as objects about which assertion or denial may take place in the particular discourse. This collection we call our universe of discourse, leaving out of consideration, for the time, every thing not belonging to it. Thus, in talking of geometry, our terms have no reference
to perfumes.
Within the universe of discourse, whether large or x and non-jr are still mutually contradictory, and every thing is in one or the other e.g., within the universe 16.
small, the classes
;
mammals, every thing
is
man
or brute.
17. The exhaustive division into x and non-;ir is applicable any universe, and so is of particular importance in logic. But a special universe of discourse may be capable of some
to
entirely different division into contradictories, equally exhaustive. Thus, with reference to any particular magnitude, all
THE ELEMENTS OF GEOMETRY. magnitudes
of that
kind
may be
exhaustively divided into the
contradictories,
Greater than, equal
to, less
than.
IV. Contranominal, Converse, Inverse, Obverse. i8. if
If
X and y are
classes, our typical statement x \s, y means, it also belongs to class y]
a thing belongs to class x, then
e.g.,
Man
is
mortal, means, to be in the class men,
is
to be in
the class mortals. the typical statement is true, then every individual x belongs to the class y hence no x belongs to the class non-j^, or no thing not j/ is a thing x that is, every non-y is nonvr If
:
:
;
e.g.,
the immortals are not-human.
The statements x is y, and non-y the contranominal form of the other.
is
non-;ir,
are called each
Though both forms express the same fact, it is, nevertheless, often of importance to consider both. One form may more connect the fact with others naturally already in our mind, and so
show us an unexpected depth and importance of meaning. 19. Since x is y means all the x's are ys, the class y thus
contains
all
the individuals of the class x, and may contain Some of the ys, then, must be xs. Thus,
others, besides.
"
from "a crystal is solid we infer "some solids are crystals." This guarded statement, some y is x, is called the logical converse of x is y. It is of no importance in geometry. 20. If, in the true statement x is y, we simply interchange the subject and predicate, without any restriction, we get the inverse statement y is x, which may be false. In geometry it often happens that inverses are true and When the inverse is not true, this arises from the important. circumstance that the subject of the direct statement has been more closely limited than was requisite for the truth of the statement.
ON 21.
The contranominal is
non-^,
Of
LOGIC.
the
of
inverse,
namely,
non-;t;
is
called the obverse of the original proposition.
course,
if
the inverse
is
true, the obverse is true,
and
To
prove the obverse, amounts to the same thing as proving the inverse. They are the same statement, but may the meaning expressed, in a different light to our minds. put vice versa.
22. If the original
non-y
The
is non-x, its
statement
inverse
is
y
is
x
is Xy its
is y, its
obverse
contranominal is is non-x is iton-y.
two are equivalent, and the last two are equivalent. Thus, of four such associated theorems it will never be necessary to demonstrate more than two, care being taken that the two selected are not contranominal. 23. From the truth of either of two inverses, that of the other cannot be inferred. If, however, we can prove them both, then the classes x and y are identical. first
A perfect definition is always invertable. V.
On Theorems.
A
theorem is a statement usually capable of being inferred from other statements previously accepted as true. 25. The process by which we show that it may be so in24.
ferred 26.
is
called the proof or the demonstration of the theorem. corollary to a theorem is a statement whose truth
A
follows at once from that of the theorem, or from what has
been given in the demonstration of the theorem. theorem consists of two parts, the hypothesis, or 27. that which is assumed, and the conclusion, or that which is
—
A
asserted to follow therefrom. 28. A geometric theorem usually relates to some figure, and says that a figure which has a certain property has of necessity also another property or, stating it in our typical form, X is j, "a figure which has a certain property" is "a ;
figure having another specified property."
THE ELEMENTS OF GEOMETRY. The
first
names or
part
theorem relates The theorem is
defines
the figure to which the
the last part contains an additional property. ; first stated in general terms, but in the proof we
usually help the mind by a particular figure actually drawn on the page so that, before beginning the demonstration, the the;
orem
restated with special reference to the figure to be used.
is
29.
Type.
— Beginners
in
geometry sometimes find it diffiwhat is assumed and what
cult to distinguish clearly between has to be proved in a theorem.
has been found to help them here,
It
tion
what
of
is
given
is
what and the demonstration, in special statement of
if
the special enuncia-
one kind of type; the in another sort of type required,
printed in is
still
;
another.
In the course of the
proof, the reason for any step may be indicated in smaller type between that step and the next. 30. When the hypothesis of a theorem is composite, that is, consists of several distinct hypotheses, every theorem formed by interchanging the conclusion and one of the hypotheses is an inverse of the original theorem.
VI. 31.
Often
in
On
Proving Inverses.
geometry when the inverse, or
its
equivalent,
the obverse, of a theorem, is true, it has to be proved geometBut if we have rically quite apart from the original theorem.
x is y, and also that there is but one individual in the class y^ then we infer that y is x. The extra-logical to an is establish here contained in the inverse proof required
proved that every
proof that there
is
but one y.
Rule of
it has been proved that x is y, that no two ;i:*s are the and that there are as many individuals in class x as in
32. If
same
^,
Identity.
class yy then
we
infer
y
is
x.
ON
LOGIC,
Rule of 33. If the
Inversion.
hypotheses of a group of demonstrated theorems
exhaustively divide the universe of discourse into contradictories, so that one must be true, though we do not know which,
and the conclusions are also contradictories, then the inverse theorem of the group will necessarily be true. Examples occur in geometry of the following type
of every
:
If
a
is
greater than
If
a
is
equal to
If
a
is less
than
by 3,
b^
then
then
c is greater
than
—
d.
c is equal to d.
then c
is
less
than
d.
Three such theorems having been demonstrated geometritrue. cally, the inverse of each is always and necessarily Take, for instance, the inverse of the first ; namely, when than dy then a is greater than b. for the second and third theorems This must be true
c is greater
;
imply than d.
that'
if
a
is
not greater than
b,
then
c is
not greater
THE ELEMENTS OF GEOMETRY,
CHAPTER
IL
THE PRIMARY CONCEPTS OF GEOMETRY. I.
34.
Definitions of Geometric Magnitudes.
Geometry
is
the science which treats of the properties
of space.
A
35. part of space occupied by a physical body, or marked out in any other way, is called a Solid.
A
/I-
THE PRIMARY CONCEPTS OF GEOMETRY. 43.
9
A surface on a moving body may generate a solid. We
cannot picture any motion of a solid which will generate any thing else than a solid. Thus, in our space experience, we have three steps down 44.
from a solid to a point which has no magnitude, or three steps up from a point to a solid so our space is said to have three ;
dimensions.
A
45. Straight Line is a line which pierces space evenly, so that a piece of space from along one side of it will fit any side of any other portion. ;
46.
A
Curve
is
a line no part of which
Take notice: the word mean "straight line."
47.
48.
\
^^ a
straight.
forth
is
A Sect is the part of
curve.
"line," unqualified, will hence-
a line between two definite points.
A Plane
Surfacey or a Planey is a surface which divides so that a piece of space from along one side of it space evenly, will fit either side of any other portion. 49.
50.
A
plane
51.
A
Figure
is generated by the motion of a line always a fixed point and leaning on a fixed line. passing through
is
any
definite combination of points, lines,
curves, or surfaces. 52.
A Plane Figure is in one plane.
THE ELEMENTS OF GEOMETRY.
10
53. If a sect turns about one of its end points, the other end point describes a curve which is called a Circle, The fixed end point is called the Center of the Circle,
c
54.
The Radius of a
^
3
Circle is a sect
drawn from the center
to the circle.
A
Diameter of a Circle is a sect drawn through the center, and terminated both ways by the circle. 56. An Arcis a. part of a circle. 55.
Lines are such as are in the same plane, and which, being produced ever so far both ways, do not meet. 57. Parallel
C 58.
When two
J) drawn from the same point, they make with each other, an Angle,
lines are
are said to contain, or to
THE PRIMARY CONCEPTS OF GEOMETRY.
II
The
point is called the Vertex, and the lines are called the line drawn from the vertex, and turning Arms, of the angle.
A
about the vertex in the plane of the angle from the position of coincidence with one arm to that of coincidence with the other, said to turn through the angle the quantity of turning is greater. is
;
and the angle
is
greater as
Since the line can turn from the one position to the
59.
other in either of two ways, two angles are formed by two lines drawn from a point.
Each If
we
of these angles is called the Explement of the other. say two lines going out from a point form an angle, we
\oy are fixing the attention upon one of the two explemental angles really form ; and usually we mean the smaller angle.
which they 60.
Two
angles are called
Equal
if
they can be placed so
arms coincide, and that both are described simultaneously by the turning of the same line about their common that their
vertex. 61. If two angles have the vertex and an arm in common, and do not overlap, they are called Adjacent Angles ; and the angle made by the other two arms on the side toward the common arm is called the Sum of the Adjacent Angles, Thus,
THE ELEMENTS OF GEOMETRY.
12
for the word "angle," the sign of equality using the sign addition (+, plus), and of the sign (=), "4-
%.
62.
A
AOC =
4.
Straight Angle has
AOB + its
arms
t ^OC. in the
same
line,
and on
different sides of the vertex.
The sum
63.
arms
exterior is
of
in the
two adjacent angles which have their line on different sides of the vertex
same
a straight angle.
64.
each
is
When said to
O 65.
sum
any two angles is a straight angle, be the Supplement of the other.
the
of
O2
If
Oi
two supplemental angles be added,
^ their exterior
THE PRIMARY CONCEPTS OF GEOMETRY. arms
will
form one line
;
13
and then the two angles are called
Supplemental Adjacent Angles,
66. 67.
A Right Angle is half a straight angle. A Perpendicular to a line is a line that
angle with
68.
makes a
right
it.
When
the
sum
of
two angles
is
a right angle, each
is
said to be the Co7nplement of the other.
An An
Acute Angle is one which is less than a right angle. Obtuse Angle is one which is greater than a right 70. but less than a straight angle. angle, 71. The whole angle which a sect must turn through, about one of its end points, to take it all around into its first position, 69.
THE ELEMENTS OF GEOMETRY.
14
one plane, the called a Perigon.
or, in
whole amount
of angle
\
round a point,
is
^
72. Since the angular magnitude about a point increased nor diminished by the number of lines which radiate from the point, the sum
is
neither
of all the angles about a point in a plane is
a perigon.
A
Reflex Angle is one which is greater than a straight angle, but less than a perigon. 74. Acute, obtuse, and reflex angles, in distinction from 73.
right angles, straight angles, and perigons, are called Oblique Angles ; and intersecting lines which are not perpendicular to
each other are called Oblique Lines. 75. When two lines intersect, a pair of angles contained by the same two lines on different sides of the vertex, having no
arm
in
76.
common,
are called Vertical Angles.
That which divides a magnitude into two equal parts is and is called a Bisector.
said to halve or bisect the magnitude,
77. If we imagine a figure moved, we may also suppose to leave its outline, or Trace^ in the first position.
it
THE PRIMARY CONCEPTS OF GEOMETRY. 78.
A
79.
The
15
Triangle is a figure formed by three lines, each intersecting the other two.
three points of intersection are the three Vertices
of the triangle.
The three sects joining the vertices are the Sides of the The side opposite A is named a the side opposite triangle. 80.
;
^is^. 81.
An
Interior Angle oi a triangle
is
one between two of
the sects.
82.
An Exterior
Angle of a triangle is one between either which is a continuation of another side. 83. Magnitudes which are identical in every respect except the place in space where they may be, are called Congruent. 84. Two magnitudes are Equivalent which can be cut into sect
and a
line
parts congruent in pairs. II.
Properties of Distinct Things.
The whole The whole
greater than its part. equal to the sum of all its parts. 87. Things which are equal to the same thing are equal to one another. 85. 86.
is
is
1
THE ELEMENTS OF GEOMETRY,
6
be added to equals, the sums are equal. equals be taken from equals, the remainders are
88. If equals
89.
If
equal. 90. If to equals
unequals are added, the sums are unequal,
and the greater sum comes from adding the greater magnitude. 91. If equals are taken from unequals, the remainders are unequal, and the greater remainder is obtained from the greater magnitude.
Things that are double of the same thing, or of equal one another. 93. Things which are halves of the same thing, or of equal things, are equal to one another. 92.
things, are equal to
III.
Some Geometrical Assumptions about
94.
A
solid or a figure
may be imagined
Euclid's Space. to
move about
in
Magnitudes which will coinspace without any other change. cide with one another after any motion in space, are congruent ;
and congruent magnitudes can, after proper turning, be made to coincide, point for point, by superposition. that is, if two lines have 95. Two lines cannot meet twice two points in common, the two sects between those points ;
coincide.
two lines have a common sect, they coincide throughTherefore through two points, only one distinct line can
96. If out.
pass. 97. If
two points
of a line are in a plane, the line lies wholly
in the plane. 98. All straight angles are equal to lines which intersect one 99.
Two
parallel to the
same
line.
one another. another cannot both be
TABLE OF SYMBOLS.
IV.
The Assumed
17
Constructions.
100. Let it be granted that a line may be drawn from any one point to any other point. loi. Let it be granted that a sect or a terminated line may be produced indefinitely in a line. 102. Let it be granted that a circle may be described around
any point as a center, with a radius equal
—
to a given sect.
Here we are allowed the use of a 103. Remark. with marked not divisions, and the use of a pair edge,
straight of
com-
passes the edge being used for drawing and producing lines, the compasses for describing circles and for the transferrence ;
of sects.
But
it
is
more important
to note the
implied restriction,
namely, that no construction is allowable in elementary geometry which cannot be effected by combinations of these primary constructions.
SYMBOLS USED.
THE ELEMENTS OF GEOMETRY,
i8
CHAPTER
III.
PRIMARY RELATIONS OF LINES, ANGLES, AND TRIANGLES. I.
Angles about a Point.
Theorem
I.
104. All right angles are equal. Such a proposition is proved in geometry by showing be true of any two right angles we choose to take.
^
G The hypothesis
it
to
3
J£
J)
will be, that
C
O
we have any two
right angles
;
A OB
and right ^ FHG. as, for instance, right t a of definition the right angle (^6)^ the hypothesis will By is half is half a mean, ^ straight angle at O, and ^
FUG
A OB
a straight angle at H. The conclusion is, that
^
AOB and
^
FUG are equal.
The
proof consists in stating the equality of the two straight are halves, referring and which the angles of angles to the assumption (in 98) that all straight angles are equal,
AOB
FUG
and then stating that therefore the right ^ AOB equals the right ^ FUG, because, by 93, things which are halves of equal things are equal.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. Now, we may the abbreviations
symbol
.'.
for the
Hypothesis.
2^
A OB \s>
A
(66.
By
is
^
9
and condense this as follows, using and st. for " straight," and the
right,"
word "therefore"
AOB == ^ AOB half a
Conclusion.
Proof.
restate " for
rt.
1
:
^ also ^ FUG is rt ^ ^ FHG. st. ^ also 4. FHG is half a st. ^.
rt
.
;
;
right angle
is
half a straight angle.)
98, all straight angles are equal,
^AOB =
.-.
4.
FHG.
Halves of equals are equal.)
(93.
From
a point on a line, there cannot be more than one perpendicular to that line. 105.
Corollary.
Theorem 106.
II.
All perigons are equal.
Xbj,
V^^
G Hypothesis. Conclusion.
Proof.
Any
4-
^ ^^
'^
4 AOA = line
>»
^ perigon ; also
4
DHD
is
a perigon.
4 DHD.
through the vertex of a perigon divides
it
into
two
straight angles.
By
98,
all
straight angles are equal, .'.
Perigon at (92.
Corollary.
O
equals perigon at
Doubles of equals are
From
H.
equal.)
the preceding demonstration, 107. follows that half a perigon is a straight angle.
it
THE ELEMENTS OF GEOMETRY.
20
Theorem io8.
III.
intersect each other^ the vertical angles are
If two lines
equal.
AC
Hypothesis. Conclusions.
and
^A0B= %.BOC
Proof. angle
AOC,
BD =:.
are two lines infersecf/ng at O.
^ COD. y^DOA.
AOB and BOC AO and DC are in one line, BOC = 4..
Because the sum of the angles
and by hypothesis .-.
^
AOB +
4.
is
the
St.
In the same way,
^BOC -^ ^COD = .-.
^AOB+^ BOC =
Take away from
4. (89. If equals
angle
BOC,
and we
AOB = 4COD.
may prove
4
side of a line.
common
^ COD.
be taken from equals, the remainders are equal.)
In the same way we
i.
^,
^ BOC +
these equal sums the
have
Exercises,
St.
BOC = 4DOA.
Tv^to angles are
Show
contain a right angle.
that the lines
formed at a point on one which bisect these angles
RELATIONS OF LINES, ANGLES, AND TRIANGLES.
Theorem log.
angle equal 07tly
two
to the
i.
IV.
go out from a point
one not adjacent
to
it,
the
so as to
four
make
each
lines will form
ijitersecting lities.
Let
OA, OB, OC, OD, be four lines, and let ^ AOB = ^ COD, and point O;
Hypothesis.
common
lines
If four
21
with
the
^BOC
==
DOA.
AO and OC are in one line. BO and OD are in one line. AOB BOC + ^ COD +
CoNCXUSiONS.
Proof. (71.
By
4.
-\-
The sum
hypothesis, .'.
4
twice .-.
4.
AOB =
4 COD, bhA
DOA = is
4BOC =^
a perigon,
a perigon.)
4 DOA,
BOC = a perigon; 4. 4 AOB + 4 BOC = a 4 A OB +
twice
4.
st.
(107. Half a perigon .'.
(65. If
4.
of all the angles about a point in a plane
is
a straight angle.)
AOand OC form
two supplemental angles be added,
one
line.
their exterior
arms
will
form one
4 AOB and 4 DOA are BO and OD form one line.
In the same way we may prove that plemental adjacent angles, and so that
line.)
sup-
no. Corollary. The four bisectors of the four angles formed by two lines intersecting, form a pair of lines perpendicular to each other.
THE ELEMENTS OF GEOMETRY.
22
II.
111.
Angles about
Two
A line cutting across other lines
Points. is
called a Transversal.
plane two lines are cut in two distinct points at each of the points four angles are detera transversal, by 112. If in a
mined.
Of these eight
angles, four are between the two given and are called Interior Angles ; the
lines (namely, ^, J, a^ b)y
other four
lie
outside the two lines, and are
called Exterior
Angles.
Angles, one at each point, which lie on the same side of the transversal, the one exterior, the other interior, like / and a, are called Corresponding Angles.
Two
angles on opposite sides of the transversal, and both interior or both exterior, like J and a, are called Alternate Angles.
Theorem
V.
113. If two corresponding or two alternate angles are equal or if two interior or two exterior angles 07i the same side of the transversal are supplemental^ then every angle is eqtial to its, cory
responding and to its alternate angle, and is supplemental to the angle on the same side of the transversal which is interior or exterior according as the first is interior or exterior.
RELATIONS OF LINES, ANGLES, AND TRIANGLES.
— Hypothesis,
First Case.
Conclusions.
2
4.b
4.4
\-
%.a
=
4.1
= — '\-
— i. -^.a = 2
4a
=z
(108. If
4c,
two
4.
2^tr.
:4.
-{-
4
^
—
4-
St.
4.
d
4h = 4d, 41 — 43, 42 = 44,
lines intersect, the vertical angles are equal.)
Moreover, since, by hypothesis, equal, ox
}^
-2^
z=z
c
Proof.
23
2j^
=^
as
supplements are
7, their
^• (98. All straight angles are equal.)
(Eg. If .-.
and
St.
2^
equals be taken from equals, the remainders are equal.)
4a = 41 = 4c == 43, 4b = 42 = 4d=4 4, = 4 ^ + 4b — 4 a \- 4 4 = 4 i 4 d = 4 2 + 4c = 4b-^4-3> -\-
(88. If
equals be added to equals, the sums are equal.)
Second Case.
If,
alternate angles equal, First
Case again.
instead of two corresponding,
we
substitute for
—
Third Case. Hypothesis. But 4 4 -\- 4 I =^s\.4, :.
which gives again the
Fourth Case. But
^
^
+
^ «
+
4a =
^
;^
we have
its vertical,
=
5f.
^
St.
4 ^
.
41, "^
4-
^, .-.
4a =
which gives again the First Case.
41,
^
=
given two
which gives the
First Case.
— Hypothesis.
^ ^=
one
^^- 4-'
THE ELEMENTS OF GEOMETRY.
24
III. Triangles.
114.
An Equilateral
Triangle
is
one in which the three sides
are equal.
115.
equal. 116.
An
A
Isosceles
Triangle
is
one which has two sides
Scalene Triangle has no two
sides equal. 117. When one side of a triangle has to be distinguished from the other two, it may be called the Base ; then that one
of the vertices opposite the base is called the Vertex. 118. we speak of the angles of a triangle, the three interior angles.
When
119.
angle. nuse.
A Right-angled The
Triangle has one of
side opposite the right angle
is
we mean
angles a right called the Hypotheits
121.
An An
Obtuse-angled Triangle has one of its angles obtuse. Acute-angled Triangle has all three angles acute.
122.
An
Equiangular Triangle
120.
is
one which has
all
three
angles equal. 123. When two triangles have three angles of the one equal respectively to the three angles of the other, a pair of equal The pair of sides oppoangles are called Homologous Angles. site
homologous angles are called Homologotis Sides.
RELATIONS OF LINES, ANGLES, AND TRIANGLES.
Theorem
2$
VI.
124. Two triangles are congruent if tzuo sides and the inclnded angle in the one a7'e equal respectively to two sides and the included a7igle in the other,
M
Z ABC
Hypothesis.
BC = MN,
4.B
triangle,"
and
and
=
for
LMN
AB = LM,
two iriangles, with
^M.
=^
The two
Conclusion. "
3L
"
triangles
are
congruent;
or,
using
A
for
congruent,"
h^ABC^ t^LMN. Proof. the vertex
Apply the triangle
M
shall rest
N on the same
side of
LMN to A ABC
on B, and the
BA as
;
Then, because the side equals because ^ J/ ^ ^, the side
point C.
upon (95. If
side
Now,
MN
equals since the point
C, therefore the side
two
lines
BA,
the point
MN
=
because the
ML
C.
ML
A
in such a manner that on BA, and the point
side
BC,
L
the
rests
will rest
point
N
L
will rest
upon
the line
will
rest
upon
BC
;
upon the
upon A, and the point
N
rests
LN coincides with the side A C.
have two points in common, the two sects between those points coincide.)
Therefore every part of one triangle will coincide with the corresponding part of the other, and the two are congruent. (94.
125. In
Magnitudes which
any pair
sides are equal.
of
will coincide are congruent.)
congruent triangles, the homologous
THE ELEMENTS OF GEOMETRY.
26
Theorem
VII.
In an isosceles triangle the angles opposite the equal sides are equal. 126.
ABC a triangle, with AB = BC. ^ A = :^C. Imagine the triangle ABC to be taken up, turned over, and
Hypothesis.
Conclusion.
Proof. put
down
ABC
left
Then,
since,
and now designate the angular points ; to distinguish the triangle from its trace
in a reversed position
A
A' (read
prime), behind.
B\ C,
in the triangles
ABC, C'B'A\
AB = C'B' by hypothesis, AB = CB,
Also
4.B :.
(124. Triangles are
But
^
C"
congruent
if
127.
=^
^A =
^
B'A\
B',
^ a.
two sides and the included angle are equal
in each.)
= ^ C, .-.
(87.
BC =
and
Things
equal to the
Corollary.
^A = ^C same thing
Every
are equal to one another.)
equilateral triangle
is
also equi-
angular.
Exercises.
same
line.
2.
The
bisectors of vertical angles are in the
RELATIONS OF LINES, ANGLES, AND TRIANGLES.
Theorem 128.
Two
2/
VIII.
triangles are coitgruent if two angles
cluded side in the one are equal respectively the included side in the other.
to
two
and the inaiigles and
TVf
ABC and LMN ^ LN. AC ^C^^N, Hypothesis.
Conclusion.
The two
Apply the
ABC ^
triangle
LMN to
M
Because
2j^
N
2&
lie
so that the
along the side
A C,
and
B.
LN^
will rest
upon point C.
Z = ^ ^, .-.
Because
point
AC =^
ABC
the triangle
LN
AC
.*.
A = ^ Z,
A LMN.
point L shall rest upon A, and the side the point lie on the same side oi
Then, because the side
4.
triangles are congruent.
A Proof.
two triangles, with
^ iV .'.
side
=
^
will rest
upon the
line
AB»
upon the
line
CB.
C,
side
Because the sides
LM
NM
will rest
LM and NM rest respectively upon the lines AB
and CB, .'.
the vertex
M,
resting
upon the vertex B,
upon both the the only point
Therefore the triangles coincide in
lines
AB
common
all
and CB, must
to the
their parts,
two
rest
lines.
and are congruent
..J
THE ELEMENTS OF GEOMETRY.
28
Theorem
IX.
129. Two triangles are congruent if the three sides of the one are equal respectively to the three sides of the other.
a
Imagine
opposite to the side on which
Case
Then
in
A
ABM,
And
and
^ ABM =
join
A
AB = LM,
LMN in
falls
on the
such a way side oi
AC
BM.
AB = AM,
^ AMB.
an isosceles triangle the angles opposite the equal sides are equal.)
for the
same
reason, in
Therefore
A BCM,
^ CBM = ^ ABM + ^ CBM = .'.
(88. If
that
falls,
to
M
because, by hypothesis, .'.
(126. In
B
the vertex
having
BM cuts the sect AC.
When
I.
LMN,
ABC ^ a LMN. A ABC to be applied
LN coincides with A C, and
that
and
CA = NL.
and
Conclusion,
Proof.
ABC
Triangles
BC = MN,
because
BC =
4.
CMB.
^
AMB
-f-
CM,
^ CMB',
equals be added to equals, the sums are equal.)
is, .'.
^ABC^i. AMC, A ABC ^ A LMJV.
{124. Triangles are congruent
if
two sides and the included angle are equal
in each.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. Case IL When AC, as C.
BM passes
29
through one extremity of the
sect
Then, by the
first
step in Case
I.,
^B ^^.M', and
same reason
for the
as before,
A
Case AC.
III.
When
ABC ^
BM
falls
A LMN, beyond the extremity
sect
AB = AM, = :. t ^^B^ ABM = And in A BCM, because BC CM, CMB = 4 CBM. :. Therefore the remainders, ^ ABC = AMC; Then
in
A ABM^
because, by hypothesis,
4.
:^
(89. If
and
so, as in
equals be taken from equals, the remainders are equal.)
Case
I.,
A
ABC ^
A LMN,
of
the
THE ELEMENTS OF GEOMETRY.
30
CHAPTER
IV.
PROBLEMS.
A
Problem is a proposition in which something 130. required to be done by a process of construction, which termed the Solution. 131.
consists,
The
—
solution
of
is
is
a problem in elementary geometry
(i) In indicating how the ruler and compasses are to be used in effecting the construction required. (2) In proving that the construction so given is correct. (3) In discussing the limitations, which sometimes exist, within which alone the solution is possible.
Problem 132.
I.
To describe an equilateral triangle upon a given (T,.
,.
.
Jl'
Given, the sect
Required,
AB.
to describe
an equilateral triangle on AB.
I".
sect.
PROBLEMS. With center
Construction.
A
3
and radius AB^ describe the
1
circle
BCD. With center (using O
B and radius BA^ describe O A CF
" circle ").
for
(102.
A
circle
may be
described with any center and radius.)
Join a point C, at which the circles cut one another, to the points
^and^. (100.
Then
A line may be
will
ABC
be an equilateral
A AB =
Because
Proof.
.'.
and because
B
is
drawn from any one point
is
triangle.
the center of
O BCD,
being radii of the same circle the center of the O CF,
AC,
BA = BC, being radii of Therefore AC = AB = BC, and an
Things equal
to the
equilateral triangle has
same thing
On a give7i
line, to
-
Given,
AB, a
lino ; a,
the
same
mark
circle.
are equal to one another.)
been described on
Problem 133*
;
A
.-.
(87.
to any other point.)
AB,
II.
off
a
sect
equal
to
a given
sect.
r
a sect
Required, to mark off a sect on AB equal to a. Construction. From any point (9 as a center, on the arc of a circle with a radius equal to a.
D
AB,
CD
describe
If be the point in which the arc intersects AB, then will be the required sect. Proof. All the radii of the circle around O are, by construction,
equal to a.
OD
is
one of these
radii, therefore it is
equal to a.
THE ELEMENTS OF GEOMETRY.
Problem 134.
To
bisect
III.
a given angle.
Given, ihe angle AOD. Required, to bisect it. Construction. From (9 as a center, with any radius, OA, describe the arc of a circle, cutting the arms of the angle in the points A and B. Join
equilateral
Join
On AB, on triangle, ABC.
AB.
OC.
Proof.
The
line
the side remote from
side
describe,
by 132, an
OC will bisect the given angle AOD. CAO and CBO we have
In the triangles
OA = OB, CA = CB by and the
(9,
CO
construction,
common. .-.
A
CAO^
A CBO,
(129. Triangles with the three sides respectively equal are congruent.) .-.
.'.
Exercises,
i.
^ COA = ^ COB,
CO is
ihe bisector of
^ A OB.
Describe an isosceles triangle having each
of the equal sides double the base. 2.
Having given the hypothenuse and one
right-angled triangle, construct the triangle.
of the sides of a
PROBLEMS.
33
Problem IV. Through a given point on a given
135.
line, to
draw a perpen-
dicular to this line.
c
^\
:b
i
AB, and the point C in it draw from C a perpendicular to AB.
Given, fhe fine
Required, Solution.
to
By
134, bisect the
st.
^ ACB.
Exercises. 5. Solve 134 without an equilateral triangle. Divide a given angle into four equal parts. 3. Find a point in a line at a given distance from a givea 6.
When
the problem impossible } in a line equally distant from two given without the line. points 5. From a given point without a given line, draw a line
point. 4.
is
Find a point
making with the given
line
an angle equal to one-half a right
angle. 6.
/4
is
a point without a given
line,
EC.
Construct an
A
as vertex, whose base shall lie isosceles triangle, having to a given sect. along EC, and be equal II.
If
a right-angle triangle have one acute angle double
the other, the hypothenuse
is
double one
side.
THE ELEMENTS OF GEOMETRY.
34
Problem V. To
136.
a given
bisect
sect.
#
•
/
*
•
/
*« %
I
»
t
/
Given, the sect
AB.
to bisect
Required, Construction.
•
'
On
it.
AB
describe an equilateral triangle
ABCy by
132.
By
AB
134, bisect the angle be bisected in D.
Proof.
;and
ACB by the line meeting AB in D,
Then
shall
AC
•=
In the triangles
BC
by
A CD and BCD
4.ACD= ^BCD, common. construction, and CD AD = BD, is
.-.
(124. Triangles are congruent
if
two sides and the included angle are equal
AB
is
bisected at
in each.)
D.
The line drawn to bisect the angle at 137. Corollary I. the vertex of an isosceles triangle, also bisects the base, and is perpendicular to 138.
it.
Corollary
II.
The
line
drawn from the vertex
isosceles triangle to bisect the base, also bisects the vertical angle.
Exercises.
12.
is
perpendicular to
of an
it,
and
Construct a right-angled isosceles triangle
on a given sect as hypothenuse.
PROBLEMS.
35
Problem VI. 139'
From a point without a given
line^ to
drop a perpendicu-
lar upon the line.
?€
i^ Given, the line
AB, and
C
the point
without
it.
Required, to drop from C a perpendicular upon AB. Construction. Take any point, Z>, on the other side of from C, and, by 102, from the center C, with radius CZ?, describe the arc and G, at FDG, meeting
AB
By
AB F FG at H.
136, bisect
CH.
Join
C^ shall be ±y^^ (using
± for
Proof.
CF =
the words "perpendicular to").
Because in the triangles
CG, and
HF =
and
HG,
.-.
^
CHF and CHG^
CH
CBF =
is
by construction,
common,
^ CHG.
(129. Triangles with three sides respectively equal are congruent.) .-.
^ FHC,
perpendicular to 140.
being half the
st.
^ FUG,
is
a
rt.
^
;
and
CH
is
AB.
The
Corollary.
line
drawn from the vertex of an it, and also
isosceles triangle perpendicular to the base, bisects bisects the vertical angle.
Exercises. bisect the angle
13.
Instead of bisecting FG, would
FCG
}
it
do to
THE ELEMENTS OF GEOMETRY.
36
CHAPTER
V.
INEQUALITIES.
The symbol
141 •
a> b means less
than
that a
is
>
is
called the
Sign of Inequality and
greater than b\ so a
d.
Theorem X. 142.
the
An
exterior angle of a triangle is greater than either of
two opposite interior
Hypothesis.
angles.
Zef ihe side
AB of
the triangle
ABC
be produced
to £>.
Conclusions.
Proof.
By
By 100 and
HF =
By
^
^ BCA.
^CBD> ^BAC. BC in ZT.
136, bisect loi, join
AH, lOQ, join
CBD >
BF,
AH, and
produce
it
to
F)
making, by 133,
INEQUALITIES. Then, in the triangles
CH =
AHC
AH =
HB,
and
HF,
37
FHB,
by construction,
and
^ ^iYC
= ^ FEB',
(io8. Vertical angles are equal.)
A
/. (124. Triangles are congruent
/.
Now,
AHC ^
A FHB,
two sides and the included angle are equal
if
i.
HCA =
4.
in each.)
HBF.
4. CBD > 4. HBF,
The whole
(85.
4
.-.
if
Similarly,
CB be
greater than
CBD >
produced
the vertical
.-.
is
to
4.
G,
its part.)
HCA,
it
may be shown
that
4.ABG>4BAC, 4. CBD > 4 BAC. Theorem XL
143.
Any two
angles of a triangle are together less than a js
straight angle.
ABC
Hypothesis.
Let
Conclusion.
4A
Proof.
Produce
CA
be any a.
^^<
-f
to
Then 4 BAD > 4 B, (142. An exterior angle of a triangle
144.
.-.
4
.-.
St.
BAD + ^>^
Corollary
I.
st.
^
.
D.
4
^+ No
is
greater than either opposite interior angle.)
BAC>4 B
-\-
4 BAC,
^ A. triangle can have
more than one
right angle or obtuse angle. 145.
Corollary
from a point
II.
to a line.
There can be only one perpendicular
THE ELEMENTS OF GEOMETRY.
38
Theorem
XII.
If one side of a triangle be greater than a second^ the angle opposite the first must be greater than the angle opposite^ 146.
the second.
^
a ABC, with side a > ^ BA C> :^C.
Hypothesis,
Conclusion.
Proof.
By
100, join
BD =
^
C,
Still
BD —
c.
^
:.
c,
BDA =
^ BAD.
an isosceles triangle the angles opposite the equal sides are equal.)
And, by 142, the exterior rior
off
c.
AD.
Then, because (126. In
a cut
133, from
By
side
more
^
also
.-.
BDA of A CDA > the ^ BAD >^C.
opposite inte-
is^BAC>)^C.
Corollary.
one side of a triangle be less than the second, the angle opposite the first will be less than the angle 147.
If
opposite the second. For,
iia
148.
:.
b
Inverses.
By By By Therefore, by
> a\
We
146, if 126,
:.
4.
b,
A,
.'.
:4
A
<:^B.
^B;
^ A =: ^ B ^ A < ^ B.
.'.
^A
If
lateral.
4.
;
.*. b, 147, if 33, the inverses are necessarily true, namely, a > b; If .'. > ^B,
If
149.
B>
have now proved. a > b, .'. ^A >
a = a <
if
by 146,
4 A = 4 B, 4 A < ^B,
Corollary.
An
.'. .-.
a = b; a < b.
equiangular triangle
is
also
equi-
INEQUALITIES.
Theorem 150.
The perpendicular
point and a given
lei
Hypothesis,
100, join
A
151.
^
sect
he a given point, and
by construction,
between a given
BC
a given
line.
4.
^
ADF >
ADF 4.
is rt.
^
,
AFD,
angles of a triangle are together less than a straight angle.)
.-.
(148. If angle
Jl
least
BC
Since,
Any two
the
AD < AF. .-.
{143.
is
line.
AD
A
Conclusion.
Proof.
XIII.
A.BC, By 139, from A drop to F^ any point o( except £>.
Construction.
By
39
ADF
is
AF>AD.
greater than angle AFD, therefore side side AD:\
AF is greater than
Except the perpendicular, any sect from a point an Oblique.
to a
line is called
Exercises.
14.
Determine the
least path from one point it shall meet a given
to another, subject to the condition that
straight line. 15.
The
greater than
four sides of a quadrilateral figure are together its
two diagonals.
THE ELEMENTS OF GEOMETRY.
40
Theorem XIV.
Two
152.
obliques from a point,
foot of the {)erpendiculary are equal,
making equal sects from the and make equal angles with
the line.
AC 1. BD. BC = DC. Conclusions. AB = AD. ABC = ADC. a ACB ^ ACD. Proof, Hypothesis.
4.
4.
t^
(124. Triangles are congruent
if
two sides and the included angle are equal
in each.)
For every oblique, there can be drawn 153- Corollary. one equal, and on the other side of the perpendicular.
Exercises.
16.
ABC
bisected by a line meeting
BD, and 17.
A C greater
is
BC
a triangle whose angle at
D.
Prove
AB
A
is
greater than
than CD.
Prove that the sum of the sects from any point to the is greater than one-half the perime-
three angles of a triangle ter of the triangle.
18. The two sides of a triangle are together greater than twice the line drawn from the vertical angle to the middle point of the base.
INEQUALITIES.
41
Theorem XV. 154.
Of any two makes
that which
obliques between a given point and line^ from the foot of the perpen-
the greater sect
dicular is the greater.
Hypothesis.
AC 1. BD,
Conclusion.
AF
CB
B
'j
Proof.
By
then, in
A AFCy
> CF. < AB.
153, take
F on /.
(143.
Any two angles of a
^ AFC + ^ AFB = .-.
(143.
Any two angles of a
155'
side of the perpendicular as
A
is less
Corollary.
^
AFC <
rt.
^
ACF
is rt.,
^,
^ AFB > st.
^
rt.
^,
^.
ABF <
i AFB,
triangle are together less than a straight angle.)
,\ (148. If angle
same
triangle are together less than a straight angle.)
.-.
since
the
because, by hypothesis,
AF
No more
than two of
is
less than
all
b.)
the sects can
be equal. Proof.
For no two
can be equal.
sects
on the same
side of the perpendicular
THE ELEMENTS OF GEOMETRY.
42
Theorem XVI. 156.
Any two
a triangle are together greater than
sides of
the third side.
Ir/
D
'"'
Hypothesis.
Let
Conclusions,
a a
By
By
be a triangle.
4- l?>c.
b
Proof.
ABC \-
c>b.
-\-
oa.
loi, produce
BC to Z>
;
making, by 133,
CD =
b.
100, join A£>.
Then, because
AC = ,'.
Now,
i.
BAD >
4.
(85.
148,
But
BD =
The whole
157.
= ^ CAD,
is
also
:2^
greater than
its part.)
BAD > BDA. %.
BD>BA. + b, and BA is c, .-.
a
.*.
Similarly
^ CDA
CAD,
.-.
By
CD,
it
a
may be shown
Corollary.
-{-
b>c.
that
Any
a
+
side of
c>b, and
that b
a triangle
the difference between the other two sides.
is
-\-
c>a.
greater than
INEQUALITIES.
43
Theorem XVII. If from the ends of one of the sides of a triangle two drawn to a point within the triangle^ these will be less than the other two sides of the triangle, but will together contain a greater angle. 158.
sects
be
M
D
Hypothesis. Conclusions.
is
(I.)
(11.)
Proof.
Produce
(I.)
Join
a po/nf within A
ABC.
AB + BC>AD + DC.
ADC >B. AD and CD. 4.
CD to meet AB in F. + BF) > (CF), and [DF +
Then {CB
(156. In a triangle any
/.
FA;\ > [DA-].
two sides are together greater than the
BA = (CB -^ BF) FA>(CF) -^ FA = CD IDF + FAI >CD + Next, in A AFD
CB
-{-
-\-
-\-
(II.)
third.)
iDA'].
%.ADC>i.AFD. (142. Exterior angle of a triangle is greater than opposite interior angle.)
And
for the
same reason,
in
4AFD>4.By
sum
.-.
i.ADC>:4.B.
perimeter of a triangle is greater than of sects from a point within to the vertices.
Exercises. the
The
A CFB
19.
^-^^^
44
ELEMENTS OF GEOMETRY. Theorem XVIII.
If two triangles have two sides of the one equal respecto two sides of the other, but the included angle of the
159.
tively first greater than the included angle of the second^ then the third side of the first will be greater than the third side of the second.
ABC
Hypothesis.
and
DEF
are two
iriangles, in
which
AB = DE, BC = EF, DEF, i. ABC > 4.
AC> DF.
Conclusion.
Proof. the point
D
By 134, AC.
EF shall coincide with BCy and ^C as A.
Place the triangles so that on the same side of
fall
bisect
^ ABD^ and
let
G be
the point in which the bisector
meets
By
100, join
^ DBG',
DG.
ABG and BDG, by construction, ^ ABG = common. by hypothesis, AB = BD and BG A ABG ^ A jDBG,
Then, in the
triangles
]
is
.-.
(124.
Two
triangles are congruent
if
two sides and the included angle are equal in each.)
/.
AG =
/.
CA = CG + GA = CG
Z>G, -\-
GD,
INEQUALITIES. But
CG + GD>
(156.
Any two
CD,
sides of a triangle are together greater than the third.) .-.
160. 4.
45
Corollary. \i c and so, by 1 59,
CA>FD,
— fy
a
:=•
d,
and
^B<^ Ey
then
E>^B\
e>d,
161. Inverses.
By By By
We
.;
have
159,
a 4.B
124,
if
160,
\i^B
b<e.
now
> 4.E,
^^ =
^^,
< 4.E,
Therefore, by 33, Rule of Inversion,
lib >
e,
proved,
when a
h >
e-,
b=e; b <
e.
=^ d,
and
THE ELEMENTS OF GEOMETRY,
46
Problem VI I. 162. to three
To construct a triangle of which the sides shall be equal given sects, but any two whatever of these sects must be
greater than the third.
\
INEQUALITIES.
47
would be on OGy and the triangle would become a sect. But as we have proved, in 156, that any two sides of a triangle are together greater than the third side, our solution of Problem VII will apply to any three sects that could possibly form a triangle. If a tri163. Corollary. of made be hinged rods, angle though the hinges be entirely free, the rods cannot turn upon
them
the triangle
;
is rigid.
Problem VIII. 164. to
At a given point
a given
ift
a given
line, to
make an angle equal
angle.
AB
and ^ C, Me poini A, and the line Required, at A, with arm AB, to niake an angle ^ C. Construction. By loo, join any two points in the arms of ^ C, thus forming a A DCE, By 162, make a triangle whose three sides Given,
shall
CD
=
be equal to the three sides oi and CE intersect at A, :,
DCE\
^FAG =
making the
sides equal to
:i^DCE,
(zag. Triangles with three sides respectively equal are congruent.)
THE ELEMENTS OF GEOMETRY.
48
CHAPTER
VI.
PARALLELS.
Theorem XIX. ^^5*
-^
t'^o lines cut by
a transversal make alternate angles
egualy the lines are parallel.
/ / the contranominal of 142, part of which may be stated thus If two lines which meet are cut by a transversal,
This
is
:
their alternate angles are unequal. 166. Corollary. If two lines cut
by a transversal make
corresponding angles equal, or interior or exterior angles on the same side of the transversal supplemental, the lines are parallel.
For we know, by 113, that either makes also the alternate angles equal. Exercises.
draw a
line
given angle.
21.
From
of these
suppositions
a given point without a given
line,
making an angle with the given line equal to a
PARALLELS.
49
Problem IX. Through a given pointy
167.
to
draw a
line parallel to a
given
line.
/-?
Given, a line
Required, (using
to
for the
II
AB
Proof. ;
4.
By
I|
AB
AB take any point, as C. By 100, join PC, CP, by 164, make ^ CPD = 4. PCB.
In
P in the line
PD equal
P
line through
words "parallel to").
Construction.
At
and a point P.
draw a
is
II
AB.
construction, the transversal
PC
makes
CPD = ^ PCB, .-.
by 165,
PD
is
11
alternate angles
AB,
Exercises. 22. Draw, through a given point between two^ which are not parallel, a sect, which shall be terminated the given lines, and bisected at the given point. by
lines
23.
If
a line bisecting the exterior angle of a triangle beshow that the triangle is isosceles.
parallel to the base,
Sects from the middle point of the hypothenuse of right-angled triangle to the three vertices are equal. 24.
25.
Through two given points draw two
with a given 26.
bisects
lines, forming,, a triangle equiangular with a given triangle. are equal sects, drawn from the extremities of.
line,
AC, BD,
AB
on opposite sides of it, such that the angles BAC,. are together equal to a straight angle. Prove that
the sect
ABD,
a,
AB
CD.
THE ELEMENTS OF GEOMETRY,
50
Theorem XX. 1
68.
If a transversal cuts two parallels the alternate angles y
are equal.
/L
AB CD
Hypothesis.
Proof.
HKD =
be
parallel to
by
CD
H and K by transversal FG.
^ KHB.
through H, making alternate angles equal,
CD, by 165. By our assumption,
to
•*•
4.
A line
lel
cut at
H
Conclusion.
99, two different lines through
is
paral-
H cannot
both
;
H
31, the line through
makes But, by hypothesis,
|I
CD
is
identical with the line
which
alternate angles equal.
AB
is
parallel to
.-.
^
KHB =
CD through i7,
^ HKD,
If a transversal cuts two parallels, it 169. Corollary I. makes the alternate angles equal; and therefore, by 113, the corresponding angles are equal, and the two interior or two exterior angles on the same side of the transversal are supple-
mental. 170.
two
parallels,
171.
a line be perpendicular to one of will be perpendicular to the other also.
Corollary it
II.
If
CoNTRANOMiNAL OF
equal, the two lines meet.
i68.
If alternate
angles are un-
PARALLELS.
51
So, if the interior angles on the same side of the transversal are not supplemental, the two lines meet ; and as, by 143, they cannot meet on that side of the transversal where the two interior angles are greater than a straight angle, therefore they must meet on the side where the two interior angles are together less
than a straight angle. 172.
allel to
CoNTRANOMiNAL OF 99. Lines in the same plane parthe same line cannot intersect, and so are parallel to
one another.
Theorem XXI. 173.
the
Each
ABC
Conclusion.
^ CBD = ^A
CBF, and by 169, .'.
any A, wih side
From B, by
by adding, ^
Exercises.
167,
^A = A
2y,
-{-
sum of
angles.
Hypothesis.
Proof.
^
exterior angle of a triafigle is equal to the
two interior opposite
produced
io
D,
^C.
}-
draw
AB
BF
\\
AC.
Then, by 168,
^C
^ DBF;
^C
=
^
Each angle
two-thirds of a right angle.
FBD of
-}-
^
CBF =
4.
CBD.
an equilateral triangle
Hence show how
is
to trisect a right
angle. 28.
If
any
of the angles of
an isosceles triangle be twomust be equilateral.
thirds of a right angle, the triangle
THE ELEMENTS OF GEOMETRY.
52
Theorem XXII. 174.
equal
to
The sum of the three a straight angle.
Hypothesis.
Conclusion.
Proof.
By
173,
Add .-.
175.
to
ABC 4.
i.B
is
+^^+^C=st. ^.
side, as
CA,
to
D.
^-tC = ^DAB.
both sides
^ CAB
a triangle
any A.
CAB
Produce a
interior angles of
^-
4.
BAC.
4.B
Corollary.
+ ^C =
4.
DAB +
4.
BAC =
st.
^.
In any right-angled triangle the two
acute angles are complemental.
TRIANGLES.
CHAPTER
53
VII.
TRIANGLES.
Theorem XXIII. 176. site side
Two ifz
triangles are congruent if two angles and the one are equal respectively to two angles
an
oppothe
and
corresponding side in the other.
ABC
Hypothesis.
and
DFG
As, with
^A=
^C = AB = Conclusion.
Proof.
By
By
^ Z>, ^G, DF.
t^ABC^ t. DFG. ^^-f-^^ + ^C =
174,
hypothesis,
4.
(89. If equals
A
-^
4.C
=^
4
D
-\-
st.
^ =
4.
D^
-4.
F
-\-
4.G,
be taken from equals, the remainders are equal.) .-.
(128. Triangles are congruent
A if
ABC ^
A DFG,
two angles and the included side are equal
in each.)
THE ELEMENTS OF GEOMETRY.
54
Theorem XXIV. 177'
tively to
If i'^o triangles have two sides of the one equal two sides of the other^ and the angles opposite
respecto
one
pair of equal sides equals theti the angles opposite to the other pair of equal sides are either equal or supplemental.
The angles included by the equal sides must be either equal or unequal. Case I. If they are equal, the third angles are equal. {174.
The sum
of the angles of a triangle
is
a straight angle.)
Case II. If the angles included by the equal sides are unequal, one must be the greater. Hypothesis.
ABC
FGH as,
and
= AB = BC =
¥^
^ABC Conclusion.
Proof.
By
^C 164,
-f
^
make /.
(128. Triangles are congruent
/.
^=
with
^
^,
FG, GJI,
> 4.G. st.
^
.
ABD = ^.G, A ABD ^ A FGH, the
if
-^
two angles and the included side are equal in each.)
4.
BDA =
^
ZT,
and
BD =
GH,
TRIANGLES. But, by hypothesis,
BC — .'.
55
GH,
BC = ^ BDC =
BD, 4.C.
{126. In an isosceles triangle the angles opposite equal sides are equal.)
But
4.
BDA +
^
BDC =
st.
4,
^H
^
4.C
.'.
Corollary
^
St.
^.
two triangles have two sides of the 178. one respectively equal to two sides of the other, and the angles opposite to one pair of equal sides equal, then, if one of the I.
If
angles opposite the other pair of equal sides they are oblique but not supplemental, or
if
is if
a right angle, or the side opposite
the given angle is not less than the other given side, the angles are congruent.
tri-
Two right-angled triangles are conII. the gruent hypothenuse and one side of the one are equal respectively to the hypothenuse and one side of the other. Corollary
179.
if
On
the Conditions of Congruence of
180.
A triangle
Two
Triangles.
has three sides and three angles.
The
three angles are not all independent, since, whenever two of them are given, the third may be determined by taking their sum from a straight angle. In four cases
we have
proved, that,
if
three independent
parts of a triangle are given, the other parts are determined ; in other words, that there is only one triangle having those parts
:
—
(124) (128)
(129)
Two Two
sides
The
three sides.
Two
and the angle between them. angles and the side between them.
angles and the side opposite one of them. In the only other case, two sides and the angle opposite one (176)
THE ELEMENTS OF GEOMETRY.
56
if the side opposite the given angle is shorter than the other given side, two different triangles may be formed, each of which will have the given parts. This is called the
of them,
ambiguous case. Suppose that the side a and side r, and a> c, then, making the :i C, and cutting off
25^
C, are given.
CB
as center, and describing an arc with radius equal to c, it cut in two points and the two unequal triangles
CD
;
A^BC
If
= a, taking B
may
ABC
and
required conditions. In these the angles opposite the side are supplemental,
by
177.
will satisfy the
BC
Loci.
The aggregate
of all points and only those points a given condition, is called the Locz^s of points satisfy that condition. satisfying 181.
which
Hence, in order that an aggregate of points, L, may be properly termed the locus of a point satisfying an assigned condition, C, it is necessary and sufficient to demonstrate the fol182.
lowing pair of inverses (i)
If
(2)
If
We
know from If
—
a point satisfies C, a point is upon Z,
contranominal (3)
:
:
—
a point
is
18,
it is
it
upon L.
satisfies
C
that, instead of (i),
not upon Z,
it
we may prove
does not satisfy
C.
its
TRIANGLES. Also, that, instead of
(2),
57
we may prove
the obverse of
(I): (4)
If
a point does not satisfy Cy
it is
not upon L,
Theorem XXV. locus of the point to which sects from two given points are equal is the perpendicular bisector of the sect joining 183.
The
them*
ii
Hypothesis.
AB
Conclusion.
PC ±
Proof.
Then A
—7
a sect,
^5
C its mid-point PA = PB,
AB.
Join PC.
ACP
is
^
A BCP,
(129. Triangles with the three sides respectively equal are congruent.) /.
4.
ACP =
CP 184.
Conclusion.
Proof,
Two
4.BCP,
the perpendicular bisector of
AB.
Inverse.
Hypothesis.
(124.
is
a
P, any point on
PA —
CP 1. AB
at
its
mid-point C.
PB.
ACP ^ aBCP.
triangles are congruent
if
two sides and the included angle are equal each.)
in
THE ELEMENTS OF GEOMETRY.
58
Theorem XXVI. 185. The locus of points from which perpendiculars on two given intersecting lines are equals consists of the two bisectors of the angles betweeit the given lines.
Hypothesis.
Given
AB
and
Proof,
CD
Therefore
all
lines are equal, lie
and
P
any
1.
(179. Right triangles having hypothenuse
186.
intersecting at O,
PM A. AB ^ PN CD. ^ POM = ^ PON, a POM ^ A PON.
point such that Conclusion.
and one side equal are congruent.)
points from which perpendiculars on two intersecting on the bisectors of the angles between them.
Inverse.
Hypothesis.
P, any point on bisector.
PM = PN. POM ^ PON.
Conclusion. Proof,
a
(176. Triangles
having two angles and a corresponding side equal in each are congruent.)
Hence, by 110, the two bisectors of the four angles between AB and CD are the locus of a point from which perpendiculars on AB and
CD are
equal.
Intersection of Loci. 187. ditions,
Where it is we leave
if
required to find points satisfying two conout one condition, we may find a locus of
points satisfying the other condition.
TRIANGLES.
59
Thus, for each condition we may construct the corresponding locus and, if these two loci have points in common, these points, and these only, satisfy both conditions. ;
Theorem XXVII. 188. tJiree
There
and
is one^
givcfz points not on a
only onCy point are equal.
from which
sects to
lifie
Given A, B, and C, not in a line. Construction and Proof. By 183, every point to which sects from are equal, lies on DL/, the perpendicular bisector of AB. A and and C are equal is the locus of points to which sects from Again
Hypothesis.
B
B
:
the perpendicular bisector
FF\
Now, Diy and FF' intersect, since, if they were parallel, AB, which perpendicular to one of them, would be perpendicular to the other also, would be one line. Let them intersect in O. By by 170, and is
ABC
95, they cannot intersect again.
189. Thus,
and /.
Therefore, by 183,
O
is
OB =
OA,
OC = OB =
OA, OC.
on the perpendicular bisector of
^C
There-
fore
Corollary.
The
three perpendicular bisectors of the sides which sects to the three ver-
of a triangle meet in a point from tices of the triangle are equal.
THE ELEMENTS OF GEOMETRY.
60
Problem X. To find points from which perpe7tdiculars on three given which form a triangle are equal.
190.
lines
Given, a,
b,
c,
three lines interseciing
in
the
three distinct
points A, By C.
The locus of points from which perpendiculars on the a and b are equal consists of the two bisectors of the angles
Solution.
two
lines
between the
lines.
The
locus of points from which perpendiculars on the two lines b and c are equal, similarly consists of the two lines bisecting the angles between b and c. Our two loci consist thus of two pairs of Hnes. Each
one pair cuts each Hne of the second pair in one point ; so that four get points, O, O^, O2, O^, common to the two loci. 191. By 185, the four intersection points of the bisectors of the
line of
we
angles between a and the angles between a
Corollary.
b, and between b and and c. Therefore
The
c, lie
on the
six bisectors of the interior
bisectors of
and exterior
angles of a triangle meet four times, by threes, in a point.
POLYGONS.
CHAPTER
61
VIII.
POLYGONS. I.
Definitions.
A
Polygon is a figure formed by a number of lines of which each is cut by the following one, and the last by the 192.
first.
193.
The common
points
of
the
consecutive
lines
are
called the Vertices of the polygon. 194. The sects between the consecutive vertices are called
the Sides of the polygon.
195.
The sum
of the sides, a
eter of the polygon.
broken
line,
makes the Perim-
THE ELEMENTS OF GEOMETRY.
62
196. The angles between the consecutive sides and towards the enclosed surface are called the Interior Angles of the poly-
gon.
Every polygon has
197.
A
polygon
is
as
many
interior angles as sides. when no one of
said to be Convex
interior angles is reflex. 198. The sects joining
its
the vertices not consecutive are
called Diagonals of the polygon.
199.
another,
200.
another,
When it is
When it is
the sides of a polygon are
all
equal to one
called Equilateral,
the angles of a polygon are
called Equiangular,
all
equal to one
POL YGONS. 201. is
A
polygon which
is
63
both equilateral and equiangular
called Regular,
202. Two polygons are Mutually Equilateral if the sides of the one are equal respectively to the sides of the other taken the same order.
m
203. of the
Two polygons are Mutually Equiangular if the angles one are equal respectively to the angles of the other
taken in the same order.
204.
Two
polygons
may be mutually
equiangular without
being mutually equilateral. 205. Except in the case of triangles, two polygons may be mutually equilateral without being mutually equiangular.
THE ELEMENTS OF GEOMETRY.
64 '
2o5.
A
polygon of three sides
is
a Trigon or Triangle ; one
of four sides is a Tetragon or Quadrilateral ; one of five sides is a Pentagon ; one of six sides is a Hexagon; one of seven
a Heptagon ; one of eight, an Octagon ; of nine, a Nonagon of ten, a Decagon; of twelve, a Dodecagon ; of fifteen, a Quindecagon. sides
is
;
207.
The Surface
enclosed by 208.
A
its
of a polygon
is
that part
of the
plane
perimeter.
Parallelogram
is
a quadrilateral whose opposite
sides are parallel.
/ 209.
A
Trapezoid
is
II.
a quadrilateral with two sides parallel.
General Properties.
Theorem XXVIII. 210. If
two polygons be mutually equilateral and mutually
equiangular^ they are congruent. Proof. Superposition they may be applied, the one to the other, :
so as to coincide.
Exercises. 29. Is a parallelogram a trapezoid could a triangle be considered a trapezoid }
}
How
POL YGONS.
65
Theorem XXIX. The sum of the
211.
straight angles than
it
interior angles of
has
A polygon of n sides. Sum of ^^s = (n — 2)
Hypothesis. Conclusion.
we can draw
a polygon
is
two
less
sides.
st. 2^'s.
the diagonals from any one vertex with out cutting the perimeter, then we have a triangle for every side of the polygon, except the two which make our chosen vertex. Thus, we have
Proof.
(«
—
If
all
whose angles make the by 174, the sum of the angles
2) triangles,
But,
interior angles of the polygon. in
each triangle
angle,
Sum
.*.
212.
Corollary
sides are {n 213.
of ^'s in polygon
—
3)
=
From each
I.
(n
—
2)
st.
is
a straight
^'s.
vertex of a polygon of n
diagonals.
Corollary
II.
The sum
of the angles in a quadrilat-
eral is a perigon.
214
gon of
Corollary 71
sides is
(n
III.
—
Each angle
2) St.
^'s
of an equiangular poly-
THE ELEMENTS OF GEOMETRY.
66
Theorem XXX. 215.
hi a convex polygo7tf the sum of the exterior angle s, one made by producing each side i7t order, is a perigon.
at each vertex^
Hypothesis.
A convex polygon of n
sides,
each
in
order pro-
duced at one end. Conclusion.
Proof. angle, as
^
Every
by
=
of exterior ^^s angle, as
interior
^, together
=
st.
^
2^
perigon.
A, and
its
+ all exterior ^ = = (n — 2) 2^'s. 211, all interior = a perigon. sum of exterior ^ 's
all
Exercises. 30. polygon of n sides ?
How many
n
St.
^ 's.
st.
2j^'s
.*.
adjacent exterior
.
all interior 2^ 's
.*.
But,
Sum
's
diagonals can be drawn in a
31. The exterior angle of a regular polygon is one-third of a right angle find the number of sides in the polygon. 32. The four bisectors of the angles of a quadrilateral form :
a second quadrilateral whose opposite angles are supplemental. 33. Divide a right-angled triangle into two isosceles triangles. 34.
of the
In a right-angled triangle, the sect from the mid-point
hypothenuse to the right angle
is
half the hypothenuse.
POL YGONS.
III.
67
Parallelograms.
Theorem XXXI. If two opposite sides of a quadrilateral are equal and parallel^ it is a parallelogram. 216.
Hypothesis.
ABCD
Conclusion.
AjD
Proof.
AC.
Join
a quadrilateral, with
Then
(168. If a transversal cuts
By
hypothesis,
AB =
and
\\
CD,
BC.
\\
AB =
^
A CD.
parallels, the alternate angles are equal.)
CD, and ACis common,
/. (124. Triangles are congruent
^BAC =
two
if
.-.
^
BCA =
^ CAD,
two sides and the included angle are equal
BC
I!
in each.)
AD.
(165. Lines making alternate angles equal are parallel.)
Exercises. 35. Find the number of elements required to determine a parallelogram.
The
four sects which connect the mid-points of the consecutive sides of any quadrilateral, form a parallelogram. 36.
37. The perpendiculars let fall from the extremities of the base of an isosceles triangle on the opposite sides will include
an angle supplemental to the vertical angle
BE
^
of the triangle.
CE
bisects the angle of a triangle ABC, and bisects the exterior angle the is equal to one-half CD, angle 38.
If
A
the angle A.
E
THE ELEMENTS OF GEOMETRY.
68
Theorem XXXII. The
217.
equal
and angles of a parallelogram and each diagonal bisects it.
opposite sides
one another^
to
AC
Hypothesis. Lei be a diagonal of n? £7 for the word " parallelogram "). Conclusions. CD.
4.
AB = BC = DAB =
^B
=:
ABC ^ ^BAC = A
Proof.
CAD=
4
ABCD
are
(using the sign
DA. 4.
BCD.
4.D.
A CDA. 4. ACD, and -^BCA)
(168. If a transversal cuts parallels, the alternate angles are equal.)
therefore, adding,
4
BAC +
CAD = 4 BAD = 4 ACD + 4 ACB = 4 BCD. side A C included between the equal angles common,
4.
Again, the
is
aABC^
.-.
(128. Triangles are congruent
AB =
.-.
Exercises.
If
a CDA,
two angles and the included side are equal
BC =
CD,
39.
if
DA,
and
4
B^
one angle of a parallelogram be
in each.)
4 D. right, all
the angles are right. of the one equal are to one angle of the other, the parallelograms mutually equi40.
If
angular.
two parallelograms have one angle
POLYGONS. 218.
Corollary
I.
Any
69
pair of parallels intercept equal
sects of parallel transversals.
219.
Corollary
II.
If
two
lines
be respectively
parallel
lines, any angle made by the first pair is equal or supplemental to any angle made by the second pair. If two angles have their arms respec220. Corollary III.
to
two other
or supplemental. tively perpendicular, they are equal
For, revolving one of the angles through a right angle
around its vertex, its arms become perpendicular to their traces, and therefore parallel to the arms of the other given angle.
THE ELEMENTS OF GEOMETRY.
70
Corollary
221.
gram
IV.
a right angle,
is
If
all
its
one of the angles of a paralleloangles are right angles, and it is
called a Rectangle,
Corollary V. are equal,
gram Rhombus.
223.
all
its
If
two consecutive sides of a paralleloand it is called a
sides are equal,
A Square is a rectangle having consecutive sides equal.
224. Both diagonals, ^Cand BD, being drawn, it may with a few exceptions be proved that a quadrilateral which has any two of the following properties will also have the others :
1.
2.
3.
AB CD. BC r>A. AB = CD. BC = DA. DAB = ^ BCD. ABC = 4 CDA. The bisection of ^C by BD. II
II
4. 4.
10.
The The The
bisection of
BD by AC.
bisection of the
^hy AC.
bisection of the
/07
by
BD.
2>
POLYGONS. These
ten
71
in pairs will give forty-five pairs ; with may be required to establish any of the
combined
each of these pairs
it
eight other properties, and thus three hundred and sixty questions respecting such quadrilaterals may be raised.
For example, from
i
/.
Two
(128.
and
A
triangles are congruent
2,
217 proves
ABE ^ if
3,
A CDE,
two angles and the included side are equal
in
each.)
the diagonals
.*.
The
of a parallelogram
bisect each other.
inverse is to prove i and 2 from 7 and 8 the diagonals of a quadrilateral bisect each other, If :
the
a parallelogram. figure Since 225. by 170 two lines perpendicular to one of two parallels are perpendicular to the other, and by 166 are parallel, and so by 218 the sects intercepted on them are equal, thereis
fore
all
perpendicular sects between two parallels are equal.
Exercises. 41. If the mid-points of the three sides of a triangle be joined, the four resulting triangles are equal. 42. If the diagonals of a parallelogram be equal, the parallelogram 43. it is
is
a rectangle.
the diagonals of a parallelogram cut at right angles,
a rhombus.
44. is
If
If
the diagonals of a parallelogram bisect the angles,
it
a rhombus. 45.
If
a square.
the diagonals of a rectangle cut at right angles,
it is
THE ELEMENTS OF GEOMETRY.
72
Theorem XXXIII. 226.
The three perpendiculars from a point.
the vertices of
a triangle
to the opposite sides vieet in
Hypothesis.
Lei DLf, EE',
from the vertices D, E, F,
DU
be the three perpendiculars
FF', meet in a point O. By 167, through D, E, F, draw AB, BC,
Conclusion.
Proof.
EE'
FF\
,
A DEF.
to the opposite sides of
,
CA
\\
FE, DF,
DE. Then
By
the figures
In same way, But, since
AD = FE =
.'.
D
E and F
is
DB, mid-point of AB.
are mid-points oi
AC
are respectively
they must be
A line perpendicular to .'.
are parallelograms.
..
DI/, EE', FF', :.
(170.
ADFE, DBFE,
217,
± AB,
and BC.
EF, FD, DE,
AC, and BC,
one of two parallels
by 189, they meet
±
is
perpendicular to the other.)
in a point.
46. The intersection of the sects joining the of opposite sides of any quadrilateral is the midmid-points sect of the point joining the mid-points of the diagonals.
Exercises.
POL YGONS.
73
Theorem XXXIV. If three or more parallels intercept equal sects on one tratisversal^ they intercept equal sects on every transversal. 227.
^—Z
Hypothesis.
AB = BC = CD
=,
etc.,
are sects on the trans-
AC, intercepted by the parallels a, b, c, d, etc. FG, GBy BK, etc., are corresponding sects on any
versa/
other trans-
versal.
Conclusion.
Proof. parallel to
AD
=,
etc.
draw, by 167,
FL,
GM, and HN,
all
;
they are
.-.
FG = GH = HK
From F, G, and Hy all
equal, because
AB
z=z
BC — CD
=,
etc.
(217. Opposite sides of a parallelogram are equal.)
But
= ^ HGM = ^ KHN, and = 4 GHM = t ^KN, 4. FGL
^ GFL
(169. If a transversal cuts .-.
A
two
parallels, the
FGL ^
(176. Triangles are congruent
if
A
corresponding angles are equal.)
GHM ^
A HKN,
they have two angles and a corresponding side
equal in each.) /.
HK.
Corollary. The intercepted part of each parallel from the neighboring intercepts by equal sects.
228. differ
FG= GH ^
will
THE ELEMENTS OF GEOMETRY,
74
Theorem XXXV. 229. The line drawn through the mid-point of one of the nonparallel sides of a trapezoid parallel to the parallel sidesj bisects the remaining side.
Hypothesis.
In the trapezoid
ABCD, BF = FC,
and AB
II
CD
WFG, Conclusion.
AG —
GD.
intercept equal sects on any transversal, they intercept equal sects on every transversal.
Proof.
230.
By
227,
if parallels
Corollary
I.
The
line joining the mid-points of the
non-parallel sides of a trapezoid is parallel to the parallel sides, for we have just proved it identical with a line drawn parallel to
them through one mid-point. 231.
Corollary
of a triangle a line bisect the third side
If through the mid-point of one side II. be drawn parallel to a second side, it will and, inversely, the sect joining the mid-
;
points of any two sides of a triangle
and equal 232.
A
to half of
sect from
is
parallel to the third side
it.
any vertex of a triangle Medial of the
of the opposite side is called a
to the mid-point triangle.
POLYGONS. 233.
Three or more
lines
75
which intersect
in the
same point
are said to be Concurrent. 234.
Three or more points which
lie
on the same
line are
said to be Collinear,
Theorem XXXVI. 235.
The
three medials of of each.
a triangle are concurrent in a
trisection point
Proof. mid-point of
By in
Let two medials,
OA, and
231, in
Hoi
A ABO,
AD
OB,
GH
is
and Join
BE, meet in O, Take G GH, HD, DE, EG.
equal and parallel to
\AB
\
so
is
the
DE
A ABC', :.
But,
by 216,
GHDE
is
a parallelogram.
by 224, the diagonals of a parallelogram /.
AG= GO ^
OD,
So any medial cuts any other
and
bisect each other,
BH ^ HO =^
at its point of trisection
vertex, /,
the three are concurrent.
OE.
remote from
its
THE ELEMENTS OF GEOMETRY.
76
The
intersection point of medials is called the Centroid The intersection point of perpendiculars from of the triangle. 236.
the vertices to the opposite sides is called the Orthocenter of the The intersection point of perpendiculars erected at
triangle.
the mid-points of the sides
is
called the Circumcenter of the
triangle.
IV. Equivalence.
Two plane figures
237.
are called equivalent if
we can prove
must contain the same extent of surface^ even if we do how to cut them into parts congruent in pairs. The base of a figure is that one of its sides on which
that they not show 238.
we imagine
it
to rest.
Any side of a figure may be taken as its base. 239. The altitude of a figure is the perpendicular highest point to the line of its base. So the altitude of a parallelogram
dropped from any point of one side
is
from
its
the perpendicular
to the line of the opposite
side.
240.
The words
a triangle
"
may have
altitude
"
" and " base are correlative
three different altitudes.
;
thus,
POL YGONS.
77
Problem XI. 241.
To describe a square upon a given
Given,
M©
seci
AB.
Required, to describe a square on AB. Construction. By 135, from A draw
By By By
133, in
AC
make
AC 1. AB. AD = AB. B>F
||
AB.
draw BI^
\\
AD.
167, through Z> draw
167, through
AF will be the required square. Proof, By construction, AF AB = FD, = ^ ^, 7^
is
4^:
B
a parallelogram,
and
and
AD = ^^ =
(217. In a parallelogram the opposite sides
But,
by construction,
-\Again, ^ two parallel
(169. If
^B
^i?;
^
Z).
and angles are equal.)
AB — AD, AF equilateral. is
.'.
A
sect.
=^
st. •^.
lines are cut
by a transversal, the two
interior angles
are
supplemental.)
But,
by construction,
^^ .-.
and
all
four angles are
242. it is
4.A
= ^B,
rt.
Corollary.
a rectangle.
is rt.,
If
a parallelogram have one angle right,
THE ELEMENTS OF GEOMETRY,
78
Theorem XXXVII. 243.
nuse
is
In any right-angled triangle the square on the hypotheequivalent to the sum of the squares of the other two y
sides.
\J^
ya Hypothesis,
Conclusion.
a ABC,
right-angled at B, -\- square on Square on
AB
^C =
By 241, on hypothenuse A C, on the
Proof.
describe the square ADFC. On the greater of the other two sides, as
= 4.
AB.
Join
Translate the
on
position of
sect
A
ABC
CF,
call
CG
until
=
CAB =
upward, keeping point
A
AD, and
A
F\
A
ABC ^ is
\
A CGF,
on D, and
C
oxi
on
sect
then
call
B' the
B.
Likewise translate
then
133, lay off
CG, and ^
,\
C
BC, by
FG.
Then, by construction, CA = FC, and AB FCG, since each is the complement of A CB
point
AC. A ABC,
square on
side toward the
CGF
to the right until
G^ the position of G,
C
is
on A^ and
F on £>
j
POL YGONS. The
79
AG'DB'FGBA,
resulting figure,
be the squares on the
will
AB and BC.
other two sides, For, since the
sum of
.*.
the angles at Z>
G'D and DB'
=
st. :^,
are in one line.
GB to meet this line at H. Then GF = FB' = BC, and
Produce
.*.
Again,
G'A
GFB'H
=
AB,
.*.
AG'HB
.-.
sq.
B'
=
=
equals square
^ GFB' on BC,
and
=
4.
4 ABH,
is
rt.
t\. 4.
^
6^'
the square
oiAC=
sq. of
G'AB
=
4.
FGB,
on ^^,
^^ +
Corollary.
sq.
of
^C
Given any two squares placed side by and BC in line to cut this figure into three pieces, two of which being translated without rotation, the figure shall be transformed into one square. 244.
side,
with bases
AB
;
K
THE ELEMENTS OF GEOMETRY.
80
Theorem XXXVIII.
Two
245. a^igular^
triangles are congruent if they are mutually equialtitudes equaL
and have corresponding
Hypothesis, tude
BH =
a5
altitude
a
Conclusion,
Proof.
Rt.
A
ABC
and
BDF mutually equiangular,
and
alti-
GB,
ABC ^
BDG ^
A BDF.
rt.
a BCH,
when two
angles and a side in one are equal to two angles and a corresponding side in the other.)
(176. Triangles are congruent
.'.
(124. Triangles are congruent
A
BD =
BC,
ABC ^
A BBF.
when two
angles and the included side are equal in
each.)
Exercises. 47. If from the vertex of any triangle a perpendicular be drawn to the base, the difference of the squares
on the two sides of the triangle is equal to the difference squares on the parts of the base.
Show how
of the
to find a square triple a given square. 49. Five times the square on the hypothenuse of a rightangled triangle is equivalent to four times the sum of the 48.
squares on the mcdials to the other two sides.
POL YGONS,
8l
Theorem XXXIX. 246.
equal
to
If two consecutive sides of one rectangle be respectively two consecutive sides of another, the rectangles are con-
gruent.
Two
Hypothesis.
^s rt
c
a
3£
ABCD
quadrilaterals
AB = FG, BC = ABCD ^ FGHK. AB CD, and AD \\BC,
js
and
FGHK,
wiih
all
GH,
Conclusion.
Proof. (166.
K
\\
interior angles
on same side of transversal are supplemental,
lines are
parallel.)
/.
a rectangle
/.
AB =
CD,
is
a parallelogram,
and
AD =
BC.
(217. Opposite sides of a parallelogram are equal.)
In the same way, .-.
(210. If
FG = BK, and GH = ABCD ^ FGHK,
FK,
two polygons be mutually equilateral and mutually equiangular, they are congruent.)
247.
Corollary.
two consecutive sides
A rectangle is ;
so
if
two
completely determined by sects, a and b, are given, we
Ctr
of the rectangle of a and b, or we may call it the It can be constructed by the method given in rectangle ab. 241 to describe a square.
may speak
THE ELEMENTS OF GEOMETRY.
82
Theorem XL.
A parallelogram
248.
and
is
equivalent to the rectangle of its base
altitude.
c r
Hypothesis.
AF =
ABCD any £y,
of which side
3
AD is taken as
base;
altitude of £y.
Conclusion.
Proof.
^ABCD =
rt.
^AFGD.
AF ^ DG, and AB =
DC.
(217. Opposite sides of a parallelogram are equal.)
By
^
construction,
.-.
(179.
G and ^ F are t.ABF^ aDCG. rt.,
Right triangles are congruent
if
hypothenuse and one side are respectively
equal in each.)
ABGD
the trapezoid take away A the same trapezoid take the equal isleftthert. z:7^i^6^Z>.
From
£7
DCG,
ABCD. From
/. (89. If
^ABCD ^
rt.
and then
is
left
A ABFj and there
^ AFGD.
equals be taken from equals, the remainders are equal.)
All parallelograms having equal bases 249. Corollary. and equal altitudes are equivalent, because they are all equivalent to the same rectangle.
Exercises.
50.
base and on the parallels.
Equivalent parallelograms on the
same
side
of
it
are
same
between the same
POL YGONS. Exercises.
51.
83
Prove 248 for the case when
(7
and
F coin-
cide.
through the vertices of a triangle lines be drawn and produced until they meet,
If
52.
parallel to the opposite sides,
the resulting figure will
contain
three
equivalent
parallelo-
grams.
On
53.
the
same base and between the same
parallels as a
given parallelogram, construct a rhombus equivalent to the parallelogram. 54.
Divide a given parallelogram into two equivalent paral-
lelograms. 55. Of two parallelograms between the same parallels, that the greater which stands on the greater base. Prove also an inverse of this. is
Equivalent parallelograms situated between the same have equal bases.
56.
parallels
Of parallelograms on equal
57.
which has the greater 58.
A trapezoid sum
is
bases, that
the greater
is
altitude.
equivalent to a rectangle whose base is sides, and whose altitude is the
two parallel perpendicular between them. half the
59.
The
of the
sect joining the mid-points of the non-parallel sides
of a trapezoid is half their 60.
BC,
If
sum.
E and F are the mid-points of the
of a parallelogram
ABCD,
the lines
opposite sides,
BEy DFy
ADy
trisect the
diagonal AC. 61.
Any
line
drawn through the intersection
of the diag-
onals of a parallelogram to meet the sides, bisects the surface. 62. The squares described on the two diagonals of a rhom-
bus are together equivalent to the squares on the four sides. 63. Bisect a given parallelogram by a line passing through any given point. 64. In 244, what two rotations might be substituted for the
two translations
?
THE ELEMENTS OF GEOMETRY.
84
Theorem XLI. Alternative Proof. 250. All parallelogra7ns having equal bases tudes are equivalent.
and equal
alti-
zrC
2).
Two ^s with equal bases and equal aliiiudes. They can be cut into parts congruent in pairs.
Hypothesis.
Conclusion.
Construction. Place the parallelograms on opposite sides of their coincident equal bases, AB. Produce a side, as FB, which when continued will enter the other parallelogram. If it cuts out of the parallel-
H
CD
at before reaching the side opposite AB, then will the other cutting side, as CB, when produced, also leave the £j before reaching the side that is, the point, K, where ; opposite
ogram
ABFG
AB
FG
CB
cuts the line
For, through
Then,
As
in
A G will
and side
A G.
\\
AB,
ABH and ABK 2j:
(i68. If
be on the sect
H and K draw HL and KM = ^
I
I,
2^ 2
=
2j:
2,
a transversal cuts two parallels, the alternate angles are equal.)
AB
is
common
;
.-.
A
ABH ^
A ABK,
POL YGONS.
85
AHB < altitude of ^yABCD, altitude of A ABK < altitude of £j ABFG, and K Now, A ABH ^ A BBL, and A ABK ^ A BKM. But altitude of A
.*.
is
(217.
The diagonal
of a parallelogram
makes two congruent
on
A G.
sect
triangles.)
Taking away these four congruent triangles, we have left two paraland KMFG, with equal bases, and^ equal but diminTreat these in the same way as the parallelograms first
HLCD
lelograms, ished altitudes.
given
and so continue
;
CRN,
until a
produced
side, as
FRQ^BXid
so the other,
also, reaches the side opposite the base before leaving the paral-
lelogram.
Then, as before, the As but
now we know
FRN and QRC
by 241, they are congruent,
.-.
FN = GN =
.-.
.-.
Also,
are mutually equiangular;
their corresponding altitudes are equal.
QR ^ RF ^
PG,
CQ,
DQ.
NR = RC
and
=^
PD,
= ^FRP = ^RQD, 4.G = ^ CRP = RNG, 4.D 4 DPR = ^ PRN, 4.
4.GPR = 2fPRQ; and therefore the remaining trapezoids, (210.
If
PRQD ^ PRNG.
two polygons be mutually equilateral and mutually equiangular, they are congruent.)
251.
Corollary.
Since a rectangle
therefore a parallelogram
is
is a parallelogram, the equivalent to rectangle of its
base and altitude.
KMFG
Exercises. 65. How do you know that HLCD and have equal altitudes } 66. How do you know, that, if Q is on the sect CD, on the sect FG ?
N
is
THE ELEMENTS OF GEOMETRY.
86
Theorem XLII.
A
252.
base triangle is equivalent to half the rectangle of its
and altitude.
a ABC, a AB C
Hypothesis,
Proof.
in
.'.
^AFBC =
altitude
BD. \\
CA,
aACB ^ aBFA.
The diagonal
rectangle of
A parallelogram is
.*.
and
F;
(217.
(248.
AC
half the rectangle oi \\
meeting AF
But
with base
—
AC and BD, and draw AF A CB, through B draw BF Through
Conclusion,
of a parallelogram bisects
it.)
AC and BD,
equivalent to the rectangle of
AABC=^i ^AFBC =
\
base and altitude.)
its
rectangle oi
AC
and
BD,
All triangles on the same base having 253. Corollary I. their vertices in the same line parallel to the base, are equivalent. 254. Corollary II. Triangles having their vertices in the
same
point,
and
for their bases equal sects of the
same
line,
are equivalent.
Corollary III. If a parallelogram and a triangle be the same base and between the same parallels, the paralupon lelogram is double the triangle. 255.
Exercises. equal altitudes.
6^.
Equivalent triangles on equal bases have
POL YGONS.
Zj
Theorem XLIII. If through any point on the diagonal of a parallelogram drawn parallel to the sides, the two parallelograms one on each side of the diagonal, will be equivalent. 256.
two
lines be
y
P
Hypothesis.
and
HK lines
P
BC
||
the four sides in the points F,
n?
The diagonal
(217.
From A
G, H, K,
AKPG ^ a PFCH, a ABD - a BCD. A KBP = A BFP, A GPD = A PHD,
Conclusion, Proof,
ABD
take away
of a parallelogram bisects
A
KBP
.-.
(89. If equals
it.)
and A
AKPG. From the equal A B CD take PBD, and we have left the £y PFCH,
cj
GPD, and we have left away the equal A s BFP and
^AKPG=^PFCH.
be taken from equals, the remainders
will
be equal.)
KBFP BD
257. In figures like the preceding, parallelograms like are called parallelograms about the diagonal
GPHD while AKPG and
^
BD
of ABCD; FG any point on diagonal AB and respectively, and meeting
ih rough
z:7S
and
PFCH
allelograms about the diagonal
are called
complements
;
of par-
BD.
68. When are the complements of the paralabout a lelograms diagonal of a parallelogram congruent ?
Exercises.
THE ELEMENTS OF GEOMETRY.
SS
V. Problems.
Problem XII. 258.
atighy
To describe a parallelogram equivalent
and having an angle equal
JO
Given,
a
ABC
Required, angle = }f. G.
a given
tri-
JL
=
a parallelogram
Bisect
to
a given angle.
and ^ G.
to describe
Construction.
to
AC mD. ADF =
At Dy by 164, make ^ By 167, through A draw By 167, through B draw
AH
:^
\\
BFH
A ABC,
while having an
G.
DF. \\
CA,
AHFD will be the parallelogram required. Proof. Join DB, Then A ABD = aBCD, same
(254. Triangles having their vertices in
the
/.
But
cjAHFD is also
(255. If a parallelogram
same
same
A
line,
ABC
double
point,
is
double
sects of
A ABD,
A ABD,
and a triangle be upon the same base and between the
parallels, the parallelogram is
n7AHFD = ^ ADF = ^ 6^.
/.
and, by construction,
and for bases equal
are equivalent.)
double the triangle.)
t^ABC,
POL YGONS.
89
Problem XIII. a given
259. Oft
sect as base, to
describe a parallelogram
and having an
equivalent to a givejz triangle^
aiigle
equal
to
a
given angle.
X
AB, a C£>F, ^
Given, ihe sect
Required,
an angle
=
^
to describe
HBL =
}^
on
G.
AB a parallelogram = A CDF, while having
G.
Construction.
^
/V
Q
.^.
By
G, and place
Produce KL.
Draw
^ (169. If a transversal cuts
make nj
258, it
so that
AM HKM \\
two
BL. -f-
4.
parallels,
it
BHKL —
BH
is
Join
on
A CDF^ and
line
having
AB produced.
MB.
KMA =
St. 4.,
makes the two
interior angles supple-
mental.) .-.
Therefore
T^HKM + ^KMBKst.^.
KH and MB meet
(171. If a transversal cuts
two
if
lines,
produced through and the
H and B.
interior angles are not supplemental,
the lines meet.)
Let them meet in Q.
LB and MA
to
meet
Through Q draw QN'm F and N. .'.
(256.
Complements
/.
260.
QN
\\
HA, and
produce
^ NPBA = ^ BHLK,
of parallelograms about the diagonal are equivalent.)
^NPBA =
Corollary.
A CFD,
and
4.
ABP =
4 G.
Thus we can describe on a given base a
rectangle equivalent to a given triangle.
THE ELEMENTS OF GEOMETRY.
90
Problem XIV. 261.
To
describe a triangle equivalent to a given polygon.
D
Given, a polygon ABCDFG. Required, to constrtcct an equivalent triangle. Construction. Join the ends of any pair of adjacent
and BC, by the sect CA. Through the intermediate produced in H. Join HC. Polygon
vertex,
ABCDFGA =
B, draw a hne polygon
y
sides, as
AB
CA^ meeting
GA
BCVFGB,
and we have obtained an equivalent polygon having fewer Proof.
aABC=
(253. Triangles
Add
to
sides.
a ABC.
having the same base and equal altitudes are equivalent.)
each the polygon .'.
ACDFGA,
ABCDFGA = BCDFGB.
In the same way the number of sides until reduced to three.
may be
still
further diminished
by one
262. Corollary I. Hence we can describe on a given base a parallelogram equivalent to a given polygon, and having an angle equal to a given angle.
Thus we can describe on a given 263. Corollary II. base a rectangle equivalent to a given polygon. To compare the surfaces of different poly264. Remark.
we need only to construct rectangles equivalent to the given polygons, and all on the same base. Then, by comparing the altitudes, we are enabled to judge
gons,
of the surfaces.
POL YGONS.
91
VI. Axial and Central Symmetry.
two figures coincide, every point A in the one coinThese points are said to cides with a point A' in the other. 265. If
correspond.
two congruent figures there and one, one, point in the other those points only corresponds " " which coincide if one of the two called corresponding being
Hence
to every point in one of
;
superimposed upon the other. Hence, calling those parts corresponding which coincide if the whole figures are made to coincide, it follows, that corresponding parts of congruent figures are themselves congruent.
figures
is
Symmetry with Regard to an
Axis.
we start with two figures in the position of coinciand take in the common plane any line x^ we may turn dence, the plane of one figure about this line x until its plane, after 266. If
half a revolution, coincides again with the plane of the other figure.
The two in the
figures themselves will then have distinct positions but they will have this property, that they ;
same plane
THE ELEMENTS OF GEOMETRY.
92
can be made to coincide by folding the plane over along the line X.
Two figures in the same plane which have this property are said to be symmetrical with regard to the line x as an axis of symmetry.
Symmetry with Regard to a Center. 267. If
common plane of two coincident we X^ may turn the one figure about this
we take
in the
figures any point point so that its plane slides over the plane of the other figure without ever separating from it.
turning be continued until one line to JT, and therefore the whole moving figure, has been turned through a straight angle about X.
Let
this
Then the two congruent figures still lie in the same plane, and have such positions that one can be made to coincide with the other by turning it in the plane through a straight angle about the fixed point X.
Two
figures
which have
this property are said to be symas center of symmetry.
metrical with regard to the point
268.
Any
X
single figure has axial
symmetry when
it
can be
divided by an axis into two figures symmetrical with respect to that axis, or has central symmetry when it has a center such
drawn through
cuts the figure in two points symmetrical with respect to this center.
that every line
it
POL YGONS.
93
Theorem XLIV. 269. If a figure has two axes of symmetry perpeiidicidar to each othery then their intersection is a center of symmetry.
^ 'O
^:
For,
if
X and y be two
axes at right angles, then to a point
correspond a point A' with regard to
To
these will correspond points
These points A^ and A/j
To
see
this, let
us
first
will
x
A^ and
correspond to
fold over along
A will
as axis.
y;
A/
with regard to
j
as axis.
each other with regard to x. then A falls on A,, and A'
on A/.
we now, without
folding back, fold over along x, A, and with it on which coincides with A/. Ai, A\ At the same time OA and OA/ coincide, so that the angles A Ox and A/ Ox' are equal, where x' denotes the continuation of x beyond O. It follows, that A OA/ are in a line, and that the sect AA/ is bisected at O, or 6> is a center of symmetry for AA/, and similarly for A^ and A\ If
will fall
BOOK
II.
RECTANGLES.
A Contimions Aggregate is
an assemblage in which two same have the boundary. adjacent parts Discrete or Discontinuous Aggregate is one in which 271. two adjacent parts have different boundaries. 270.
A
A pile of
cannon
We
balls is a discrete aggregate.
know
that any adjacent two could be painted different colors, and so they have direct independent boundaries.
Our
fingers are a discontinuous aggregate. 272. All counting belongs first to the fingers.
" " implied and bound up in the word number the assumption that a group of things comes ultimately to the same finger, in whatever order they are counted. 273.
There
is
This proposition *•
is
involved in the meaning of the phrase
distinct things."
Any
one and any other of them make two.
attached to two of
be
If
they are
fingers in a certain order, they can also attached to the same fingers in the other order. Thus, one
my
order of a group of three distinct things can be changed into any other order while using the same fingers, and so on with a
group of
four, etc. 95
THE ELEMENTS OF GEOMETRY.
96
By generalizing the use of the fingers in counting, has made for himself a counting apparatus, which each one carries around in his mind. This counting apparatus, the natural series of numbers, was made from a discrete aggregate, 274.
man
and so
only correspond exactly to discrete aggregates. In row of shot, we can find between any two, only a a 275. will
number
of others, and sometimes none at all. in so row of six shot Just regard to any two numbers. can be divided into two equal parts but the half, which is finite
A
;
three,
we cannot
divide into two equal parts
:
and so
in a series
of numbers. 276. But in 136 we have shown how any sect whatever may be bisected, and the bisection point is the boundary of both
So a
parts.
something numbers. 277.
line is not a discrete aggregate of points. It is in kind from the natural different series of totally
The
science of numbers
is
founded on the hypothesis
The science of space is founded of the distinctness of things. on the entirely different hypothesis of continuity. 278. Numbers are essentially discontinuous, and therefore unsuited to express the relations of continuous magnitudes. 279. In arithmetic we are taught to add and multiply numbers
:
we
will
now show how
the laws for the addition and
these discrete aggregates are applicable to sects, which are continuous aggregates. multiplication of
THE COMMUTATIVE LAW FOR ADDITION. 280. In a sum of numbers we may change the order in which the numbers are added. If X and y represent numbers, this law is expressed by the
equation
X
-\-
y
=
y
\' X,
RECTANGLES.
97
It depends entirely on the interchangeability of of the units of numeration.
any pair
281. The sum of two sects is the sect obtained by placing them on the same line so as not to overlap, with onfe end point in common.
-I
1
t
Thus, the sum of the sects a and b means the sect AC^ which can be divided into two parts,
AB = 282.
BC =
and
a,
The commutative law holds a
,
-^^
b
—
f
3L
for the
b,
summation
of sects.
b -^ a,
.±^ S C
A AC=
CA';
AC
revolved through a straight angle may be superimposed upon C^A\ and will coincide point for point. The more general case, where three or more sects are
for
added, follows from a repetition of the above. Thus, the commutative law for addition in geometry depends entirely on the possibility of motion without deformation. .
THE ELEMENTS OF GEOMETRY.
98 283.
The sum
of
superimposing two
two rectangles is the hexagon formed by and bringing the bases into the same
sides,
line.
Thus,
if
two adjacent sides
the other c and
the sum
one are a and
and of of the rectangles ab and cd is of
b,
ABCDFGA, 284.
angles
;
The commutative law holds for the addition of rectis, the sum is independent of the order of summa-
that
tion.
C,
ab for
ABCDFGA
AB'CD'F'CA\
^
•\-
turned
and
cd
=.
over
cd
-\-
ab\
may be superimposed upon
will coincide
we can
C
3",
,J3
with
it.
describe on a given base a rectangle Since, by 263, to a equivalent given polygon, the more general case, where three or more rectangles are added, follows from a repetition of the above.
RECTANGLES.
99
THE ASSOCIATIVE LAW. 285. In getting a sum of numbers, we may add the numbers together in groups, and then add these groups. If we use parentheses to mean that the terms enclosed have
been added together before they are added to another term, this law may be expressed symbolically by the equation
X 286.
The
-\-
{y
\-
z)=z X
^y
z.
-^
associative law holds for the
summation
of sects.
a-\-{b-\'C)=^a-\-b-\-c=^ AD. c
287.
The
associative law holds for the
angles.
ad
-f {df 4" eg)
=
ad
-\-
df
+
summation eg.
d
WmmHk
d
i^
of rect-
THE ELEMENTS OF GEOMETRY.
lOO
THE COMMUTATIVE LAW FOR MULTIPLICATION. 288.
The product
of
numbers remains unaltered
factors be interchanged.
xy 289.
=
The commutative law
if
the
yx.
holds for the rectangle of two
sects.
a.
If
a and b are any two
sects, rectangle
or ab for rectangle ab cide with it.
may be
—
ab
= rectangle
ba,
ba,
so applied to rectangle ba as to coin-
THE DISTRIBUTIVE LAW. 290.
To
multiply a
multiply each term obtained.
x{y 291.
ucts
we
The
numbers by a number, we may the sum, and add the products thus
sum
of
+
of
z)
—
xy
distributive law holds
-\-
xz.
when for numbers and prod-
substitute sects and rectangles.
a{b \-
c)
—
ab -^ ac
\
|
f
i^twi^vv
RECTANGLES.
lOI
if we add the rectangle ab to the rectangle aCy so that a side a in the one shall coincide with an equal side a in the r and other, the sum makes a rectangle whose base is ^
for
+
whose
altitude
is
In the same altitude,
we
a
may
is,
,>,""''''
get
+
c
state this in
If there be
any two
-j-
a)=
ab
+
ac
4
words as follows one of which
^«V\
f,.
J
'
"
"»'>'
>'"{'* }l' ]\
:
sects
is
divided into any
number of
partSy the rectangle contained by the two sects is equivalent to the rectangles co7ttained by the undivided sect and the several parts of the divided sect. 292. \i b
-\-
c =. a, then
ab
-\-
ac =. a(b
-{-
c)
=
aa
=
a^.
& Therefore If a sect be divided into any two parts^ the rectangles contained by the whole a^td each of the parts are together equivalent to the
,,,
.
a(b
We
that
the rectangle a{b -\- c). way, by adding three rectangli^s (?f; the sanje ;
square on the whole
sect.
:
;
J
THE ELEMENTS OF GEOMETRY.
102
293. If ^
= ^,
a{b
then
\- c)
—
a{b \- a)
—
ab
•\-
^
aa
ab -{
a^.
amC
Therefore
If a
sect he
divided into any two partSy the rectangle con-
tained by the whole and one of the parts is equivalent to the rectangle contained by the two parts together with the square on ,
the aforesaid part.
The
rectangle of two equal sects is a square, and b){a •\- b). only a condensed way of writing {a But, by the distributive law, i^6t.
(a
+
by
+
is
{a
By
-{-
b){a
b)
=
{a
=
a{a
=
aa
+
b)a
-\-
{a
+
b)b.
the commutative law, {a
By
+
+
b)a
+
b).
the distributive law,
a{a
-i-
b)
^
ab
-\-
=
a^
-\-
ab.
a
In the same way,
+ h)b = (a + hy =
(a
ab a^
-\-
b^,
+
ab
+
(ab
+
b^),
RECTANGLES,
By a:"
103
the associative law,
\- ab
{ab -f b')
-\-
,\
{a
=
+ {ab + ab) + b^ = + by = «2 + 2ab + b^.
a^ -^ 2ab
^=»
+
b^,
Therefore If a sect be divided into any two parts, the square on the whole sect is equivalent to the squares on the two parts together y
with twice the rectangle contained by the two parts. The square on a sect is four times the 295. Corollary. square on half the sect.
*
t
J)
CL
JB
a
I
296.
the distributive law, and 294,
By
a(a
From
-{-
b
+
this, if
b)
-\-
b^
=
a""
-^ ab -^
ab -^
AB at for BD put
we
bisect the sect
b^
C,
=
(a -^ by.
and divide it into and for DC put by
two unequal parts at Z>, and a, we get the theorem If a sect is divided into two equal parts and also into two unequal parts, the rectangle contained by the U7iequal partSy :
y
together with the square on the line betweeti the points of sectiony is equivalent to the square on half the sect. 297. If
CD
==
by
and
AC =.
a
-\- by
then their sum
AD =
a
-^ b -\- by and their difference is a therefore the rectangle contained by their sum and difference equals a(a 2b), which, by 296, is the difference between (a by and ^^ ;
+
+
THE ELEMENTS OF GEOMETRY.
104-
Therefore
The rectangle contained by the sum and difference of any two sects is
equivalent to the differeftce between the squares on those
sects.
we
298. If
and for
bisect the sect
BD put a,
and for =^ {a
BC
AB put
and produce it to Z?, then the above equation,
at Cy b,
by, gives us the theorem a sect be bisectedy and.produced to any pointy the rectangle If contained by the whole line thus produced a^id the part of it produced^ together with the square on half the sect bisected^ is equiv-
a{a \- b
'\-
b) -\-
b'^
:
•\-
alent to the square on the line the part produced.
299.
By
294,
+
{a
By
^
2ab -^
a^
=
a^
+
2ab
+
b^
-{-
the half
and
a".
b"
-\-
a" =z 20"
+
2ab
+
^^
the distributive law, 20^
By
-{-
made up of
is
the associative law, a"
By
by
which
-f-
2ab
-\-
b^
=
2a{a
-^ b) \- b^.
the commutative law,
2a{a /.
(a
-^ b) -\'
by
-{-
b''
-{-
a^
= =
2{a
+
b)a
2(a
-{-
b)a
+ +
^, b\
Therefore If a sect be divided into any two partSy the squares on the whole line and on one of the parts are equivalent to twice the
RECTANGLES.
•105
rectangle C07ttained by the whole a7id that party together with the square on the other part.
«
Z-
300.
By
the commutative and distributive laws, and 294,
4(^
By
+
b)a
+
b^
=
4a^
+
4ab
+
^'
=
(2a
+
by.
the associative and commutative laws, {2a -\-by=^(a
+
a-\-
•
.-.
4{a
4-
b)a 4-
€i
a
^'
=
(la
([^
+
by
=
+ ^]
^]
+
+
«)%
ay.
^
a a
I
L-
Therefore If a sect be divided into any two parts, four times the rectangle contained by the whole sect and one of the parts together ^
THE ELEMENTS OF GEOMETRY.
I06
with the square on the other part, is equivalent to the square on which is made up of the whole sect and the first part. 301. By the associative and commutative laws, and 294,
the line
{{a
•\-
f\
ay
-\-
=
4- ^2
=^+ By
2ay
(^ 4-
^ab
+
4^5^
+
h'
-\-
h^
2«2
•=.
+
/^ab 4-
2^
+
2«^
by
+
2«%
the distributive law, and 294,
20" 4- Acib 4-
2b'-
.-.
4- 20"
([«
+
—
^]
2(^2 4- 2ab
+ ay
-f-
4- ^^
b^)
=
»
20"
—
2(« 4- ^)^
2{a
+
-\-
2^^
^
^
^^^^^ ,
•\-
.
.
Therefore If a sect be divided into two equal and also two unequal partSy the squares on the two unequal parts are together double
and on
the squares on half the sect
of
{a
the line between the points
section.
302.
By
+ by
4-
294,
/^^
and the associative and
_, ^2
=
_|_
2^3 4.
distributive laws,
^2 4- ^2
J + ^J +
,ab 4-
2b^
=
2
2(^Y+
4-
(^
^y.
Therefore If a sect be bisected and prodticed to any point, the square on the whole line thus produced and the square on the part of it produced are together double the squares on half the sect and on the line 303.
made up of the half and the part produced. The projection of a point on a line is the
perpendicular from the point
a4- --^-^r^^^C^
to the line.
foot of the
RECTANGLES.
107
304. The projection of a sect upon a line is the part between the perpendiculars dropped upon the line from the ends of the sect.
For example, A'B' line
is
the projection of the sect
AB
on the
c.
Theorem
I.
In an obtuse-angled triangle the square on the side oppoangle is greater than the sum of the squares on sides the other two by twice the rectangle contained by either of 305.
y
site the obtuse
those sides
and the projection of
Hypothesis,
a ABC,
Conclusion,
a""
Proof.
By
=
b^
c"
^ CAB -{-
it.
obtuse.
2bj\
294, (b
Adding
with -\-
the other side upon
+ yy =
b'
+
2bj
+ y^
b^
+
2b/
+ y^ +
h^ to both sides,
+ JY + h^ = + j'Y + h^ = a\ {p
But
h^.
^
and (b j' -f /^^ ^, (243. In a right triangle, the square of the hypothenuse equals the squares of the other sides.) .-.
1-
^
^=^
=
^2 -f 2bj
+
c^.
sum
of the
THE ELEMENTS OF GEOMETRY.
io8
Theorem
II.
any triangle the square on a side opposite any acute less than the sum of the squares on the other two sides
306. In
^
angle is by twice the rectangle contained by either of those sides projection of the other side upon it.
Hypothesis,
a ABC,
Conclusion,
c^
Proof.
Adding
By
with
=
4- 2bJ
^ C
+
a^
and
the
acute.
b^.
295,
h^ to both sides, ^^
+P
=
-h h^
2bj
+ Alf +
h^.
But
j^
-{-
h""
=
and
a^,
Alf
+ k^=
c^,
(243. In a right triangle, the square of the hypothenuse equals the
squares of the other sides.) b""
.*.
307.
+
a""
=
Having now proved that By243,
if
By By
305,
if
306,
if
;^^
=
^ A > ^^ <
2b/
+
^.
in a triangle,
=
<5-
+
^;
> <
b^
-{-
c^
b^
+
c\
^^
=
rt.
^ ^
rt. ;^,
/.
«^
rt.
^,
.'.
a^
rt.
^,
/.
a^
;
Therefore, by 33, Rule of Inversion, If a^ If a"
If a^
= > <
b^
+
c^,
.-.
b^ \- c^,
.'.
+
.\
b^
c%
^A> ^A<
rt.
rt.
;
;
^.
sum
of the
RECTANGLES.
Theorem
109
III.
a medial be drawn from the vertex of the sqicares on the two sides is equivalent to twice the square on half the base, together with twice the square on the medial. 308.
In any
triangle^ if
to base^ the stun
Hypothesis.
i^ABC,
Conclusion,
a^
Proof.
— Case a"
=
-{•
I.
with medial
4.
{\bY
BM
=•
i.
=
^{hby + 22^. If BMA = BMC, and c' = + /^ c^
4.
then
{\by
+
/^
{243. In a right triangle, the square of the hypothenuse equals the sum of the squares of the other sides.)
Case be the
If
II.
greater.
Then, in the In
A BMA, by
BMA
^ BMC, one of BMC. obtuse-angled triangle BMC, by 305, a-= (1^)^ + /^+ 2{\bj), 2(^
does not equal
them must
Call the greater
306,
Adding, a-
+
c^
/.
+
2{\bj) a^
+
=
c^ =.
2{iby
+
2{^by
-f 2i\
2/^
+
2{lbj),
The difference of the squares on the 309. Corollary. two sides is equivalent to twice the rectangle of the base and the projection of the medial on it.
no
THE ELEMENTS OF GEOMETRY.
Theorem 310.
IV.
The sum of the squares on the four sides of any quad-
rilateral is greater than the sum of the squares on the diagonals by four times the square on the sect joining the mid-points of the
^ -q
diagonals.
Hypothesis. onals
AC
and
EF is the sect joining BD of the quadrilateral ~A&
Conclusion.
Proof.
By
Draw
+
'BC''
+
'C&
+
the mid-points of the diag-
ABCD, 'DA^
BE and VE.
308,
AB'
+
BC = 2f,^^V3 ^j +
2^:^,
and
C&^DA'^
2(^ + 2DB^.
Adding,
AB^^ BC^ But,
-^
C& + DA^= 4^^Y + 2^" +
2DE'.
by 308,
BE"
+ DE" = sj^^Y +
2.5^,
= AC^ + BD" +
^EF".
RECTANGLES,
The sum
Corollary.
311.
of
III
on
the squares
sides of a parallelogram is equivalent to the
sum
the four
of the squares
on the diagonals, because the diagonals of a parallelogram bisect each other.
Problem To square
312.
I.
a7iy polygon.
H
.
\ '
I
\
/
I
I
\K
*i?
\
u.-
i
GrvEN, any polygon N. Required, to describe a square equivalent to N. Construction. Describe, by 263, the rectangle
Then,
if
AB —
BC^
the required square
is
ABCD =
N,
ABCD.
AB be not equal to BC, produce BAy and cut off AF = AD. BF in G, and, with center G and radius GB^ describe FLB. Produce DA to meet the circle in H. If
Bisect
The square on
AH shall be equal to N.
GH.
Proof.
Then, by 296, because the sect Join equally in G and unequally in A^ rectangle
by 243
BF
is
BA,AF + AG^ = GF" = GH^ = AH' +
;
AH""
=
rectangle
BA,AF =
rectangle
BA,AD =
JV,
divided
AG",
THE ELEMENTS OF GEOMETRY.
112
Problem To divide a given
313.
II.
sect into
two parts so that the
rect-
angle contained by the whole and one of the parts shall be equal to the square on the other part.
Z Given, the sect
J)
AB.
=
AC*. Required, to find a point C such that rectangle AB,BC Construction. On AB^ by 241, describe the square ABDF,
AF
in G. 136, bisect Produce FA, and make
By
On
AH
BG. GB.
AHKC, AB,BC = JC^
describe the square
Then
KC
Produce to Z. Then, by 298, because FA is bisected and produced to H, rectangle FB,BA 'A& = 'GB'' = 'GB"
Proof. in
G
=
GA^
+
+ AB\
But, since
BK =
/.
FB,HA =
AB",
.-.
FB,BK =
AB",
BA,
Take from each the common /.
that
Join
GH =
part
AL,
BC =
CD)
is,
AC^ = CB,BD = CB,BA,
RECTANGLES.
Problem
113
III.
To describe an isosceles triangle having each of the at the base double the a?igle at the vertex. angles 314.
Take any
Construction. that rectangle
AB,AC = BC^.
sect,
AB,
and, by 313, divide it in C so C and radius CB, describe
With center
a circle.
With same
radius, but center
preceding circle in D.
Join
A, describe a
ABD will be the Proof.
By
By 134, AC. Then
construction,
bisect
^
triangle required.
CB = CD = DA,
ADC by DF.
AB'
circle intersecting the
AD, BD, CD.
By
+ AD" = BD^
137,
-f
DF ± AC, and bisects is
2AB,AK
The square on
a side opposite an acute angle is less than the squares on the (306. other two sides by twice the rectangle of either and the projection of the other
on
But
it.)
^C =
2AF, .'.
.'.
by our
initial
AB,AC = AB,2AF = 2AB,AF\ construction, BC^ == 2AB,AF,
+ AD" = BD" + BC\ AB = BD, 'AB' = 'BD', But, by construction, AD = BC, = = 2 ^ B. = = BDC ADB DAB ACD ^ ^ ^ ^B + ^ .-.
AB"
.*.
.'.
.'.
(173.
The
exterior angle of a triangle equals the angles.)
sum
of the
two opposite
interior
V
BOOK THE I.
III.
CIRCLE.
Primary Properties.
315. If a sect turns about one of its end points, the other end point describes a curve called the Circle. 316. The fixed end point is called the Center of the circle. 317. The moving sect in any position is called a Radius of
the circle. 318. it, all
As
the motion of a sect does not enlarge or diminish
radii are equal.
319. Since the
moving sect, after revolving through a perireturns to its gon, original position, therefore the moving end describes a closed curve. point This divides the plane into two surfaces, one of which sect. This finite plane surface
swept over by the moving
called the surface of the circle.
Any
part of the circle
is
called an Arc,
x«5
is
is
THE ELEMENTS OF GEOMETRY.
Il6
Theorem 320. tha?t,
The
equal
is
Any
or without the
If a point
Proof. center
a point fro^n the center of a chxle is less or greater thaji, the radiusy according as the point
sect to
to,
is within, on,
a radius, for
is
it is
S
circle.
on the
point, Q, within the circle
is
circle,
the sect drawn to
it
from the
one of the positions of the describing
/.
If
I.
lies
OQ <
on some
OQR,
OR,
without the circle, then the sect :,
radius,
sect.
OS contains
a radius ORy
OS > OR,
321. By 33, Rule of Inversion, a point is within, on, or without the circle according as its sect from the center is less than, equal to, or greater than, the radius.
322.
the
A
circle.
Secant
is
a line which passes through two points on
THE CIRCLE.
Theqrem 323.
A
11/
II.
secant can meet the circle in only two points.
Proof.
By
definition, all sects joining the center to points
on the
but from a point to a line there can be only two equal
circle are equal, sects. (155.
324.
where
No more
A Chord is the part
and one 326.
of a secant
line.)
between the two points
intersects the circle.
it
325.
than two equal sects can be drawn from a point to a
A
Segment of a circle is the figure made by a chord two arcs into which the chord divides the circle. When two arcs together make of the
an entire
circle,
each
is
said to be the
Explemettt of the other. 327.
When two
equal, each 328.
is
explemental arcs are a Semicircle.
When two
explemental arcs are
unequal, the lesser is called the Arcy and the greater is called the Arc. 329.
A
cording as
segment its
arc
is
is
called a
Minor Major Major or Minor Segment
a major or minor arc.
ac-
THE ELEMENTS OF GEOMETRY.
Il8
Theorem 330.
A
circle
has only one center.
Hypothesis.
Let F, G, and
Conclusion.
O
Proof.
Join
III.
FGH has
FG and
H be points
on a O.
only one center.
GH.
Since, by definition, a center is a point from which all sects to the circle are equal, therefore any center of a circle through and is in
F
G
the perpendicular bisector of FG, and any center of a circle through and is in the perpendicular bisector of GH.
H
G
(183.
The
locus of the point to which sects from two given points are equal perpendicular bisector of the sect joining them.)
But these two perpendicular bisectors can point, /.
O FGH
intersect
in
is
the
only one
has only one center.
The perpendicular bisector of any 331. Corollary I. chord passes through the center. 332. Corollary II. cle, or of any given arc
To of
find the center of
a
circle,
any given cirtwo draw non-parallel
chords and their perpendicular bisectors. point where these bisectors intersect. 333. 334.
The
is
the
A Diameter is a chord through the center. A diameter is equal to two radii so all diameters
are
center
:
bisected by the center of the circle, and are equal.
THE CIRCLE.
Theorem 335
•
Circles
CD —
C
and
O
aro the centers,
circles are congruent.
Apply one
Proof.
of which
circles
radius OF.
The
Conclusion.
IV.
of equal radii are congruent.
Two
Hypothesis.
and radius
II9
circle to the other so that the center
O
shall
OF
fall upon line CD. coincide with center C, and sect Then, because will coincide with the point D. Then every CD, the point
OF =
F
one
particular point in the
circle
must coincide with some point
in the
other circle, because of the equality of radii. (321.
A
point
is
on the
circle
when
Corollary.
sect
from the center
o c^ o
/.
336.
its
is
equal to the radius.)
a
After being applied, as above, the second
its center and still it will coincide though the point F no longer falls upon D. Hence, considering one circle as the trace of the other,
circle
may be
with the
A about
turned about
first,
circle
can be
;
—
made
to slide
along itself by being turned
its center.
This fundamental property of this curve allows us to turn any figure connected with the circle about the center without
changing
its
relation to the circle.
THE ELEMENTS OF GEOMETRY.
120
337. Circles
which have the same center are called Con-
centric.
Theorem V. a point in com-
338. Different concentric circles cannot have
mon.
Proof.
The
points of the circle with the lesser radius are
all
within
the larger circle. (321.
A
point
is
within the circle
if
its
distance from the center
is
less than the
radius.)
339.
First Contranominal of 338.
with a point in 340.
common
Second Contranominal of
cles with a point in
341.
The
Two
different circles
are not concentric.
common
338.
Two
concentric
cir-
coincide.
center of a circle
is
a Center of Symmetry^ the
end points of any diameter being corresponding points. This follows at once from the definition of Central Symmetry, and the fundamental property that the circle slides along itself when turned about its center, and so coincides after turning about the center through aity angle. curve which will slide upon its
with
itself
The
circle is the only closed
trace.
THE
CIRCLE.
121
Theorem VI. 342.
The perpendicular from the center of a
circle to
a secant
chord ; and, if a line through the center bisect a chord not passing through the center^ it cuts, it at right angles. bisects the
Proof. center, only
For any chord, there one
line
is only one perpendicular from the through the mid-point and center, only one per-
pendicular bisector; and these, by 331, are identical. (331.
343.
The perpendicular
bisector of any chord passes through the center.)
Every diameter
is
an Axis of Symmetry,
For if we fold over along a diameter, every point on the part of the circle turned over must fall on some point on the other part, since its sect from the center, which remains fixed, is a radius. Every line which is an axis of symmetry of a a diameter. if For, not, there would be a point symmetrical to the center, and this, too, would again be a center the circle would thus have two centers. Inverse.
circle contaiits
:
(330.
Each an
A
circle
has only one center.)
circle has therefore, besides its center of symmetry,, number of axes of symmetry.
infinite
THE ELEMENTS OF GEOMETRY.
122
Theorem 344.
Every chord lies wholly within the
A
circle.
B
be any two points in ABC. on chord AB and between A B is within Every point
Hypothesis.
Lei
Conclusion.
and
O ABC.
the
Proof.
By •
VII.
Take any
point, Z>, in
Join
Then
OA, 0Z>, OB. OB makes a greater
pendicular from (342.
The
O
on the
line
sect than
The
from the foot of the per-
OB >
AB bisects chord AB.)
OD,
oblique which makes the greater sect from the foot of the perpendicular
.*.
A point is within the
345*
OD
perpendicular from center on line
is
^320.
AB.
AB,
.-.
(r54.
chord
332, find O, the center of the circle.
Corollary.
D
circle
If
is
the greater.)
within the circle.
if its
sect
from the center
is less
than the radius.)
a line has a point within a
circle, it is
a
secant, for the radius is greater than the perpendicular from the center to this line so there will be two sects from the :
each equal to the radius will pass through two points on the circle. Thus, again, the circle is a closed curve.
center to the
line,
;
that
is,
the line
THE
CIRCLE.
Theorem 346.
In a
circle,
1
23
VIII.
two chords which are not both diameters do
not mutually bisect each other.
Hypothesis.
Lei ihe chords
AB, CD,
which do not both pass
through the center, cut one another in the point F, in the O A CBD. Conclusion. and do not mutually bisect each other. Proof. If one of them pass through the center, it is not bisected
AB
by the
other,
CD
which does not pass through the center.
If neither pass through the center, find the center O, and join If is the bisection point of one of the chords, as AB, then
OF.
F
OFB =
:^
(342. If a line
rt.
^,
through the center bisect a chord not passing through the center, cuts it at right angles.) .*.
.*.
Exercises. chords ?
69.
OFD OF does
^
What
is
is
it
oblique,
not bisect
CD.
the locus of mid points of parallel
70. Prove by symmetry that the diameter perpendicular to a chord bisects that chord, bisects the two arcs into which this chord divides the circle, and bisects the angles at the center
subtended by these
arcs.
THE ELEMENTS OF GEOMETRY.
124
Theorem
IX.
any point not the center^ sects be drawn to the is that which meets different points of a circle, the greatest the center the least is part of the the circle after passing through ; same line. 347. If from
From any
Hypothesis.
whose center
drawn
to
a
Q DBHK,
A CB > AD.
First Conclusion.
Proof.
point A, sects are
is C.
Join C£>.
CB = CD, AB = AC -^ CB = AC
Then, because .'.
(156.
Any two
When A
is
within the circle,
BC ^ KC
ixom both Taking away When A is on the circle, is
is
156.
AH < AK.
.*.
a point.
without the circle,
AC = AH Taking away HC = 348.
AC, by
sides,
AH
When A
CD> AD.
AH < AK.
Second Conclusion. Proof.
-j-
sides of a triangle are greater than the third.)
KC,
Corollary.
any other chord.
-{-
HC
KC, by
AH < AK.
The diameter
of
a circle
156.
is
greater than
THE CIRCLE.
1
25
Theorem X. 349*
If from any point three
sects
drawn
to
a
circle
are equal,
that point is the center.
the point O to O center of O ABC.
Hypothesis.
From
Conclusion.
O
Proof.
O
is
of chord
Join
is
AB, and
ABC,
lei
OA = OB =
OC,
B C.
on the perpendicular bisector of chord
AB, and
also
on
that
BC,
(183.
The
locus of a point to which sects from two given points are equal perpendicular bisector of the sect joining them.)
(332.
The
center
.-.
is
O is
center of
is
the
O ABC.
the intersection of perpendicular bisectors of two non-parallel chords.)
350.
From any point not the 349. drawn more than two equal sects to a
CoNTRANOMiNAL OF
center, there cannot be circle.
351. Corollary I. common, they coincide.
If
two
Because from the center of one to points .*.
on the other
circles
have three points in
circle three equal sects
circle,
they are concentric, and they have a point in .*.
(340.
Two
can be drawn
common,
they coincide.
concentric circles with a point in
common
coincide.)
THE ELEMENTS OF GEOMETRY.
126
352.
one
Corollary
circle
353-
Through three
II.
points not
more than
can pass.
Corollary III. Two different in more than two points.
circles
cannot meet
one another 354.
A
circle
is
circumscribed about a polygon
when
it
the vertices of the polygon. passes through Then the polygon is said to be inscribed in the circle. all
Problem 355. scribe
a
I.
Through any three points not in the same
lincy to de-
circle.
yo
c
Given, three points, A, B, and C, not in the
Required,
to
describe
a
circle
same
fine.
which shall pass through all of
them.
Construction.
By
188, find the point O, such that
OA = OB = 356.
Corollary.
To
OC,
circumscribe a circle about any given
triangle, by 355, pass a circle through its three vertices. 357. Four or more points which lie on the same circle are
called Concyclic,
THE
CIRCLE.
12/
Theorem XI. from the center on equal chords are on andy unequal chords ^ that on the greater is the
358. Perpendiculars
equal ; lesser,
C
From center
Hypothesis. Off,
Conclusion.
Proof.
By
Draw
hypothesis,
O,
OK, OL, JL chords Off = OK < OL.
AB = CD> FG,
OA, OB, OC, OB, OF. CD,
the radii
AB = .-.
A
OAB^
A 0C£>,
(129. Triangles with three sides respectively equal are congruent.)
the altitude
Off
Again, because CZ> >
FG,
,*.
.-.
But ~CK'
ZX'
But
+
Z^ = 'OC^
=
corresponding altitude
OK,
CK > FL, = OF" = FL' + T&.
_
> 'FL\ .-.
:,
K(f
OK
< OL,
By 33, Rule of Inversion, Chords having equal perpendiculars from the center are equal and, of chords having unequal perpendiculars, the one 359«
;
with the lesser
is
the greater.
THE ELEMENTS OF GEOMETRY.
128
Angles at the Center.
II.
The explemental
angles at the center of a circle, whose are said to be subtended by^ or to stand radii, arcs the opposite them intercepted by the explemental upony the major arc. radii, the reflex angle upon 360.
arms are the same
361.
A
Sector
is
the figure formed by two radii and the arc
included between them. 362.
The angle
of
-^
the sector
is
the
angle at the center which stands upon the arc of the sector. 363.
A given
sect
is
said to subtend a
certain angle from a given point when the lines drawn from the point to the ends of the sect form that angle.
An
inscribed angle is formed by two chords from the same point on the circle, and is said to stand upon the arc between its arms. 364.
t^.
THE CIRCLE.
Theorem
129
XII.
365. In the same or equal circles^ equal arcs are congruent, subtend equal angles at the center^ determine equal sectors^ and are subtended by equal chords.
Hypothesis. Conclusions.
OA = radius CF, and arc AB = arc FG, AOB = ^ FCG. Sector AOB = sector FCG, Chord AB = chord FG. sector A OB over the sector FCG so that the
Radius I.
4.
II.
III.
Proof.
Place the
O shall fall on because OA = CF, center
C,
,'.
and the radius
point
A
falls
OA
on the
line
CF.
on point F.
Again, because the radii are equal, every point of the arc fall
AB
will
on some part of the circle FG, But arc AB = arc FG^ ,',
,',
and
Then,
B on G, AB = chord FG, ^AOB = 4 FCG, sector AOB = sector FCG, point
falls
chord
366. In the same or equal circles the sum of two minor arcs the arc obtained by placing them on the same circle so as not to overlap, with one end point in common.
is
THE ELEMENTS OF GEOMETRY,
I30
Theorem XIIL
A
367.
sum of two
an angle at
arcs of the
the center equal to the
same or equal circles subtends
sum of
which each
the angles
arc subtends separately.
o
Proof.
Placing the arcs with two end points in common at B, join and C, to the center O,
B and the other end points, A Then
the angles are in such a position, that, by definition 61, 4.
But
4 AOC
subtending 368.
may
AOB +
t
BOC =
subtended by arc
is
^ AOB,
and arc
BC subtending
Two
Corollary.
^ AOC,
AC, which
is
the
sectors of the
same or equal
be so placed as to form a sector whose arc
their arcs,
surface
is
whose angle the
sum
is
the
sum
sum of
arc
AB
^ BOC. is
the
of their angles,
circles
sum
of
and whose
of their surfaces.
Theorem XIV. 369. In the
same or equal
circles,
of two unequal
greater sicbtends the greater angle at the center, the greater sector. Proof.
If the first arc
second plus a third arc tends
;
is
and
greater than the second, it is equal to the so, by 367, the angle which the first sub-
greater than the angle which the second subtends which the third arc subtends at the center. is
And 370.
like is true
From
arcs, ths
and determines
by the angle
of the sectors.
365 and 369, by 33, Rule of Inversion,
THE
CIRCLE.
I31
In the same or equal circles, equal angles at the center intercept equal arcs, and determine equal sectors and, of two ;
unequal angles at the center, the greater intercepts the greater arc, and determines the greater sector. 371.
Corollary.
A
diameter of a circle divides the en-
closed surface into two equal parts.
from 365 and 369, by 33, Rule of Inversion, In the same or equal circles, equal sectors have equal arcs and equal angles and, of two unequal sectors, the greater has the greater arc and the greater angle. 372. Again,
;
Theorem XV. 373. In the same or equal circles^ of two Ufiequal minor arcs, the greater is subtended by the greater chord ; of two unequal major arcSy the greater is subtended by the lesser chord, J9.
AB > arc CD, Chord AB > chord CD. AOB > 4. COD, Minor arc
Hypothesis.
Conclusion.
Proof.
4.
(369. .-.
(159.
Two
The
m
greater arc subtends the greater angle at the center.)
t.'s^AOB and
COD,
side
AB > side
CD.
two sides equal, but the included angle greater have the third side greater in the first.)
triangles with
in the
first,
Secondly, because the minor arc with its major arc together make entire circle, therefore to a greater major arc will correspond a
up the lesser
minor
arc,
and therefore a
lesser chord.
THE ELEMENTS OF GEOMETRY,
132
From 365 and 373, by 33, Rule of Inversion, In the same or equal circles, equal chords subtend equal major and minor arcs and, of two unequal chords, the greater 374.
;
subtends the greater minor arc and the lesser major
arc.
Theorem XVI.
An
375.
angle at the center of a circle
same
angle standing upon the
Fig.
Hypothesis.
is
Fig.
I.
Lei
double the inscribed
arc.
AB
be any arc,
O
the
2.
center,
C
any point
AB.
on the circle not on arc
A OB =
2:^ A CB. CO, and produce it to D. Because, being radii, OA = OC,
Conclusion.
Proof.
^
Join
^
.'.
(126.
The
^
.-.
(173.
The
OCA = ^ OAC,
angles at the base of an isosceles triangle are equal.)
AOD =
4.
OCA
4-
4.
OAC =
exterior angle of a triangle equals the
sum
2^ OCA.
of the interior opposite
angles.)
Similarly,
Hence angles
OCB;
^
DOB =
(in Fig.
i)
AOD, DOB, that
is,
^
is
2^ OCB.
the
sum
or (in Fig. 2) the difference sum or difference of
double the
AOB =
2^ ACB.
of the
OCA
and
THE
III.
CIRCLE.
133
Angles in Segments.
376. An angle made by two lines drawn from a point in the arc of a segment to the extremities of the chord is said to be
inscribed in the segment, and
is
called the angle in the seg-
ment.
377.
Corollary
I.
Angles inscribed
in the
same segment
of a circle are equal.
ADB
For each of the angles ACBy subtended at the center by the arc ARB, 378.
Corollary
II.
If
ments by a chord, any pair
a circle
of angles,
is
one
is
half of the angle
divided into two segin
each segment,
will
be supplemental.
For they are halves of the explemental angles standing on the same explemental arcs.
at the center
THE ELEMENTS OF GEOMETRY.
134
379.
Corollary
rilateral inscribed in
The opposite angles of every quadIII. a circle are supplemental.
For they stand on explemental explemental
arcs,
and so are halves
of
angles at the center.
Theorem XVII. 380. From a point on the side toward the segment, its chord subtends an angle less than, equal to, or greater than, the angle in the segment, according as the point is without, on, or within
the arc of the segment.
Hypothesis.
AC
the chord of
without the segment on the within the segment.
Conclusions.
segment are equal,
Since,
same
any segment, side of
by 377, we know
we have I.
II.
AC
all
only to prove
^
AFC
<
4AQC>
ABC. 4 ABC. 4.
ABC P any point B Q any point
as
;
y
angles inscribed in the
THE CIRCLE. Proof.
I.
Let
R be
RA, making A APR) An
(142.
II.
1
PC meets APC
the point where
then
4.
exterior angle of a triangle
is
the circle,
and
35
join
greater than either interior opposite angle.)
For the same reason, if AQvs, produced to meet the circumferand SC joined, making A CSQ, 4. > ^ CSA.
AQC
'ence at S,
By 33, Rule of Inversion, the side toward a segment, the vertex of a triangle, with its chord as base, will lie without, on, or within the arc of the segment, according as the vertical angle is less than, equal to, 381.
On
or greater than, an angle in the segment.
Theorem XVIII. If two opposite angles of a quadrilateral are supplea circle passing through any three of its vertices will mental, 382.
contain the fourth.
Hypothesis.
ABCD is a quadrilateral, with ^ A
Conclusion.
The
four vertices
lie
on the
circle
=
st. ^. -\- ^ C determined by any
three of them.
Proof.
By
Take any side of
355, pass a circle through the three points B, C, D. on the arc of the segment, on the same
DB as A.
{379.
The
From On
DFB
point, F,
Join
FB,FD.
Then^
i^
+
;^
C =
st.
^.
opposite angles of an inscribed quadrilateral are supplemental.)
hypothesis,
.-.
%.
A =
-^^
F,
.*.
A
is
on the arc
DFB.
the side toward a segment, the vertex of a triangle, with its chord as base, will lie on its arc if the vertical angle is equal to an angle in the segment.)
(381.
THE ELEMENTS OF GEOMETRY.
136
Theorem XIX. 383. The angle i7t a segment is gi^eater thany equal to, or less than, a right angle, accordifig as the arc of the segment is less
than, equal
to,
or greater than, a semicircle.
Hypothesis. is
Take
O.
B
Conclusions.
AD
Lei
and
C
I.
^
be
a diameter of a circle whose center
points on the
ACD =
rt.
same 2^,
circle.
being half the straight angle
AOD. An
(375.
angle at the center II.
(143.
Any two
.-.
is
double the inscribed angle on the same
arc.)
^ADC
angles of a triangle are together less than a straight angle.) III.
.-.
i.
ABC >
It.
4.,
since 4.
ADC +
(379. Opposite angles of
384.
A circle,
By
33,
segment
Rule is
i.
ABC =
St.
4..
an inscribed quadrilateral are supplemental.)
of Inversion,
less than, equal to, or greater than, a semi-
according as the angle in
less than, a right angle.
it is
greater than, equal
to,
or
THE CIRCLE.
I37
Theorem XX. intersect within a circley an angle formed are each equal to half the angle at the center standing on the sum of the arcs they intercept.
If two chords
385.
and
its vertical
Let the chords
Hypothesis.
AC,BD
intersect at
F
within
the
circle.
Conclusion.
^
BFC =
half
on (arc
at center standing
^
BC
}-
arc £>A) .
Proof.
Join CB>.
^ (173.
The
BFC =
^
exterior angle of a triangle
BDC + is
^ DCA,
equal to the
sum
of the opposite interior
angles.)
ter
But 2 ^ on DA ;
(375.
An
BDC =
at center
angle at the center .-.
{367.
^
A sum of
2
^
BFC :=
is
on BC, and
2 4.
DCA =
^
at cen-
double the inscribed angle upon the same
^
at center
two arcs subtends an angle
on arc
{BC
-\-
arc.)
DA).
at the center equal to the
sum
of the
angles subtended by the arcs.)
Exercises.
71.
The end
circle are the vertices of a 72.
points of two equal chords of a
symmetrical trapezoid.
Every trapezoid inscribed
in a circle is symmetrical.
THE ELEMENTS OF GEOMETRY.
138
Theorem XXI.
An
386.
angle
formed
by two secants
is
half the angle at the
center standing on the differeftce of the arcs they intercept.
Lei iwo lines from
Hypothesis.
F
cut the circle whose center
O, in the points A, B, C, and D. =^ Conclusion. zX O qh difference between arc :4.
is
F
arc
\^
Proof.
Join^C 4.
Doubling both arc
AB
and
CD.
AB .*.
CAD =
sides, :^ at
(9
i.F^- 4.C. on arc
CD =2^i^-f^at0on
;
twice
F=
^
difference of
2^ s
at
O
on
arcs
CD and AB.
IV. Tangents. 387. is
A
line
said to be a 388.
The
called the
which
Tangent
will
meet the
circle in
one point only
to the circle.
point at which a tangent touches the Point of Contact.
circle is
THE
CIRCLE.
1
39
Theorem XXII.
Of Imes
passiiig through the perpendicMlar is a tangent to the circle, 389.
end of any
and every
radius, the
a
other line
is
oblique to
PC.
secant.
A
Hypothesis.
Conclusions.
OP perpendicular PB is a tangent at P, PC is a secant.
radius I.
II.
Proof. is
The
(I.)
from
sect
O
to
to
PB,
any point on PB, except P,
> OP, (150.
The perpendicular
.'.
{321.
A
point
is
every point of is
the least sect between a point and a line.)
PB
without a circle
if
except its
sect
P
is
outside the circle.
from the center
is
greater than the
radius.)
Proof.
(II.)
A
sect perpendicular to
PC
is
less
than the oblique
OP, .*.
a point of .*.
390.
Corollary
I.
PC
PC
is
is
within the circle
;
a secant.
One and
only one tangent can be
to a circle at a given point on the circle. To draw a tangent to a circle 391. Corollary II.
drawn
point on the point.
circle,
draw the perpendicular
at
a
to the radius at the
THE ELEMENTS OF GEOMETRY.
140
Corollary
392.
III.
The
radius to the point of contact
perpendicular to the tangent. The perpendicular to a tangent from IV. the point of tangency passes through the center of the circle. The perpendicular drawn from the 394. Corollary V. of
any tangent
is
Corollary
393«
center to the tangent passes through the point of contact.
On the Three Relative
Positions of a Line and a
Circle.
Corollary VI.
395'
A line will be a secant,
a tangent, or
not meet the center
is
396.
circle, according as its perpendicular from the less than, equal to, or greater than, the radius.
By
33,
Rule
of Inversion,
The
perpendicular on a line from the center will be less than, equal to, or greater than, the radius, according as it is a secant, tangent, or non-meeter.
Theorem XXIII. 397.
An
ajigle
point of contact
is
formed by a tangent and a chord from
y
intercepted arc.
Hypothesis,
the
half the angle at the center standing on the
AB
with center at O.
is
tangent at
C,
and
CD
is
a chord of
THE Conclusions.
Proof.
At
I4I
DCB = \t DOC, ^DCA = \ explement DOC.
^
C erect
chord
CF (393.
CIRCLE.
The perpendicular
is
CF ± AB,
a diameter of the
to a tangent
circle.
from the point of contact passes through the
center of the circle.)
Join
OD.
Then
OCA =
^st.
^at
^ OCD =
\4.
FOD,
tX.%.
O,
also
(375.
The
angle at the center is
double the inscribed angle on the same
arc.)
Therefore, adding
DCA = ^reflex ^ DOC, therefore the supplement of ^ DCA, which ^ DCB, ^
is
ment of
reflex
which
^ DOC,
398. Inverse.
If
is 4.
is
half the exple-
DOC
the angle at the center standing on the
arc intercepted by a chord equals twice the angle made by that chord and a line from its extremity on the same side as the arc, this line is a tangent.
There is but one line which will make from 397, that a tangent makes it. know, already Proof.
Exercises.
73.
The chord which
tact of parallel tangents to a circle 74. 75.
How may
is
this angle
;
and we
joins the points of con-
a diameter.
A
397 be considered as a special case of 375 parallelogram inscribed in a circle must be a rect-
If
a series of circles touch a given line at a given point,
.?
angle. y6.
where
will their centers all lie
The
}
angle of two tangents is double that of the chord of contact and the diameter through either point of contact. yy.
THE ELEMENTS OF GEOMETRY.
142
Theorem XXIV. The angle formed by a tangent and a secant is half the at the center standing on the difference of the intercepted angle 399.
arcs.
Hypothesis.
whose center
Of iwo
lines
from D, one cuts at
Conclusion.
O, the other is tangent to the 2 :^ difference between ^
Proof.
AB.
is
Join
D= ^
(173.
The
exterior angle of a triangle
^
.-.
(375.
ABC = ^D +
An
is
and C
O
the
O
at A.
AOC and ^ AOB.
^ DAB,
equal to the two interior opposite angles.)
AOC = 2^D
angle at the center is
B
same
-{-
:^
AOB,
double the inscribed angle on the same
arc.)
and (397.
An
.-.
400.
angle formed by a tangent and chord is half the angle at the center on the intercepted arc.)
twice
7^
Dis
the difference between
Corollary.
AOC and ^ AOB.
The angle formed by two tangents
half the angle at the center standing
intercepted arcs.
^
is
on the difference of the
THE
CIRCLE.
Problem
I43
II.
given point without a
401. Frofn a
circle to
draw a tangent
to the circle.
Given, a
center O, and a point P outside draw through P a tangent to the circle.
O wth to
it
Required, Construction.
Join OP. Bisect OP in C. With center C and radius CP describe a circle cutting the given and G. circle in Join PF and PG.
F
These are tangents Proof.
to
O FGB.
Join OF.
^ OFF (383.
.-.
(389.
The
is
a
rt.
angle in a semicircle
PF
is
^, is
tangent at i^ to
A line perpendicular to a radius at its
Two
a right angle.)
O BFG.
extremity
is
a tangent to the
circle.)
tangents drawn to a circle from the same external point are equal, and make equal angles with the line joining that point to the center. 402.
Corollary.
THE ELEMENTS OF GEOMETRY.
144
V.
Two
Circles.
Theorem XXV. 403. The line joining the centers of two circles which meet in two points is identical with the perpendicular bisector of the common chord.
Proof. For the perpendicular bisector of the pass through the centers of the two circles. (331.
404.
The perpendicular
Corollary.
common chord must
bisector of any chord passes through the center.)
Since any
common chord
is
bisected by
the line joining the centers, therefore if the two circles meet at a point on the line of centers, there is no common chord,
and these
have no second point in common. which meet in one point only are said to touch each other, or to be tangent to one another, and the point at which they meet is called their point of contact. circles
405.
Two
406.
By
Two
not concentric, have always one, and only one, axis of symmetry ; namely, their line of centers. this is the only line which contains a diameter of
common For each.
circles
343,
circles,
THE
CIRCLE.
1 45
Theorem XXVI. Obverse of 407.
If two
404.
have one common point not on the
circles
through their centers ^ they have also another
O
Hypothesis.
common
point
B
with center
O
and
O
common
line
point.
with center C, having a
OC.
not on
They have another common point. Join OC, and from B drop a line perpendicular
Conclusion.
Proof.
Dy and prolong
it,
to
OC
2X
making DF = BD. F the second common point. is
A
For (124.
ODB ^
A OFD,
A
and
CBD ^
Triangles having two sides and the included angle equal
A in
CFD
;
each are con-
gruent.) .-.
OF =
(321.
A
point
408.
is
on the
F
is
-,
on both Os.
circle if its distance
CoNTRANOMiNAL OF
CF = CB
and
OB,
.*.
from the center
407.
If
two
is
equal to the radius.)
circlcs
touch One
another, the line through their centers passes through the point of contact.
Two circles which touch one another tangent at their point of contact namely, the perpendicular through that point to the line joining their 409.
have a
centers.
Corollary.
common
;
THE ELEMENTS OF GEOMETRY.
146
410. Calling the sect joining the centers of two circles their center-sect Cy and calling their radii r^ and ^2, we have, in
—
regard to the relative positions of two circles, I. If ^>ri ^2, therefore the Os are wholly exterior.
+
the
2.
If ^
=f +
3.
\i c
Os
J
therefore the
r^,
r^,
but c
cut each other.
> the
Os touch
externally.
difference of radii, therefore
THE
=
^
If
4.
CIRCLE.
the difference of
radii,
1
therefore the
47
Os touch
internally.
If ^
5.
< the
difference of radii, therefore one
O
is
wholly
interior to the other.
411.
By
above are
33,
Rule
of
Inversion, the
five
inverses to the
true.
78. How must a line through one of the comtwo intersecting circles be drawn in order that the two circles may intercept equal chords on it ? 79. Through one of the points of intersection of two circles draw the line on which the two circles intercept the greatest
Exercises.
mon
points of
sect.
80.
of
two
If
lines be drawn through the point of contact the lines joining their second intersections with
any two
circles,
each circle will be parallel.
THE ELEMENTS OF GEOMETRY,
148
VI. Problems.
Problem To
412.
bisect
Given, ihe arc
III.
a given
BD.
to bisect
Required, Construction.
m F;
dX
C is Proof.
arc.
it.
Join
BD, and
bisect the sect
BD
F erect a perpendicular cutting the arc in
the
mid point of
C.
the arc.
]omBC,CD.
aBCF^ aDFC, (124. Triangles having
two sides and the included angle in each equal are congruent.)
chord
.-.
BC =
aicBC = (374. In the
413.
when
A
same
polygon
all its
circle,
is
chord CD, arc
CD.
equal chords subtend equal minor arcs.)
said to be circtimscribed about a circle
sides are tangents to the circle.
The
circle is
414.
A circle which touches one
then said to be inscribed
other two sides produced
is
i7i
the polygon.
side of a triangle
called an Escribed Circle.
and the
THE
CIRCLE,
149
Problem IV. 415.
To describe a
not all parallel^
circle
and do
touching three given lines which are
not all pass through the
same
point.
Given, three lines intersecting in A, B, and C.
Required, to describe a circle touching them. Construction. Draw the bisectors of the angles
These four bisectors
A
at
A
and C.
will intersect in four points, (9, (9i, O2, O^.
circle described with
any one of these points as center, and
its
perpendicular on any one of the three given lines as radius, will touch all
three.
Proof. distant from
186, every point in the bisector of an angle
By its
arms
Therefore, since
from
O
on
AB
is
equally
;
O
is
on the
bisector of
^
equals the perpendicular from the equals perpendicular from O on CB, since O of ^ C.
Ay the perpendicular on AC, which also is also on the bisector
O
THE ELEMENTS OF GEOMETRY.
150
four tangents common to two circles occur in two pairs intersecting on the common axis of symmetry. 416.
The
Problem V. 417.
In a given
circle to inscribe
a triangle equiangular
to
a
given triangle.
Given, a
O
and A ABC.
to describe in the
Required, Construction. point D. Make ^
Draw a
HVK ^
^
C,
Q
a
/^
tangent
and
4.
equiangular
GH
to
A ABC.
touching the
GDL =
circle
at
the
^ A.
K and L being on the circumference, join KL. Proof. (174.
^
DKL KDL =
The
4.
^K {397.
the required triangle.
three angles of a triangle are equal to a straight angle.)
=
An inscribed angle is An angle formed by a
{375.
is
B.
}^A,
and
^L
=
half the angle at the center
tangent and a chord
on the intercepted
is
arc.)
^
C.
on the intercepted
arc.)
half the angle at the center
BOOK
IV.
REGULAR POLYGONS. I.
Partition of a Perigon.
Problem 418.
To
bisect
I.
a perigon.
Solution. To bisect the perigon at the point O, draw any line through O. This divides the perigon into two straight angles, and all straight angles are equal.
419.
Corollary.
By drawing
a second line through
O
at
right angles to the first, we cut the perigon into four equal parts ; and as we can bisect any angle, so we can cut the perigon into 8, 16, 32, 64, etc., equal parts.
THE ELEMENTS OF GEOMETRY.
152
Problem To
420.
trisect
II.
a perigon.
';d
\
\
Solution. line
BO;
on
an equilateral
To
trisect the
^ For
^ Z>(9C is (174.
The
may
sect
OC;
on
OC
<9,
to
O
draw any by 132,
construct,
CDO.
DOB
one-third of a perigon.
is
one-third of a
st.
^,
three angles of a triangle are equal to a straight angle.) .*.
421.
perigon at the point
BO produced take a triangle
V
^
DOB
Corollary.
Since
cut the perigon into
Remark.
To
is
6,
two-thirds of a
we can
st.
bisect
12, 24, 48, etc.,
^.
any angle, so we
equal parts.
any given angle is a problem the of strict power beyond Elementary Geometry, which allows the use of only the compasses and an unmarked ruler. There is an easy solution of it, which oversteps these limits only by using two marks on the straight-edge. The trisection of the angle, the duplication of the cube, and the quadrature of the circle, are the three famous problems of antiquity. 422.
trisect
REGULAR POLYGONS:
Problem 423.
To cut a perigon
Solution.
Then ^ (174.
^
By
st.
^
ABChss\ng
.
three angles of any triangle are equal to a straight angle.) fifth
of a perigon at a "point O, construct, by 164,
^ Corollary.
GOB =
Since
etc.,
Problem To cut a perigon
4.A.
we can
cut a perigon into 10, 20, 40, 80,
425.
III.
314, describe an isosceles triangle
Therefore, to get a
424.
53
into five equal parts.
two-fifths of a
is
The
1
bisect
any angle, we may-
equal parts.
IV.
into fifteen equal parts, 'J3
=
Solution. At the perigon point O, by 420, construct the one-third of a perigon. one-fifth of a perigon. By 423, make ^
^
AOC
AOB =
Then five,
fifteen, ^ AOC contains BOC contains two.
of such parts, as a perigon contains and ^ contains three, therefore ^
AOB
So bisecting
^
BOC gives one-fifteenth of
a perigon.
THE ELEMENTS OF GEOMETRY.
154
Hence a perigon may be
Corollary.
426.
divided into 30,
60, 120, etc., equal parts.
II.
Regular Polygons and Circles.
Problem V. To inscribe
427.
number of
i?t
a
circle
a regular polygon having a given
c
sides.
This problem can be solved
if
a perigon can be divided into the
given number of equal parts.
For
let the
perigon at O, the center of the
circle,
be divided into a
number of equal parts, and extend their arms to meet B, C, D, etc. Draw the chords, AB, BC, CD, etc. Then shall ABCD, etc., be a regular polygon. For
if
the figure be turned about
its
center O, until
the circle in A,
OA
coincides
OB, therefore, because the angles are all equal, OB coincide with the trace of OC, and 6>C with the trace of OJD, etc. ;
with the trace of will
then
AB will coincide with the
CD,
etc.
;
.-.
.*»
BC, and
AB = BC ^ CD ^ the polygon
Moreover, since then .'.
trace of
is
BC with the
trace of
etc.,
equilateral.
^^C will coincide with the BCD = etc., t ^BC =
trace of
BDCj
-4.
.*.
Therefore
ABCD,
the given circle.
the polygon etc., is
is
equiangular.
a regular polygon, and
it
is
inscribed in
REGULAR POLYGONS. From
Remark.
428.
no advance was made
1
55
the time of Euclid, about 300 B.C.,
in the inscription of
regular polygons found that a regular polygon of 17 sides was inscriptible, and in his abstruse Arithmetic, published in 1 80 1, gave the following In order that the geometric division of the circle into n parts may be possible, n must be 2, or a higher power of 2, or else a prime number of the form 2^ i, or a product of two until
Gauss,
in 1796,
—
:
+
prime numbers of that form, or else the a of power of 2 by one or more different prime numproduct or
more
different
bers of that form.
In other words,
it is
necessary that 7t should contain no odd -\- i, nor contain the same divisor of
divisor not of the form 2"'
more than once. Below 300, the following 38 are the only possible values
that form
«:
2,
4,
3,
5,
6,
8,
of
10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48,
51, 60, 64, 6%, 80, 85, 96, 102, 120, 128, 136, 160, 170, 192, 204,
240, 255, 256, 257, 272.
Exercises. 81. The square inscribed in a circle is double the square on the radius, and half the square on the diameter.
Prove that each diagonal regular pentagon. 82.
is
parallel to a side of
the
An
inscribed equilateral triangle is equivalent to half a regular hexagon inscribed in the same circle. 84. An equilateral triangle described on a given sect is 83.
equivalent to one-sixth of a regular hexagon described on the
same
sect. If
85.
a triangle
is
the circumscribed circle
show that the radius of double that of the inscribed and the
equilateral, is
;
radius of an escribed, triple.
The end
%6,
angles
:
points of a sect slide on two lines at right
find the locus of its mid-point.
THE ELEMENTS OF GEOMETRY.
156
Problem VI. To circumscribe about a given having a given number of sides. 429.
C
jm:
This problem can be solved given number of equal parts.
For
let the
a regular polygon
circle
X'
a perigon can be divided into the
if
perigon at O, the center of the
circle,
be divided into a
number of equal angles, and extend their arms to meet the Bj C, D, etc. Draw perpendiculars to these arms at A, B,
circle in
D,
C,
A,
etc.
These
will be tangents. Call their points of intersection
Then For
if
shall
KLM,
the figure
with the trace of
K, L, M,
etc.
be a regular polygon. be turned about its center etc.,
OB,
O
until
OA
all
equal,
then, because the angles are
coincide with the trace of
OC, and
6^
C with
the trace of
coincides
OD,
OB
will
etc.
Therefore the tangents at A, B, C, etc., will coincide with the traces of the tangents at B, C, D, etc. Hence the polygon will coincide with its trace ; .*.
KL =
LM =
etc.,
also
^K therefore the polygon circle.
is
=^
:^L
regular,
=
and
^M =
it is
etc.
;
circumscribed about the given
REGULAR POLYGONS, 430.
Hence we can circumscribe about
Corollary.
cle regular
1
polygons of
3,
4,
5,
6,
10,
8,
12,
15,
16,
a
5/
cir-
17, etc.,
sides.
Problem VII. 431.
To circumscribe a
circle
about a given regular polygon.
Given, a regular polygon, as ABCDE. Required, to describe a circle about it. and ^ Construction. Bisect :4-
ABC by lines
£AB
O and
With center
O.
radius
OA
This shall be the required
Proof.
Join
OC,
circle.
Then
A (124.
intersecting in
describe a circle.
OBC^
A OBA,
Triangles having two sides and the included angle
in
each equal are con-
gruent.) .*.
^
OCB = ^ OAB = .-.
Similarly prove each .-.
^
half one
i.BCD\%
-4.
of the regular polygon,
bisected.
of the polygon bisected,
OA = OB = OC = OD = OE,
(148. In a triangle, sides opposite equal angles are equal.)
therefore a circle with radius
OA
passes through
circumscribed about the given polygon.
B^ C, D, Ey and
is
THE ELEMENTS OF GEOMETRY,
158
Problem VIII. To
432.
inscribe
a
circle in
a given regular polygon.
T C
Given, a regular polygon, ABODE, Required, to inscribe a circle in it. Construction. Bisect 2< and ^
EAB
ABC by lines
From O drop OP perpendicular to AB. The O with center O and radius OP shall be the O Proof. Join 00, OD, OE, and draw 0Q1.B0, X VE, and OT J. EA. Then
intersecting in
O.
A (124.
OBC^
required.
OR 1.
OD, OS
A OBA,
Triangles having two sides and the included angle
in
each equal are con-
gruent.)
/.
^ OCB
=
2j^
.*.-
Similarly prove each
OAB = ^
^
half
BCD
is
one
;^
of the regular polygon,
bisected.
of the polygon bisected.
Again,
A (176. Triangles
OBP ^
A OBQ,
having two angles and a corresponding side in each equal are congruent.)
.-.
OP =
OQ.
REGULAR POLYGONS.
1
59
Similarly,
OQ
OR = OS
=z
therefore a circle with radius at points F, Q,
R,
433.
it is
Corollary
cles of a regular 434.
OT,
touch
AB, BC, CD, BE, EA,
T;
S,
.*.
OP will
=^
inscribed in the given polygon.
I.
The
inscribed and circumscribed
cir-
polygon are concentric.
Corollary
II.
The
bisectors of the angles of a reg-
meet in a point which is the center both of the circumscribed and inscribed circles, and is called the center
ular polygon
all
of the regular polygon. 435.
Corollary
The
III.
perpendicular bisectors of the
sides of a regular polygon all pass through its center. 436. The radius of its circumscribed circle is called the
radius of a regular polygon.
The
radius of
its
inscribed circle
called its apothem. 437. The side of a regular hexagon inscribed in a circle is For the sects from the center to the ends equal to the radius.
is
of a side
make an
isosceles triangle,
one-third a straight angle
III.
;
Least Perimeter in Surface
438-
Any two
in
therefore
one
it is
of
whose angles
Equivalent
Figures. Isoperimetric Figures.
figures are called Isoperimetric
perimeters are equal.
is
equilateral.
— Greatest when
their
THE ELEMENTS OF GEOMETRY.
l60
Theorem
Of
439*
which
I.
all equivalefit triangles having the has the least perimeter.
same
base^ that
is isosceles
Lei
Hypothesis.
ABC
be an isosceles triangle, and same base.
ABC
any
equivalent triangle having the
AB + AC
Conclusion.
A A'
Proof.
BC.
\\
Equivalent triangles on the same base, and on the same side of
(253. Inverse.
are between the
Draw
CND
J_
same
BA produced in D. 4.NAC ^ )^ACB,
AA' meeting ^
(168. If a transversal cuts
two
it,
parallels.)
Join
AD.
parallels, the alternate angles are equal.)
^ACB = ^ABC, (126. In
an isosceles triangle the angles opposite the equal sides are equal.)
^ (169. If a transversal cuts
.-.
(128. Triangles having
ABC =
two
^ DAN,
parallels, the corresponding angles are equal.)
t.ACN^
t.ADN,
two angles and the included side equal
in
each are con-
gruent.)
(183.
The
.*.
AN
.'.
AD =
is
the perpendicular bisector of
AC,
and
AD =
CD,
AC.
locus of the point to which sects from two given points are equal, perpendicular bisector of the sect joining them.)
is
the
REGULAR POLYGONS.
l6l
But
BD < A'B + (156.
Any
A'D,
two sides of a triangle are together greater than the
AB
/.
-ir
AC
third.)
A'C.
Of all equivalent triangles, that which has the least perimeter. equilateral For the triangle having the least perimeter enclosing a given surface must be isosceles whichever side is taken as the 440.
Corollary.
is
base.
Theorem
II.
441. Of all isoperimetric triangles having the that which is isosceles has the greatest surface.
same
base,
MAS
Lei ABC be an isosceles triangle; and lei A'BC same base BC, have an equal perimeier; thai is, A'B + A'C = AB + AC. a ABC > A A'BC. Conclusion, Proof. The vertex A' must fall between BC and the parallel AJV; since, if it fell upon AN, by the preceding proof, A'B -\- A' C > AB + AC', and, if it fell beyond AN, the sum A'B + A'C would be still Hypothesis.
standing on ihe
greater.
Therefore the altitude of
A A'BC, and hence 442. is
also
Corollary.
equilateral
is
A
ABC
is
greater than the altitude of
surface.
Of
all
isoperimetric triangles, that which
the greatest.
For the greatest isosceles
its
triangle having a given perimeter is taken as the base.
whichever side
must be
THE ELEMENTS OF GEOMETRY.
l62
Theorem
III.
443* Of all triangles formed with the same two given sides that in which these sides are perpendicular to each other has the y
greatest surface.
Lei
Hypothesis.
AB, BC,
Conclusion,
Proof.
A
ABC
ABC, A'BC,
be two triangles having the sides be right BC\ and let ^
ABC
respectively equal to A'B,
is
a
Taking
ABC > ^C as
A A'BC. the
greater than the altitude
common base, the A'D of A A'BC.
Theorem
altitude
AB
of
IV.
444. Of all isoperimetric plane figures^ the circle contains the greatest surface.
Proof. With a given perimeter, there may be an indefinite number of figures differing in form and size. The surface may be as small as we please, but cannot be increased indefinitely. Therefore,
be one greatest forms.
the figures of the same perimeter, there must or several equivalent greatest figures of different figure,
among
all
REGULAR POLYGONS.
1
63
Every closed figure, if the greatest of a given perimeter, must be convex; that is, such that any sect joining two points of the perimeter Hes wholly within the figure. For let ACBNA be a non-convex figure, the sect
AB
line
joining two of the points in its perimeter, lying without if the re-entrant portion be revolved about the ; into the position AC'B, the figure has the same
AB,
the figure
perimeter as the
Now
let
face of part,
A
first figure,
AFBCA
given perimeter
AB to
A CB AC'BNA
then
;
but a greater surface.
be a figure of greatest surface formed with a
then, taking any point
A
in
its
perimeter, and drawing
bisect the perimeter, it also bisects the surface. For if the surone of the parts, as AFB, were greater than that of the other were revolved upon the line into CB, then if the part
AFB
AB
AF'BFA
the position AF'B, the surface of the figure would be greater than that of the figure and yet would have the same perim-
AFBCA,
eter.
Now
the angles
AFB
angles the chords
and
AFB
and
AF'B
must be right angles,
else the tri-
AF'B
AF,FB,
could be increased, by 443, without varying AF',F'B, and then (the segments AGF, FEB,
still standing on these chords) the whole figure would have increased without changing its perimeter. But is any point in the curve AFB therefore this curve is a \
AG'F', F'E'B,
F
semicircle. (384.
The
arc of a segment
Therefore the whole figure
is
is
a semicircle
a
circle.
if
the angle in
it is
right.)
THE ELEMENTS OF GEOMETRY.
164
Theorem 445*
Of
V.
all equivalent plane figures^ the circle has the least
peri^neter.
Hypothesis. file
Let
Proof.
A
;
C
be a circle, and
same surface as C. Conclusion. The perimeter of then,
Suppose by 444,
B
a
A
than that of A, same perimeter as the
figure
C
:.
has the less surface has the less
;
perimeter of
.*.
C<
perimeter of B, or of A.
Theorem 446.
any ofher figure having
less
circle with the
But, of two circles, that which
perimeter
C is
A
Of
VI.
all the polygons constructed with the same given which can be inscribed in a circle.
sides, that is the greatest
Hypothesis. a, b, c, d, e,
Lei
P
be
and inscribed
a polygon constructed with the sides a circle S, and let P' be any other
in
REGULAR POLYGONS. polygon constructed with the same sides, and not inscriptible
1
65
in
a
circle.
F > P'.
Conclusion. Proof. circular
P.
Upon
segments
The whole
the sides a,
b, c, etc.,
of the polygon
P'
construct
on the corresponding sides of S' thus formed has the same perimeter as the
equal to those standing
figure
circle S, therefore,
by 444, surface oi
S>
S'
\
subtracting the circular
segments from both, we have
P>P',
Theorem
VII.
447. Of all isoperimetric polygons having the the regular polygon is the greatest. sides of
samt number
i
Proof.
I.
The
greatest polygon P, of
all
the isoperimetric poly-
gons of the same number of sides, must have its sides equal ; for if two of its sides, as AB' B'C, were unequal, we could, by 439, increase its ,
substituting for the triangle
surface
by
triangle
ABC.
II.
The
AB' C
the isoperimetric isosceles
polygon constructed with the same number of by 446, be inscriptible in a circle. Therefore it is a
greatest
equal sides must, regular polygon.
^y. Of all triangles that can be inscribed in a whose vertices are the feet of the altitudes that given triangle, of the original triangle has the least perimeter.
Exercises,
THE ELEMENTS OF GEOMETRY.
i66
Theorem
VIII.
448. Of all equivalefit polygons having the same sides ^ the regular polyg07i has the least perimeter.
Hypothesis.
Let
P
be
M
a regular polygon, and any equivasame number of sides as P.
lent irregular polygon having the
Conclusion.
Proof.
number of
The perimeter of
P
is less
than that of
M.
Let TV be a regular polygon having the same perimeter
and the same number of
sides as
M
M
;
then,
by 447,
P
or
But, of two regular polygons having the same number of sides, that which has the less surface has the less perimeter ; therefore the perimeter of
P
is
less
than that of
-A^
or of
Theorem 449. eter, its
sides.
M.
IX.
If a regular polygon be constructed with a given perimsurface will be the greater, the greater the fiumber of its
REGULAR POLYGONS.
P be
Let
Proof.
167
the regular polygon of three sides,
and
Q
the
regular polygon of four sides, constructed with the same given perimeter.
AB
In any side
of
P take
D
P
the polygon ; polygon of four sides, in which the straight angle with each other ; then, by 447, of four sides is less than the regular isoperi-
may be regarded as an sides AD, DB, make a
any arbitrary point
irregular
P
the irregular polygon of four sides. metric polygon
Q
In the same manner
it
follows that
imetric polygon of five sides,
Q is less
than the regular isoper-
and so on.
Theorem X. 450.
the
lesSy
equivalent 7'egular polygonSy the perimeter will be the greater the number of sides.
Of
p
R
P
and Q be equivalent regular polygons, and Let have ihe Q greater number of sides. will be greater than that of Q. Conclusion. The perimeter of Proof. Let i? be a regular polygon having the same perimeter as Hypothesis.
lei
P
Q and
the
same number of
sides as P-, then,
Q>Ry therefore the perimeter of
P
is
or
by 449,
P>R',
greater than that of
R or of
Q.
BOOK
V.
RATIO AND PROPORTION. Multiples. 451.
In Book V., capital letters denote magni-
Notation.
tudes.
Magnitudes which are or may be of different kinds are denoted by letters taken from different alphabets.
^ +^
is
abbreviated into 2A.
A The
-h
A +A=
small Italic letters m^ n, /,
mA
g,
sA,
denote whole numbers.
A taken m times nA means A taken n times mA + nA = (m + n)A. nA taken m times mnA. means
;
;
.*.
is
452.
A greater magnitude is
magnitude when the
greater
is
said to be a Multiple of a lesser the sum of a number of parts
each equal to the less that is, when the greater contains the less an exact number of times. ;
Z69
THE ELEMENTS OF GEOMETRY.
I/O
A
lesser magnitude is a Submultiple^ or Aliquot Party 453. of a greater magnitude when the less is contained an exact number of times in the greater. 454. When each of two magnitudes is a multiple exactly contains, a third magnitude, they are said to be mensurable.
no magnitude which each
455. If there is
number
nitudes will contain an exact
of
of,
or
Com-
two given mag-
of times, they are called
Incommensurable.
Remark.
456.
important the student should know,
It is
that of two magnitudes of the same kind taken at hazard, or one being given, and the other deduced by a geometrical construction, it is very much more likely that the two should be
incommensurable than that they should be commensurable. To treat continuous magnitudes as commensurable would be to omit the normal, and give only the exceptional case. This makes the arithmetical treatment of ratio and proportion radically incomplete and inadequate for geometry.
Problem 457.
common
To find the greatest common submultiple or greatest divisor of two given magnitudes, if any exists.
1
I
Let
to as
I.
^
1
1
1
1
AB
and
CD be
^
Ic^n
the two magnitudes.
From AB, the greater, cut off as many parts as possible, each equal CD, the less. If there be a remainder FB, set it off in like manner often as possible upon CD. Should there be a second remainder
HD,
set
it
off in like
manner upon the
first
remainder, and so on.
RATIO AND PROPORTION. The process
will terminate
only
if
a remainder
I71
is
obtained which
is
an aliquot part of the preceding one ; and, should it so terminate, the two given magnitudes will be commensurable, and have the last remainder for their greatest
For suppose
common
HD
divisor.
the last remainder.
Then HD is an aliquot part of FB, and so of CH, and therefore CD, and therefore of AF. Thus being a submultiple of AF and FB, it is contained exactly in AB. And, moreover, it is the greatest common divisor of AB and CD. For since every divisor of CD and AB must divide AF, it must divide FB or CH, and therefore also HD. Hence the common divisor cannot be greater than HD. of
458.
Inverse of 457.
If
two magnitudes be commensura-
ble, the above process will terminate.
For now, by hypothesis, we have a greatest common
divisor
G.
But G is contained exactly in every remainder. For G, being a submultiple of CD, is also an aliquot part of AF, a multiple of CD and therefore, to be a submultiple of AB, it must be an aliquot part of FB the first remainder. SimFB being divisible by G, ilarly, CD and the first remainder must be so, and in the same way the the second remainder third and every subsequent remainder. But the alternate remainders decrease by more than half, and so the process must terminate at G for otherwise a remainder would be reached which, being less than G, could not be divisible by G. The above process applied 459* CoNTRANOMiNAL OF 457. to incommensurable magnitudes is interminable. If the above process be intermin460. Obverse of 457. ;
HD
;
magnitudes are incommensurable. On this depends the demonstration, given in 461, of a remarkable theorem proved in the tenth book of Euclid's " Elements."
able, the
THE ELEMENTS OF GEOMETRY.
1/2
Theorem 461.
I.
The side and diagonal of a square are incommensurable.
F
Hypothesis.
Lei
ABCD
,
be a square;
AB, a
side;
AC, a
diagonal.
Then
Conclusion.
AC> AB,
Proof.
will
AB and AC he incommensurable.
but <
2AB,
EC obtained by setting off on AC = AE AB. part Erect EE perpendicular to AC and meeting CE> in E. Join AE. Therefore a
first
remainder
is
di
A AE>E ^ A AEE, (179.
Right triangles are congruent when the hypothenuse and one side are equal respectively in each.)
.-.
Again, angles,
rt.
ECE,
A is
CEE half a
is rt.
E>E
=
isosceles,
EE. because one of the complemental
^, .;.
CE = EE = ED.
Hence a common divisor of EC and EC.
EC and DC would be also a
common
divisor of
But
EC and EC are again the side and diagonal of
fore the process
is
interminable.
a square, there-
RATIO AND PROPORTION. 462.
any
1
73
Another demonstration that the side a7id diagonal of
sqtiare are incommensurable.
F
If you suppose them commensurable, let represent their common measure ; that is, the aliquot part which is contained an exact number of times in the side, and also an exact number of times in the diagonal.
Let
m
represent the
number of times
P
is
contained in the side S,
and n the number of times F is contained in the diagonal I) ; so — nF, where m and n are whole numbers. S = mF, and Then, by 243, the square on Z> = 2 sq. on S, and
that
D
n^
therefore n^
is
therefore n
is
odd j
=
2m^,
an even number, an even number, since the square of an odd number
therefore n
may be represented by .*.
4^2
.*.
2^^
_ =
is
2^,
2m^, m^,
m
must be even, and can be represented by 2t, and therefore must be even, and so on forever. ^ That is, m and n must be such whole numbers that they will divide by 2, and give quotients which will again divide by a, and so on forever. This is impossible ; for even if m and n were high powers of 2, they therefore
would by constant
division reduce to
i,
an odd number.
Scales of Multiples. 463.
By
taking magnitudes each equal to A, one, and two, them together we obtain a set of magni-
three, four, etc., of
tudes depending upon A, and all known when A is known ;' and so on, each being namely. A, 2^, 3^, 4A, 5^, 6A, obtained by putting A to the preceding one. .
.
.
This we shall call the scale of multiples of ^. 464. If in be a whole number, mA and mBa.re multiples of
A
and B, or the same multiples
of
ca;lled equi-
^and
B.
THE ELEMENTS OF GEOMETRY.
1/4
465.
We
assume: As
than By so
is
mA
A
greater than, equal to, or less to, or less than
is
mB
greater than, equal
respectively. 466.
As
By
mA
33,
Rule
of Inversion,
greater than, equal than, equal to, or less than greater is
Theorem 467.
or less than
to,
mB^
so
is
A
B respectively. II.
Commensurable magnitudes have also a com7non mul-
tiple.
of
If A and B are commensurable A which is also a multiple of B.
Proof.
The
Let
C be
a
common
scale of multiples of
C,
is
magnitudes, there
divisor of
A
is
some multiple
and B.
C is 2C,3C
Now, by hypothesis, one of the multiples of this scale, suppose pC, equal to A, and one, suppose qC, is equal to B. Hence, by 465, the multiple pqC is equal to qA^ and the same
multiple
is
equal to
pB
;
therefore,
qA 468.
Inverse of 467.
=
pB.
Magnitudes which have a common
multiple are commensurable. Proof. into
If
pA =
qB, then
—
will
go
p
times into B, and q times
A.
Any whole number or fraction is commensurable with whole number and fraction, being each divisible by unity every 469.
over the product of the denominators. To find a common multiple, we have only to multiply together the whole numbers and the numerators of the fractions.
RATIO AND PROPORTION. Multiplication by a whole
470.
number
1
or fraction
is
75
dis-
tributive,
m{A +
jB
+ ...) = mA
-{-
mB +
..
,
471. Multiplication being commutative for whole or fractions, .*.
= mnA = nmA =
m(nA)
numbers
n{mA).
Magnitudes which are of the same kind can, being
472.
multiplied, exceed each the other.
Scale of Relation.
The
473.
kind
is
a
list
Scale of Relation of two magnitudes of the same of the multiples of both, all arranged in ascending
order of magnitude so that, any multiple of either magnitude being assigned, the scale of relation points out between which ;
multiples of the other it lies. 474. If we call the side of any square 5, and
Dy their scale of relation S,
n,
\oSy 116",
2S,
ZDy
2D, 3^, 4^, 126",
186*, 13Z), 196", 14Z), 141.S',
looD,
9Z>,
will
commence
3A 5^>4A
thus,
66", 76*, 5
—
A
its
86*,
diagonal
6A
9^.
7^
135, 146", ioZ>, 156", iiDy i65, i2Z>, 176",
20^, 216", i5Z>, etc.
1426*, etc.
14142136', iooooooZ>, 14142146', etc.
1414213566', iooooooooZ>, 1414213576', etc.
And we
have proved that no multiple of 5 will ever equal any multiple of D. 475. If A be less than B, one multiple at least of the scale of A will lie between each two consecutive multiples of the scale of B.
Moreover,
if
A
and
B are
two
finite
magnitudes of the same
THE ELEMENTS OF GEOMETRY.
176 kind,
however small
A may
be,
we may, by continuing
the scale
A
of multiples of sufficiently far, at length obtain a multiple of than B. greater
A
So we are
We We
I.
II.
justified in saying,
can always take can always take
mA nA
such that
A
but not greater than qB, provided that than q.
The
476. if
one
is
altered
scale of relation of
B ox pB.
greater than
it is
is
greater than/^,
less
two magnitudes
altered in size ever so little
;
for
than By and will
be changed
some multiple
magnitude can be found which will exceed, or
same multiple
/
fall
of the
short
before alteration, by more than the of, other original magnitude, and consequently the interdistribution of the multiples of the two original magnitudes will differ the
of
it
from the interdistribution of the multiples of one of the original magnitudes and the second altered. Hence, when two magnitudes are known, the order of their multiples is fixed and known. Inversely, if by any means the order of the multiples known, and also one of the original magnitudes, the other is fixed size, even though we may not yet be in a condition find
is
of to
it.
477*
The
order in which the multiples of
multiples of B,
when
all
A
lie
among
the
are arranged in ascending order of
magnitude, and the series of multiples continued indefinitely, determines what is called the Ratio of A to B, and written
A
:B,in which the first magnitude is called the Afitecedenty and the second the Consequent. 478. Hitherto we have compared one magnitude to another, with respect to quantity, only in general, according to the logical division greater than, equal to, less than.
But double, cial
cases
of
triple,
a
more
every two magnitudes
quadruple, quintuple, sextuple, are spesubtile relation of the
same
kind.
which
exists
between
RATIO AND PROPORTION. Ratio
is
1
77
the relation of one magnitude to another with
respect to quantuplicity.
and B interlie in the same order the ratio ^ to ^ is the same then Dy and the four magnitudes are said to be
479. If the multiples of
A
C and
as the multiples of to D, as the ratio
C
Proportio7taly or to
This
same
written
is
form a Proportion. A B w C D, and read \
:
''
A
to
B
is
the
C to
D!' The second pair of magnitudes may be of a different kind from the first pair. A and are called the Extremes, B and C the Means and is said to be a Fourth Proportional to A, By and C. 480. In the special case when A and B are commensurable, we can estimate their quantuple relation by considering what multiples they are of some common standard and so we can get two numbers whose ratio will be the same as the ratio of A to B. 481. The ratio of two magnitudes is the same as the ratio of two other magnitudes, when, any equimultiples whatsoever of the antecedents being taken, and likewise any equimultiples whatsoever of the consequents, the multiple of one antecedent as
D
^
D
;
is greater than, equal to, or less than, that of its consequent, according as the multiple of the other antecedent is greater
than, equal to, or less than, that of its consequent. to is the same as the ratio of 482. Thus, the ratio of
^
^
to
D
mA as mC
when
according
is
greater than, equal
to,
is
greater than, equal
to,
C
or less than nBy or less than nBy
whatever whole numbers nt and 71 may be. 483. Three magnitudes (A, By C) of the same kind are said to be proportionals when the ratio of the first to the second is the same as the ratio of the second to the third that is, when ;
A
:
B
:.
B
:
C.
In this case, C is said to be the Third Proportional to B the Mean Proportional between A and C.
By and
A
and
THE ELEMENTS OF GEOMETRY.
178
Theorem 484. Ratios
same as one
III.
which are the same as the same ratio are the
another.
Hypothesis.
A
Conclusion.
C
B w C
\
D
:
X
::
\
:
D^zh^
A
\
B w X
\
Y.
Y.
C
D
X
Y
the inverse of 479, the multiples of and have the and as those of the same and, being true of the multiples of and Y, therefore the multiples of and have the same interorder as those of and D.
Proof.
By
B
A
same interorder
-,
X
C
Thus ratios come under the statement, 485. Remark. "Things equal to the same thing are equal to each other;" and a proportion may be spoken of as an equality of ratios. 486. Since
therefore
pA
is
if
the
inverse of a
a pair of
numbers
greater than, equal
to,
/
complete definition is true, and q can be found such that
gB when pC is not Or less than qD, then the
or less than
respectively greater than, equal to. ratio oi to is unequal to the ratio of
A
B
The
first
C to
Z>.
to be to the second in a greater (or less) 487. ratio than the third to the fourth when a multiple of the first is
takes a more (or less) advanced position of the second than the
same multiple
among the
multiples
of the third takes
among
those of the fourth. 488. The ratio of ^ to ^ is greater than that oi C to D when two whole numbers m and n can be found such that mA
greater than nB, while 7nC is not greater than that is equal to nB, while mCis less than nD.
is
nD
;
or such
mA
Theorem
IV.
489. Equal magnitudes have the same ratio to the same magnitudey and the same has the same ratio to equal magnitudes.
RATIO AND PROPORTION. For So
and
if
A
if
if
mA
==
B, /.
mA =
/.
mB
> nC,
equal, equal
;
if less, less
A
/.
And
also if
if
79
mJB,
> nC;
;
C
:
B
::
C.
:
pT > mA, /7'> mB;
.-.
and
1
equal, equal
;
if less, less
;
T A
/.
T
::
:
B.
:
Theorem V. magnitudes^ the greater has a greater other any magnitude than the less has ; and the same has a greater ratio to the less, of two other magnimagnitude 490.
Of two unequal
ratio to
tudeSy than it has to the greater. If
A
> B,
m
mB
can be found such that
is
less
than
mA
by a
greater magnitude than C.
will
Hence, if mA = nC, or if mA is between be less than nC ; therefore, by 488,
A Also, since
C
.-.
491. li
From 489 and
:
490,
:
B
by :
A :C>B',C,
or
\i
B
>
C
>
33, ::
A=
:
mA,
;
of Inversion,
B,
B.
T :A B. T '.A>T A < B, /.
:B,
.-.
If
A
:
C
:
C, or ii
+
A,
Rule
T
{n
C.
:
not >
is
A'.C v,B:CyOr\i T A .-.
li
C
while « C
nC > mB,
nC and
-.B,
i)C,
mB
THE ELEMENTS OF GEOMETRY.
l80
Theorem
VI.
492. If any number of magnitudes be proportionals^ as one the antecedents is to its consequent^ so will all the antecedents of taken together be to all the consequents.
Let
A
'.
B
C
'.'.
D
'.
E
'.'.
F,
\
then
A
B
\
A
'.'.
E B
C^
-^
',
^-
D -^ F.
For as mA >, =, or < nB, so, by inverse of 482, < nZ> ; and so also is 7nE >, =, or < nF; so also
.'.
mA + mC
is
and therefore so
is
-{-
mE >,
w(^
+ C+
B
A+ C+E
E)
=,
or
=,
>,
or
<
is
mC
>,
nB + nD +
< n(B
-{-
D
-\-
=,
or
nF,
F);
whence
A
:
i:
Theorem 493.
same as
but
as
mpA
pA
=
>,
=,
or
is
Z>
+
F,
VII.
pmAj and mqB
pmA >,
=,
or
mB w A
\
< qB,
.*.
so
-\-
The ratio of the equimultiples of two magnittides is the the ratio of the magnitudes themselves.
mA For
B
:
as
so
=
mpA
is
A
pA
:
B.
>,
=,
or
<
mqB
',
qmB^ >,
=,
< qmB, whatever be /.
\
B
::
or
< qB,
the values of
mA
:
mB.
p
and q,
RATIO AND PROPORTION-.
l8l
Theorem VIIL If four magnitudes be proportionals then, if the first be greater than the third, the second will be greater than the fourth; 494*
and if
y
A
Given, li
and if
equal, equal ;
A
>
\
B w C
\
C,
therefore,
from our
therefore,
by 491,
D.
A
.-.
'.
A=
C, ,',
A
,\
C
i
:
:
.-.
li
A
<
B
>
C
n
>
C
B
> D.
B
::
£>
::
B
=Zf,
B
<
:
B;
'.
B;
h)rpothesis,
C li
less, less.
C, .-.
A
.-.
C
:
:
.-.
C C
£> <
B
C C
<
:
B,
:
B,
:
B,
:
B,
Z>.
Theorem IX. If four magnitudes of the same kind be proportionals
495.
they will also be proportionals
Let
A
\
B
\\
C
:
2?, the four
then alternately,
A
For, by 493,
mA: mB
'.'.
/.
therefore,
by 494,
being true for
all
mA >,
C
'.
magnitudes being of the same kind,
'.'.
B
'.
D.
\
:
\
:
=,
m A
< «C, and «,
or
:
C
y
alternately.
A B :: C D w nC mA mB \\ nC nD
values of /.
when taken
i:
mB >,
as
B
'.
D.
'.
nD,
\
=,
or
< nD; and,
this
1
THE ELEMENTS OF GEOMETRY.
82
Theorem
X.
If two ratios are equals the sum of the antecedent and consequent of the first has to the consequent the same ratio as the sum of the antecedent and consequent of the other has to its 496.
consequent. \i
A
'.
B
'.'.
C
'.
D,
then
A
-^
B B w C+D \
:
I?.
Proof. Whatever multiple oi A -{- B we choose to examine, take same multiple of A, say 1 7^, and let it lie between some two multiples of B, say 23^ and 24B ; then, by hypothesis, 17C lies between 2$Z> and 24D. the
Add B)
lies
I
'jB to all the
between
first,
and
i
yZ) to
all
40B and 41B, and i7(C
the second
+
D)
;
then
1
7
(^
+
between 4oZ> and
412? ; and in the same manner for any other multiples.
BOOK VL RATIO APPLIED. I.
Fundamental Geometric Proportions.
Theorem
I.
497. If two lines are cut by three parallel lines the intercepts on the one ai-e to 07ie another in the same ratio as the corresponding intercepts on the other. ^
Hypothesis.
Lei the three parallels AA',
other lines in A, B, C, and
A\ B\ C,
BB\ CC,
respectively. 183
cut two
1
THE ELEMENTS OF GEOMETRY,
84
AB
Conclusion.
On
Proof.
BC
:
the line
::
ABC,
A'B'
= mAB, and, in the same way, BN = From
same side of B. in J/' and iV^'.
^
B'M'
.-.
(227. If three or
B'C,
:
by laying
off
n.BC,
M and N draw
=
BM
AA\
A'B'C
sects
taking
lines
1|
AB, take J/ and JV on
on one parallels intercept equal intercept equal sects on every transversal. )
more
sects
m
But whatever be the numbers
cutting
the
B'N'^n.B'C
and
m.A'B\
m
and
n, as
transversal, they
BM (or m
.
AB)
.
.5'C")
is
>,
= ox
so
is
B'M'
(or
w
.
A'B') respectively .-.
498.
whether
AB
:
BC
::
>,
=,
A'B'
or :
< B'JST
(or n
;
^'C.
Remark.
B is
Observe that the reasoning holds good, between A and C, or beyond A, or beyond C,
If the points A and ^' coincide, the I. be a triangle therefore a line parallel to one side of a triangle divides the other two sides proportionally. If two lines are cut 500. Corollary II. by four parallel on the one are to one another in the same lines, the intercepts
499.
Corollary
figure ACC will
;
corresponding intercepts on the other. is produced, and the line cut at a point outside the sect AB, the sect is said to be divided exter-
ratio as the
501. If a sect
AB
P
AB
nally at P,
AB,
and
AP
and
BP
are called External Segments of
RATIO APPLIED. In distinction,
if
P
the point
is
1
on the sect AB^
it
85
said
is
to be divided internally.
Theorem
II.
A
given sect can be divided internally into two segments having the same ratio as any two given sectSj and also 502.
externally unless the ratio be 07ie of equality ; and^ in each casCj there is only one such point of division.
R
B
G
J"
C
w
\
\\ \ JI\
X
Given, ihe sect On a line from
AB.
A
making any angle with AB, take
equal to the two given sects.
Draw If
it
drawn
||
could be divided internally at
A G would (476.
The
\\
CG to
meet
be to
AC
and
CD
BD.
DB and meeting AB F in the given ratio.
CF
internally at
Join
in F.
G in
By the
497,
same
AB
ratio,
is
divided
BH being
AD in H,
GB as ^ C to
scale of relation of
CH, and
two magnitudes
therefore not as
will
be changed
if
^ C to
one
is
CD.
altered in
size ever so little.)
Hence ratio.
F
is
the only point which divides
AB internally in the
given
THE ELEMENTS OF GEOMETRY,
i86
D
A
CD
and are on the same side of C, be taken so that Again, if the like construction will determine the external point of division. would In this case the construction will fail if AC, for
D
CD =
coincide with A.
As above, we may
also prove that there
can be only one point of
external division in the given ratio.
CN. 503.
Inverse of 499.
triangle proportionally
For a
parallel
is
from
second side in the same
A line which
divides two sides of a
parallel to the third. one of the points would divide the ratio,
but there
only one point of
is
division of a given sect in a given ratio.
Theorem
III.
504. Rectangles of equal altitude are^ to one another in the ratio as their bases.
same
^
:
B Let
AC, BC, be two
their bases
In the
OA, OB, on line
OAB
JV
rectangles having the the same side of OC,
OM = m ,OA, MC and NC.
take
complete the rectangles
M
common
and
side
ON =
OC, and
n.OB, and
RATIO APPLIED. Then < ON, so .*.
MC = m.AC, and NC = n.BC; MC respectively >, =, or < NC\
and
OM
as
is >,
1
87
=,
or
is
^C
rectangle
^C
rectangle
:
::
base
OA
:
OB.
base
Corollary.
Parallelograms or triangles of equal tude are to one another as their bases. 505.
Theorem ter
Let
O
and
C
AM =
m AB .
Also take an arc
and
n
AOM>,
=,
But as arc
XN =
CN equals
.
n
angles at the cen-
circles
;
AB, XL, any
;
(365. In equal circles, equal arcs
CK
circles,
arcs on which they stand.
be the centers of two equal
OA
then the angle or the sector between
<
IV.
506. In the same circley or in equal and sectors are to one another as the
two arcs in them. Take an arc
alti-
.
and
OM equals m A OB. .
subtend equal angles at the center.)
KL
;
then the angle or sector between
KCL. or
<
KCN,
so respectively
is
arc
AM
>,
=,
or
KN\
(370 and 372. In equal circles, equal angles at the center or equal sectors intercept equal arcs, and of two unequal angles or sectors the greater has the greater arc.)
/.
AOB
:
KCL
:
:
arc
AB
:
arc
XZ,
THE ELEMENTS OF GEOMETRY,
1 88
II.
Similar Figures.
Similar figures are those of which the angles taken in the same order are equals and the sides between the equal angles 507.
proportional
The 4.
ABCD
figure
B'y etc.,
and
AB
is
similar to
A'B'C'iy
:
A'B'
BC
::
:
B'
C
::
Theorem 508.
\i 4.
A ^
A\
4.
4.
B^
also
CD
i
CI/
:
:
etc.
V.
Mutually equiangular triangles are similar,
3.
Hypothesis, at F, G, and H. Conclusion.
Proof.
ABC, FGH,
AB
FG w BC GH FGH to A ABC so that
Apply A
GH
having ^.s at A, B, and C,
i^s
.
\
falls on with B, and Then, because ^ Cr
:
BC.
= ^ ^, GF .-.
falls
on BA.
:
CA
:
=
^.s
HF,
the point
G
coincides
RATIO APPLIED. Now,
since
BF'H'
2j^
= ^A
89
by hypothesis,
F'H'
/.
1
AC,
II
(166. If corresponding angles are equal, the lines are parallel.)
therefore,
AB
by 498,
:
BF'
CB
-.:
:
BH\
FGH so that the at ZT and CA HF. C coincide, we may prove that BC GH A triangle is similar to any triangle cut 509. Corollary. In the same way, by applying the
A
2< s
\
off
by a
line parallel to
one of
:
:
:
its sides.
Theorem
VI.
510. Triangles having their sides taken are similar,
i7t
order proportional
B
J£
jF
FG w BC GH :: CA HF, = ^ H, and ^ A = ^ F. On BA take BF" = GF, and draw F'H CA AB F'B :: BC BH' CA H'F'.
Hypothesis.
AB
Conclusion.
^ C
Proof.
:
\
-.
\\
.-.
:
:
:
:
:
(508. Mutually equiangular triangles are similar.)
Since
FG =
F'B,
/.
BC
:
(484. Ratios equal to the
therefore,
by 491,
In the same way .',
But
F'BH
is
GH same
i:
BC BH\
GH = BH, HF — H'F', aFGH^aF'BH\
similar to
ABC.
:
ratio are equal.)
;
I
THE ELEMENTS OF GEOMETRY.
go
Theorem
VII.
Two
triangles having one angle of the one equal to one the other^ and the sides about these angles proportional, angle of are similar. 511.
Hypothesis.
Conclusion,
Proof.
In
^ B = ^ G, and AB BC w FG GH, a AB C '^ t^ FGH (using ~ for the word " similar ") BA take BF' = GF, and draw F'L AC, :
(508.
A F'BL
AB
:
BC
FG by construction AB BC ',
/.
therefore,
- A ABC,
Equiangular triangles are similar.)
.-.
F'B —
.
\\
:.
But
:
FB BL X
F'B
',:
;
:
BL.
therefore,
from our hypothesis,
','.
F'B
:
GH,
i:
F'B
:
GH,
by 491,
.-.
(124. Triangles having
BL = GH, A F'BL ^ A FGH,
two sides and the included angle respectively equal are congruent.)
.-.
A
FGH -
A ABC.
RATIO APPLIED.
Theorem
I91
VIII.
512. If two triangles have two sides of the one proportional two sides of the other^ and an angle in each opposite one corresponding pair of these sides equal, the angles opposite the other pair are either equal or supplemental. to
The
angles included by the proportional sides are either or unequal. equal Case I. If they are equal, then the third angles are equal. (174.
The sum
of the angles of a triangle
is
a straight angle.)
Case II. If the angles included by the proportional sides are unequal, one must be the greater. Hypothesis.
4.A=^^F,
^B>^G.
Conclusion.
Proof.
ABC and FGH ^ C
Make ^
4- .^ ZT =^
ABD
A
.-.
(508. .-.
=
as
7^
^
st.
AB
with
\
.
G',
ABD -
A
FGH;
Equiangular triangles are similar.)
AB
'.
BD
'.'.
FG
GH'y
\
therefore, from our hypothesis,
AB therefore,
by 491,
BD =
\
BD
','.
AB
BC;
'.
BC, .*.
^ C
But ^ BDC + ^ BDA = /.
BC
st.
= ^ BDC. ^,
^ C+ ^
Zr
=
St.
^.
w
FG
\
GH,
THE ELEMENTS OF GEOMETRY.
192
Corollary. If two triangles have two sides of the one proportional to two sides of the other, and an angle in each 513.
opposite one corresponding pair of these sides equal, then if one of the angles opposite the other pair is right, or if they are oblique, but not supplemental, or if the side opposite the given angle in each triangle is not less than the other proportional
the triangles are similar. 514. In similar figures, sides between equal angles are called The ratio of a side of one Homologous^ or corresponding. side,
polygon to
its
homologous side
in a similar
polygon
is
called
the Ratio of Similitude of the polygons. Similar figures are said to be similarly placed vih^n each side of the one is parallel to the corresponding side of the other.
Theorem
IX.
If two unequal similar figures are similarly placed^ all joining a vertex of one to the corresponding vertex of the
515. lines
other are concurrent.
Hypothesis.
ABCD
and A'B'C'U two
similar figures
simi-
larly placed.
Conclusion.
Proof.
Since
The
lines
AA\ BB\ CC,
.-.
AA' and BB'
Call their point of intersection P,
AP
etc.,
meet
AB and A'B' are unequal,
:
A'P
'.:
BP
:
are not
||
in a point P.
.
Then
B'P
::
AB
:
A'B,
(508. Equiangular triangles are similar.)
RATIO APPLIED. In the same way,
if
BB' and
BQ
CC meet in
B'Q
::
:
A'B'
::
:
B'P
::
'.
I93
BC
:
Q, then
B'C,
But, by hypothesis,
/. .'.
{502.
A line
AB BP
the point
can be cut only
Q
BC BQ
'.
:
B'
C
,
B'Q,
coincides with P.
at a single point into external
segments having a given
ratio.)
Corollary. Similar polygons may be divided into the same number of triangles similar and similarly placed. For if, with their corresponding sides parallel, one of the polygons were placed inside the other, the lines joining corresponding vertices would so divide them. 516.
517.
The
point of concurrence of the lines joining the equal is called the
two similar and similarly placed figures Center of Similitude of the two figures.
angles of
518.
Corollary.
The
sects from the center of similitude
where it meets corresponding sides along any of the similar figures are in the ratio of those sides. line to the points
Exercises,
^j.
Construct a polygon similar to a given
polygon, the ratio of similitude of given.
the two polygons being
THE ELEMENTS OF GEOMETRY.
194
Theorem X.
A
perpendicular front the right angle to the hypothenuse divides a right-angled triangle into two other triangles similar 519.
to the
whole and
to
one another.
Hypothesis,
a
ABC
Conclusion,
t.
ACD ^
right-angled af C.
CD ± AB, Proof.
ABC -^
^ CAD = ^BAC, .-.
(174.
t.
The
and
t^
rt. 2^
CBD.
ADC =
rt.
^ ACB,
^ACD = ^ABC,
three angles of any triangle are equal to a straight angle.) ..
In the same way we
aACD ~ hABC. CBD ~ A ABC; ~ aACD A CBD.
may
.-.
prove
A
Corollary I. Each side of the right triangle is a proportional between the hypothenuse and its adjacent
520.
mean
segment. For since A
A CD ~ A ABC, /.
521. tional
Corollary
II.
AB
:
AC
::
AC
:
AD.
The perpendicular
between the segments
of the
is
a
mean
propor-
hypothenuse. 522. Corollary III. Since, by 383, lines from any point in a circle to the ends of a diameter form a right angle, therefore, if from any point of a circle a perpendicular be dropped
RATIO APPLIED.
195
upon a diameter, it will be a mean proportional between the segments of the diameter.
XL
Theorem
523. The bisector of an interior or exterior angle of a triangle divides the opposite side internally or externally in the ratio of the other two sides of the triangle.
ABC any A. BD the bisector of ^ at B. AB BC :: AD DC. Proof. Draw AF BD. Then, of the two angles at B given equal by hypothesis, one Hypothesis.
Conclusion.
\
:
\\
the corresponding interior angle at F, alternate angle at (168.
A
line cutting
A, two parallels makes alternate angles equal, and (169) sponding angles equal.) .-.
(126. If
two angles of a
AB =
'.BC
'.'.
be parallel to a side of a
BF.
AD
'.
triangle,
DC, it
cuts the other sides propor-
tionally.) .-.
corre-
triangle are equal, the sides opposite are equal.)
BF
But (499. If a line
equals
and the other the corresponding
AB :BC::AD:
DC,
THE ELEMENTS OF GEOMETRY.
196
Since, by 502, a sect can be cut at only one and at only one point exter-
524. Inverse.
point internally in a given ratio,
nally in a given ratio, therefore, by 32, Rule of Identity, if one side of a triangle is divided internally or externally in the ratio of the other sides, the line drawn from the point of division to
the opposite vertex bisects the interior or exterior angle. 525. When a sect is divided internally and externally into segments having the same ratio, it is said to be divided harmonically. 526.
angle
at
The bisectors of an interior and exterior one vertex of a triangle divide the opposite side har-
Corollary.
monically.
Theorem 527.
If a
and Q, the A and B.
sect,
sect
is divided harmonically at the points P will be divided harmonically at the points
AB,
PQ
!:b
.
I
3L
Hypothesis.
XII.
The sect
AB
)
,
-n
a
divided internally at F,
and exter-
nally at Q, so that
AF therefore,
by
by
BF
::
AQ BQ;
AF
::
BQ AQ;
BQ
::
AF
:
inversion,
BF therefore,
:
:
:
alternation,
BF The
:
:
AQ.
points A, B, and F, Q, of which each pair divide the sect terminated by the other pair, are called harmonically four Harmo7iic Points, 528.
RATIO APPLIED.
197
Rectangles and Polygons.
III.
Theorem
XIII.
529' If four sects are proportional^ the rectangle contained by the extremes is equivalent to the rectangle contained by the
means.
ad
ac
Let the four sects
a, b,
Cj
Then
d,
be proportional.
rectangle
ad
=
be.
On a and on b construct rectangles with and on d construct rectangles of altitude a.
Proof.
On
c
altitude
=
c.
Then a
:
b
::
ac
:
and
be,
c
d
by hypothesis, a
/.
therefore,
ac
b
'.
:
be
::
c
::
ac
by 491, be
=
ad.
::
ac
:
ad.
each other as their bases.)
(504. Rectangles of equal altitudes are to
But,
',
d,
:
:
ad]
THE ELEMENTS OF GEOMETRY.
198
If two rectangles are equivalent, the sides form the extremes, and the sides of the other
530. Inverse. of the
one
will
the means, of a proportion.
a^.
^c
Recfangle ad b w c a
Hypothesis.
Conclusion, Proof. but ac
:
be
:
a
—
ad
Since :
\
-.
:
.*.
531.
Corollary.
angle of the
extremes
a
If is
be.
therefore,
be,
and ac
b,
= d.
:
ad
:
b
:
::
:
c
by 489, ac c :
:
be
ac
:
ad\
d,
:
d.
three sects are proportional, the rectequivalent to the square on the mean.
Theorem XIV. 532'
If two chords
intersect either
within or without the
circle, the rectangle co7ttained by the segments of the ofie is equivalent to the rectangle contained by the segments of the other.
RATIO APPLIED.
AB and CD intersect in AP PB = rectangle CP PD.
Hypothesis.
Let the chords
Conclusion.
Rectangle
Proof.
^PAC = {376.
Angles
199 P.
.
.
^ PDB,
in the
same segment
of a circle are equal.)
and
APC = A APC -
^ .-.
^ BPD, A BPD,
(508. Equiangular triangles are similar.)
.-.
AP
.-.
AP.PB = CP.PD.
'.
CP
'.'.PD
\
PB,
Let the point P be without the circle, 533. Corollary. coincide and suppose DCP to revolve about P until C and then the secant DCP becomes a tangent, and the rectangle
D
PD
;
becomes the square on PC. Therefore, if from a point CP. without a circle a secant and tangent be drawn, the rectangle of the whole secant and part outside the circle is equivalent to the square of the tangent.
THE ELEMENTS OF GEOMETRY.
200
Theorem XV. The rectangle of two sides of a triangle is equivalent to two sects drawn from that vertex so as to make the two sides and produced^ one to the base, with equal angles 534.
the rectangle of
^
the other to the circle circumscribing the triangle.
ABE = ^ CBD. Rectangle AB .BC =z DB
Hypothesis.
^
Conclusion.
Proof.
Join
(376.
4.
CBD =
4.
AE.
Angles in the same segment of a
ABE,
CBD ~ A ABE, AB BE DB BC, AB.BC = DB,BE,
Corollary
DB(BD 532).
;
Therefore,
BD'
when
BD
'.
BE coincide, they bisect AB BC ^ DB BE = + BD DE = BD' CD DA (by
If
I.
therefore
+ DE) =
'.'.
'.
.-.
B
circle are equal.)
A
/.
535.
BE,
by hypothesis, .-.
the angle
.
Then
and
rectangle .
.
.
-\-
.
the bisector of an angle of a triangle
RATIO APPLIED.
201
meets the base, the rectangle of the two sides is equivalent to the rectangle of the segments of the base, together with the square of the bisector.
536.
If BD be a perpendicular, BE is a II. BAE is then right therefore in any triangle
Corollary
diameter, for angle
;
^
C
the rectangle of two sides is equivalent to the rectangle of the diameter of the circumscribed circle by the perpendicular to the base from the vertex.
Prove that the inverse of 535 does not BC. Prove 535 when it is an exterior angle which is bi-
Exercises. hold
when
89. sected.
ZZ.
AB =
THE ELEMENTS OF GEOMETRY.
202
Theorem XVI. 537. The rectangle of the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles
of
its opposite sides.
Hypothesis.
ABCD
Conclusion.
Rectangle
Proof.
By
164,
an inscribed quadrilateral.
A C BD = AB .CD + BC. DA,
make ^
.
DAF =
^ BAC,
Toeachadd^i^^C. Then, in ^^
A CD and ABF,
^DAC =
4.BAF;
also
4.ACD = ^ABD, (376.
Angles inscribed in the same segment are equal.)
.-.
(508.
.-.
therefore,
l\ACD^
A ABF,
Equiangular triangles are
AC AB :
::
CD
similar.)
:
BF;
by 529,
AC BF ^ AB CD, Again, by construction, ^ DAF = ^ BAC, .
.
RATIO APPLIED. Moreover,
^
ADF =
%.
ACB,
(376. Angles inscribed in the
.-.
203
same segment are
equal.)
t.ADF'^ ^ABCy
(508. Equiangular triangles are similar.)
.-.
therefore,
,\
by 529,
AB .CD
-{-
AC AD :
::
To
alter
3L
:
DF',
AC ,FD = AD.BC, BC .BA = AC.BF-^ AC.FZ> = AC{BF + FD) = AC. BD. Problem
538.
CB
a given
sect in
I.
a given
ratio.
^-^^
204 539. This
is
ELEMENTS OF GEOMETRY.
the same as finding a fourth proportional to
three given sects. To find a third proportional to a and the above construction.
Every
540.
some
alteration of a
by
make
is
magnitude
AB =
^ in
an alteration in
ratio.
Two
more alterations are jointly equivalent to some one and then this single alteration which produces the alteration, of two is said to be compounded of those two. effect joint The composition of the ratios of ^ to ^ and (7 to Z> is performed by assuming F, altering it into Gy so that F G :\ a by then altering G into H, so that G -.: C D. The joint effect turns F into H) and the ratio oi F to is the ratio compounded of the two ratios, a b and C D. 541. A ratio arising from the composition of two equal or
:
:
:
H
:
H
:
ratios is called the Duplicate
:
Ratio of either.
Theorem XVII. 542. Mutually equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
and
Hypothesis.
In
^ AC, ^ BCD =
Conclusion.
£:7
AC
BC
:
CH,
-.
^y
CG
^
=z ratio
HCF of C7 CG. compounded of DC
:
CF,
RATIO APPLIED.
205
Place the £j^ so that HC and CB are in one line DC and CF are in one line. Complete the BF.
Proof.
;
then,
/:7
by log,
Then
£yAC
:
^BF
.:
DC
\'.
BC
:
CF,
'.
CH,
and
^BF
'.
C7CG
(505. Parallelograms of equal altitude are as their bases.)
c7
.'.
BC
AC
has to
z:7
CG
the ratio
DC
compounded of
:
CF
and
CH,
:
Corollary
Triangles which have one angle of the to one angle of the other, being halves of equiangular parallelograms, are to one another in the ratio compounded of the ratios of the sides about those angles. 543.
I.
one equal or supplemental
Corollary
II. Since all rectangles are equiangular therefore the ratio compounded of two ratios parallelograms, between sects is the same as the ratio of the rectangle con-
544.
tained by the antecedents to the rectangle of the consequents. If the ratio compounded of a b\ be written a'y and b :
— ^ b a .
,
this corollary proves
of ratios obeys the
Thus
7
b
.
:
^ b— = a-^ b .
a
same laws
;
and the composition
as the multiplication of fractions.
? =: ^, and so the duplicate ratio of two sects b
is
the
b^
same
as the ratio of the squares on those sects. be seen hereafter that the special case of 542, when the parallelograms are rectangles, is made the foundation of all It will
mensuration
of surfaces.
Exercises.
90.
If
mutually equiangular parallelograms are
equivalent, so are rectangles with the 91.
same
sides.
Equivalent parallelograms having the same sides as
equivalent rectangles are mutually equiangular.
THE ELEMENTS OF GEOMETRY.
206
Theorem XVIII. 545. Similar triangles are to one another as the squares on their corresponding sides.
Let the similar triangles
AB, AC,
sides
BC
II
HK.
Since,
Join
by 505,
AHC
:
Exercises. is
as to
AH, AK, and
have the therefore
CH, triangles of equal altitude are as their bases,
A
\
:
:
:
triangles
be placed so
ABC A AHC w AB AH, A AHX :: AC AK = AB AH, by hypothesis AB AB AB"A ABC A AHK = .-.
and A
ABC^ AHK,
along the corresponding sides
92.
The
:
AH AH
ratio of the surfaces
;
AH^ of
two similar
the square of the ratio of similitude of the
tri-
angles.
an angle of a triangle also bisect a the is isosceles. side, triangle 94. In every quadrilateral which cannot be inscribed in a 93. If the bisector of
circle,
sum
the rectangle contained by the diagonals is less than the two rectangles contained by the opposite sides.
of the
95. The rectangles contained by any twp sides of triangles inscribed in equal circles are proportional to the perpendiculars on the third sides.
96.
Squares are to one another in the duplicate ratio of
their sides.
RATIO APPLIED.
207
Theorem XIX. 546. Similar polygons are to each other as the squares on their corresponding sides, ^ j^
J2)'
ABCD
Hypothesis.
which
BC
and B'
By
By
and
AE'CU
two
similar polygons,
of
are corresponding sides.
ABCD
Conclusion.
Proof.
C
:
A'B'C'iy
516, the polygons
may
:
:
AB^
:
A'B'\
be divided into similar
triangles.
545, any pair of corresponding triangles are as the squares on
corresponding sides,
therefore,
t.ABD
BI>
t^BCD
BC^
t.A'B'D
B'Lf^
/^B'C'jy
B'C'^
by 496,
ABD + A A'B'D^ + A
A BCD A B'C'D^
In the same way for a third pair of similar
BC B'C^
triangles, etc.
THE ELEMENTS OF GEOMETRY.
208
Corollary. If three sects form a proportion, a polyfirst is to a similar polygon similarly described on the on gon the second as the first sect is to the third. 547.
Theorem XX. 548. The perimeters of any two regular polygons of the same number of sides have the same ratio as the radii of their circum-
scribed circles.
Proof. of sides are
The all
angles of two regular polygons of the same number equal, and the ratio between any pair of sides is the
same, therefore the polygons are similar. But lines drawn from the center to the extremities of any pair of sides are radii of the circumscribed circles, and make similar triangles,
AB therefore,
by 496,
RATIO APPLIED.
Problem
209
II.
To find a mean proportional between two
549.
give?t sects.
J)
Let line,
AC
{383.
BD
B
^C BD
perpendicular to
(521. If
the two given sects. Place AB, BC, in the same describe the semicircle ADC, From draw
AB^ BC, be
and on
An
is
the
mean
proportional.
angle inscribed in a semicircle
is
a right angle.)
from the right angle a perpendicular be drawn to the hypothenuse, be a mean proportional between the segments of the hypothenuse.)
Exercises.
97. If the given sects
as in the above figure,
between them
how would you
were
^C and ^6",
find a
mean
it
will
placed
proportional
1
sum of two sects is greater than the mean between them. proportional The 99. rectangle contained by two sects is a mean proportional between their squares. 100. The sum of perpendiculars drawn from any point within an equilateral triangle to the three sides equals its 98. Half the
altitude.
The bisector of an angle of a triangle divides the into two others, which are proportional to the sides triangle of the bisected angle. loi.
102.
Lines which trisect a side of a triangle do not
the opposite angle. 103. In the above figure, F, then
a
ABD =
a FCB.
if
trisect
AF _L AD meet DB produced at
THE ELEMENTS OF GEOMETRY.
2IO
Problem 550.
On a given
III.
a polygon similar
sect to describe
to
a given
polygon.
Let
ABODE be the given polygon,
and A'B' the given
sect.
Join^Z>, AC.
Make
= ABC, ^B'A'C = ^BAC, ^B'C'A' = ^BCA,
4.
.'.
A'B'C
i.
and
A A'B'C
A ABC.
4.A'Ciy
= i.ACD, = 4 CAD, = 4. CDA,
Then make C'A'iy
4. .-.
4 C'lyA'
and
AA'C'iy ~ A Therefore, from the
first
A'B'
:
A CD,
etc.
pair of similar triangles,
B'C
::
AB
C'A'
::
BC
:
BC,
and
B'C From
:
:
CA.
the second pair of similar triangles,
C'A' .\
B'C
:
:
CD' CD'
::
CA
::
BC
:
:
CD, CD,
and so on. Thus all the ^s in one polygon are = the corresponding ^s in the other, and the sides about the corresponding ^s are proportional. /.
the polygons are similar.
RATIO APPLIED.
211
Theorem XXI. 551. In
enuse
is
a right-angled
equivalent
to
triangle,
the
any polygon upon the hypoth-
sum of
the similar
described polygons on the other two
Draw
and
similarly
sides.
CD ± AB.
Therefore, by 520,
AB therefore,
:
AC
::
AD
::
Q
:
S.
BD
::
Q
:
R,
AC AD; :
by 547,
AB
:
Similarly,
AB
:
Therefore, by 496,
AB
:
AD + DB
::
Q
.
R+
S,
Exercises. 104. If 551 applies to semicircles, show the triangle equivalent to two crescent-shaped figures, called the lunes of Hippocrates of Chios (about 450 B.C.).
THE ELEMENTS OF GEOMETRY.
212
Problem
IV.
To describe a polygon equivalent
552.
to oney
and- similar
to
another^ given polygon.
K A. JiXL
Let
D be the one, and ABC the other, given polygon.
By
262,
AC describe any
on
Z37
C£ = ABC,
^AM = D, and having ^ AEM = ^ 549, between
By ^y
^^
55°> o^
construct the figure
also
describe
KGH
similar
GH.
and
similarly
ABC.
KGH may be
AB
AC and AF find a mean proportional
described to the figure
It
and on
CAE.
is
the figure required.
proved, as in 542, that
^C and AF are in one line, and
LE and EM\
therefore,
by 505,
AC AF '.
'.'.
^7
CE
'.
,17
AM
:
:
ABC
:
Z>.
But
AC AF :
(547. If three sects
.-.
therefore,
::
ABC KGH, :
form a proportion, a polygon on the first is to a similar polygon on the second as the first sect is to the third.)
by 491,
ABC
:£>.'.
ABC KGH;
KGH =
:
£>,
GEOMETRY OF THREE DIMENSIONS.
BOOK
VII.
OF PLANES AND
LINES.
553. Already, in 50, a plane has been defined as the surface generated by the motion of a line always passing through a fixed point while it slides along a fixed line.
554. Already, in 97, the theorem has been assumed, that, if of a line are in a plane, the whole line lies in that
two points plane.
Two
other assumptions will
555.
Any number
of planes
—
now be made may be passed through any line. :
'L
2 556.
^
A plane may be revolved on any line lying in
it.
ax3
THE ELEMENTS OF GEOMETRY.
214
Theorem Through two intersecting
557.
I.
lines^ one
plane and only y
one,
passes.
Two One
Hypothesis.
Conclusion.
in B. and plane, only one, passes through them.
lines
AB, BC, meeting
FG
Proof.
be passed through AB, and, by By 555, let any plane be revolved around on an as axis until it meets any point C 556,
AB
of the line
The
BC.
BC
then has two points in the plane FG) and therefore, by 554, the whole line BC\%'m. this plane. Also, any plane containing AB and BC must coincide with FG. For let Q be any point in a plane containing AB and BC. line
Draw since
QMN in
M and JV
point in the plane Similarly,
this
plane to cut
AB, BC, inM and N. Then, FF, therefore, by 554, Q is a
are points in the plane
FF.
any point
in a plane containing
therefore any plane containing
AB, BC, must lie in must coincide with FF. AB, BC,
FF;
Corollary I. Two lines v^rhich intersect lie in one and a plane is completely determined by the condition plane, that it passes through two intersecting lines. 559' Corollary II. Any number of lines, each of which 558.
intersects all the others at different points, lie in the
same
but a line may pass through the intersection of two plane others without being in their plane. ;
OF PLANES AND LINES. Corollary
560.
III.
A
line^
21 5
and a point without
that line^
determine a plane.
3
Proof.
AB the line,
Suppose
CD
and
C the
D
point without
AB.
in AB. Then one plane conany point tains and CD, therefore one plane contains and C. must contain ; therefore, any Again, any plane containing
Draw
the Hne
to
AB
AB
D
AB
AB and C must contain CD also. only one plane that can contain AB and CD. there only one plane that can contain AB and
plane containing
But there Therefore
Hence
561.
is
is
the plane
is
C.
completely determined.
Corollary IV.
Three points not
in the
same
line
determine a plane.
For
let
A, B,
C
be three such points.
Draw
the line
AB.
AB
plane which contains A, B, and C must contain and C\ and a plane which contains and C must contain A, and C are contained by one plane, and one By and C. Now,
Then a
AB
AB
only
one
;
.therefore A, B,
only.
and
C
Hence the plane
is
are contained by one plane, and
completely determined.
THE ELEMENTS OF GEOMETRY.
2l6
562.
Two
determine a plane.
parallel lines
For, by the definition of parallel lines, the two lines are in the same plane and, as only one plane can be drawn to contain one of the lines and any point in the other line, it follows ;
that only one plane can be
drawn
to contain both lines.
Theorem
II.
563. If two planes cut one a7iotheri their
common
section inust
be a straight line, iL
AB
Hypothesis.
Lei
Conclusion.
Their
M
and
CD
common
N MN
be two intersecting planes. is a straight line.
section
Let and be two points common to both planes. the straight line and are in Therefore, by 554, since both planes, the straight line lies in both planes. And no point out of this line can be in both planes ; because then Proof.
MN.
Draw
two planes would each contain the same it,
which, by 560,
Hence
M
line
and the same point without
is
impossible. every point in the common section of the planes
straight line
MN.
N
lies
in the
OF PLANES AND LINES. CoNTRANOMiNAL OF
564.
A
554.
line
217
which does not
lie
may have no point, and cannot have more common with the plane.
altogether in a plane
than one point, in Therefore three planes which do not pass through the same line cannot have more than one point in common for, by 563, ;
the points
common
two planes
to
have only one point in
on a
lie
line,
and
this line
can
common
with the third plane. All are 565. planes congruent hence properties proved for one plane hold for all. plane will slide upon its trace. ;
A
Principle of Duality.
When any figure
is given, we may construct a reciprocal of points, and points instead of instead figure by taking planes
566.
planes, but lines
where we had
lines.
The
figure reciprocal to four points which do not lie in a will consist of four planes which do not meet in a point. plane From any theorem we may infer a reciprocal theorem.
Two a
line
A
Two
points determine a line.
Three points which are not determine a plane. line and a point without lines in a plane
line.
Three planes which do not pass
it
through a line determine a point. A line and a plane not through it determine a point.
determine a plane.
Two
planes determine a
in
determine
a point.
Two lines through a point determine a plane.
There is also a more special principle of duality, which, in for the plane, takes points and lines as reciprocal elements ;
they have this fundamental property in common, that two elements of one kind determine one of the other. Thus, from a proposition relating to lines or angles in axial symmetry, we get a proposition relating to points or sects in central symmetry.
The
angle between two correlines is bisected by the
The
sect
between
sponding
sponding points
axis.
center.
is
tvvo
corre-
bisected by the
THE ELEMENTS OF GEOMETRY.
2l8
Theorem
it
III.
567. If a line be perpendicular to two lines lying in a plane, will be perpendicidar to every other line lying in the plane a7td
passing through
Hypothesis.
its foot.
Let the line
EF be
perpendicular to each of the
at E, the point of their intersection. Conclusion. J_ GH, any other line lying in the plane
lines
AB, CD,
EF
ABCD,
and passing through E. Proof. Join
AD
Take and
AE = BE, and CE = BE. B C, and let G and H be the
joining lines intersect the third line
Take any point
F
in
EF.
points in which the
GH.
Join
FA, FB, EC, FD, EG, EH.
Then
A
AED ^
A BEC,
(124. Triangles having two sides and the included angle equal are congruent.)
.-.
AD =
BC,
and
^
DAE =
^ CBE.
Then
A
AEG ^
A BEH,
(128. Triangles having two angles and the included side equal are congruent.)
/.
GE = BE,
and
AG =
BH.
OF PLANES AND LINES.
219
Then
A (124. Triangles having
AEF ^
A BEF,
two sides and the included angle equal are congruent.)
AF = BF In the same way
CF = DF.
Then
A
ADF ^
A BCF,
(129. Triangles having three sides in each respectively equal are congruent.)
^ Z^AF = ^
.-.
CBF
Then
A
AFG ^
A BFH,
(124. Triangles having two sides and the included angle equal are congruent.)
FG =
FH.
Then
A
FFG ^
A FFJI,
(129. Triangles having three sides in each respectively equal are congruent)
.-.
As
GH
through
is
FFG = FE _L
^ FFH, GH.
any hne whatever lying in the plane
ABCDy
and passing
-£", .*.
568.
^
A
line
EF J_ every such line.
meeting a plane so as to be perpendicular to
every line lying in the plane and passing through the point of Then intersection, is said to be perpendicular to the plane. said to be perpendicular to the line. At a given point in a plane, only one 569. to the perpendicular plane can be erected ; and, from a point without a plane, only one perpendicular can be drawn to the also the plane
is
Corollary.
plane.
THE ELEMENTS OF GEOMETRY.
220
570.
Inverse of 567. same point lie
line at the
AB
Hypothesis. fine
any
All lines perpendicular to another in the
line
same
J.
BD
plane.
BC
and BE.
any other
± AB.
^C
Conclusion.
For
Proof. plane
BDE
one plane
;
in
if
in the plane
is
BDE.
not, let the plane passing through
the
BE.
AB, BC,
Then AB, BC, and
BE
cut the
are
all
in
AB 1.BC', ABE we have two lines BCy BE, both ± AB
at
line
and because
AB
J_
BD and BE, therefore, by 567,
AB ± BE. But, by hypothesis, therefore, in the plane
B, which, by 105,
is
impossible .-.
;
BC lies in the plane DBE.
Corollary. If a right angle be turned round one of arms as an axis, the other arm will generate a plane and when this second arm has described a perigon, it will have 571.
its
;
passed through every point of this plane. without a given line to pass 572. Through any point a plane perpendicular to AB. In the plane determined by
D
AB and the point D, draw DB J_ AB ABD about AB,
the line rt. 4.
AB
;
then revolve the
OF PLANES AND LINES.
Theorem
221
IV.
same plane are parallel
573. Lines perpendicular to the
to
each other.
Hypothesis.
AB,
Conclusion.
AB
Proof.
BD,
Make
MN at
CD L plane \\
DE =
A (124. Triangles having
B, D.
DE X BD in the plane MN,
and draw and AB, join BE,
Join
the points
CD.
AD, AE.
ABD ^
Then
A EDB,
two sides and the included angle equal are congruent.)
AD = BE. A ABE ^ A EDA, /.
Then
(129. Triangles having three sides in each respectively equal are congruent.)
^
.-.
But,
by hypothesis,
ABE
is
a
rt.
EDA
.-.
.*.
ABE =
AD, BD, CD,
^, is
a
are
rt.
all
(570. All lines perpendicular to another at the
But
AB
is
^, in
and they are both
AB and CD He X BD, .'.
two
AB
in the
II
one plane.
same point
lie
B
in the
A and are same plane ;
in this plane, since the points .*.
(166. If
EDA,
i.
same
in
plane.)
it,
CD.
interior angles are supplemental, the lines are parallel.)
THE ELEMENTS OF GEOMETRY.
222
If one of two parallels is perpen574. Inverse of 573. dicular to a plane, the other is also perpendicular to that plane.
7 AB AB
Hypothesis.
Conclusion.
Proof.
For
if
CD ± plane MN,
CD,
\\
_L plane
MN.
AB, meeting
the plane
MN
at
B,
is
not
±
it,
let
^i^beXit; therefore,
But
by 573,
BF
CD,
\\
and, by hypothesis,
BA
\\
CD.
this is impossible,
(99.
Two
intersecting lines cannot both be parallel to the .-.
575.
Corollary
two intersecting
I.
lines,
same
line.)
AB ± plane MN.
If one plane be perpendicular to one of and a second plane perpendicular to the
second, their intersection is perpendicular to the plane of
the two
For is
lines.
-
their intersection
perpendicular
a
to
DF line
D
through parallel to AB, and also perpendicular to a line through parallel to BC.
D
(574. If, of
576. line,
two
parallels,
Corollary
one be perpendicular to a plane, the other
II.
Two
lines,
is also.)
each parallel to the same
are parallel to each other, even though the three be not in
OF PLANES AND LINES.
223
one plane. For a plane perpendicular to the third line will, by 574, be perpendicular to each of the others; and therefore,
by
573, they are parallel.
Problem To draw a
577.
line perpendicular to
given point without
GrvTEN, ihe
plane to
A^x2Ni
a giveft platie from a
it.
BH and
draw from
Required, Construction.
I.
A
the point
a
line _L
A
without
plane
it.
BH.
In the plane draw any line
BC,
and, by 139, from
AD LBC.
and from A, draw, by In the plane, by 135, draw DF 1. BC AF ± DF. AF will be the required perpendicular to the plane. Proof. Through F draw GH BC. Then, since ^C ± the plane ADF, \
139,
\\
A
line
(574. If
two
(567.
perpendicular to any two lines of a plane at their intersection perpendicular to the plane.) /.
GH ± the plane ADF,
lines are parallel,
and one
is
perpendicular to a plane, the other
also.)
/.
GH ± the line AF of the plane ADF,
AF±GH. AF ± DF, But, by construction, AF _L the plane passing through .'.
,'.
is
GH, DF,
is
THE ELEMENTS OF GEOMETRY.
224
Problem 578.
To
erect
a perpendicular
to
II.
a given plane from a given
point in the plane.
1 A
be the given point. point B, without the plane, draw, by 577, and from A, by 167, draw BC^ plane, Let
From any
AD
.'.
{574. If one of
two
parallels
is
BC ±
the
||
AD JL the plane. perpendicular to a given plane, the other
is also.)
579. The projection of a point upon a plane is the foot of the perpendicular drawn from the point to the plane.
580. The projection of a line upon a plane is the locus of the feet of the perpendiculars dropped from every point of the line upon the plane.
OF PLANES AND LINES.
225
These perpendiculars are all in the same plane, since any two are parallel, and any third is parallel to either, and has a point in their common plane, and therefore lies wholly in that plane
therefore the projection of a line
;
Theorem 581.
A
line
makes with
its
own
is
a line.
V. projection upon
a plane a
less
angle than with any other line in the plane.
meet
the plane
MN
project/on upon the plane in the plane through
MJV, and
BC
Hypothesis. be
Let
BA
its
B
and
let
Take
Then, in As
4.
ABC.
BC =
ABA'
BA'. Join and ABC,
AB = BA' =
A C and A' C.
AB, BC,
AA' < AC, (150.
The
perpendicular
is
the least sect from a point to a line.)
^ ABA' <^ABC. In two triangles, a = a', d = b\ c
(161.
ii
BA'
any other line drawn
MN.
^ ABA' <
Conclusion.
Proof.
at B,
therefore
C<
C.)
THE ELEMENTS OF GEOMETRY,
226
is
582. The angle between a line and its projection on a plane called the Inclination of the line to the plane. 583. Parallel planes are such as never meet, how far soever
they
may be
584.
produced.
A line is parallel
far soever they
may
to a plane
Theorem 585. Planes to
when they never meet, how
be produced.
VI.
which the same line
is
perpendicular are par-
allel.
For,
if not,
they intersect.
Call any point of their intersection
X,
Draw from X a line in each plane to the foot of the common perpenThen from this point X we would have two perpendiculars to dicular. the
same
line,
(145.
which
is
impossible,
There can be only one perpendicular from a point to a /.
the planes cannot intersect, .*.
line.)
and
are parallel.
If two
lines are parallel^ every plane through one of the them, except plane of the parallels, is parallel to the other.
586.
AB
GH
DEF
and be the parallels, and any plane then the line AB and the plane DEF are parthrough GH\ For the plane of the parallels ABHG intersects the allel. plane DEF in the line GH\ and, if AB could meet the plane DEFy it could meet it only in some point of GH', but AB cannot meet GH, since they are parallel by hypothesis. Therefore AB cannot meet the plane DEF. Applications, (i) Through any given line a plane can be Let
passed parallel to any other given line. (2) Through any given point a plane can be passed parallel to
any two given
lines in space.
OF PLANES AND LINES.
Theorem
227
VII.
587 • If a pair of mtersecting lines be parallel to another pairy but not in the same plane with them, the plane of the first pair is parallel to the plafie of the second pair.
Hypothesis.
AB
Conclusion.
Plane
Proof.
Through
From
DE, and
ABC
Two
lines
GH
II
DEy
ABy
and
EF.
HK
\\
meeting
\\
in
H,
BC,
ABH +
St.
^
it
EF,
HK
and
t ^IIG =
each
other.'
^.
BH was drawn X plane DEF, 4.BHG =
i\,^.
':\
:^ABH^ and, in
\\
DEF. plane DEFy
line are parallel to
.-.
So
\\
BC
plane
each parallel to the same /.
But because
\\
B draw BH JL
H draw GH /.
(576.
\\
same way,
rt.
^;
CBH = ^ HB \_^\^XiQ ABCy plane DEF. plane ABC rt.
i.
;
.-.
:. (585. Planes to
which the same
\\
line is perpendicular are parallel.)
THE ELEMENTS OF GEOMETRY.
228
Theorem If a pair of intersecting angle made by the first pair
588. ajty
angle
made
is
equal ,or supplemental
In
BE. DE, AB, BE, and ED,
Join
AB
\\
this plane,
from
must meet the
A
line
are in
.-.
In same way,
since,
AD
11
line
Call the intersection point
I|
ABED
is
any two
a
ziz
and
AD =
BC =
and
CF = BE,
EF,
AD = construction, AD
by CF,
ACFD
D.
;
AB = DE,
,'.
(216. If
BE.
draw a
BE.
CF.
.-.
But
one plane.
DE. .*.
hysi6,
any
\\
\\
.'.
It
to
AB DE, and BC EF. ^ ABC = or supplemental to ^ DEF.
Conclusion. Since
lines be parallel to another pair^
by the second pair.
Hypothesis.
Proof.
VIII.
\\
BE, and CF
||
BE,
therefore,
isazij,
opposite sides of a quadrilateral are equal parallelogram.)
and
parallel,
it
is
a
OF PLANES AND LINES. .-.
A
:,
229
AC = DF, ABC ^ A DBF,
(129. Triangles with three sides of the
one equal to three of the other are con-
gruent.)
/.
4.
ABC =
^ DBF.
Theorem IX. 589.
If
tivo parallel
cornnio7t sections
with
it
planes be cut by another plane^ their are parallel.
A
and B, and the third plane X. Call the parallel planes the lines of intersection are in one plane, since they both in the plane X.
Then
lie
Again, because one of these lines is in the plane A, and the other B, they can never meet.
in the parallel plane
Therefore the two lines are in one plane, and can never meet is,
;
that
between two
par-
they are parallel. 590.
allel
Corollary.
planes are equal.
Parallel sects included
THE ELEMENTS OF GEOM^^TRY.
230
Theorem X. 591.
Parallel lines intersectijig the same plane are equally
inclined to
it.
Hypothesis.
AB
I|
BB'
1.
CD, plane
ACF, ACF.
Djy JL plane Conclusion. ^ BAB' = '^ DCU, Proof. ^ ABB' = or supplemental
to
2^
CDL/.
a pair of intersecting lines be parallel to another pair, any angle made by the first pair is equal or supplemental to any angle made by the second pair.)
(588. If
But
%.
ABB' cannot
be supplemental to
.-.
and therefore
their
-4.
;^
CDIf
,
since each
is
acute,
ABB' = 4 CDiy;
complements are equal.
Two lines not in the same plane are regarded as makwith one another the angles included by two intersecting ing lines drawn parallel respectively to them. We know, from 588, that these angles will always be the 592.
same, whatever the position of the point of intersection.
OF PLANES AND LINES,
23 1
Theorem XL
V
^^^ /^'^^j be cut by three parallel planesy the corre593* spondi7ig sects are proportio7ial.
Let the
lines
AB, CD, be
cut by the
||
MN^ FQ,
planes
RS,
in the
points Aj Ej By and C, F, D.
Conclusion.
AE
\
EB
Join
AD,
Join
A C, BD, EGy EG.
w CF
cutting the plane
FQ
G,
Then
EG (589. If parallel planes be cut
FD.
:
in
II
BD.
by a third plane,
their
common
sections with
it
are
parallel.)
In the same way, /.
AC AE EB :
\\
GF) ::
AG
::
CF
:
GDy
and
AG (499.
A
line parallel to the
/.
'.
GD
'.
FD
'y
base of a triangle divides the sides proportionally.)
AE
\
EB
\\
CF
\
FD,
THE ELEMENTS OF GEOMETRY.
232
Theorem
Two
594.
common
XII.
same plane have
lines not in tJie
one,
and
only one^
perpendicular.
AB
and CD, two fines not in the same plane, and Hypothesis. therefore neither parallel nor intersecting each other. Conclusion. There is one line, and no more, perpendicular to both
AB and
CD.
Proof. revolve on
Through one of the
CD
as axis until
Let A^B^ be the projection of
which
point in
AB.
the plane
with
;
AB on
the plane
CD.
it
MN,
and
Then O,
O
let
like
be the
every point
the foot of a perpendicular to the plane from some is perpendicular to Then, since
PO
Call this point P.
perpendicular to
is
CD
and A'B'.
But A'B'
is
par-
AB,
AB. .-.
(170.
parallel to
because, being in a plane parallel to AB, it can never meet and, being the projection of AB, it is, by 580, in the same plaijc
allel to
AB
MN,
is
CD, pass a plane, and let it AB. Call this plane MN.
lines, as
this projection intersects
in the projection,
point of
it is
A
PO ± A'B',
-since
line perpendicular to
it is
also _L
one of two parallels
is
AB,
perpendicular to the other
also.)
/.
If there could be
OP ± both AB and
any other
common
CD.
perpendicular, call
it
P'Q.
OF PLANES AND LINES, Through
Q
draw
MN, QR
in the plane
FQ ± AB, CD, but, by hypothesis, FQ Since
.',
\\
233
AB.
FQ A-QR;
1.
/. /.
^
Q is
,\
is
X plane
O, the only point
At a given point
J/W;
a point in A'B', the projection of
common
to
AB,
A'B' and CI>,
F Q coincides with PO.
.'.
(569.
i^'e
in a given plane, only
one perpendicular to the plane can be
erected.)
Theorem The smallest
595.
sect
XIII.
between two lines not
m
the
same
pla7ie is their commojz perpe^tdictdar.
For
if
any sect drawn from one
to the other
is
not perpendicular to
both, by dropping a perpendicular to the hne it cuts obliquely from the point where it meets the other line, we get a smaller sect. (150.
596.
The perpendicular
Remark.
lines are drawn,
one
If
is
the smallest sect from a point to a line.)
two planes
in
and two intersecting each plane, these lines may, for the same intersect,
\ww
THE ELEMENTS OF GEOMETRY.
234
Theorem XIV. 597.
The smallest
sect
from a point
to
a plane
is the
perpen-
dicular.
Conclusion.
AD ± plane MN. B AD < AB.
Proof.
AB, BD.
Hypothesis.
Join
In .-.
(150.
The perpendicular
is
A
any oiher point of
ABDy^ ADB is
AD < AB.
a
rt.
MN.
^,
the least sect from a point to a line.)
Theorem XV. 598. Eqtial obliques from a point to a plane meet the pla7ie whose center is tJie foot of the perpendicular from the point to the plane.
in a circle
OF PLANES AND LINES. Hypothesis.
AB, AC, equal
sects
from
A
235 to the
plane
MN.
AD ± plane MN, Conclusion. DB = DC. A ^ CD. Proof. Rt. A ABD ^ rt.
(179. If
two right triangles have the hypothenuse and one side respectively equal, they are congruent.)
Corollary I. To draw a perpendicular from a point a plane, draw any oblique from the point to the plane ; revolve this sect about the point, tracing a circle on the plane ; 599.
to
find the center of this circle,
Corollary
and join
it
to the point.
through the center of a circle a line be passed perpendicular to its plane, the sects from any point of this line to points on the circle are equal. 600.
II.
If
Theorem XVI. 601. The locus of all points from which the two sects drawn two fixed points are equal, is the plane bisecting at right angles the sect joining the two given poifits. to
For the sect
is
line
from any such point to the mid point of the joining therefore all such lines form a plane ;
perpendicular to that sect
perpendicular to that sect at
its
mid
point.
(570. All lines perpendicular to another line at the
same point
lie in
the
same
plane.)
THE ELEMENTS OF GEOMETRY.
2z6
Theorem XVII. 602.
The
lars onto the
locus of all points from, which the two perpendicusides of two fixed planes are equal, is a plane
same
determined by
07ie
such point aiid the intersection line of the tzvo
given planes.
Hypothesis.
Lei
AB
and
CD
be the two given planes, and
one such point. Conclusion.
K
AB
and CZ> from any Perpendiculars dropped on to and the intersection line are point in the plane determined by
X
BC
equal.
Proof.
AB, and
cuts the line
Draw
J_
BC.
KC
i|
PH ±
P
any point in the plane XCB. Draw plane the point where the plane BBIf Call plane CB>.
Take
BB
G
Join
BG.
KC
BG, BG, GH.
Because the perpendiculars from equally inclined to the two planes.
is equal, therefore the same inclination to each as
KC
(591. Parallels intersecting the
/. /.
^ A
K
are given
But
BG
;
same plane are equally
BGB = BGB ^ BB =
^ BGH, A BGH,
BH,
inclined to
it.)
has
OF PLANES AND LINES.
237
Theorem XVIII. If three lines not in the same plane meet at one pointy two of the angles formed are together greater than the third. any 603.
Proof.
The theorem
requires proof only greater than each of the others.
considered
is
greatest
BAC, make
4.
F draw
Through and DC.
BAF = ^ BAD. BFC cutting AB in B, %.
a line
DB
A (124. Triangles having
when
BAD ^
the third angle
In the plane of the
AF — AC in C.
Make
AD.
and
Join
A BAFy
two sides and the included angle respectively equal are congruent.) .-.
But from A {156.
BCD we
Any two
BD = BF. BD + DC> BC.
have
sides of a triangle are together greater than the third.)
BD and BF^ DC > FC,
And, taking away the equals
.'.
(161. If
in
As
CAD and
CAF, we have ^
CAD
> ^ CAF.
two triangles have two sides respectively equal, but the third side greater in the first, its opposite angle is greater in the first.)
BAD and BAF gives ^ BAD + 4 CAD > BAG.
Adding the equal ^ s
4.
THE ELEMENTS OF GEOMETRY.
238
Theorem XIX. 604.
not in
made
vertices of a convex polygon be joined to a point plane^ the snin of the vertical angles of the triangles so
If the
its
is less
than a perigon. :
s
The sum
of the angles of the triangles which have the equal to the sum of the angles of the same number of triangles having their vertices at (9 in the plane of the polygon.
Proof.
common
vertex
S
is
But :^ %.
SAB + 2^ SAB SBA 4- t SBC
(603. If three lines not in a plane
meet
> ^
BAE,
> ^
ABC,
at a point,
etc.
any two of the angles formed
are together greater than the third.)
lience, summing all these inequalities, the sum of the angles at the bases of the triangles whose vertex is S, is greater than the sum of the angles at the bases of the triangles whose vertex is O therefore ;
the is,
sum of
less
the angles at
than a perigon.
S
is less
than the
sum
of the angles at O, that
BOOK
VIII.
TRI-DIMENSIONAL SPHERICS. 605.
If
one end point of a sect
other end point
is
606.
The
607.
The moving
fixed
is
fixed,
the locus of the
a Sphere.
end point
sect in
is
called the Ce7iter of the sphere.
any position
is
called the
Radius
of
the sphere. 608.
As
the motion of a sect does not change
it,
all radii
are
equal. 609. The sphere is a closed surface for it has two points on every line passing through the center, and the center is midway between them. ;
«39
THE ELEMENTS OF GEOMETRY.
240
Two such points are called Opposite Points of the sphere, and the sect between them is called a Diameter. 6ii. The sect from a point to the center is less than, equal to, or greater than, the radius, according as the point is within, 6io.
on, or without, the sphere. For, if a point is on the sphere, the sect drawn to it from the center is a radius if the point is within the sphere, it lies on some radius if without, it lies on the extension of some ;
;
radius.
612.
By
33,
Rule of Inversion, a point
is
within, on, or with-
from the center according as the sect to less than, equal to, or greater than, the radius. out, the sphere,
it
Theorem 613.
The common
section of
I.
a sphere and a plane
Take any sphere with center O. Let A, B, C, etc., be points common and
is
to the sphere
is
a
circle.
and a plane,
OD the perpendicular from O to the plane. Then OA = OB = OC = etc.,
being radii of the sphere, .*.
(598.
ABC,
etc., is
a circle with center
D.
Equal obliques from a point to a plane meet the plane in a circle whose center is the foot of the perpendicular from the point to the plane.)
TRI-DIMENSIONAL SPHERICS. Corollary.
614.
The
line
24 1
through the center of any circle through the cen-
of a sphere, perpendicular to its plane, passes ter of the sphere.
A Great Circle of
615.
a sphere
is
any section
of the sphere
made by
a plane which passes through the center. All other circles on the sphere are called Small Circles.
616.
Corollary.
All great circles of the sphere are equal, radius the radius of the sphere. 617. The two points in which a perpendicular to its plane, through the center of a great or small circle of the sphere, intersects the sphere, are called the Poles of that circle. since each has for
its
618. Corollary. Since the perpendicular passes through the center of the sphere, the two poles of any circle are opposite points, and the diameter between them is called the Axis of that circle.
Theorem 619.
Every great
II.
circle divides the sphere into
two congruent
hemispheres.
For
if
one hemisphere be turned about the fixed center of the
sphere so that its plane returns to the great circle will coincide with its will coincide.
its
own
former position, but inverted, trace, and the two hemispheres
THE ELEMENTS OF GEOMETRY.
242 620.
Any two
great circles of a sphere bisect each other.
Since the planes of these circles both pass through the center of the sphere, their line of intersection is a diameter of the sphere, and therefore of each
they will
622.
Through any two points
point,
in a sphere, not the extremi-
and only one, great circle can be passed two given points and the center of the sphere determine
ties of a diameter, one,
for the
circle.
any number of great circles pass through a also pass through the opposite point.
621. If
;
TRI-DIMENSIONAL SPHERICS. plane. Through opposite points, an indefinite can be passed. circles great its
243
number
of
Through any three points in a sphere, a plane can be and but one therefore three points in a sphere deterpassed, mine a circle of the sphere. 624. A small circle is the less the greater the sect from its center to the center of the sphere. For, with the same hypothenuse, one side of a right-angled triangle decreases as the other 623.
;
^_^
increases.
625.
A Zone
is
parallel planes.
a portion of a sphere included between two circles made by the parallel planes are
The
the Bases of the zone. 626.
point,
A
line or plane is tangent to a sphere
and only one,
627.
Two
when
it
has one
common
with the sphere. spheres are tangent to each other when they have in
one point, and only one, Exercises.
105.
If
in
common.
through a fixed point, within or with-
out a sphere, three lines are drawn perpendicular to each other, intersecting the sphere, the sum of the squares of the three Also the sum of the squares intercepted chords is constant. of the six
segments
of these chords is constant.
three radii of a sphere, perpendicular to each other, are projected upon any plane, the sum of the squares of the 106.
If
three projections the sphere.
is
equal to twice the square of the radius of
^-^^
244
ELEMENTS OF GEOMETRY. Theorem
628.
A
extremity
plane perpendicular is
tangent
to
III.
a radius of a sphere at
its
to the sphere.
For, by 597, this radius, being perpendicular to the plane, is the from the center to the plane ; therefore every point of
smallest sect
the plane
is
without the sphere except the foot of
this radius.
Corollary.
Every line perpendicular to a radius at tangent to the sphere. 630. Inverse of 628. Every plane or line tangent to the to is the radius drawn to the point of perpendicular sphere 629.
its
extremity
is
For since every point of the plane or line, except the point of contact, is without the sphere, the radius drawn to the point of contact is the smallest sect from the center of contact.
the sphere to the plane or line
;
therefore,
by
597,
it
is
per-
pendicular.
Exercises. 107. Cut a given sphere by a plane passing through a given line, so that the section shall have a given radius. 108. Find the locus and from point B is b.
of points
whose
sect from point
A
is ^,
TRI-DIMENSIONAL SPHERICS.
Theorem
cle
IV.
631. If two spheres cut one another, their intersection is a cirwhose plane is perpendicular to the line jointing the centers of
the spheres,
and
wJiose cejiter is in that
Hypothesis.
and
C
Let
and
O
be the
liiie.
centers of the spheres,
A
B
any two points in their intersection. and are on a circle having Conclusion.
B
A
OC, and
its
Proof.
its
center on the line
plane perpendicular to that line. Join CA,
because they have
Then
CB, OA, OB.
A the
245
OC
CAO^
common,
A CBO,
CA =
CB, and
OA =
OB,
radii of
same sphere.
As are =, therefore perpendiculars from A and B upon and meet OC aX. the same point, D. and are in a plane ± OC and, being equal sects,
Since these
OC are
equal,
Then
AD
their extremities
632.
A
DB B are in a circle having
and
Corollary.
By moving
',
its
center at
D.
the centers of the two inter-
secting spheres toward or away from each other, we can make their circle of intersection decrease indefinitely toward its therefore, if two spheres are tangent, either internally or externally, their centers and point of contact lie in the same
center
line.
;
THE ELEMENTS OF GEOMETRY.
246
Theorem
Through any four points not in
633. spJiere^
as
V.
and only
the
same plane^ one
one^ ca7i be passed.
Let A, B, C, D, be the four points. Join them by any three AB, BC, CD. Bisect each at right angles by a plane.
The plane bisecting
CD.
bisecting
AB, and
Moreover,
(575. If
BC has the line EH'vn common with the plane FO in common with the plane bisecting
has the line
EH
i.
plane
ABC, and
FO ± plane BCD,
one plane be perpendicular to one of two intersecting
plane perpendicular to the second, their intersection
plane of the two .-.
.'.
sects,
is
lines,
and a second
perpendicular to the
lines.)
EH\.EG,
and
FO JL FG,
EU and FO meet, since, by hypothesis,
Call their point of meeting O. containing A, B, C, and D,
O
shall
^
at (9
> o and <
st.
^.
be the center of the sphere
TRI-DIMENSIONAL SPHERICS. For
O
is
in three planes bisecting at right angles the sects
247
AB, BC,
CD, (601.
The
all points from which the two sects drawn to two fixed points the plane bisecting at right angles the sect joining the two given
locus of
are equal,
is
points.)
634.
A
Tetrahedro7t
is
a solid bounded by four triangular The sides of the triangles are
plane surfaces called its faces. called its edges.
635.
Corollary
I.
A
sphere
may be
circumscribed about
any tetrahedron. 636.
Corollary
of a tetrahedron
common
II.
The
lines perpendicular to the faces
through their circumcenters, intersect at a
point.
III. The six planes which bisect at right the six of tetrahedron a all pass through a comangles edges
637.
mon
Corollary
point.
THE ELEMENTS OF GEOMETRY.
248
Problem 638.
To
inscribe
I.
a sphere in a given tetrahedron.
Through any edge and any point from which pertwo faces are equal, pass a plane. the same with two other edges in the same face as the first, and
Construction. pendiculars to
Do
its
the three planes so determined intersect at O. of the required sphere.
let
(602.
The
locus of
all
O
shall
be the center
points from which the perpendiculars on the same sides of is a plane determined by one such point and the intersec-
two planes are equal,
tion line of the given planes.)
639.
Corollary.
In the same way, four spheres
may be
escribed, each touching a face of the tetrahedron externally, and the other three faces produced. 640.
The
641.
A
solid
bounded by a sphere
globe-segment
is
is
called a Globe.
a portion of a globe included be-
tween two
The
parallel planes. sections of the globe
made by
the parallel planes are
the bases of the segment.
Any sect perpendicular to, and terminated by, the bases, the altitude of the segment. 642.
A
revolving
figure with
it
two points
fixed can
still
is
be moved by
about the line determined by the two points.
TR /-DIMENSIONAL SPHERICS.
249
This revolution can be performed in either of two senses, and continued until the figure returns to its original position. The fixed line is called the Axis of Revoliction. If the axis of revolution slides
is
any
upon by a plane is a
through the center, a sphere because every section of a sphere
line passing
its trace.
This
is
circle.
Any figure has central symmetry if bisects all sects through it terminated
it
has a center which
by the surface. The and coincides with its trace throughout any motion during which the center remains fixed. Thus any figure drawn on a sphere may be moved about on the sphere has central symmetry,
sphere without deformation. But, unlike planes, all spheres are not congruent. Only those with equal radii will coincide. In general, a figure drawn upon one sphere will not fit upon
So we cannot apply the test of superposition, except on the same sphere or spheres whose radii are equal. Again, if we wish the angles of a figure on a sphere to remain the same while the sides increase, we must magnify the whole on the same sphere similar figures cannot exist. sphere another.
:
Two
643.
when
Two
points are symmetrical with respect to a plane
this plane bisects at right angles the sect joining them. figures are symmetrical with respect to a plane when
every point of one figure has its symmetrical point in the other. Any figure has planar symmetry if it can be divided by a plane into
two
sphere
is
The figures symmetrical with respect to that plane. symmetrical with respect to every plane through its
Any two spheres are symmetrical with respect to plane through their line of centers. every such plane cuts the spheres in two great circles and Every center.
;
the five relations between the center-sect, radii, and relative position of these circles given in 410, with their inverses, hold for the
two spheres.
Any three spheres are symmetrical with respect to the plane determined by their centers.
THE ELEMENTS OF GEOMETRY,
250
Theorem
VI.
If a sphere he tangent to the parallel planes containing opposite edges of a tetrahedrony and sections made in the globe 644.
and
tetrahedron by one plane parallel to these are equivalenty made by any parallel plane are equivalent.
sectio7is
Hypothesis. the
KJ
Lei
be the sect _L the edges
EF and GH in
tangent planes.
II
Then
Kf = DT,
Let sections
KR =
the diameter.
made by plane
± Kf
R
at
Conclusion. Proof. .'.
^ LN = A
Since
MW MZ
(593. If lines be cut
\
'.
± DT
and
^ MO =
DI, be equivalent; that is, Draw any parallel plane ACBLSN.
at I, where
O PQ.
O AB.
LEU ^ A MEW, and A LHV ^ A MBZ, LU w EM EL :: fR fS; LV HM HL KR KS :
'.
:
'.'.
:
:
-.
;
by three parallel planes, the corresponding sects are proportional.)
.-.
.'.
{542.
WM. MZ
^MO
:
:
UL .LV
ziyLN
::
:
:
fR
RK
.
:
fS SK;
fR .RK fS .SK :
.
(i)
Mutually equiangular parallelograms have to one another the ratio which
compounded
of the ratios of their sides.)
is
TRI-DIMENSIONAL SPHERICS.
251
But
QFQ (546.
OAB
'.
'.:
PI^
'.
AC^
::
TI ID .
:
TC
,
CD.
(2)
Similar figures are to each other as the squares on their corresponding sects.)
(522. If from any point in a circle a perpendicular be dropped upon a diameter, will be a mean proportional between the segments of the diameter.)
it
hypothesis and construction, in proportions (i) and (2), the third, and fourth terms are respectively equal,
By first,
^ LN ^O AB,
.-.
Theorem 645.
The
sects
joining
its
pole
VII. to
points on any circle of a
sphere are equal.
Proof.
(600.
perpendicular to
on the
its
If through the center of a circle a line
be passed
plane, the sects from any point of this line to points
circle are equal.)
646.
Corollary.
Since chord
fore the arc subtended
by chord the subtended arc equals by chord
Hence the
PA PA PB
equals chord PB, therein the great circle in the great circle PB.
great-circle-arcs joining a pole to points
circle are equal.
PA
on
its
THE ELEMENTS OF GEOMETRY.
252 So,
one of
if
an arc of a great
its
circle
be revolved in a sphere about
extremities, its other extremity will describe a circle
of the sphere. 647. One-fourth a great circle 648.
with
its
is
called a Quadrant.
The
great-circle-arc joining
pole
is
a quadrant.
any point
in a great circle
TRI-DIMENSIONAL SPHERICS, 651. If
from the
vertices,
A and BC
F, of
253
any two angles
in a
and GH^ be described, the sphere, as poles, great circles, will to be one another in the ratio of the arcs of these angles circles intercepted
For the angles and GOH.
between their sides (produced
A
and
F are
if
necessary).
equal respectively to the angles
BOC
(591. Parallels intersecting the
same plane
are equally inclined to
it.)
GOH
But BOC and are angles at the centers of equal cirand cles, therefore, by 506, are to one another in the ratio of the arcs BC and GH. 652. Corollary. Any great-circle-arc drawn through the pole of a given great circle
perpendicular to that circle. FG is a quadrant therefore the great circle described with G as pole passes through F, and so the arc intercepted on it between and is, by
FG
is
± GH.
is
For, by hypothesis,
;
GF
GH
But, by 651, the angle at G is to this as a quadrant straight angle is to a half-line. Inversely, any great-circle-arc perpendicular to a great cir-
648, also a quadrant.
cle will pass
through its pole. For if we use G as pole when the angle 'at 6^ is a right angle, then FH, its corresponding arc, is a quadrant when GF is
a quadrant
;
therefore,
by 649,
F is
the pole of
GH,
THE ELEMENTS OF GEOMETRY.
254
Theorem The smallest
653.
line in
VIII.
a sphere^ between two points,
is the
great-circle-arc not greater than a semicircle^ which joins them.
AB
a greaf-c/rch-arc, not greater than a on a sphere. two semicircle, joining any points A and let the A and be First, joined by the broken line A CB, points which consists of the two great-circle-arcs A C and CB. Hypothesis.
is
B
B
AC
Conclusion.
-it
CB > AB.
Join O, the center of the sphere, with A, Bj and C.
Proof.
^
AOC
COB >
^
-I-
same plane meet
^ AOB.
one point, any two formed are together greater than the third.)
(603. If three lines not in the
But the corresponding arcs are
The smallest
let
F be
line
in the
same
of the angles
ratio as these angles,
AC + CB>AB.
.-.
Second,
at
any point whatever on the great-circle-arc AB. A to must pass through B.
on the sphere from
For by revolving the
great-circle-arcs
B AB and BB
about
A
and
B
as poles, describe circles.
These
F be by
circles
touch at P, and
any other point in the
great-circle-arcs, then,
by our
FA /.
FA
> FA, and
lie
circle
wholly without each other
whose pole
is
B, and
join
;
for let
FA,
First,
-\-
FB> AB,
F lies without the circle whose pole
is
A.
FB
TRI-DIMENSIONAL SPHERICS.
255
ADEB
be any line on the sphere from A to B not passing and therefore cutting the two circles in different points, one A portion of the line ADEB, namely, DE, lies in Z>, the other in £. be revolved about between the two circles. Hence if the portion be revolved A until it takes the position AGP, and the portion
Now
through
let
/*,
AD
BE
B
AGPHB will be less than BHP, ADEB. Hence the smallest line from A to B passes through P, that must be the arc through any or every point in AB consequently AB about
the line
into the position
is,
;
it
itself.
654. Corollary. between two points.
A
sect
is
the smallest line in a plane
BOOK
IX.
TWO-DIMENSIONAL SPHERICS. INTRODUCTION.
.
655. Book IX. will develop the Geometry of from theorems and problems almost identical with assumption gave us Plane Geometry. In Book have been demonstrated by considering the sphere in ordinary tri-dimensional space.
But,
if
we
the Sphere, those whose VIII., these as contained
really confine our-
they do not admit of demonstration, except by making some more difficult assumption and so they are the most fundamental properties of this surface and its selves to the sphere
itself,
:
characteristic line, the great circle
our
first
plane and
;
just as the assumptions in of the
book were the most fundamental properties its
characteristic line.
So now we whenever
and,
spherical line.
will call a great circle
in this
Sect
simply the spherical line ;
book the word
now means
line is used, it means a part of a line less than a
half-line. 257
THE ELEMENTS OF GEOMETRY.
258
Two-Dimensional Definition
of the Sphere.
656. Suppose a closed line, such that any portion of it may be moved about through every portion of it without any other Suppose a portion of this line is such, that, when change. moved on the line until its first end point comes to the trace of its second end point, that second end point will have moved to Call such a portion a halfthe trace of its first end point. a sect. and lesser line, Suppose, that, while the portion any extremities of a half-line are kept fixed, the whole line can be so moved that the slightest motion takes it completely out of its trace, except in the two fixed points. Such motion would generate a surface which we will call the Sphere.
Fundamental Properties
of the Sphere.
ASSUMPTIONS. 657.
A
figure may be moved about in a sphere without any that is, figures are independent of their place on ;
other change the sphere.
Through any two points
in the sphere can be passed a with the congruent generating line of the sphere. In Book IX., the word line will always mean such a line, and sect
658.
line
mean
a portion of it less than half. sects cannot meet twice on the sphere that is, if 659. two sects have two points in common, the two sects coincide
will
Two
;
between those two points. 660. If two lines have a common sect, they coincide Therefore through two points, not end points of throughout. a half-line, only one distinct line can pass. 661. A sect is the smallest path between its end points in the sphere.
TWO-DIMENSIONAL SPHERICS. 662. will
fit
259
A
piece of the sphere from along one side of a line either side of any other portion of the line.
DEFINITIOXS. of a sect is kept fixed, the other end the point moving sphere describes what is called an arc, while the sect describes at the fixed point what is called a The angle and arc are greater as the amount spherical angle.
663. If
one end point in
of turning in the sphere is greater.
When
a sect has turned sufficiently to fall again into but on the other side of the fixed point or verline, is called a straight angle, and the arc the described tex, angle described is called a semicircle. 664.
the same
665. Half a straight angle is called a right angle. 666. The whole angle about a point in a sphere perigon the whole arc is called a circle.
is
called a
:
The point is called the pole of sects are called its spherical radii.
the circle, and the equal
ASSUMPTIONS. 667.
A circle
can be described from any pole, with any sect
as spherical radius. 668. All straight angles are equal.
THE ELEMENTS OF GEOMETRY.
26o
669.
Corollary Corollary
All perigons are equal. If one extremity of a sect
I.
is in a line, the two angles on the same side of the line as the sect are
670.
II.
together a straight angle.
671.
Corollary
angles having a
III.
common
Defining adjacent angles as two common arm, and not over-
vertex, a
it follows, that, if two adjacent angles together equal a straight angle, their two exterior arms fall into the same
lapping, line.
672.
Corollary IV.
If
two sects cut one another, the
vertical angles are equal.
673. Any line turning in the sphere about one of its points, through a straight angle, comes to coincidence with its trace, and has described the sphere. 674.
Corollary
I.
The sphere
is
a closed surface.
TWO-DIMENSIONAL SPHERICS. 675. it
Corollary
II.
Every
26 1
line bisects the sphere,
— cuts
into hemispheres.
Reduced Properties of the Sphere.
Theorem 676.
All
lines in
a sphere
I.
intersect,
and
bisect each other at
their poi7its of intersection.
BB' and
CC
be any two lines. by 675, each of them bisects the sphere, therefore the second cannot lie wholly in one of the hemispheres made by the first, therefore they intersect at two points let them intersect at A and A' If BA C be revolved in the sphere about A until some point in the Let
Since,
.
;
sect
AB coincides with
some point
in the sect
AB\
then
ABA'
will lie
along AB'A'. (660.
Through two
points, not
end points of a
half-line,
only one distinct line can
pass.)
Then,
also, since the angles
BA C
and B'A C", being
by 672, equal, ACA' will along AC A'. Therefore of intersection of AB and A C must coincide with the lie
intersection of
AB' and
A C,
and
ABA'
be a
vertical, are,
the second point
second point of half-line, as also A CA'.
THE ELEMENTS OF GEOMETRY.
262
677.
by two
A spherical
figure such as
half-lines, is called
ABA'CA^ which
is
contained
a Lime.
Theorem
II.
The angle contained by
the sides of a bine at one of their points of intersection equals the angle co7itained at the other.
678.
For,
pose
4-
if
the two angles are not equal, one must be the greater. 4- A'.
Sup-
A>
Move the lune in the sphere until point A' coincides with the trace of point A, and half-line A'BA coincides with the trace of half-line CA' Then, since we have supposed 4 4- ^' ^^ half-Hne A' CA
A
would
^>
.
start
trace of
ABA'
.
A
^
AC
A' and the between the trace of neither between could meet it again them, Starting
from the trace of
TWO-DIMENSIONAL SPHERICS. until
it
reached the trace of A'
lune less than
its trace, .*.
which
;
is
263
and so we would have the surface of the contrary to our
first
assumption (657).
the two angles cannot be unequal.
DEFINITIONS. 679.
from a
The
Stipplement of a Sect
is
the sect by which
it
differs
half-line.
680. One-quarter of a line is called a Quadrant. 618.
A
formed by
682.
Spherical Polygon
is
a closed figure in the sphere
is
a convex spherical polygon of
sects.
A
Spherical Triangle
three sides.
Symmetrical Spherical Polygons are those in which the and angles of the one are respectively equal to those of
683.
sides
THE ELEMENTS OF GEOMETRY.
264
If one end of a one polygon, and one end of another sect pivoted within the symmetrical polygon, and the two sects
the Other, but arranged in the reverse order. sect
were pivoted
w^ithin
revolved so as to pass over the equal parts at the same time, one sect would move clockwise, while the other moved counterclockwise.
Two
points are symmetrical with respect to a fixed line, called the axis of symmetry, when this axis bisects at right angles the sect joining the two points. 684.
Any two when every
figures are symmetrical with respect to an axis
point of
one has
its
symmetrical point on the
other.
Theorem
III.
685. The perimeter of a convex spherical polygon wholly contained within a second spherical polygon is less than the perimeter
of the second.
Let ABFGA ABCDBA.
be a convex spherical polygon wholly contained
in
TWO-DIMENSIONAL SPHERICS. Produce the sides
A G and GF
to
K and L respectively. By 66 1,
a sect
is
26$
meet the containing perimeter
the smallest path between
its
end
points,
BF < BCLF, and LG < LKG, and KA < KDEA BFGA < BCLGA < BCKA < BCDEA.
.-.
at
;
.-.
Theorem The swn of
686. less
than a
the sides
IV.
of a convex spherical polygon
is
line.
^C
ABODEA
be the polygon, and let its sides BA and be meet again at M. Since the polygon is convex, the ^ ^CZ> < St. ^, and also ^ BAE < st. ^ ; therefore ABCDEA lies wholly within ABCMA therefore, by 685, the perimeter oi ABCDEA Let
produced
to
;
<
the perimeter of
ABCMA,
that
is,
less
than a
line.
SYMMETRY WITHOUT CONGRUENCE. 687. If
two figures have central symmetry
in a plane, either
can be made to coincide with the other by turning it in the This holds good when for plane through a straight angle. *'
plane" we substitute "sphere." If two figures have axial symmetry
made
in a plane,
they can be
by folding the plane over along the axis, but not by any sliding in their plane. That is, we must use the to coincide
THE ELEMENTS OF GEOMETRY.
266
third dimension of space,
and then their congruence depends
of the plane that its two sides are indistinthat so guishable, any piece will fit its trace after being turned This procedure, folding along a line, can have no place over.
on the property
in
a strictly two-dimensional geometry
;
and, were
we
in tri-
dimensional spherics, we could say, that from the outside a sphere is convex, while from the inside it is concave, and that a piece of it, after being turned over, will not fit its trace, but only touch
it
at
one point.
So
figures with axial
symmetry,
according to 684, on a sphere cannot be made to coincide and the word symmetrical is henceforth devoted entirely to such. ;
Theorem
Two
688.
V.
spherical tria7igles having two sides
and
the in-
cluded a7tgle of one equal respectively to two sides and the inchtded angle of the other, are either congruent or symmetrical
(i)
If the parts given equal are arranged in the
DEF and until
it
(2)
DEF
ABC,
then the triangle coincides with ABC.
DEF can
same
order, as in
be moved in the sphere
If the parts given equal are arranged in reverse order, as in
and A'B' C\ by making a
ABC, we
triangle symmetrical to
get the equal parts arranged in the
which proves
same order
A'B'C, as in
as
DEF,
DEF congruent to any triangle symmetrical to A'B' C
TWO-DIMENSIONAL SPHERICS.
Theorem
26y
VI.
If two spherical triaiigles have two sides of the one to two sides of the other, but the included angle
689.
equal respectively
tJie first greater than the included angle of the second, tJien the thi7'd side of the first will be greater than the third side of
of
the second.
Hypothesis.
AB = DE,
Conclusion.
BC
When
Proof.
on the same side
Z)F
will
the triangle
hut
AC =
EF,
AB coincides with DE,
if
but
4.
BAC>i. EDF,
DF
-4.
BAC
F are
C and > ^ EZ>F,
the points
hne AB, then, since he between BA and A C, and side of the
is
must stop
the
either within
on BC, or after cutting BC. within the triangle, by 685,
^C + CA > EF + FD
DF,
;
BC>EF.
.-.
cuts
AC ^
ABC,
F
When
and
> EF.
When F Hes on BC, then EF is but a part of BC. In case DF BC, call their intersection point G. Then BG -{ GF > EF, and
GC + GD>AC, hut FD = AC,
.-.
BC
-ir
.'.
FD >AC +
EF',
BC>EF.
when one
pair of equal sides coincide, the other pair he on oppoof the line of coincidence, the above proof will show the third side of a triangle symmetrical to the first to be greater than the third If,
site sides
side of the second triangle,
and therefore the third side of the
greater than the third side of the second.
first
THE ELEMENTS OF GEOMETRY.
268
From 6^Z and
689, by 33, Rule of Inversion, if two two have sides of the one equal respectively spherical triangles to two sides of the other, the included angle of the first is 690.
greater than, equal to, or less than, the included angle of the second, according as the third side of the first is greater than, equal to, or less than, the third side of the second.
Corollary.
Therefore, by 688, two spherical triangles having three sides of the one equal respectively to three sides of the other are either congruent or symmetrical. 691.
Problem To
692.
Let In
BA C be
its
Join
bisect
a given spherical angle.
the given
2^
.
arms take equal sects
BC.
I.
With
B as pole,
AB and A C each less and any
than a quadrant. BC, but
sect greater than half
not greater than a quadrant, as a spherical radius, describe the circle EDF. With an equal spherical radius from C as pole describe an arc intersecting the circle
DC, DA.
Z>^
For the "as
BD = DC
EDF
at
D
within the lune
BAC.
Join
^^C having AD common and AB =
DB,
shall bisect the angle
ABD, A CD,
(spherical radii of equal circles), are, .-.
^ BAD =
^ CAD.
AC, and
by 691, symmetrical,
TWO-DIMENSIONAL SPHERICS. 693.
Corollary
To
I.
bisect the re-entrant angle
produce the bisector of the angle 694. line
Corollary
To
II.
269
from a given point in the
BAC,
BAC.
erect a perpendicular to a given line, bisect the straight angle at
that point.
Problem 695
•
With
To
A
bisect
a given
1 1.
sect.
and B, the extremities of the given
spherical radii greater than half arcs intersecting at C.
AB,
sect, as poles, and equal but less than a quadrant, describe
^C
and BC\ and, by 692, bisect the angle Join the bisector to meet at D.
ACBy and
produce
AB
D
is
the
mid point of
For, by 688,
696.
a"
ACD
Corollary.
the sect is
AB.
symmetrical to '^ BCD,
The
spherical triangle are equal.
angles at the base of an isosceles
THE ELEMENTS OF GEOMETRY.
270
Problem 697. To draw a perpendicidar point in the sphere not in the line.
III. to
a given line from a given
Given, Me line AB^ and point C. on the other side of the line AB, and with C as Take a point and CD as spherical radius, describe an arc cutting AB in F pole, and G. Bisect the sect FG at B, and join CH. CH shall be ± AB. and FCH are symmetrical. For, by 691, 'as
D
GCH
698.
If two
lines be
Theorem
VII.
drawn
a sphere^ at right angles from which all
in
given line, they will intersect at a point drawn to the given line are eqnal.
AB and CB, drawn at right angles to AC, intersect at B, ^C again at A' and C respectively.
Let
meet
to
a
sects
and
TWO-DIMENSIONAL SPHERICS. Then ^ (678.
BA'C = ^ BAC,
The angles contained by
AC =
on the
line
AC,
common
slide
^ ^C^';
AC
supplement
Hence, keeping
.
each of the
.-.
A
into coincidence with
A' C, being all right, AB will and hence the figures ABC and A' along C'B,
CB
of intersection,
A'C,
ABC until AC comes
Then, the angles at A, C,
and
=
two points
the sides of a lune, at their are equal.)
moreover,
for they have the
^ BCA'
and
27 1
,
lie
and A'
C
C
.
along A'B,
BC coincide.
half-lines
ABA'
and
CBC
is
bisected at B.
In like manner, any other line drawn at right angles to through B, the mid point of ABA'
^ C passes
.
Hence every
Corollary
699.
radius
Corollary
two points 701.
the line 702.
AC to B
I.
A
is
line
a.
is
quadrant.
a circle whose
spherical
a quadrant.
is
700.
sect from
in a line,
Corollary is
II.
A
and not III.
perpendicular to
Corollary
IV.
point which
a quadrant from
is
in the line, is its pole.
Any
sect from the pole of a line to
it.
Equal angles
at
the poles of lines
intercept equal sects on those lines. If be the pole, and 703. Corollary V.
K
any other line, the angles and semilunes one another 3.s AC to FG.
ABC
FG
and
a sect of
FKG
are to
THE ELEMENTS OF GEOMETRY.
2/2 704.
We
see,
from 698, that a spherical triangle may have
two or even three right angles.
ABC
B
has two right angles, a spherical triangle and C^ called a bi-rectangular triangle. is By 698, the vertex and are quadrants. the pole of BC, and therefore If
it
A
is
AB
AC
The Polar
of a given spherical triangle is a spherical of whose sides are respectively the vertices the poles triangle, of the given triangle, and its vertices each on the same side of
705.
a side of the given triangle as a given vertex.
Theorem
VIII.
706. If^ of two spherical triangles^ the first is the polar of the second^ tJien the second is the polar of the first.
A'B'C
Hypothesis.
Lei
Conclusion.
Then
ABC
be the polar of ABC, the polar of A'B'C
is
TWO-DIMENSIONAL SPHERICS, Join A'B and A'C. therefore the pole of A' ,
Proof. Since
C is
B
C
is
the pole of
A'B\ .*.
In
like
therefore
by 700, A'
is
BA'
a quadrant
is
a quadrant
is
the pole of
;
and since
\
BC,
manner, B' is the pole of A C, and C" of ^^. A and ^' are on the same side oi B'C'y and B' on the
B
Moreover,
same
CA'
2/3
C
^'C, and
side of
.'.
and
ABC
is
C on the same side oi A'B',
the polar of
A'B'C
Theorem IX. 707. In a pair of polar triangles, any angle of either intercepts^ on the side of the other which lies opposite to it^ a sect which is the
supplement of that
side.
ABC
and A'B^C be two polar Produce A'B' and B' to meet is the pole of B' therefore Since Let
C C
B
is
=
,
CD CD = BC
the pole of A'B' ^ therefore half-line,
but
BE
-^
of -5 C which A' intercepts,
Exercises. as
its
angle
is
109.
is
Any
triangles.
BC BE
is
at Z> is
a quadrant -h
DE.
is
to half a right angle.
^
;
;
respectively.
and since
therefore
Therefore
the supplement of
lune
and
a quadrant
BE
DE,
-h
C
CD
the sect
BC.
to a tri-rectangular triangle
THE ELEMENTS OF GEOMETRY.
274
Theorem X. 708. If two angles of a spherical triangle be equals the sides which subtend them are equal.
Hypothesis.
Conclusion.
In ^ ABC let BC — AB.
4.
A^
4.
C,
For draw A'B'C, the polar of ABC. and A'B' the equal 4% A and C intercept equal sects. and A'B', being, by 707, the supplements of these Therefore B' Proof.
Now, on B'
C C
equal sects, are equal, (696.
The
.*.
angles at the base of
.'.
—
'^' 4- ^'> an isosceles spherical triangle are equal.)
"4-
BC and AB are equal, BC = AB.
the supplements oi /.
Theorem XI. 709.
If one angle of a spherical triangle be greater than a must be greater than the side
second, the side opposite the first opposite the second.
Make 4 by 708, AZf
In^ABC, 4 C>4A. .'.
ACD =4A; = CD.
TWO-DIMENSIONAL SPHERICS. But, by 66i,
CD
-\-
2^$
DB> BC, AD \- DB>BC.
.'.
From 708 and 709, by 33, Rule of Inversion, one side of a spherical triangle be greater than a second, the angle opposite the first must be greater than the angle 710. If
opposite the second.
Theorem
XII.
711. Two spherical triangles having two angles and the included side of the one equal respectively to two angles and the included side of the other^ are either congruent or symmetrical.
For the
first
triangle
can be moved
with the second, or with a triangle
in the sphere into coincidence
made symmetrical
Theorem
to the second.
XIII.
Two
spherical triangles having three angles of the one equal respectively to three angles of the other^ are either cojigruent 712.
or symmetrical. Since the given triangles are respectively equiangular, their polars are respectively equilateral. (702.
Equal angles at the poles of lines intercept equal sects on those lines; and, by 707, these equal sects are the supplements of corresponding sides.)
THE ELEMENTS OF GEOMETRY.
2/6
Hence
these polars, being, by 691, congruent or symmetrical, are respectively equiangular, and therefore the original spherical triangles are respectively equilateral.
Theorem XIV.
An
exterior angle of a spherical triangle is greater than, or less than, either of the interior opposite angles, accordeqnal as the medial ing from the other interior opposite angle is less than, equal to, or greater than, a qtiadrant. 713.
to,
A CD be
Let at
E.
Join
an exterior angle of the
BE, and
a"
produce to F, making
A (688. Spherical triangles having
ABE ^
ABC.
EF =
By
695, bisect
BE.
Join
AC
FC.
A CFE,
two sides and the included angle equal are congru-
ent or symmetrical.)
^ BAE = ^ FCE. BE be a quadrant, BEE
.-.
If,
now, the medial
F lies
676,
on
is
a half-line, and, by
BD;
^DCE coincides with ^ FCE, DCE = ^ BAE. the medial BE be less than a quadrant, BEF less than a .'.
.'.
If line,
and
is
F lies
A C and CD DCA > :. ^ DCA >
between
.-.
And Hne, and
^
4.
4.
FCE,
^ BAC.
BE be greater than a quadrant, BEF F lies between CD and A C produced
if
;
/.
^
.-.
4.
DCA DCA
half-
;
< 4. FCE, < 4 BAC.
is
greater than a half-
TWO-DIMENSIONAL SPHERICS. Thus, according as quadrant, the exterior interior opposite
2^
BE
greater than, equal to, or less than, a less than, equal to, or greater than, the
is
^ A CD
2//
is
BA C,
From
713, by 33, Rule of Inversion, as the exterior angle is greater than, equal According or less the interior to, than, opposite angle BAC, the medial is less than, equal to, or greater than, a quadrant. 714.
ACD
B£
Theorem XV. 715. Two spherical triangles having the angles at the base of the one equal to the angles at the base of the other, and the sides opposite one
pair of eqnal ajigles eqnal, ai'e eitJier congruent or symmetrical, provided that in neitJier triangle is the third angular point a quadrant from any point in that half of its base not adjacent to
if
First,
as
triangles, 4.
C=
will lie
the parts given equal in
ABC
F,2iXidAB =z along BC, and 4.
lie
clockwise in the two spherical
DEF, where we suppose 4- ^ — 4- ^^ DE, make BE coincide with AB then EF
and
;
VF must A G,
coincide with
if it
could
AGC
it
if the equal parts lie in one spherical triangle clockwise, other counter-clockwise, as in a" A'B' and 'a DEF, then
Second,
A
For
AGB
AH in
AC.
with exterior would make 2^^ = interior opposite ^ C, and therefore, by 714, with medial a quadrant, which is contrary to our hypothesis.
take any other position, as •4.
of the sides equal by hypothesis.
07ie
the
DEF
^'a ABC,
C
which
is
symmetrical to
a A'B'C,
THE ELEMENTS OF GEOMETRY.
2/8
Theorem XVI. If a
716, to
sect
drawn
lute be less than
a
drawn which
in the sphere from a point perpendictilar it is the smallest which can be
a qiiadranty
the sphere from the point to the line ; of others that nearer the perpendicular is less than that which is more also to every sect drawn on one side of the perpendicidar
i7t
is
remote ;
,
there can be
Let
A
drawn
one^
be the point,
and only
one^ sect
equal on the other
side.
BD the Hne, AB J_ BD^ and AD nearer than
AE to AB. Produce
AB to
C,
BC ^ AB. AD = CD.
making
Then (688. Spherical triangles having
Join
CD, CE.
two sides and the included angle equal are con-
gruent or symmetrical.)
But,
by 661,
AD + CD /.
Also,
or
tAD > AC
AD
or
2AB,
> AB.
by 685,
AE + CE or 2AE > AD -f CD or 2AD. Again, make BE = BD. Join AE. As before, AE = AD, and we have already shown sects
on the same
side of the perpendicular can be equal.
that
no two
TWO-DIMENSIONAL SPHERICS. 717.
A
279
Corollary. The greatest sect that can be drawn from is the supplement of AB.
BD
to
Theorem XVII. 718.
equal
to
If two spherical triangles have two sides of the one two sides of the other, and the angles opposite 07ie pair
of equal sides equal, the angles opposite the other pair equal or supplemeiital.
2^
First, given the equal parts AB = DE, AC — E arranged clockwise in the two spherical triangles.
ai'e
either
DF, and ^
B=
^ = D, the spherical triangles are congruent. = :^ D, one must be the greater. :^ ^ is not Suppose ^ A > ^ D. Make DE coincide with AB. Then, since ^B = :^E, side EF will lie along BC; since A> ^ D, side DF will lie between AB and A C, as at A G. Now, the two angles at G are supplemental but one ^ F and the other = isosceles. C, because "a CFG If
}^
2s^
If
4.
is
;
is
:j^
Second, if the parts given equal lie clockwise in one spherical triand 'a DEF, angle and counter-clockwise in the other, as in "a A'B' the above proof shows then, taking the "a , symmetrical to A'B'
C
C
ABC
^
F
equal or supplemental to
^
C, that
is,
to
^
C
,
THE ELEMENTS OF GEOMETRY.
280
Theorem XVIII. 719.
Symmetrical
isosceles spherical triangles are congruent.
For, since four sides
and four angles are equal, the
between clockwise and counter-clockwise
is
distinction
obliterated.
Theorem XIX. 720. The locus of a point from which the two sects drawn to two given points are equal is the line bisecting at right angles the sect joining the two given points.
it,
For every point in this perpendicular bisector, and no point out of possesses the property.
TWO-DIMENSIONAL SPHERICS.
281
Problem IV. 721. To pass a circle through auy three points, or to find the circumcenter of any spherical triangle.
Find the intersection point of perpendiculars erected
at the
mid-
points of two sides.
Theorem XX. made with a
side of a spherical triangle by joining its extremity to the circiimc enter^ equals half the anglesum less the opposite angle of the triangle. 722.
Any
Yox^A + i.
angle
:^B
OCA = ^
+
^C=2^
\
OCA +
2
^ OCB
±2^
OAB,
U OCB ± ^ OAB) ^A
-{-
^B ^ ^C -
^B.
THE ELEMENTS OF GEOMETRY.
282
Corollary.
723.
Symmetrical spherical triangles are equiv-
alent.
For the three pairs of isosceles triangles formed by joining the vertices to the circumcenters having respectively a side and two adjacent angles equal, are congruent.
Theorem XXI. 724. When three lines mutually intersecty the two triangles on opposite sides of any vertex are together equivalent to the lune with that vertical angle.
"aABC^^ ADF = For
CH^
DF —
having the
common
lune
ABHCA.
BC, having the common supplement CD and FA = common supplement HF\ and AD = BH^ having the \
supplement
HD
ABC +
aADF^aBCBT, ADF = A ABC + a BCH =
;
.'.
.-.
A
The
A
lune
ABHCA.
Spherical Excess of a spherical triangle is the sum of its angles over a straight angle. In general, the spherical excess of a spherical polygon is the excess of the sum of its angles over as many straight angles as it has 725.
excess of the
sides, less two.
TWO-DIMENSIONAL SPHERICS.
283
Theorem XXII. 726. is
A
half the
spherical tria?tgle is equivalent to a lime whose angle t7
iangles spherical excess.
ABC ABC
until they meet again, Proof. Produce the sides of the 'a now forms part of three two and two, at D^ F^ and H. The "a lunes, whose angles are A, Bj and C respectively.
But,
by 724, lune with ^
^ =
a
ABC +
a ADF.
Therefore the lunes whose angles are A, B^ and C, are together equal to a hemisphere plus twice a" ABC.
But a hemisphere .-.
2^ ABC =
727.
triangle
is
a lune whose angle
lune whose
^
is
(^
is
a straight angle,
+^+ C=
st.
^)
lune whose
^
is e,
The sum of the angles of a spherical I. greater than a straight angle and less than 3 straight
Corollary is
angles. 728.
Corollary
greater than
II.
Every angle
of a spherical triangle
is
^e.
Corollary III. A spherical polygon is equivalent to a lune whose angle is half the polygon's spherical excess. 730. Corollary IV. Spherical polygons are to each other 729.
as their spherical excesses, since, angles.
by 703, lunes are as their
THE ELEMENTS OF GEOMETRY.
284
Corollary V.
731.
To
construct a lune equivalent to any 2) straight angles, subtract {71
—
spherical polygon, add its angles, halve the remainder, and produce the arms of a half until they
meet
again.
Theorem XXIII. If the line be completed of which the base of a given spherical triangle is a secty and the other two sides of the triangle be produced to meet this line^ afid a circle be passed through these 732.
tzvo points of intersectio7Z and the vertex of the triangle^ the locus of the vertices of all triangles equivalent to the given triangle^ and on the same base with it^ and on the same side of that base,
arc of this circle termifiated by the intersection points containing the vertex. is the
Let
AB
ABC
and
CB
be the given spherical
to
AC
meet
pass a circle through
AP
produced Let
triangle,
in
B,D,E.
AC
F be
base.
Produce
By
any point in the arc
AP
and CP.
its
D and E respectively.
and
719,
EBD,
produced passes through the opposite point the a PDE. through E, forming with Join F, its circumcenter, with P, D, and E. Since, by 678, the two angles of a
Join
D, and
DE
CP
lune are equal, /.
= PCA = ^AFC ^
^ FAC -4.
St.
St.
^ ^
- PDE, - ^ FED, :^
4.DFE',
TWO-DIMENSIONAL SPHERICS. /.
^{PAC^- PCA
=
4-
2 St.
APC) = ^s —
2St.
^s
^ i^Z?^
2
-
=
^ {PDE
285
+ PED - DPE)
a constant.
Any angle made with a
side of a spherical triangle by joining its extremity (722. to the circumcenter, equals half the angle-sum less the opposite angle of the
spherical triangle.)
733. In a sphere a line is said to touch a circle circle, but will not cut it.
when
it
meets the
Theorem XXIV. The Ime drawn at right angles
734'
a
circle at its
Let
BD
any point
By
7
1
extremity touches the
be perpendicular
C in BD. C > AB, 6,
And no
-(4
therefore
C is
AE
_L
BF. .*.
By
^
716,
is
radius of
circle.
to the spherical radius
other line through B, as
For draw
to the spherical
AB.
without the circle.
BF, can be
^^ > AE,
within the circle.
tangent.
Join
A with
THE ELEMENTS OF GEOMETRY,
286
Theorem XXV. In a sphere the sum of one pair of opposite angles of a qtiadiilateral inscribed in a circle equals the sum of the other 735.
Join E^ the circumcenter, with A, B^ inscribed quadrilateral.
By
+
696,
4.
ABC =
4.
BAE +
C, Z>, the
4 BCE, and
4.
vertices of the
ADC =
4.
DAE
4 DCE, .\
4
ABC 4-
4
ADC =
4
BAD +
4 BCD,
Theorem XXVI. 736. In equal circles^ equal angles at the pole stand on equal arcs.
For the
figures
may be
slidden into coincidence.
TWO-DIMENSIONAL SPHERICS.
287
Theorem XXVII. Equg,l spherical chords cut equal circles into the same
737.
two
arcs.
Theorem XXVIII. 738. In equal circles^ a?tgles at the corresponding poles have the same ratio as their arcs.
Theorem XXIX. 739'
Four pairs of equal
circles
non-concurrent lines in a sphere.
can be drawn
to touch three
BOOK
X.
POLYHEDRONS. 740.
A Polyhedron
741.
The bounding
is
a solid bounded by planes.
planes,
by their
intersections, determine
the Faces of the polyhedron, which are polygons. 742. The Edges of a polyhedron are the sects in which faces meet. 289
its
THE ELEMENTS OF GEOMETRY.
290 743. its
The Summits
of a
polyhedron are the points in which
edges meet.
744. A Plane Section of a polyhedron is the polygon in which a plane passing through it cuts its faces. 745. A Pyramid is a polyhedron of which all the faces, except one, meet in a point.
746.
The
point of meeting
not passing through the apex 747.
The
faces
Latei'al Faces 748.
the one
called the Apex^ and the face taken as the Base.
is
is
and edges which meet
at the
apex are called
and Edges.
Two is
polygons are said to be parallel when each side of parallel to a corresponding side of the other.
A
Prism is a polyhedron two of whose faces are congruent parallel polygons, A»d the other faces are parallelo749.
grams. 750.
The Bases
of a
prism are the congruent parallel poly-
gons. 751.
752.
The Lateral Faces of a prism are all except its bases. The Lateral Edges are the intersections of the lateral
faces.
753.
A Right
pendicular to
Section of a prism
its lateral
edges.
is
a section by a plane per-
POL YHEDRONS. 754.
The
Altitude of a
Prism
Altitude of a
Pyramid
is
291
any sect perpendicular
to
both bases.
755. its
The
vertex to the plane of
A
756. Right Prism dicular to its bases.
its is
is
the perpendicular from
base.
one whose
lateral
edges are perpen-
N 757. Prisms not right are oblique.
758.
A Parallelopiped
is
a prism whose bases are parallelo-
grams. 759.
A
Quader
is
A
Cube
a quader
a parallelopiped whose six faces are rect-
angles.
760.
is
whose
six faces are squares.
THE ELEMENTS OF GEOMETRY.
292
Theorem
I.
761. All the summits of any polyhedron may he joined by one closed line breaking only in them^ and lying wholly on the surface.
For, starting from one face,
ABC
.
.
.
,
each side belongs also to a
neighboring polygon.
A
and B, we may omit AB, and use the remainTherefore, to join der of the perimeter of the neighboring polygon a. In the same way, and C, we may omit BC, and use the remainder of the to join
B
perimeter of the neighboring polygon b, unless the polygons a and b have in common an edge from B. In such a case, draw from in ^
^
the diagonal nearest the edge common to a and b take this diagonal and the perimeter of b beyond it around to C, as continuing the broken ;
Hne
;
and proceed
polygon
in the
same way from
C
around the neighboring
c.
When this procedure has edge in common with ABC
taken in .
.
.
all
summits
in faces having
an
we may, by proceeding from the the same way take in the summits
,
closed broken line so obtained, in on the next series of contiguous faces, etc.
So continue
until the single closed
once, through every summit.
broken Hne goes once, and only
POL YHEDRONS.
Theorem
593
II.
762. Cutting by diagonals the faces not triangles into triangleSj the whole sinface of a7iy polyhedron contains four less
triangles than double its
number of summits.
r=
2(^-2).
For, joining all the summits by a single closed broken line, this cuts 2 trithe surface into two bent polygons, each of which contains 5
—
angles, where
763.
S is
the
Corollary.
of
any polyhedron summits, less two. 764.
and
number of summits.
Remark.
is
The sum of all the as many perigons as
Theorem
II. is
angles in the faces the polyhedron has
called Descartes'
Theorem,
really the fundamental theorem on polyhedrons, though this place has long been held by Theorem III., called Euler's is
Theorem, which follows from
it
with remarkable elegance.
THE ELEMENTS OF GEOMETRY.
294
Theorem
III.
765. The number of faces and summits in any polyhedron^ taken together^ exceeds by two the number of its edges.
Case
I.
If all the faces are triangles.
7^
But
2(6"-
Then, by 762,
2).
also
2E each edge belongs to two time 3 is contained in 2E.
for
By
=
adding,
Case
II.
== 2>F, faces,
and so we get a
— 2F + 2(6" — F + S = E -^ 2,
we have 2E
2); that
triangle for every
is,
If not all the faces are triangles.
Since to any pyramidal summit go as many faces as edges, we may replace any polygonal face by a pyramidal summit without changing the
POL YHEDRONS.
F
-^ equality or inequality relation oi ment only adds the same number to
S
F
to a summit.
F
\-
S =
But, after
E
E
-^ 2
;
for
such replace-
E, and changes one face polygonal faces have been so replaced,
all
by Case
2,
-\-
\.o
295
as to
Therefore always the relation was
I.
equality.
Theorem 766.
IV.
Quaders having Co7igruent bases are
to
each other as
their altittcdes.
A
A-
t Hypothesis. Let a and d be and Q, having congruent bases B.
Q
Conclusion.
Proof. with altitude
Q
\
w
a
\
Q
ma
B
d.
Of a
take any multiple,
ma
mQ.
is
the altitudes of two quaders,
In the same way, take equimultiples
;
then the quader on base
nd nQ, ^
Q greater than, equal According have ma greater than, equal to, or less than as 7n
is
to,
nd
or less than, n Q^ ;
therefore,
by
we
defini-
tion,
Q
',
Q
Not all the faces eight. five sides. There
'.
d.
no. In no polyhedron can triangles and three-
Exercises,
faced summits both be absent
III.
a
\\
is
nor
;
together are present at least the summits have more than
all
no seven-edged polyhedron.
296
THE ELEMENTS OF GEOMETRY.
Theorem 767.
V.
Quaders having equal altitudes are
bases.
2C71
/
to
each other as their
POL YHEDRONS.
Theorem 768.
Two
quaders are
of the ratios of their bases
y\
to
297
VI.
each other in the ratio compounded
and altitudes.
THE ELEMENTS OF GEOMETRY.
298
Theorem
VII.
a quader of equiva-
769. Afiy parallelopiped is equivalent to and equal altitude.
lent base
AB
an oblique parallelepiped For, supposing the four to AB^ take sect parallel edges prolong
ee J_ Now
DF
CD, and
the solids
\\
DEB
CE. Through
CEA
and
and edges respectively equal.
DEA,
on an oblique AB, and
CD —
CE
DE
and
base, draw-
parallel planes.
ipsiss
are congruent, having all their angles in turn from the whole solid
Taking each
AB — CD. GH = CD, and through G and H pass planes
leaves parallelopiped In the line take
CD
perpendicular to
The
solids
GH.
DEH and CEG are congruent, therefore
parallelopiped
CD = GH. Now prolong the four HK, and through L and
edges not parallel to
M
As
GH, and
pass planes perpendicular to
GH — LM;
but before, parallelopiped equivalent base and equal altitude to parallelopiped
LM
AB.
take
LM —
LM.
is
a quader of
POL YHEDRONS.
Theorem
299
VIII.
770. A plane passed through two diagonally opposite edges of a parallelopiped divides it into two equivalent triangular prisms.
f=a>\
THE ELEMENTS OF GEOMETRY.
300
Theorem 772.
IX.
If a pyrajnid be cut by a plane parallel
section is to the base as the square of
from
to its base, the
perpe7idiciLlar on
it,
the vertex, is to the square of the altitude of the pyramid.
The cut
tJie
section
them
and base are
similar, since
corresponding diagonals
into triangles similar in pairs because having all their sides
respectively proportional.
For
A'C
VC B'C BC AC AC
AC
:
and
.-.
and
in the
::
VC
:
:
VC
:
::
:
::
B'
:-.
A'B'
VO"
:
C
:
:
FC,
:
BC;
:
AB,
VO,
same way,
B'C
BC
t.A'B'C'^
.-.
section
:
:
base
::
ABC,
t^
A' C'^
:
AC^
tic, ::
VO^
:
VC
In pyramids of equivalent bases and 773. Corollary. equal altitudes, two sections having equal perpendiculars from the vertices are equivalent.
POL YHEDRONS.
301
Theorem X. 774.
bases
Tetrahedra
and equal
{triangular pyramids)
having equivalent
altitudes are equivalefit.
Divide the equal altitudes a into n equal parts, and through each point of division pass a plane parallel to the base.
By 773, all sections in the first tetrahedron are triangles equivalent to the corresponding sections in the second. Beginning with the base of the first tetrahedron, construct on each a
section, as lower base, a prism high, with lateral edges parallel to
one
of the edges of the tetrahedron. In the second, similarly construct prisms on each section, as upper base.
Since the first prism-sum is greater than the first tetrahedron, and the second prism-sum is less than the second tetrahedron, therefore the difference of the tetrahedra is less than the difference of the prism-sums. But, by 771, each prism in the second tetrahedron is equivalent to the prism next above it on the first tetrahedron. So the difference of the prism-sums is simply the lowest prism of the first series. As n increases, this decreases, and can be made as small
THE ELEMENTS OF GEOMETRY.
302
we please by taking n sufficiently great ; but it the constant difference between the tetrahedra, as
difference
is
always greater than that constant
and so
must be nought.
Theorem XI.
A
775. triangular pyramid is one-third of a triangular prism the same base and altitude. of
Let
E ABC
be a triangular pyramid.
Through one edge of the
A C,
pass a plane parallel to the opposite lateral edge, EB, and pass a plane parallel to the base. The prism through the vertex has the same base and altitude as the given pyramid. The
base, as
E
ABC DEE plane AEE
cuts the part added, into two triangular pyramids, each and have the equivalent to the given pyramid; for same altitude as the prism, and its bottom and top respectively as bases ;
E ABC
while
E AFC and E AFD
A DEE
have the same altitude and equal bases.
BOOK XL MENSURATION, OR METRICAL GEOMETRY. CHAPTER THE METRIC SYSTEM.
I.
LENGTH, AREA.
776. In practical science, every quantity is expressed by a one a number, the phrase consisting of two components, other the name of a thing of the same kind as the quantity to be expressed, but agreed on among men as a standard or Unit.
—
777. this
The Measurement
of a
magnitude consists
in finding
number.
the process of ascertaining approximately the ratio a magnitude bears to another chosen as the standard and the measure of a magnitude is this ratio 778.
Measurement, then,
is
;
expressed approximately in numbers. 779. For the continuous-quantity space, the fundamental unit actually adopted is the Meter, which is a bar of platinum preserved at Paris, the bar supposed to be taken at the temperature of melting ice. 780. This material meter sally chosen,
is
the ultimate standard univer-
because of the advantages of the metric system 303
THE ELEMENTS OF GEOMETRY,
304
it, which uses only decimal and sub-multiples, being thus in harmony with the multiples
of subsidiary units connected with
decimal nature of the notation of arithmetic. 781. The metric system designates multiples by prefixes derived from the Greek numerals, and sub-multiples by prefixes from the Latin numerals.
Prefix.
MENSURATION-, OR METRICAL GEOMETRY. 784.
An
305
may be practically measured by the known sect, such as the edge of a ruler
accessible sect
direct application of a suitably divided.
But because
incommensurability, any description of a
of
must be usually imperfect and merely approximate. Moreover, in few physical measurements of any kird are more than six figures of such approximations accurate and that degree of exactness is very seldom obtainable, even by the most delicate instruments. 785. For the measurement of surfaces the standard is the square on the linear unit. 786. The Area of any surface is its ratio to this square. 787. If the unit for length be a meter, the unit for area, a square on the meter, is called a square meter (m."" better, "^•''). sect in terms of the standard sect
;
;
In science the Square Centivteter primary unit of surface. 788.
i^^-"^)
is
adopted as the
To find the area of a rectangle.
Rule. Multiply Formula. R = Proof.
the base by the altitude. ab.
— Special
Case
:
When
the base and altitude of
the rectangle are commensurable. In this case, there is always a sect which will divide both base and altitude exactly. If this sect be
assumed as
linear unit, the lengths a
and b are integral
THE ELEMENTS OF GEOMETRY.
306 numbers.
In the rectangle
ABCD,
divide
AD
into ^,
and
AB
b, equal parts. Through the points of division draw lines These lines divide the parallel to the sides of the rectangle.
into
rectangle into a
assumed unit
number
of squares, each of
which equals the
In the bottom row, there are b such and, since there are a rows, we have b squares resquares peated a times, which gives, in all, ab squares. of surface.
;
The composition of
Note.
ratios includes
numerical multiplication
as a particular case.
But ordinary multiplication
is
also
an independent growth from
addition.
In
of view, the multiplier indicates the number of the multiplicand indicates the thing This is not a mutual operation, and the product
this latter point
additions
or repetitions, while
added or repeated. is
always in terms of the unit of the multiplicand. The multiplicand the multiplier is an aggregate of repetitions. ]
may be any aggregate To repeat a thing does gate of the
same
not change
it
in kind, so the result
is
an aggre-
sort exactly as the multiplicand
But if the multiplicand itself is also an aggregate of repetitions, the two factors are the same in kind, and the multiplication is commutative. is the only sort of multiplication needed in mensuration ; for are supposed to be expressed exactly or approximately in numbers, and in our rules it is only of these numbers that we speak. Thus,
This
all ratios
"
when
the rule says, Multiply the base by the altitude," it means, Multiply the number taken as the length of the base, by the number which is the measure of the altitude in terms of the same linear unit. The
product that
is
is, its
a number, which we prove to be the area of the rectangle ; numerical measure in terms of the superficial unit. This is
the meaning to be assigned whenever in mensuration product of one sect by another.
General Proof.
5
L
we speak of
the
represent the unit for length, and the unit of surface, and £7 ab the rectangle, the length of If
MENSURATION, OR METRICAL GEOMETRY. whose base
is
b
and the length
of
whose
altitude
is
307
a^
then,
by
542,
£y ab
aL
bL
But the first member of this equation is the area of the rectangle, which number we may represent by R and the second member is equal to the product of the numbers a and b ;
;
.-.
R=
ab.
olJL
Example
i.
Find the area of a ribbon
i™-
long and
Answer.
Since a square
is
a rectangle having
its
i<^"^
^-^^•'^
wide.
=
loo'^'^-^.
length and breadth
equal, therefore
789.
To
Rule.
area of a square. Ta^e tJie second power of the number denoting the find the
its side.
length of
Note.
This
is
why
the product of a
number
into itself
is
called
the square of that number.
790.
Given, fhe area of a square, to find
Rule. area.
ifte
length of a side.
Extract the square root of the number denoting the
THE ELEMENTS OF GEOMETRY.
308 791.
Metric Units of Surface.
I
hectar
I
dekar
= =
I
ar
=1
I
deciar
I
centiar
I
milliar
(^^•)
(a-)
Example
2.
= = =
hectometer
i sq.
i
dekameter
sq.
meter
sq.
I sq.
decimeter
I sq.
centimeter
I sq.
millimeter
How many square
Remark.
loooo
sq. meters.
1000
sq. meters.
100
sq. meters.
10
sq. meters.
i
sq.
meter.
.1
sq. meter.
.01
sq. meter.
.0001 sq. meter. .000001 sq. meter.
centimeters in 10 millimeters square?
Answer,
792.
= = = = = = = = =
(lo'"'"-)^
=
loo"^"^-^
=
i<^™-^,
Distinguish carefully between square meters
and meters square.
We say 10 square kilometers (lo^"'-'^), meaning a surface which would contain 10 others, each a square kilometer while " the expression '* 5 kilometers square (5^™-)^ means a square whose sides are each 5 kilometers long, so that the figure con;
tains 25^"'•^
Example
3.
A
square
Find
is looo'"-^.
its
side.
Answer. Viooo"^-
793. of
=
31.623™*
Because the sum of the squares on the two sides
a right-angled triangle
is
the square of the hypothenuse,
therefore, also,
>
Given, the hypothenuse
Rule.
and one
Multiply their
the square root.
Formula,
c^
—
a"^
=•
sum {c -\-
side, to find the other side.
by their difference^
a) {c
—
a)
=
b^.
and
extract
MENSURATION, OR METRICAL GEOMETRY.
From
this
are given two
follows, that, in an acute-angled triangle, if we sides and the projection of one on the other, or
it
two sides and an Exercises.
309
altitude,
112.
we can
What must
find the third side.
be given in order to find the
medials of a triangle } 113. If on the three sides of
any triangle squares are described outward, the sects joining their outer corners are twice the medials of the triangle, and perpendicular to them.
THE ELEMENTS OF GEOMETRY.
310
CHAPTER
II.
RATIO OF ANY CIRCLE TO ITS DIAMETER.
Problem 794.
I.
Given^ the perimeters of a regular inscribed
ilar circumscribed polygon^
and a
sim-
compute the perimeters of the regular inscribed a7id circumscribed polygons of double the numto
ber of sides.
a
Take
AB a side
JO
point
side of its
mid-
BG\
then
E.
A
AE
CD a AB at
of the given inscribed polygon, and
the similar circumscribed polygon, tangent to the arc
B
AF
and draw the tangents and Join AE, and at is a side of the regular inscribed polygon of double the
sides, and number of
FG
is
number of
a side of the circumscribed polygon of double the
sides.
Denote the perimeters of the given inscribed and circumscribed polygons by / and q respectively, and the required perimeters of the inscribed and circumscribed polygons of double the number of sides \>y p' and / respectively.
MENSURATION, OR METRICAL GEOMETRY. Since
OC
the radius of the circle circumscribed about the poly-
\s
gon whose perimeter
is q,
/. (548.
The perimeters
q
p
I
OC
:-.
OF
OC
The
radii of their cir-
circles.)
^ COE,
bisects the .'.
(523.
OE.
:
two similar regular polygons are as the
of
cumscribed
But, since
311
OE
:
CE
::
:
EE,
bisector of an angle of a triangle divides the opposite side in the ratio of the other two sides of the triangle.)
q
.-.
p
\
CE
\\
EE,
\
whence, by composition,
p
•\-
EG
q
'.
2p
'.:
CE
-^
EE
2EE = CE
:
:
EG w
a side of the polygon whose perimeter is contained in q, tained as many times in / as
since
is
is
q
\
/, and
(f, is
con-
CE
p
:, If,
-^
q
::
2p
:
q
:
q'.
now, the letters be taken to represent lengths in terms of the unit
sect Z, this proportion
is
pL
-t-
qL
::
2pL
:
which gives the number
qL
:
q'L,
,
^ = 7+7 Again, right
A
AEH ~ EEN, since acute EAH = AH AE EN EE, / /, p :/ 4.
.',
:-.
'.
.'.
AJI and
<')
AE are
:
EEN,
:
:
:
contained the same number of times in
p and /' and and are contained twice that EJV number of respectively, the letters be taken times in /' and / respectively. to If, now, represent length in terms of the same unit sect Z, this proportion is since
EE
pL which gives the number
:
/Z ,
:
:
/Z ,
—
:
/Z,
;
Therefore from the given lengths p and q we compute q' by equaand then with / and / we compute /' by equation (2).
tion (i),
THE ELEMENTS OF GEOMETRY.
312
Theorem The length of a 3.141592+. 795*
The
circle
I.
whose diameter
is
unity
length of the perimeter of the circumscribed square
side of the inscribed square
Now,
-|V2, therefore
its
=
=
perimeter
is
is
4.
A
2V 2.
2V2 and ^ 4 in 794, we find, for the perimep circumscribed and inscribed octagons,
putting
ters of the
is
is
/=
2pq
/ = V>/=
=
3-3137085,
3.0614675.
=
q
—
3.0614675, and Then, taking these as given quantities, we put / for the find the same and formulae, polygons of by 3.3137085,
=
/ 3.1825979, and p' Continuing the process, the results
sixteen sides,
table
:
—
Number Sides.
of
—
3.1 2
will
1445 2. be found as
in the following
MENSURATION, OR METRICAL GEOMETRY. But since, by 654, a chord
shorter than
is
its
313
arc, therefore
the
than any inscribed perimeter ; and, assuming that two tangents from an external point cannot be together shorter than the included arc, it follows that the circle is not longer than a circumscribed circle is longer
therefore the circle whose diameter and not longer than 3.1415928. 3.1415926
perimeter
;
is
longer than
is
unity
A
Variable is a quantity which may have successively 796. an indefinite number of different values. 797* If ^ variable which changes its value according to some law ca7z be made to approach some fixed value as nearly as we please but can never become equal to it, the constant is called the Limit of the variable. y
Example zero
is
;
for,
4.
The
limit of the fraction -, as
X
by taking x
assigned quantity, but
sufficiently great,
we can never make
x increases
we can make -
X
indefinitelyy
less
than any
zero.
it
PRINCIPLE OF LIMITS. 798. Ifiy while tending toward their respective limits, two variables are always in the same ratio, their limits will have that ratio. I
1
^
M—L
h-M
i
'
y
C'
I
C
x'
Z'
y
AC
and AL represent the limits of any two which are always in the same ratio, and Ax, Ay, represent two corresponding values of the variables Let the sects
variable magnitudes let
H^
1
themselves
;
then
Ax For
if
not,
'.Ay
'.'.AC
:
AL,
then
Ax
:
Ay
::
AC
:
to
some
sect
> or < AL,
THE ELEMENTS OF GEOMETRY,
314
Suppose, in the
Ax where AL'
By
that
first place,
is less
'.Ay
::
AC
AL',
:
than AL.
hypothesis, the variable
and may be made to
differ
Ay
from
it
continually approaches
by
less
ALy
than any given quan-
tity.
Let
Ax
and Ay, then, continue to increase, always remaining
same ratio, until Ay differs from AL by less than the or, in other words, until the point y passes quantity L'L the point Z', and reaches some point, as y\ between and Z, and X reaches the corresponding point x' on the sect AC, Then, since the ratio of the two variables is always the same, in the
;
V
Ax
:
Ay
::
Ax'
;
Ay
::
AC
:
A/
::
AC
:
A/,
:
AL\
But, by hypothesis.
Ax .-.
Ax'
AL',
:
But
Ax'
Ay'
Hence the supposition that Ax or to any quantity less than AL, is absurd. Suppose, then, in the second place, that
which
AC
.-.
:
absurd.
is
AL'
:
Ay
:
:
y
Ax
:
Ay
::
AC
:
AL",
where AL" > AL, Now, there is some AC, which is to AL 2is AC is to AL", Substituting this ratio for that oi
Ax
:
Ay
::
AC
AC
:
sect, as
to
AL,
AC,
less
AL'\ we have
than
MENSURATION, OR METRICAL GEOMETRY.
315
which, by a process of reasoning similar to the above, may be shown to be absurd. Hence, since the fourth term of the proportion can be neither greater nor less than that is,
AL
ALy
it
must equal
;
Ax 799.
Corollary.
Ay
\
:
AC
:
:
AL,
two variables are always equal, their
If
limits are equal.
Theorem 800.
Any two
Proof.
By
II.
circles are to each other
as their radii.
548, the perimeters of any two regular polygons of the sides have the same ratio as the radii of their circum-
same number of scribed circles.
The inscribed regular polygons remaining similar to each other when number of sides is doubled, their perimeters continue to have the same ratio. Assuming the circle to be the limit toward which the the
perimeter of the
inscribed
Limits, the circles have the
801. Since c
\
c'
w
polygon increases, by 798, Principle of
same
r
\
ratio as their radii.
r'
w
2r
—
• •
2r that
is,
\
2r
2r'y
f\
the ratio of any one circle to its diameter its diameter.
as the ratio of any other circle to
is
the same
THE ELEMENTS OF GEOMETRY.
3l6
This constant ratio
is
denoted by the Greek letter it. But, with unit diameter, and there-
by 795, this ratio for the circle fore for every circle, is
=
TT
3.141592+.
For any circle. Formula, c =1 2rir.
802.
CIRCULAR MEASURE OF AN ANGLE. 803.
When
its
vertex
any St.
-2^
_
is at
its
^
the center of the
intercepted arc semicircle
any
^
_
_
circle,
by 506,
arc rir^
arc
TT
So,
if
we adopt
as unit angle the radian^ or that part of a peri-
gon denoted by
-—^
st
^,
that
is,
the angle subtended at the center
z:^ of every circle
by an arc equal
to its radius,
and hence named
a radian, then
The number which expresses any angle in radians^ also expresses its intercepted arc in terms of the radius. If u denote the number of radians in any angle,
length of its intercepted arc, then u
~
/ -,
r
and
/ the
MENSURATION^ OR METRICAL GEOMETRY,
The
fraction arc divided by radius^ or Uy
is
317
called the circu-
measure of an angle. 804. Arcs are said to measure the angles at the center which include them, because these arcs contain their radius as often In this sense an as the including angle contains the radian. at is the center measured the arc angle intercepted between by lar
its sides.
THE ELEMENTS OF GEOMETRY,
3i8
CHAPTER
III.
MEASUREMENT OF SURFACES. 805.
angle of
By
any parallelogram is equivalent to the base and altitude therefore,
248,
its
rect-
;
ziy To
find the
area of any parallelogram.
Rule. Multiply Formula, hj =. 806. Corollary.
the base by the altitude. ab.
The
area of a parallelogram divided by
the base gives the altitude. 807.
By
252,
angle of its base
any triangle is equivalent to one-half the and altitude therefore,
Given, one side
rect-
;
and
the perpendicular upon
it
from the opposite
vertex, fo find the area of a triangle.
Rule. Take half the product of Formula, a '^ib.
=
the base into the altitude.
MENSURATION, OR METRICAL GEOMETRY.
319
Given, the three sides, to find the area of a triangle.
808.
Fivm half
Rule.
sum of
the
the three sides subtract each
side separately ; multiply together the half-sum and the The square root of this product is the area. remainders.
Formula, Proof.
\s
z=i
t.
{s
—
a) {s
b) {s
—
the projection of c on
Calling j
=
a"
Calling the altitude
—
b^ -^ €"
tJiree
c).
b,
by 306,
2bj\
b^ 4- c^
•••
J
//,
this gives
2b
^ Ab' .\
.-.
2/lb
2hb
=
=
4h^b^
=
4/5V
2/lb
=
V/4^2^2
y/{2bc
,
,
,
+
_ -
c^
-\-
c){b
-\-
c
—
-{ b
-\-
c) {b
-{•
c
—
.
=a /.
by 807, Ihb
4- b 4- c
\^ i/ib
=
=
A.
-
(^2 _^ ^2
a'){2bc
a^y,
_
—
^2^2^ b^
—
a){^a -{ b
—
a) {a -^ b
b
.
gives
^s{s
-
{b^ -{ c"
V(d: \- b
Writing s
But,
b^
-\'
y^o^sT
.:
—
-\-
c
—
a
_L_^
a){s
-
b) {s
-
-
c^ -\-
c)iya c)
{a
=s— c).
«^),
—
b -^ c),
—
b
a,
-\-
c)
THE ELEMENTS OF GEOMETRY.
320 809.
To find the area of a regular polygon.
Rule.
Take half
of the inscribed
Formula. Proof.
the product of its perimeter by the radius
circle.
TV
= —rp2
.
Sects from the center to the vertices divide the
polygon into congruent isosceles the radius of the inscribed circle, is
triangles whose altitude is and the sum of whose bases
the perimeter of the polygon. 810.
To find ihe area of a
circle.
Rule. Multiply its squared radius by Formula. O = /'V.
-k.
If a regular polygon be circumscribed about the area N^ by 809, is \rp.
circle, its
MENSURATION, OR METRICAL GEOMETRY.
32 1
now, the number of sides of the regular polygon be continually doubled, the perimeter / decreases toward c as limit, If,
N
N
and / are always in toward circle. But the variables and the constant ratio \r\ therefore, by 798, Principle of Limits, their limits are in the same ratio,
O =
/.
But, by
811.
802, c
=
\rc.
2r7r,
To find the area of a sector.
Rule. Multiply the length of =: Formula. 5 J//r^. J/r
=
Proof.
By
506,
5
:
O ^
:
:
/
:
the arc by half the radius.
^
:
:
/^
:
27r,
THE ELEMENTS OF GEOMETRY.
322 813.
To find the area of a sector of an annulus.
Rule.
Multiply the sum of the bounding arcs by half the
difference of their radii.
Formula. Proof.
S A .
.
=
The annular
l{r^
—
sector
=
two sectors \rj^ and \rj^. But /i and 4 are arcs subtending the same angle by 804, -^
.-.
irj^ 814.
-
\rj,
=
\rj,
=
+ W^ -
+
r^ (/, -f 4) /,). \h (/, the difference between the
is
;
therefore,
4 i^/i
—
^
1^/2
= 4(^.-^0(^ +
To find the lateral area of a prism.
/
4).
MENSURATION, OR METRICAL GEOMETRY. Rule.
323
Multiply a lateral edge by the perimeter of a right
section.
Formula.
P
The
=z
Ip.
edges of a prism are all equal. The sides of a right section, being perpendicular to the lateral edges, are the altitudes of the parallelograms which form the
Proof.
lateral
lateral surface of the prism.
815.
A
that every
Cylindric Surface
two
is
generated by a line so moving
of its positions are parallel.
J
816.
The
generatrix in any position
is
called an
Element
of
the surface. 817.
and two
A
Cylinder
is
parallel planes.
a solid bounded by a cylindric surface
THE ELEMENTS OF GEOMETRY.
324 8i8.
The Axis
centers of
819.
A
its
Truncated Cylinder
and a non-parallel
820.
of a circular cylinder is the sect joining the
bases.
is
the portion between the base
section.
To find the lateral area of a right circular cylinder.
Rule. Multiply Formula. C
=
its cl.
length by
its circle.
MENSURATION, OR METRICAL GEOMETRY.
325
Proof. Imagine the curved surface slit along an element and then spread out flat. It thus becomes a rectangle, having for one side the circle, and for the adjacent side an element. 821.
Corollary
circular cylinder
is
I.
The curved
surface of a truncated
the product of the circle of the cylinder
by
the intercepted axis.
'Jf',
For, by symmetry, substituting an oblique for the right section through the same point of the axis, alters neither the
curved surface nor the volume, since the solid between the two sections will be the same above and below the right section. 822.
Corollary
II.
The
lateral area of
any cylinder on
any base equals the length of the cylinder multiplied by the perimeter of a right section. 823.
The
lateral surface of a
prism or cylinder
is
called its
Mantel.
Exercises. 114. The mantel of a right circular cylinder equivalent to a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter of its is
base. 115.
A
plane through two elements of a right circular its base in a chord which subtends at the center
cylinder cuts
an angle whose circular measure is u find the curved surfaces of the two parts of the cylinder. ;
ratio of
the
THE ELEMENTS OF GEOMETRY,
326 824.
moving
A
Conical Surface is generated by a straight line so as always to pass through a fixed point called the
apex.
825.
A
Cone
is
a solid bounded by a conical surface and a
plane.
A right
circular cone
may be
generated by revolving a All the elements are right triangle about one pependicular. equal, and each is called the Slant Height of the cone. 826.
MENSURATION, OR METRICAL GEOMETRY. 827.
The
or cone
FricsUim of a pyramid
327
the portion
is
included between the base and a parallel section.
828.
To
Rule.
find the lateral
Multiply the
height.
area of a right circular cone.
circle
of
its
Formula.
K = \ch =
Proof.
the curved surface be
If
and spread out
flat, it
base by half the slant
irrh.
slit
becomes a sector
along a slant height, with slant
of a circle,
height as radius, and the circle of cone's base as arc by 811, its area is \ch.
;
therefore,
=
829. Corollary. Since, by 810, the base O \cr, therefore the curved surface is to the base as the slant height is to the radius of base, as the length of circle with radius h is to circle
with radius
r,
as the perigon
is
to the sector angle.
THE ELEMENTS OF GEOMETRY.
328
To find ihe lateral area of the frustum of a right circular
830. cone.
Rule.
sum of
Multiply the
slajit
height of the frustum by half the
the circles of its bases.
F
Formula.
•=
\h{c^-\' c^
=
tt// (r,
+
^2)-
Proof. Completing the cone, and slitting it along a slant height, the curved surface of the frustum develops into the difference of two similar sectors having a common angle, the arcs of the sectors being the circles of the bases of the frus-
tum.
By
813, the area of this annular sector,
F=
\h{c,
+
c).
The
axis of a right circular cone, or cone of revolution, is the line through its vertex and the center of its base.
831.
116. Given the two sides of a right-angled Find the area of the surface described when the
Exercises. triangle.
triangle revolves about its hypothenuse. Calling a and b the given altitude and base, and x the perpendicular from the right angle to the hypothenuse, by 828, the area of the surface described by a is itxa^ and by b is irxb.
Hint.
Also a
\
X w
^a" \-
b-"
\
b.
117. In the frustum of a right circular cone, on each base stands a cone with its apex in the center of the other base from the basal radii r^ and 7\ find the radius of the circle in
;
which the two cones
cut.
MENSURATION, OR METRICAL GEOMETRY.
Theorem
329
III.
832. The lateral area of a friistiim of a cone of revohttion is the product of the projection of the frustum s slant height on the axis by twice tt times a perpendicular erected at the mid-point of this slant heighty
Proof. is
PR and
term,htated by the axis.
830, the lateral area of the frustum whose slant height
By axis
and
MN
is
F= But
if
Q is
P=
PRL ~ A QCO,
27r
/.
F=
Exercises.
when
X
PM
-f
RN =
2QO,
PR X QO. one are drawn per-
;
PR X QO
27r{LP X
118.
RN)PR,
since the three sides of
pendicular to the sides of the other .-.
-}-
PR, then
the mid-point of .'.
But A
Tr{PM
=:
PL X QC,
QC) = 27r{MN X CQ).
Reckon the mantel from the two
the inclination of a slant height to one base
is
radii
half a
right angle. 119. If in the frustum of a right cone the diameter of the upper base equals the slant height, reckon the mantel from the altitude a and perimeter/ of an axial section.
THE ELEMENTS OF GEOMETRY.
330 833.
To find ihe area of a sphere.
Rule. Multiply four times = 4rV. Formula.
its
H
ir.
In a circle inscribe a regular polygon of an even Then a diameter through one vertex passes
Proof.
number
squared radius by
of sides.
through the opposite vertex, halving the polygon symmetrically. Let PR be one of its sides draw PM, RAl, perpendicular to the diameter BD. From the center C the perpendicular ;
CQ
PR.
Drop the perpendiculars PL, QO. the whole figure revolve about as axis, the semicircle will generate a sphere, while each side of the inbisects
Now,
BD
if
scribed polygon, as of a cone.
frustum
PR, will generate By 832, this
F=
2it{^MN X
the curved surface of the
C0;
and the sum
of all the frustums, that is, the surface of the solid generated by the revolving semi-polygon, equals iirCQ into the
sum
of the projections, ,\
Sum
BM, MN, NC,
of surfaces of frustums
etc.,
=
27r
which sum
CQ
X BZ>.
is
BD.
MENSURATION', OR METRICAL GEOMETRY.
As we
continually double the
polygon,
and
number
331
of sides of the inscribed
semi-perimeter approaches the semicircle as
its
limit,
surface of revolution approaches the sphere as limit, while CQ, its apothem, approaches r, the radius of the sphere, its
Representing the sum of the surfaces of the BD by 2r, we have
as limit.
frus-
tums by 2/^ and
CQ That
is,
the variable
ratio 4r7r; therefore,
have the same
sum by
is
to the variable
ratio,
H= — r
.-.
A
835.
The
To
Calot
last
find the
Rule.
is
CQ
in the constant
798, Principle of Limits, their limits
.-.
834.
^
H=
4r7r,
^r^TT,
a zone of only one base.
proof gives also the following rule
:
—
area of a zone.
Multiply the altitude of the segment by twice
the radius of the sphere.
Formula.
Z=
zarir.
ir
times
THE ELEMENTS OF GEOMETRY.
332
CHAPTER
IV.
SPACE ANGLES. 836. A Plane Angle which meet in a point. 837.
A
Space Ajigle
which meet
is
the divergence of two straight lines
is
the spread of two or more planes
in a point.
Symmetrical Space Angles are those which cut out symmetrical spherical polygons on a sphere, when their vertices 838.
are placed at
839.
A
its
center.
Steregon
is
about a point in space.
the whole amount of space angle round
MENSURATION, OR METRICAL GEOMETRY. 840.
A
Steradian
is
333
the angle subtended at the center by
that part of every sphere equal to the square of
its radius.
The space angle made by
841. only two planes corresponds to the lune intercepted on any sphere whose center is in the common section of the two planes.
A
842. Spherical Pyramid is a portion of a globe bounded a by spherical polygon and the planes of the sides of the polyThe center of the sphere is the apex of the pyramid gon. ;
the spherical polygon
is its
base.
843. Just as plane angles at the center of a circle are proportional to their intercepted arcs, and also sectors, so space
angles at the center of a sphere are proportional cepted spherical polygons^
and also
to their inter-
spheidcal pyramids.
^-^^
334
ELEMENTS OF GEOMETRY.
ratio of the space angles of two right cones but a^, having the same slant height, h. These space angles are as the corresponding calots (or zones of one base) on the sphere of radius h. Therefore, by 835, the required ratio
Example.
Find the
of altitude a^ and
is
2Trh{h
2Trh(h
— —
^i)
a^)
_
h
—
a^
MENSURATION, OR METRICAL GEOMETRY. 846.
Corollary
I.
A
lime measures twice as
335 stera-
many
angle contains radians. If two-faced space angles are equal, Corollary II. 847. so a dihedral angle may be meastheir lune angles are equal dians as
its
;
ured by the plane angle between two perpendiculars, one in each face, from any point of its edge.
848.
Suppose the vertex
of a space angle
is
put at the cen-
ter of a sphere, then the planes which form the space angle will cut the sphere in arcs of great circles, forming a spherical polygon, whose angles may be taken to measure the dihedral angles of the space angle, and whose sides measure its face angles.
Hence from any property
of spherical polygons
we may
an analogous property of space angles. For example, the following properties of trihedral angles have been proved in our treatment of spherical triangles infer
:
—
Trihedral angles are either congruent or symmetrical I. which have the following parts equal :
—
THE ELEMENTS OF GEOMETRY,
33^
Two Two
face angles and the included dihedral angle. dihedral angles and the included face angle. Three face angles.
(i) (2) (3)
Three dihedral angles. (5) Two dihedral angles and the face angle opposite one of them, provided the edge of the third dihedral angle of neither trihedral makes right angles with any line in the half of the opposite face not adjacent to one of the face angles equal by (4)
hypothesis. (6) Two face angles and the dihedral angle opposite one of them, provided the other pair of opposite dihedral angles are
not supplemental.
As one
of the face angles of a trihedral angle is greater than, equal to, or less than, another, the dihedral angle which it subtends is greater than, equal to, or less than, the dihedral II.
angle subtended by the other.
Symmetrical trihedral angles are equivalent. IV. Space angles having the same number and sum of dihedral angles are equivalent. III.
V.
The
face angles of a convex space angle are together
than a perigon.
less
To find ihe area of a spherical triangle. Rule. Multiply its spherical excess in radians by 849.
its
squared
radius.
is
=
Formula,
"a
Proof.
711, a triangle is equal to a lune
\e
;
By
therefore,
850.
its e
'a
=
A
whose angle
r'^e.
I. spherical triangle measures as contains radians.
Corollary
polygon, multiply radius.
845,
Corollary
steradians as 851.
by
er^.
its
many
714, to find the area of a spherical spherical excess in radians by its squared
II.
By
MENSURATION, OR METRICAL GEOMETRY.
CHAPTER
337
V.
THE MEASUREMENT OF VOLUMES. 852. 853.
edge
is
The Volume of a solid is its ratio to an assumed unit. The Unit for Measurement of Volume is a cube whose the unit for length.
Cubic Centimeter.
854.
Metric Units for Volume.
Solid.
Cubic.
Liquid.
I
kilostere
I
hectostere
I
dekastere
I
stere
I
kiloliter
I
decistere
I
hectoliter
I
centistere
I
dekaliter
I
liter
I
deciliter
I
=
I
cubic dekameter
cubic meter
= = = =
I
cubic decimeter
I
cubic meter.
.1
cubic meter.
=
.001 cubic meter.
.0001 cubic meter.
cubic centimeter cubic millimeter
authorized abbreviation for cubic ;
10 cubic meters.
.00001 cubic meter. I
meter
100 cubic meters.
.01 cubic meter.
=
I centiliter
The
1000 cubic meters.
is
that for stere
= =
.000001 cubic meter,
.000000001 cubic meter.
the index ^ as in icm.3 for is s., and for liter 1..
i
cubic centi-
THE ELEMENTS OF GEOMETRY,
338 855.
of
To
volume of a quader.
find the
Rule. Multiply together its lengthy breadth, and thickness. Proof. By "jG^i, two quaders have the ratio compounded the ratios of their bases and altitudes. The volume of any cube is the third 856. Corollary.
power of a
of the length of an edge. is called its cube.
This
is
why
the third power
number
857.
To
Rule. Proof.
find the volume of
By
of equivalent base
858.
To
any parcllelopiped.
altitude by the area of its base. 769, any parallelopiped is equivalent to a quader
Multiply
find the
its
and equal
altitude.
volume of any prism.
Rule. Multiply its altitude by Formula. F. P aB.
its base.
=
Proof.
By 771, any three-sided prism is half a parallelowith base twice the prism's base and the same altitude. piped, Thus the rule is true for triangular prisms, and consequently since, by passing planes through one lateral edge, and all the other lateral edges excepting the two adjacent to this one, we can divide any prism into triangular prisms of the same altitude, whose triangular bases together make the
for all prisms
;
polygonal base.
MENSURATION, OR METRICAL GEOMETRY. 859.
339
To find the volume of any cylinder.
I
J I I
^- ^--S I I
I
Rule.
its
Multiply
altitude by its base.
The cylinder is the limit of an inscribed prism number of sides of the prism is indefinitely increased,
Proof.
when
the
the base of the cylinder being the limit of the base of the prism.
But always the variable prism is to its variable base in the constant ratio a therefore, by 798, Principle of Limits, their limits will be to one another in the same ratio. ;
860.
To
find the
volume of any pyramid.
Rule. Multiply one-third of Formula. Y \aB.
=
its
altitude by its base.
'^^^
340 Proof.
By
ELEMENTS OF GEOMETRY.
775,
any triangular pyramid is one-third same base and altitude.
of a
of the
triangular prism The rule thus proved for triangular pyramids is true for all pyramids since by passing planes through one lateral edge, and all the other lateral edges excepting the two adjacent to ;
this one,
of the
we can
same
divide any pyramid into triangular pyramids
altitude
whose bases together make the polygonal
base.
861. To find the volume of any cone.
Rule.
Multiply one-third
its
altitude by its base.
Formula when Base is a Circle. V. Proof. The base of a cone is the limit
K=
^ar^ir.
of the base of
an
inscribed pyramid, and the cone is the limit of the pyramid. But always the variable pyramid is to its variable base in the
constant ratio \a. Therefore their limits are to one another in the same
K=
ratio,
and V. laB. Scholium. This applies to all solids determined by an elastic string stretching from a fixed point to a point describing any closed plane figure.
MENSURATION, OR METRICAL GEOMETRY. Corollary.
The volume
341
the solid generated by the revolution of any triangle about one of its sides as axis, is one-third the product of the triangle's area into the length of the circle described by its vertex. 862.
F=
of
\irrt.
PRISMATOID. 863.
A Prismatoid
polygons
is
a polyhedron whose bases are any two and whose lateral faces are triangles
in parallel planes,
determined by so joining the vertices of these bases that each lateral edge, with the preceding, forms a triangle with one side of either base.
THE ELEMENTS OF GEOMETRY.
342 864.
A
same two
number
of different prismatoids thus pertain to the
bases.
two basal edges which form with the same lateral of two adjoining faces are parallel, then these two sides edge 865. If
two triangular faces trapezoid.
fall in
the same plane, and together form a
MENSURATION, OR METRICAL GEOMETRY, 866.
number
867.
343
A Prismoid is a prismatoid whose bases have the same of sides,
and every corresponding pair
A frustum
of a
pyramid
is
parallel.
a prismoid whose two bases
are similar. 868.
tum
Corollary.
Every three-sided prismoid
is
the frus-
of a pyramid.
869.
If
both bases of a prismatoid become sects,
it
is
a
tetrahedron.
870.
angle,
A
Wedge
is
a prismatoid whose lower base parallel to a basal edge.
is
a rect-
and upper base a sect
871.
The
altitude of a prismatoid
is
any sect perpendicular
to both bases.
A
Cross-Sectiojt of a prismatoid 872. plane perpendicular to the altitude.
is
a section
made by a
1^^^
344 873.
To find the volume of any prismatoid.
Rule. base
from
ELEMENTS OF GEOMETRY.
Multiply one-fourth
and
a
three times
its
altitude by the
sum of
one
cross-section at two-tJiirds the altitude
that base.
Formula.
Z>
= -(^ +
3 T).
4
Proof.
prismatoid may be divided into tetrahedra, all of the same altitude as the prismatoid some, as C FGO, having their apex in the upper base of the prismatoid, and for base
Any
;
a portion of its lower base some, as O ABC, having base in the upper, and apex in the lower, base of the prismatoid and the others, as A COG, having for a pair of opposite edges a ;
;
AC
sect in the plane of each base of the prismatoid, as d.nd OG. if the formula holds for tetrahedra in these Therefore, good
three positions, it holds for the prismatoid, their sum. In (i), call T^ the section at two-thirds the altitude from the base B^ then T^ is \a from the apex. Therefore, by 772, ;
r, /.
:
^,
:
:
{\ay
:
a\
.'.
T,
= ^B„
A = -(^x + 3^0 = -(^x + -^x) = -^^, 4 4 3
3
which, by 860, equals Y, the volume of the tetrahedron.
MENSURATION, OR METRICAL GEOMETRY. In
(2),
=
B^
o,
being a point, and T^
is
\a from the apex
3
KLMN be the
(3), let
Join
;
A = ^(^, + 37;) = ^(o + ^^.) = -aB, = K 4 4
...
In
345
ANK
:
A ^6^6>
A
GNM:
A 6^^C
:
:
section T^.
Now
CK, CN, OM, ON.
A
AJV^
:
GN'
:
3
:
:
^^6^^
GA^
:
:
(^ay
:
a^
:
:
1
:
g.
;
:
{^ay
:
a^
:
:
4
:
g.
ANK
But the whole tetrahedron B^ and the pyramid C be considered as having their bases in the same plane, and the same altitude, a perpendicular from C;
CANK
:
Z>3 .-.
:
:
ANK A AGO C ANK = JZ>3. A
:
:
:
i
:
9,
may
AGO,
THE ELEMENTS OF GEOMETRY.
346 In same way,
O
GNM
:
Z>3
:
:
GNM
A
:
A
GAC
:
GNM = C KLMN + O KLMN = C ANK + O
:.
:
4
:
9,
%D^, %D^,
But, by 860,
C KLMN + O ^B,
.',
=
9
KLMN =
^aT,,
/.
\
n,
+ | |«7; = \aT,, = ^3^x = -(^x + 3^x),
.
\aT,
4
3
.
4
since here
B,
Corollary
874.
T
and
I.
w,^
875.
Since, in the frustum of a pyramid,
\
B
^i' /
and Wj are corresponding sides of B and T. For the frustum of a cone II.
Corollary
tion,
V.F =
-7rr,^
4
where
o.
are similar.
4
where
=
r^ is
the radius of T.
of revolu-
MENSURATION, OR METRICAL GEOMETRY.
347
876. To find the volume of a globe.
Rule. Multiply Formula. G
=
the cube of its radius by
|-7r.
|-7rr3.
644, a tetrahedron on edge, and a globe with the tetrahedron's altitude for diameter, have all their corre-
Proof.
By
sponding cross-sections equivalent
any one pair are equiva-
if
lent.
Hence the volumes may be proved and
G=
But a
=
2r,
and, /.
877.
by 6^
Corollary
522,
= I.
I
laT.
T=
\r
2r
|r
.
.
equivalent, as in 774,
.
.
|r fr
.
.
tt,
TT
=
%^r\
Globes are to each other as the cubes
of their radii. 878.
Corollary
II.
Similar solids are to one another as
the cubes of any two corresponding edges or sects.
DIRECTION. two points starting from a state of coincidence move along two equal sects which do not coincide, that quality of each movement which makes it differ from the other is its 879. If
Direction. 880. If two equal sects are terminated at the same point, but do not coincide, that quality of each which makes it differ from the other is its direction.
The
part of a line which could be generated by a tracing-point, starting from a given point on that line, and moving on the line without ever turning back, is called a Ray from 881.
that given point as origin.
THE ELEMENTS OF GEOMETRY.
348
Two
882.
rays from the same origin are said to have the if otherwise, they are said to they coincide
same direction
;
have different directions. 883. Two rays which have no point but the origin in common, and fall into the same line, are said to have opposite directions.
Two
884.
directions origins.
885.
origins. 886.
A
it is
A
887. its
rays lying on parallel lines have parallel-same they are on the same side of the line joining their
,
Two
directions
which
if
if
rays lying on parallel lines have parallel-opposite they are on opposite sides of the line joining their
sect
is
said to have the
same
direction as the ray of
a part. sect
is
definitely fixed
parallel-same direction, and
its
The
if
we know
its
initial point,
length. a sect could be traced
if we operation by which that the of is, point, operation carrying a in a certain direction until it parallel-same tracing-point of over a number units for is called a passes given length,
888.
knew
its
initial
Vector.
The position of B relative to A is indicated by the and length parallel-same direction of the sect AB drawn from A to B. If you start from A, and travel, in the direction indicated by the ray from A through B, and traverse the given number of units, you get to B. This parallel-same direction and length may be indicated equally well by any .other sect, such as A'B' which is equal to AB and in parallel-same direction. 890. As indicating an operation, the vector ^j5 is completely defined by the parallel-same direction and length of the transAll vectors which are of the same magnitude and ferrence. parallel-same direction, and only those, are regarded as equal. 889.
,
Thus
AB is
not equal to
BA.
MENSURATION, OR METRICAL GEOMETRY.
349
PRINCIPLE OF DUALITY. JOIN OF POINTS 891.
Two
The
Points.
line joining
The
called the Join of
AND OF
two points
point common the Two Lines.
to
is
LINES. called the Join of the lines is
two intersecting
A
fixed point, A, may be joined to 892. Pencil of Lines. other points in space. get thus all the lines which can be drawn through the The aggregate of all these lines is called a Pencil of A. point all
We
The fixed point is called the Base of the pencil. Anyone of these lines is said to be a line in the pencil, and also to be a line in the fixed point. In this sense, we say not only that a point may lie in a line, but also that a line may lie in a point, meaning that the Hne passes through the point. 893. In most cases, we can, when one figure is given, construct another such that lines take the place of points in the first, and points the place of lines. Any theorem concerning the first thus gives rise to a corresponding theorem concerning the second figure. Figures or theorems related in this manner are called Reciprocal Fignres Lines.
or Reciprocal Theorems. is
894. Small letters denote lines, and the join of two elements denoted by writing the letters indicating the elements, to-
gether.
Thus the
A
B
is the line AB, while and join of the points of of the lines a and b. or intersection, join, point
ab denotes the
ROW OF 895.
Row
A
POINTS, PENCIL OF LINES.
line contains
an
infinite
of Points, of which the line
A row is
all
points in a
line.
is
number the Base,
of points, called
a
THE ELEMENTS OF GEOMETRY.
350
The
reciprocal figure
is
all lines in
a pointy or
all
lines pass-
ing through the point. Flat Pencil is the aggregate of all lines in a plane which In plane geometry, by a pencil we pass through a given point. mean a flat pencil.
A
RECIPROCAL THEOREMS. 8961.
A
point moving along a
line describes a row.
8971,
A
sect
is
a part of a row-
described by a point moving from one position, A^ to another position,
B.
Thus
A
line turning about a a pencil. describes point 897'. An angle is a part of a
896'.
pencil described by a line turning
from one position,
a^
to
another
position, b.
to sect
AB corresponds
4-
^^•
LINKAGE. 898.
The
Peancellier Cell consists of a
rhombus movably
jointed, and two equal links movably pivoted at a fixed point, and at two opposite extremities of the rhombus.
MENSURATION, OR METRICAL GEOMETRY.
351
TO DRAW A STRAIGHT LINE. 899.
Take an
extra link, and, while one extremity is on the the other extremity to a fixed
fixed point of the cell, pivot point.
Then
pivot the
first
end to one
of the free angles of
The opposite vertex of the rhombus v/ill now describe a straight line, however the linkage be pushed or
the rhombus.
moved. is constrained to move Proof. By the bar FD, the point on the circle ADR therefore 4. ADR, being the angle in a
D
;
semicircle,
If,
now,
is
always right.
E moves on EM ± AM, ADR ~ A AME, DA AR AM AE, DA AE = RA AM. A
:. .'.
Therefore, line
EM.
if
AE AD .
But because .*.
:
:
:
,
,
is
:
E
moves on the straight constant, is a rhombus, and AC,
BDCE
AB =
D and JV are always on the variable sect AE.
THE ELEMENTS OF GEOMETRY.
352
Always AB^
AB'
.'.
= AN^ + NB% and "BE" = EN^ + NB\ - BE" = AN^ - NE" = {AN + NE) {AN - NE) = AE AD. .
900.
The
Corollary.
to keep, however varying sects a constant.
power
depends on its be deformed, the product of two
efficacy of our cell
it
Therefore a Hart four-bar cell, constructed as in the accompanying figure, may be substituted for the Peancellier six-bar cell,
since
AE AD .
Also, for the
equals a constant. cell may be substituted the quadruplane,
Hart
four pivoted planes.
CROSS-RATIO. 901. If four points are collinear, two may be taken as the extremities of a sect, which each of the others divides inter-
nally or externally in some ratio. The ratio of these two ratios
is
called the Cross-Ratio of
the four points.
The
cross-ratio
——
-—— CB DB :
is
written {ABCD).
AB
from BA^ as of opposite Distinguishing the "step" in the the two groups of two in a and sense," taking points definite order, to write out a cross-ratio, make first the two **
MENSURATION, OR METRICAL GEOMETRY. and put crossing these the
bars,
two,
A A thus —- —-
B
:
B
;
then
letters of the first
up, crosswise, the first
fill
353
group of
by the
first
second group of two, the second by the second. take the points in a different order, the value of the
letter of the If
we
cross-ratio
may change. can do this in twenty-four different ways by forming permutations of the letters.
We
But
of
all
these twenty-four cross-ratios, groups of four are
equal, so that there are really only six different ones. have the following rules
We
:
—
If in a cross-ratio the two groups be interchanged^ remains unaltered. I.
its
value
{ABCD) = {CDAB).
II. If in a cross-ratio tJie tzvo points belongifig to one of the two groups be interchanged^ the cross-ratio cJianges to its recip-
rocal.
(ABCD)^ = ^
I.
and
?
.
{ABDC)
are proved by writing out their values.
II.
it followsJ that^ if we interchange the elements in each pair^ the cross-ratio remains unaltered.
III.
From II
{ABCD) = {BADC). IV. If in a cross-ratio the tzvo middle letters be interchanged, the cross-ratio changes into its complement.
{ABCD) = This
is
I
- {ACBD).
proved by taking the step-equation for any four
linear points,
BC .AD ^ CA BD + AB CD = .
and dividing
it
by
CB AD. .
.
o,
col-
THE ELEMENTS OF GEOMETRY.
354
Theorem go2.
If any four
IV. by a transversal,
concurre^tt lines are cut
any of the four points of intersection is constant for all positions of the transversal. cross-ratio
Hypothesis. Let A, B, C, D, be the intersection points for t, the transversal in one position, and let A', B\ , jy, be the corresponding intersection points for /, the transversal in another
C
position.
{ABCD) = {A'B'Cjy);
Conclusion.
AC AD ^ CB DB , '
Proof. Through the points which cut the concurrent lines in
AACC^
A
CB
where account
is
is,
A'D'
.
'
C'B'
D'B''
B
A
and
Ca,
D2, B^, and A^, C^, D^.
BCC„
AC ^
A;_C_
that
AC^ C,B'
taken of " sense.'
and
and
on
t
draw
A ADD:,
AD DB
parallels
~ A BDD,, AD, D,B'
to
/',
MENSURATION, OR METRICAL GEOMETRY. Hence
CB
'
*
Z>B
'
D^B
C,B
AD^
D,b''
but
AD,
" .
A'ly
AC AD ^ A'C CB DB AD , '
jyB"
D,B . '
CB' ^ A'C CB'
DB'
.
A'ly
''
D'B''
355
EXERCISES. BOOK 1
20.
Show how
to
I.
make a rhombus having one
of
its
diagonals
equal to a given sect. 121. If two quadrilaterals have three consecutive sides,
and the two
contained angles in the one respectively equal to three consecutive sides and the two contained angles in the other, the quadrilaterals are congruent.
122.
Two
circles
cannot cut one another at two points on the same
side of the line joining their centers.
123. Prove, by an equilateral triangle, that, if a right-angled triangle have one of the acute angles double of the other, the hypothenuse is double of the side opposite the least angle.
Draw Draw
a perpendicular to a sect at one extremity. three figures to show that an exterior angle of a triangle 125. be may greater than, equal to, or less than, the interior adjacent angle. 126. Any two exterior angles of a triangle are together greater than1
24.
a straight angle.
The perpendicular from any
vertex of an acute-angled triangle within the triangle. 128. The perpendicular from either of the acute angles of an obtuseangled triangle on the opposite side falls outside the triangle. 127.
on the opposite
129.
and
less
side
falls
The semi-perimeter of a triangle than any two sides.
is
greater than any one sidej
357
THE ELEMENTS OF GEOMETRY.
358
130. The perimeter of a quadrilateral is greater than the sum, and than twice the sum, of the diagonals. 131. If a triangle and a quadrilateral stand on the same base, and
less
the one figure fall within the other, that which has the greater surface have the greater perimeter.
shall
132. If one angle of a triangle be equal to the sum of the other two, the triangle can be divided into two isosceles triangles.
any sect joining two parallels be bisected, this point will drawn through it and terminated by the parallels. 134. Through a given point draw a line such that the part intercepted between two given parallels may equal a given sect. 133.
If
bisect any other sect
135.
The medial from
vertex to base of a triangle bisects the inter-
cept on every parallel to the base. 136. Show that the surface of a quadrilateral equals the surface of a triangle which has two of its sides equal to the diagonals of the quadrilateral,
and the included angle equal
to either of the angles at
which the
diagonals intersect. 137. Describe a square, having given a diagonal. is the square on the is a right-angled triangle; 138. are the and squares on the other sides. hypothenuse;
BCED
ABC
ACKH
Find the center of the
ABFG square ABFG
the two diagonals), and through and the other perpendicular to
it
BC.
into four congruent quadrilaterals.
of the square until
it
(which may be done by drawing draw two lines, one parallel to BC, This divides the square
BCED draw a parallel to AB or A C.
meets the second of the other
BCED into a square and the four quadrilaterals in
ABFG
Through each mid-point of the pair,
If
sides
each be extended
they will cut the square
four quadrilaterals congruent to
ACKH and
ABFG.
139. The orthocenter, the centroid, and the circumcenter of a triangle are collinear, and the sect between the first two is double of the sect
between the 140.
triangle
last
two.
The perpendicular from is
half
the
sect
from
the circumcenter to any side of a opposite vertex to the ortho-
the
center.
141. Sects
drawn from a given point
find the locus of their mid-points.
to a given circle are bisected
;
EXERCISES.
142.
The
359
intersection of the Hnes joining the mid-points of opposite is the mid-point of the sect joining the mid-points
sides of a quadrilateral
of the diagonals. 143.
A
parallelogram has central symmetry.
SYMMETRY. 144.
symmetry
No
triangle
is
a medial.
can have a center of symmetry, and every axis of
cut by 145. Of two sides of a triangle, that is the greater which is the perpendicular bisector of the third side. 146. If a right-angled triangle is symmetrical, the axis bisects the right angle.
147.
An
angle in a triangle will be acute, right, or obtuse, according its vertex is greater than, equal to, or less than,
as the medial through
half the opposite side.
148. If a quadrilateral has axial symmetry, the
not on the axis must be even it is
two,
a
;
if
none,
it is
number of
vertices
a symmetrical trapezoid
;
149. A kite has the following seven properties ; from each prove the others by proving that a quadrilateral possessing it is a kite.
(i) other,
One
which
(2)
(3)
if
kite.
The The
diagonal, the axis, is the perpendicular bisector of be called the transverse axis.
all
tlie
will
axis bisects the angles at the vertices
which
it
joins.
angles at the end-points of the transverse axis are equal,
equally divided by the
and
latter.
(4) Adjacent sides which meet on the axis are equal. (5) The axis divides the kite into two triangles which are congruent,
with equal sides adjacent. (6)
which
is
(7) axis,
The
transverse axis divides the kite into two triangles, each of
symmetrical.
The
lines joining the mid-points of opposite sides
and are equally inclined
to
meet on the
it.
150. A symmetrical trapezoid has the following five properties ; from each prove all the others by proving that a quadrilateral possessing it is a symmetrical trapezoid.
THE ELEMENTS OF GEOMETRY.
360 (i)
Two
opposite sides are parallel, and have a
common
perpendicu-
lar bisector.
(2) The other two opposite sides are equal, either of the other sides. (3) its
Each angle
is
and equally
inclined to
equal to one, and supplemental to the other, of
two adjacent angles. (4) (5)
which
it
The diagonals are equal, and divide each other equally. One median line bisects the angle between those sides produced does not bisect, and hkewise bisects the angle between the two
diagonals.
151. Prove the properties of the parallelogram from
its
central
sym-
metry. 152. A kite with a center is a rhombus; a symmetrical trapezoid with a center is a rectangle ; if both a rhombus and a rectangle, it is a
square.
BOOK
II.
153. The perpendicular from the centroid to a line outside the triangle equals one-third the sum of the perpendiculars to that line from the vertices. 154.
If
two sects be each divided internally into any number of contained by the two sects is equivalent to the sum
parts, the rectangle
of the rectangles contained by all the parts of the other.
all
the parts of the one, taken separately,
with
155. The square on the sum of two sects is equivalent to the sum of the two rectangles contained by the sum and each of the sects. 156. The square on any sect is equivalent to four times the square
on half the 157.
grows
sect.
The
rectangle contained by two internal segments of a sect of section moves from the mid-point.
less as the point
158. The sum of when they are equal.
159.
If the
the squares on the two segments of a sect
is
least
hypothenuse of an isosceles right-angled triangle be
divided into internal or external segments, the
sum of
their squares
is
EXERCISES.
361
equivalent to twice the square on the sect joining the point of section to the right angle. 160. Describe a rectangle equivalent to a given square, and having one of its sides equal to a given sect. 161. Find the locus of the vertices of
all
triangles
on the same
base,
having the sum of the squares of their sides constant. 162. The center of a fixed circle is the point of intersection of the diagonals of a parallelogram ; prove that the sum of the squares on the sects drawn from any point on the circle to the four vertices of the is constant.
parallelogram
163. Thrice the sum of the squares on the sides of any pentagon is equivalent to the sum of the squares on the diagonals, together with four times the sum of the squares on the five sects joining, in order, the mid-
points of those diagonals. 164. The sum of the squares on the sides of a triangle is less than twice the sum of the rectangles contained by every two of the sides. 165. If from the hypothenuse of a right-angled triangle sects be cut off equal to the adjacent sides, the square of the middle sect thus formed is equivalent to twice the rectangle contained by the extreme sects.
BOOK
III.
166. From a point, two equal sects are drawn to a circle. that the bisector of their angle contains the center of the circle.
Prove
167. Describe a circle of given radius to pass through a given point its center in a given line.
and have
168. Equal chords in a circle are 169. secant.
Two
all tangent to a concentric circle. concentric circles intercept equal sects on any common
1 70. If two equal chords intersect either within or without a circle, the segments of the one equal the segments of the other.
one
171. Divide a circle into two segments such that the angle in the shall be seven times the angle in the other.
ABC
ABC
AC = AC
and bxq two triangles such that 172. prove the circle through A, B, C, equal to that through A, B,
C,
;
THE ELEMENTS OF GEOMETRY.
362
173. Pass a circle through four given points. possible
174. \i
chord
When
is
the problem
?
AB
AC
and
parallel to
BD
be two equal arcs in a
circle
ABCD,
prove
chord CD.
175. Circles described on any two sides of a triangle as diameters on the third side, or the third side produced.
intersect
176. If one circle be described on the radius of another as diameter, is bisected by the
any chord of the larger from the point of contact smaller.
177. If two circles cut,
and from one of the points of
intersection
two diameters be drawn, their extremities and the other point of
inter-
section will be collinear.
178. Find a point inside a triangle at which the three sides shall
subtend equal angles. 179.
On
the produced altitudes of a triangle the sects between the
orthocenter and the circumscribed circle are bisected by the sides of the triangle. 180. If
on the three
sides of
any triangle equilateral
triangles
be
described outwardly, the sects joining their circumcenters form an equilateral triangle.
on
181. If two chords intersect at right angles, the sum of the squares segments is equivalent to the square on the diameter.
their four
182. to
The
opposite sides of an inscribed quadrilateral are produced that the bisectors of the two angles thus formed are at
meet ; prove
right angles.
183. The feet of the perpendiculars drawn from any point in circumscribed circle to the sides of a triangle are collinear. 184.
drawn
P outside a circle, two secants, PAB, PDC, are ABCD AC, BD, are joined, and intersect at O. O lies on the chord of contact of the tangents drawn from P
From
a point
to the circle
Prove that
its
]
Hence devise a method of drawing tangents to a circle from an external point by means of a ruler only. 185. A variable chord of a given circle passes through a fixed point ;
to the circle.
find the locus of
its mid-point. 186. Find the locus of points from which tangents to a given circle contain a given angle.
EXERCISES.
363
187. The hypothenuse of a right-angled triangle is given; find the loci of the corners of the squares described outwardly on the sides of
the triangle.
BOOK
IV.
188. If a quadrilateral be circumscribed about a circle, the
sum of
two opposite sides is equal to the sum of the other two. 189. Find the center of a circle which shall cut off equal chords from the three sides of a triangle.
Draw
190.
which are not
a line which would bisect the angle between two lines but which cannot be produced to meet.
parallel,
The angle between the altitude of a triangle and the line and circumcenter equals half the difference of the angles vertex through at the base. 191.
192. The bisector of the angle at the vertex of a triangle also bisects the angle between the altitude and the line through vertex and circumcenter.
193. If an angle bisector contains the circumcenter, the triangle
is
isosceles.
194. If a circle can be inscribed in a rectangle, it must be a square. 195. If a rhombus can be inscribed in a circle, it must be a square. 196. The square on the side of a regular pentagon is greater than the square on the side of the regular decagon inscribed in the same circle
by the square on the
The
197.
radius.
intersections of the diagonals of a regular pentagon are
the vertices of another regular pentagon the alternate sides.
The sum of
198.
;
so also the intersections of
the five angles formed by producing the alternate
sides of a regular pentagon equals a straight angle.
199. The circle through the mid-points of the sides of a triangle, called the medioscribed circle, passes also through the feet of the altitudes,
and
bisects the sects
Hint.
Let
ABC
V, Z, the feet of
of
between the orthocenter and
be the triangle its
altitudes
;
;
vertices.
the mid-points of its sides ; X^ orthocenter ; C/, y, JV, the mid-points
H^ K, Z,
O
its
AO, BO, CO.
LHWU
is
a rectangle
;
so
is
HKUV\
and the angles
at
X,
V, Z, are right
THE ELEMENTS OF GEOMETRY.
364 200.
The mediocenter
is
midway between the circumcenter and
the
orthocenter.
201. The diameter of the circle inscribed in aright-angled triangle, together with the hypothenuse, equals the sum of the other two sides. 202. An equilateral polygon circumscribed about a circle is equiangular if the number of sides be odd.
203. Given the vertical angle of a triangle and the it ; find the locus of the circumcenter.
sum of
the
sides containing
204. Given locus of
the vertical
angle
and base of a
triangle;
The center of the inscribed circle. The centers of the escribed circles. (3) The orthocenter. (4) The centroid. (5) The mediocenter. 205. Of all rectangles inscribable in a circle, show
find the
(i)
(2)
that a square
is
the greatest.
BOOK
VI.
206. The four extremities of two sects are concyclic if the sects cut so that the rectangle contained by the segments of one equals the rectangle contained by the segments of the other. 207. If two circles intersect, and through any point in their common chord two other chords be drawn, one in each circle, their four extremare concyclic.
ities
Does the magnitude of the third proportional depend on the order in which the sects are taken ?
208. sects
209. shall
Through a given point
be divided
inside a circle
to
two given
draw a chord so
that
it
at the point in a given ratio.
210. If two triangles have two angles supplemental and other two equal, the sides about their third angles are proportional. their 211. Find two sects from any two of the six following data :
sum, their difference, the sum of their squares, the difference of their squares, their rectangle, their ratio.
EXERCISES.
365
212. If two sides of a triangle be cut proportionally, the lines drawn from the points of section to the opposite vertices will intersect on the medial from the third vertex. 213. Given the base of a triangle and the ratio of the two sides; find the locus of the vertex.
MISCELLANEOUS. 214. Find the locus of a point at which two adjacent sides of a rectangle subtend supplementary angles. 215.
Draw through a
two given 216.
given point a line making equal angles with
lines.
From
extremities
may
a given point place three given sects so that their be in the same line, and intercept equal sects on that
line.
217. Divide a sect into two parts such that the square of one of the on the whole sect.
parts shall be half the square
218. Find the locus of the point at which a given sect subtends a
given angle. 219. Given a curve, to ascertain whether
it
is
an arc of a
circle or
not.
220. If two opposite sides of a parallelogram be bisected, and lines be drawn from these two points of bisection to the opposite angles, these lines will be parallel, two and two, and will trisect both diagonals.
Through two given points on opposite sides of a line draw meet in it silch that the angle they form is bisected. 222. The squares on the diagonals of a quadrilateral are double of the squares on the sides of the parallelogram formed by joining the 221.
lines to
mid-points of its sides. 223. If a quadrilateral be described about a circle, the angles subtended at the center by two opposite sides are supplemental. and have a common tangent BC, 224. Two circles touch at
A
Show
that
BA C
is
a right angle.
225. Find the locus of the point of intersection of bisectors of the angles at the base of triangles on the same base, and having a given vertical angle.
THE ELEMENTS OF GEOMETRY.
366
226. Find the locus of the centers of circles which touch a given circle at
a given point.
227. Two equal given circles touch each other, and each touches one side of a right angle ; find the locus of their point of contact. 228. In a given Hne find a point at which a given sect subtends a
given angle. 229. Describe a circle of given radius to touch a given line and have its center on another given line. 230. At any point in the circle circumscribing a square, show that one of the sides subtends an angle thrice the others. 231. Divide a given arc of a circle into two parts which have their
chords in a given
ratio.
common tangent between its points of contact proportional between the diameters of two tangent circles. 233. Any regular polygon inscribed in a circle is a mean proportional between the inscribed and circumscribed regular polygons of half 232.
is
a
the
The
sect of a
mean
number of sides. 234. The circumcenter,
the
centroid, the
mediocenter, and the
orthocenter form a harmonic range.
EXERCISES IN GEOMETRY OF THREE DIMENSIONS AND IN MENSURATION. The most instructive problems in geometry of three dimensions are made by generalizing those first solved for plane geometry. This way of getting a theorem in solid geometry is often difficult, but a number of the exercises here given are specially adapted for it. In the author's "Mensuration" (published by Ginn
& Co.)
are given
one hundred and six examples in metrical geometry worked out completely, and five hundred and twenty-four exercises and problems, of which also more than twenty are solved completely, and many others have hints appended.
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