Digitized by the Internet Archive
in 2008 with funding from
Microsoft Corporation
http://www.archive.org/details/elementsofgeometOOhalsrich
o', > > > J
1 ,» • • •
THE
ELEMENTS OF GEOMETRY.
BY
GEORGE BRUCE HALSTED,
A.B., A.M., 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.
NEW YORK :
JOHN WILEY & SONS,
15 ASTOR PLACE.
1885. .
Copyright, 1885,
By JOHN WILEY & SONS.
CAJORI
ELECTROTVPED AND PRINTED
BY RAND, AVERY, AND COMPANY,
(by permission)
TO
J. J. SYLVESTER,
A.M., CAM.; F.R.S., L. AND E.; CORRESPONDING MEMBER INSTITUTE OF FRANCE; MEMBER
ACADEMY OF SCIENCES IN BERLIN, GOTTINGEN, NAPLES, MILAN, ST. PETERSBURG,
ETC.; LL.D., UNIV. OF DUBLIN, AND U. OF E.; D.C.L., OXFORD; HON.
FELLOW OF ST. JOHN'S COL., CAM.; SAVILIAN PROFESSOR
OF GEOMETRY IN THE UNIVERSITY OF OXFORD;
In (Grateful B£membrance
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
subtile term " direction ; " and teachers who know something
of the Non-Euclidian, or even the modern synthetic geom-
etries, are seeing the evils of this superficial " directional "
method.
Moreover the attempt, in these books, to take away by
definition from the familiar word "distance" its abstract
character and connection with length-units, only confuses the
ordinary student. A reference to the article Measurement
in the " Encyclopaedia Britannica " will show that around the
word "distance" centers the most abstruse advance in pure
science and philosophy. An elementary geometry has no
need of the words direction and distance.
The present work, composed with special reference to
use in teaching, yet strives to present the Elements of
Geometry in a way so absolutely logical and compact, that
they may be ready as rock-foundation for more advanced
study.
Besides the acquirement of facts, there properly belongs
to Geometry an educational value beyond any other element-
VI PREFACE.
ary subject. In it the mind first finds logic a practical
instrument of real power.
The method published in my Mensuration for the treat-
ment of solid angles, with my words steregon and sieradian,
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 I regularly gave my classes the method of
Book IX. In 1883 my pupil, H. B. Fine, at 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 graphics.
The whole work is greatly indebted to my pupil and
friend, Dr. F. A. C. Perrine. We have striven after accuracy.
Any 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 I.
CHAPTER I.
ON LOGIC.
I. Definitions. — Statements.
ARTICLE PAGE
I. A statement defined i
Immediate inference defined ... i
Equivalent statements defined . . i
A statement implied i
Statements reduced to two terms
and a copula 2
Logically simple sentence .... 2
Logically composite sentence . . 2
Conditional statements reduce to
the typical form 2
9. Subject and predicate 2
II. Definitions. — Classes.
10. Individual terms defined .... 2
11. Class terms defined 2
12. How a class is defined 3
13. Contradictory of a class .... 3
14. Contradictories include the universe, 3
III. The Universe of Discourse.
15. Universe of discourse defined . . 3
16. The classes x and non-x still con-
tradictory 3
17. Different divisions into contradic-
tories 3
IV. Contranominal, Converse, Inverse,
Obverse.
ARTICLE PAGE
18. Contranominal defined
19. Converse defined
20. Inverse defined
21. Obverse defined
22. Two theorems prove four ....
23. Classes proved identical ....
24.
25-
26.
27.
28.
29.
30-
31-
32.
V. On Theorems.
A theorem defined ....
Demonstration defined . ,
Corollary defined .....
Hypothesis and conclusion ,
A geometric theorem defined
Use of type in this work . ,
Inverses of theorem with composite
hypothesis 6
VI. On Proving Inverses.
Inverses often proved geometrically, 6
Rule of identity 6
Rule of inversion 7
vii
Vlll
CONTENTS.
CHAPTER II.
THE PRIMARY CONCEPTS OF GEOMETRY.
I. Definitions of Geometric Magnitudes.
ARTICLE PAGE
34. Geometry defined 8
A solid defined 8
A surface defined 8
A line defined 8
A point defined 8
A magnitude defined 8
35-
36.
yi-
38.
39.
40-43
44.
45-
46.
47.
48.
49.
50.
51-
52.
53-
54.
55-
56.
57-
58.
59-
60.
61.
62.
63.
64.
65.
Point, line, surface, solid, de-
fined, using motion .... 8
Why space is called tri-dimensional, 9
Straight line defined ..... 9
A curve defined 9
Line henceforth means straight
line 9
A sect defined 9
A plane defined 9
Plane defined, using motion ... 9
A figure defined 9
A plane figure defined 9
Circle and center defined .... 10
Radius defined 10
Diameter defined 10
Arc defined 10
Parallel lines defined 10
Angle, vertex, arms, defined ... 10
Explement defined 11
Equal angles defined 11
Adjacent angles and their sum de-
fined II
Straight angle defined 12
When two adjacent angles equal a
straight angle 12
Supplement defined 12
Supplemental adjacent angles . . 12
66. Right angle defined 13
67. A perpendicular defined .... 13
68. Complement defined 13
69. Acute angle defined 13
70. Obtuse angle defined 13
71. Perigon defined 13
72. Sum of angles about a point is a
perigon 14
73. Reflex angle defined 14
74. Oblique angles and lines defined . 14
75. Vertical angles defined 14
76. Bisector defined 14
Tj. A trace defined 14
•jS. A triangle defined 15
79. Vertices defined 15
80. Sides defined 15
81. Interior angle defined 15
82. Exterior angle defined 15
83. Congruent defined 15
84. Equivalent defined 15
II. Properties of Distinct Things.
85-93. The so-called axioms of arith-
metic I5> 16
III. Some Geometrical Assumptions about
Euclid'' s Space.
94-99. The so-called geometric axioms, 16
IV. The Assumed Constructions.
100-103. The so-called postulates . . . 17
Table of symbols used 17
CHAPTER III.
PRIMARY RELATIONS OF LINES, ANGLES, AND TRIANGLES,
I. Angles about a Point.
ARTICLE PAGE
104-105. Right and straight angles . 18, 19
106-107. Perigons 19
1 08-1 10. Vertical angles 20, 21
II. Angles about Two Points.
ARTICLE PAGE
III. Transversal defined .... 22
1 1 2-1 13. Corresponding and alternate
angles 22
CONTENTS.
IX
III. Triangles.
ARTICLE PAGE
114. Equilateral triangle defined ... 24
Isosceles triangle defined .... 24
Scalene triangle defined .... 24
Base and vertex defined .... 24
Angles of triangle mean interior
angles 24
[15.
16.
:i7.
ARTICLE PAGE
119. Right-angled triangle and hypoth-
enuse defined
120. Obtuse-angled triangle defined
121. Acute-angled triangle defined
122. Equiangular triangle defined
Homologous defined . . .
123.
124-129, Congruent triangles
CHAPTER IV.
PROBLEMS.
ARTICLE PAGE
130. Problem defined 30
131. Solution defined 30
ARTICLE
132-40. Deduced constructions
PAGE
30-35
CHAPTER V.
JNEQUALITIES.
ARTICLE PAGE
141. Sign of inequality explained . 36
142-150. Theorems on one triangle . 36-39
ARTICLE PAGE
151. Oblique sect defined .... 39
152-164. Obliques and triangles . . 40-47
CHAPTER VI.
PARALLELS.
ARTICLE
165-172.
Fundamental theorems
PAGE ARTICLE PAGE
48-51 I 173-175. Application to the triangle . 51,52
CHAPTER VII.
TRIANGLES.
PAGE
176-179. Congruent triangles . . . 53-55
180. Conditions of congruence of
triangles 55
181. Locus defined 56
ARTICLE PAGE
182. How to prove a locus ... 56
183-186. Loci 57,58
187-191. Intersection of loci . . . 58-60
CHAPTER VIII.
POLYGONS.
I. Definitions.
ARTICLE PAGE
192. A polygon defined 61
193. Its vertices defined 61
ARTICLE
194. Its sides defined ....
195. Perimeter defined . . .
196. Interior angles of a polygon
PAGE
. 61
. 61
. 62
CONTENTS.
ARTICLE PAGE
197. Convex defined 62
198.
199.
200.
201.
Diagonals defined 62
An equilateral polygon defined . . 62
An equiangular polygon defined . 62
A regular polygon defined ... 63
Mutually equilateral defined ... 63
Mutually equiangular defined . . 63
204, 205. Polygons may be either ... 63
206. Trigon, tetragon or quadrilateral,
pentagon, hexagon, heptagon,
octagon, nonagon, decagon, do-
decagon, quindecagon .... 64
The surface of a polygon defined . 64
Parallelogram defined 64
Trapezoid defined 64
203.
207.
208.
209.
II. General Properties.
210-215. Congruence, angles .
216-220.
221.
222.
223,
224, 225.
226.
III. Parallelograms.
Sides and angles .
A rectangle defined
A rhombus defined
A square defined .
Properties . . .
64-66
67-69
• 70
• 70
. 70
70,71
Triangle altitudes concurrent . 72
227-2:
232.
233.
234.
235-
236.
237.
238-240.
241-244.
245-255.
256, 257.
258-264.
PAGE
Intercepts 73> 74
Medial defined 74
Concurrent defined .... 75
Collinear defined 75
Medials concurrent • • . . 75
Centroid, orthocenter, circum-
center, defined 75
IV. Equivalence.
Equivalence, how proved . . 76
Base and altitude 76
Squares 77-79
Rectangles 80-86
Complements of parallelo-
grams 87
V. Problems.
Parallelograms equivalent to
polygons 88-90
VI. Axial and Central Symmetry.
265. Corresponding points defined . 91
266. Axial symmetry defined . . 91
267. Central symmetry defined . . 92
268. 269. Symmetry in one figiire . . 92, 93
BOOK II
RECTANGLES.
270. A continuous aggregate defined . 95
271. A discrete aggregate defined ... 95
272-279. Counting and number . . 95, 96
280. The commutative law for addi-
tion 96
281. The sum of two sects defined . . 97
282. The commutative law holds for
sects 97
283. The sum of two rectangles de-
fined q8
ARTICLE PAGE
284. The commutative law holds for
rectangles 98
285. The associative law 99
286. The associative law holds for sects, 99
287. The associative law holds for rect-
angles 99
288. The commutative law for multipli-
cation 100
289. The commutative law for rectangle
of sects 100
CONTENTS,
XI
ARTICLE
290.
291-293.
294, 295.
296-302.
PAGE
The distributive law . . .100
The distributive law holds for
sects and rectangles . 100-102
The square on the sum of two
sects 102, 103
Rectangles and squares . 103-106
ARTICLE PAGE
303. The projection of a point de-
fined 106
304. The projection of a sect de-
fined 107
305-311. Squares on sides . , . 107-111
312-314. Problems 111-113
BOOK III
THE CIRCLE.
ARTICLE
315-319.
320,321.
322, 323.
324-
325-
326.
327.
328.
329-
330-332.
333» 334.
335» 336.
337-340.
341-343.
344-348.
349-353.
354-356.
357.
358, 359.
360.
361, 362.
363-
364.
365.
I. Primary Properties.
PAGE
Center, radius, surface of circle, 115
Point and circle . .
16,
17
A secant
A chord defined . . .
A segment defined . .
Expleraental arcs defined
Semicircle defined .....
Major arc and minor arc de-
fined
Major segment and minor seg-
ment defined 117
Circle and center 118
Diameter 118
Congruent circles 119
Concentric circles 120
Symmetry in the circle . 120, 121
Chords 122-124
Sects 125, 126
Circumscribed circle and in-
scribed polygon 126
Coney clic defined 126
Perpendiculars on chords . .127
II. Angles at the Center.
Angles subtended by arcs . .128
Sector 128
Angle subtended by a sect . .128
Inscribed angle 128
Corresponding arcs, angles,
sectors, chords 129
PAGE
129, 130
130-132
132, 133
ARTICLE
366-368. Sum of two minor arcs .
369-374. Unequal arcs ....
375-378. Inscribed angles . , .
III. Angles in Segments.
379-386. Angles made by chords and
secants 134-138
IV. Tangents.
387, 388. Tangent and point of contact
defined 138
389-402. Properties of tangents . 139-143
V. Two Circles.
403, 404. Common chord 144
405. Tangent circles defined . . . 144
406. Common axis of symmetry . 144
407-411. Relations between the center-
sect, radii, and relative posi-
tion 145-147
VI. Problems.
412. To bisect a given arc. . . . 148
413. Circumscribed polygon and in-
scribed circle 148
414. Escribed circle defined . . .148
415-417. Problems 149, 150
Xll
CONTENTS.
BOOK IV
REGULAR POLYGONS.
I. Partition of a Perigon.
ARTICLE PAGE
418,419. Successive bisection .... 151
420-422. Trisection 152
423-426. Five and fifteen . . . 153-154
II. Regular Polygons and Circles.
427-435. To inscribe and circum-
scribe 154-159
436. Radius and apothem de-
fined 159
ARTICLE PAGE
437. Side of inscribed hexagon
equals radius 159
III. Least Perimeter in Equivalent Figures.
— Greatest Surface in Isoperimetric Fig-
ures.
438. Isoperimetric defined. . . .159
439-443. Equivalent and isoperimetric
triangles 160-162
444-446. The circle 162-164
447-450. Regular polygons . . . 165-167
BOOK V.
RATIO AND PROPORTION.
ARTICLE
451-453-
454-
455-
456.
457.
458-460.
461, 462.
463-
Multiples 169,170
Commensurable defined . . .
Incommensurable defined . .
Commensurability exceptional.
Greatest common submulti-
ple
Incommensurable magnitudes.
The side and diagonal of a
square are incommensur-
able 172,173
Scales of multiples defined
PAGE
ARTICLE
?, 170
464-472.
. 170
473-476.
. 170
477,478
, 170
479-
. 170
480-482
, 171
483.
484-491.
2,173
492-496.
. ^n
PAGE
Multiples 173-175
Scale of relation . . . 175, 176
Ratio, antecedent, consequent, 176
Proportion, extremes, means,
fourth proportional . . .177
Equality of ratios defined . .177
Third proportional, mean pro-
portional 177
Ratios compared . . . 178, 179
Ways of changing a propor-
tion 180-182
BOOK VI
RATIO APPLIED.
I. Fundamental Geometric Proportions.
ARTICLE PAGE
497-500. Intercepts by parallels . 183, 184
501-503. Segments of a sect . . 184-186
ARTICLE PAGE
504, 505. Rectangles are as their bases,
j86, 187
506. Angles are as arcs 187
CONTENTS.
Xlll
II. Similar Figures.
PAGE
Similar figures defined . . .188
Similar triangles . . . 188-192
Homologous, ratio of simili-
tude, similarly placed . .192
515,516. Similar polygons similarly
placed 192, 193
517,518. Center of similitude .... 193
519-522. The right triangle .... 194
507.
508-513.
514.
ARTICLE PAGE
523, 524. The bisector of an angle, 195, 196
525-528. Harmonic division 196
III. Rectangles and Polygons,
529-539. Rectangle of extremes equiv-
alent to rectangle of
means 197-204
540-547. Composition of ratios . 204-208
548-552. Theorems and problems, 208-212
GEOMETRY OF THREE DIMENSIONS.
BOOK VII.
OF PLANES AND LINES.
ARTICLE PAGE
553. The plane defined 213
554-556. Assumptions about the plane .213
557-562. Determination of a plane, 214-216
563-565. Intersection of planes . 216,217
566. Principle of duality . .. . .217
567-578. Perpendicular to plane . 218-224
ARTICLE
PAGE
579-581.
Projection on plane . .
224, 225
582.
Inclination defined . .
. . 226
583-593-
Parallel planes ....
226-231
594, 595-
Smallest sect between lines
,232,233
596-604.
Loci and angles made
by
planes
233-238
BOOK VIII
TRI-DIMENSIONAL SPHERICS.
ARTICLE
605-609.
610.
611,612.
613,614.
615,616.
617,618.
619.
620-624.
625.
PAGE
Sphere, center, radius, defined, 239
Opposite points and diameter
defined 240
Point and sphere 240
Plane and sphere . . . 240, 241
Great circle and small circle . 241
Poles and axis 241
Hemispheres 241
Great circles and small cir-
cles 242,243
Zone and its bases defined . . 243
ARTICLE PAGE
626-630. Tangency 243,244
631, 632. Intersecting spheres .... 245
633-639. Sphere and tetrahedron . 246-248
640, 641. Globe and globe-segment de-
fined 248
642, 643. Central and planar sym-
metry 248, 249
644. Globe equivalent to tetrahe-
dron 250
645,646. Chords and arcs 251
647-649. Quadrant 252
XIV
CONTENTS.
650-652. Spherical angles and perpen-
dicular arcs .... 252, 253
ARTICLE
653, 654.
Smallest line between two
points 254,255
BOOK IX.
TWO-DIMENSIONAL SPHERICS.
ARTICLE PAGE
655. Introduction 257
656. Two-dimensional definition of
the sphere 258
657-662. Fundamental properties, 258,259
663-666. Definitions 259
667-675. Assumptions with corolla-
ries 259-261
676. All spherical lines intersect . 261
677, 678. A lune and its angle .... 262
679-684. Definitions 263, 264
685, 686. Perimeter of polygon . 264, 265
687. Symmetry without congru-
ence 265
688-691. Congruent spherical tri-
angles 266-268
692-697. Problems 268-270
ARTICLE PAGE
698-703. The line and its pole . . 270, 271
704. Bi-rectangular triangle . . . 272
705-707. Polar triangles .... 272, 273
708-719. Spherical triangles . . 274-280
720, 721. Loci 280, 281
722, 723. Symmetrical triangles equiva-
lent 281, 282
725. Spherical excess defined . . 282
724-728. Triangles equivalent to
lunes 282,283
729-731. Polygons equivalent to
lunes 283, 284
732. Equivalent triangles on same
base 284
IZZ^TA' Tangency 285
735-739. Theorems on circles . . 286, 287
BOOK X.
POLYHEDRONS.
ARTICLE PAGE
740-744. Polyhedron, faces, edges, sum-
mits, section .... 289, 290
745-747- Pyramid, apex, base, lateral
faces and edges .... 290
748. Parallel polygons 290
749-757. Prism, altitude, right prism,
oblique 290, 291
758-760. Parallelopiped, quader, cube . 291
761. All summits joined by one line, 292
ARTICLE PAGE
762-764. Descartes' theorem .... 293
765. Euler's theorem 294
766-768. Ratio between quaders . 295-297
769. Parallelopiped equivalent to
quader 298
770, 771. Triangular prism 299
n^^ 773- Pyramid sections 300
774. Equivalent tetrahedra . . . 301
775. Triangular pyramid and prism, 302
CONTENTS.
XV
BOOK XL
MENSURATION, OR METRICAL GEOMETRY.
CHAPTER I.
THE METRIC SYSTEM. — LENGTH, AREA.
ARnCLK PAGE
776-778. Measurement 303
779-782. Meter and metric system, 303, 304
783,784. Length 304>305
785-787. Area 305
ARTICLE PAGE
788. Area of rectangle 305
789, 790. Area of square 307
791, 792. Metric units of surface . . . 308
793. Given hypothenuse and side . 308
CHAPTER II.
RATIO OF ANY CIRCLE TO ITS DIAMETER.
ARTICLE PAGE
794. To compute perimeters .... 310
795. Length of circle 312
796. Variable defined 313
797. Limit defined 313
ARTICLE PAGB
798. Principle of limits . . . .313
799-802. Value of ir 315? 316
803,804. Circular measure of an
angle 316,317
CHAPTER III.
MEASUREMENT OF SURFACES.
ARTICLE PAGE
805, 806. Area of parallelogram . . .318
807, 808. Area of triangle . . . 318, 319
809, 810. Area of regular polygon and
circle 320
811-813. Area of sector and sector of
annulus 321,322
814. Lateral area of prism . . . 322
815-822. Cylinder 323-325
823. Mantel defined 325
824. Conical surface 326
825-832. Cone and frustum . . . 326-329
Z-},1. Area of sphere 330
834, 835. Zone 331
CHAPTER IV.
SPACE ANGLES.
ARTICLE PAGE
^Z^i^yi' Plane and space angles de-
fined 332
838. Symmetrical space angles . . . 332
839. Steregon 332
840. Steradian 333
841. Space angle made by two planes . 333
842. Spherical pyramid ...... 333
ARTICLE PAGE
843. Space angles proportional . . . 333
844. To transform any polyhedral angle, 334
845. 846. Area of a lune 334
847. Plane angle as measure of dihedral, 335
848. Properties of space angles . . .335
849-851. Area of spherical triangle or
polygon 336
XVI
CONTENTS.
ARTICLE
852-854.
855,856.
857.
858, 859.
860-862.
863-865.
866-868.
CHAPTER V.
THE MEASUREMENT OF VOLUMES.
Volume and metric units . .337
Volume of quader and cube . 338
Volume of parallelopiped . . 338
Volume of prism or cyl-
inder 338,339
Volume of pyramid or cone, 339-341
Prismatoid defined . . 341, 342
Prismoid 343
ARTICLE PAGE
869,870. Tetrahedron and wedge are
prismatoids 343
871, 872. Altitude and cross-section of a
prismatoid 343
873, Volume of any prismatoid . . 344
874, 875. Frustum of pyramid or cone . 346
876. Volume of globe 347
Z^^, 878. similar solids 347
Direction 347
Principle of Duality 349
Linkage 350
Cross-ratio 352
Exercises 357-366
THE ELEMENTS OF GEOMETRY.
BOOK I.
CHAPTER I.
ON LOGIC.
I. Definitions. — Statements.
1. A statement is any combination of words that is either
true or false ; e.g. {exempli gratia^ " for example "),
All x^s are fs,
2. To pass from one statement to another, with a conscious-
ness that belief in the first warrants belief in the second, is to
infer.
3. Two statements are equivalent when one asserts just as
much as the other, neither more nor less ; e.g., A equals By and
B equals A.
4. A statement is implied in a previous statement when its
truth follows of necessity from the truth of the previous state^
ment; e.g..
All x's are ys implies some fs are x*s.
THE ELEMENTS OF GEOMETRY.
5. Any 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 equals B, x is 7, are simple examples of this typical form
of all statements. Here x and y each stand for any word or
group of words that may have the force of a substantive in
naming a class ; e.g., x\^y may mean all the x^^ are y^ ; man
is mortal, means, all men are mortals.
6. A sentence containing only one such statement is a
logically simple sentence ; e.g., Man is the only picture-making
animal.
7. A sentence that contains more than one such statement
is a logically composite sentence ; e.g., A and B are respectively
equal to C and D, is composite, containing the two statements
A equals C, and B equals D.
8. Statements that are expressly conditional, such as, if A
is B, then C is D, reduce to the typical form as soon as we see
that they mean
{The consequence of M) is N.
9. In the typical form, bird is biped, x is y, we call x and y
the terms of the statement; the first being called the subject,
and the last the predicate.
II. Definitions. — Classes.
10. Terms that name a single object are called individual
terms ; e.g., Newton.
11. Terms that name any one of a group of objects are
called class terms ; e.g., mafti crystal. 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 LOGIC,
12. A class is defined by stating enough properties to decide
whether a thing belongs to it or not.
Thus, "rational animal" was given as a definition of "man."
13. If we denote by x the class possessing any given prop-
erty, all things not possessing this property form another class,
which is called the contradictory of the first, and is denoted by
non-;ir, meaning " not x ; " e.g., the contradictory of animate is
inanimate.
14. Any one thing belongs either to the class x or to the
class non-;r, but no thing belongs to both.
It follows that X is just as much the contradictory of nonvr
as non-;r is of x. So any class y and the class non-7 ^^^ mutu-
ally the contradictories of each other, and both together include
all things in the universe ; e.g., unconscious and conscious.
III. The Universe of Discourse.
15. In most investigations, we are not really considering all
things in the world, but only the collection of all objects which
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.
16. Within the universe of discourse, whether large or
small, the classes x and non-;ir are still mutually contradictory,
and every thing is in one or the other ; e.g., within the universe
mammals, every thing is man or brute.
17. The exhaustive division into x and non-;ir is applicable
to any universe, and so is of particular importance in logic.
But a special universe of discourse may be capable of some
entirely different division into contradictories, equally exhaus-
tive. 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 X and y are classes, our typical statement x \s y means,
if a thing belongs to class x, then it also belongs to class y\
e.g., Man is mortal, means, to be in the class men, is to be in
the class mortals.
If the typical statement is true, then every individual x
belongs to the class y : hence no x belongs to the class non-y,
or no thing not y is, 3. thing x ; that is, every non-y is nonvt; :
e.g., the immortals are not-human.
The statements x is y, and non-^ is non-;ir, are called each
the contranominal form of the other.
Though both forms express the same fact, it is, nevertheless,
often of importance to consider both. One form may more
naturally connect the fact with others 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 j/'s, the class y thus
contains all the individuals of the class x, and may contain
others, besides. Some of the ys, then, must be ^s. Thus,
from **a crystal is solid " we infer "some solids are crystals."
This guarded statement, some y is Xy 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
important. When the inverse is not true, this arises from the
circumstance that the subject of the direct statement has been
more closely limited than was requisite for the truth of the
statement.
ON LOGIC.
21. The contranominal of the inverse, namely, non-;tr is
non-^, is called the obverse of the original proposition.
Of course, if the inverse is true, the obverse is true, and
vice versa. To prove the obverse, amounts to the same thing
as proving the inverse. They are the same statement, but may
put the meaning expressed, in a different light to our minds.
22. If the original statement is x is j/, its contranominal is
noii-y is non-Xy its inverse is y is Xy its obverse is non-x is non-y.
The first 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.
A perfect definition is always invertable.
V. On Theorems.
24. 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 in-
ferred is called the proof or the demonstration of the theorem.
26. A corollary to a theorem is a statement whose truth
follows at once from that of the theorem, or from what has
been given in the demonstration of the theorem.
27. A theorem consists of two parts, — the hypothesis, or
that which is assumed, and the conclusion, or that which is
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 part names or defines the figure to which the
theorem relates ; the last part contains an additional property.
The theorem is 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 is restated with special reference to the figure to be used.
29. Type. — Beginners in geometry sometimes find it diffi-
cult to distinguish clearly between what is assumed and what
has to be proved in a theorem.
It has been found to help them here, if the special enuncia-
tion of what is given is printed in one kind of type; the
special statement of what is required, in another sort of type ;
and the demonstration, in 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. On Proving Inverses.
31. Often in geometry when the inverse, or its equivalent,
the obverse, of a theorem, is true, it has to be proved geomet-
rically quite apart from the original theorem. But if we have
proved that every x is y, and also that there is but one individ-
ual in the class y^ then we infer that y is x. The extra-logical
proof required to establish an inverse is here contained in the
proof that there is but one y.
Rule of Identity.
32. If it has been proved that x is j, that no two ;tr's are the
same /, and that there are as many individuals in class x as in
class /, then we infer y is x.
ON LOGIC,
Rule of Inversion.
33. If the hypotheses of a group of demonstrated theorems
exhaustively divide the universe of discourse into contradic-
tories, so that one must be true, though we do not know which,
and the conclusions are also contradictories, then the inverse
of every theorem of the group will necessarily be true.
Examples occur in geometry of the following type : —
If a is greater than by then c is greater than d.
If a is equal to by then c is equal to d.
If a is less than by then c is less than d.
Three such theorems having been demonstrated geometri-
cally, the inverse of each is always and necessarily true.
Take, for instance, the inverse of the first ; namely, when
c is greater than dy then a is greater than b.
This must be true ; for the second and third theorems
imply that' if a is not greater than by then c is not greater
than d.
THE ELEMENTS OF GEOMETRY.
CHAPTER IL
THE PRIMARY CONCEPTS OF GEOMETRY.
I. Definitions of Geometric Magnitudes.
34. Geometry is the science which treats of the properties
of space.
35. A part of space occupied by a physical body, or marked
out in any other way, is called a Solid.
3
D
36. The common boundary of two parts of a solid, or of a
solid and the remainder of space, is a Surface.
37. The common boundary of two parts of a surface is a
Line.
38. The common boundary of two parts of a line is a Point.
39. A Magnitude is any thing which can be added to itself
so as to double.
40. A point has position without magnitude.
41. A line may be conceived of as traced or generated by
a point on a moving body. The intersection of two lines is a
point.
42. A line on a moving body may generate a surface. The
intersection of two surfaces is a line.
THE PRIMARY CONCEPTS OF GEOMETRY. 9
43. A surface on a moving body may generate a solid.
44. 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
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.
45. A 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 is ^^ "^>.
straight. a curve.
47. Take notice: the word "line," unqualified, will hence-
forth mean " straight line."
48. A Sect is the part of a line between two definite points.
49. A Plane Surfacey or a Plane, is a surface which divides
space evenly, so that a piece of space from along one side of it
will fit either side of any other portion.
50. A plane is generated by the motion of a line always
passing through a fixed point and leaning on a fixed line.
51. A Figure is any definite combination of points, lines,
curves, or surfaces.
52. A Plane Figure is in one plane.
10
THE ELEMENTS OF GEOMETRY,
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 Circle is a sect drawn from the center
to the circle.
55. A Diameter of a Circle is a sect drawn through the
center, and terminated both ways by the circle.
56. An Arc is a. part of a circle.
57. Parallel Lines are such as are in the same plane, and
which, being produced ever so far both ways, do not meet.
a J)
58. When two lines are drawn from the same point, they
are said to contain, or to make with each other, an Angle.
THE PRIMARY CONCEPTS OF GEOMETRY. II
The point is called the Vertex, and the lines are called the
Arms, of the angle. A line drawn from the vertex, and turning
about the vertex in the plane of the angle from the position of
coincidence with one arm to that of coincidence with the other,
is said to turn through the angle ; and the angle is greater as
the quantity of turning is greater.
59. Since the line can turn from the one position to the
other in either of two ways, two angles are formed by two
lines drawn from a point.
Each of these angles is called the Explement of the other.
If we say two lines going out from a point form an angle, we
\^
are fixing the attention upon one of the two explemental angles
which they really form ; and usually we mean the smaller angle.
60. Two angles are called Equal if they can be placed so
that their arms coincide, and that both are described simul-
taneously by the turning of the same line about their common
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 com-
mon arm is called the Sum of the Adjacent Angles, Thus,
12
THE ELEMENTS OF GEOMETRY.
using the sign "4- for the word "angle," the sign of equality
(=), and the sign of addition (+, plus),
%. AOC = ^ AOB + t BOC.
62. A Straight Angle has its arms in the same line, and on
different sides of the vertex.
63. The sum of two adjacent angles which have their
exterior arms in the same line on different sides of the vertex
is a straight angle.
64. When the sum of any two angles is a straight angle,
each is said to be the Supplement of the other.
O O2 Oi it
65. If two supplemental angles be added, their exterior
THE PRIMARY CONCEPTS OF GEOMETRY.
13
arms will form one line ; and then the two angles are called
Supplemental Adjacent Angles,
66. A Right Angle is half a straight angle.
67. A Perpendicular X,o a line is a line that makes a right
angle with it.
68. When the sum of two angles is a right angle, each is
said to be the Complement of the other.
69. An Acute Angle is one which is less than a right angle.
70. An Obtuse Angle is one which is greater than a right
angle, but less than a straight 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,
14
THE ELEMENTS OF GEOMETRY.
or, in one plane, the whole amount of angle round a point, is
called a Perigon.
J ^
72. Since the angular magnitude about a point is neither
increased nor diminished by the number of
lines which radiate from the point, the sum
of all the angles about a point in a plane
is a perigon.
73. 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
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 common, are called Vertical Angles.
76. That which divides a magnitude into two equal parts is
said to halve or bisect the magnitude, and is called a Bisector.
77. If we imagine a figure moved, we may also suppose it
to leave its outline, or Trace^ in the first position.
THE PRIMARY CONCEPTS OF GEOMETRY.
IS
78. A Triangle is a figure formed by three lines, each inter-
secting the other two.
79. The three points of intersection are the three Vertices
of the triangle.
80. The three sects joining the vertices are the Sides of the
triangle. The side opposite A is named a ; the side opposite
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
sect and a line 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 Equivalettt which can be cut into
parts congruent in pairs.
II. Properties of Distinct Things.
85. The whole is greater than its part.
86. The whole is equal to the sum of all its parts.
87. Things which are equal to the same thing are equal to
one another.
1 6 THE ELEMENTS OF GEOMETRY,
88. If equals be added to equals, the sums are equal.
89. If equals be taken from equals, the remainders are
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.
92. Things that are double of the same thing, or of equal
things, are equal to one another.
93. Things which are halves of the same thing, or of equal
things, are equal to one another.
III. Some Geometrical Assumptions about Euclid's Space.
94. A solid or a figure may be imagined to move about in
space without any other change. Magnitudes which will coin-
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.
95. Two lines cannot meet twice ; that is, if two lines have
two points in common, the two sects between those points
coincide.
96. If two lines have a common sect, they coincide through-
out. Therefore through two points, only one distinct line can
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 one another.
99. Two lines which intersect one another cannot both be
parallel to the same line.
TABLE OF SYMBOLS.
17
IV. The Assumed 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.
103. Remark. — Here we are allowed the use of a straight
edge, not marked with divisions, and the use of a pair 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 geom-
etry which cannot be effected by combinations of these primary
constructions.
SYMBOLS USED.
~ similar.
= equivalent.
^ congruent.
II parallel to.
_L perpendicular to.
-i- plus.
.'. therefore.
^ angle.
< less than.
> greater than-.
A triangle.
£y parallelogram*
O circle,
rt. right.
St. straight.
i8
THE ELEMENTS OF GEOMETRY.
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 it to
be true of any two right angles we choose to take.
^ 3
G
J£ D C O
The hypothesis will be, that we have any two right angles ;
as, for instance, right 4. A OB and right i. FHG.
By the definition of a right angle (^6)^ the hypothesis will
mean, :i A OB is half a straight angle at O, and ^ FHG is half
a straight angle at H.
The conclusion is, that ^ AOB and ^ FHG are equal.
The proof consists in stating the equality of the two straight
angles of which the angles AOB and FHG are halves, referring
to the assumption (in 98) that all straight angles are equal,
and then stating that therefore the right ^ AOB equals the
right 4 FHG, because, by 93, things which are halves of equal
things are equal.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. 1 9
Now, we may restate and condense this as follows, using
the abbreviations rt. for " right," and st. for " straight," and the
symbol .'. for the word "therefore" :
Hypothesis. -4. A OB is rt ^ ; also ^ FUG is rt ^,
Conclusion. ^ AOB = ^ FHG.
Proof. ^ AOB \s half a st. ^ ; also ^ FBG is half a st. ^.
(66. A right angle is half a straight angle.)
By 98, all straight angles are equal,
.-. ^AOB = ^FHG.
(93. Halves of equals are equal.)
105. Corollary. From a point on a line, there cannot be
more than one perpendicular to that line.
Theorem II.
106. All perigons are equal.
Xb^
JC i< ^
Hypothesis. 4- ^ ^^ '-^ ^ perigon ; also ^ DHD is a perigon.
Conclusion. 4A0A= 4. DHD.
Proof. Any line through the vertex of a perigon divides it into two
straight angles.
By 98, all straight angles are equal,
.*. Perigon at O equals perigon at H.
{92. Doubles of equals are equal.)
107. Corollary. From the preceding demonstration, it
follows that half a perigon is a straight angle.
20 THE ELEMENTS OF GEOMETRY.
Theorem III.
1 08. If two lines intersect each other ^ the vertical angles are
equal.
Hypothesis. AC and BD are two lines intersecting at O.
Conclusions. ^ AOB = ^ COD.
^BOC ^ y^DOA.
Proof. Because the sum of the angles AOB and BOC is the
angle AOC, and by hypothesis AO and DC are in one line,
.-. ^ AOB + 4. BOC = St. 4.
In the same way,
^BOC + 4COD = St. 2^,
.-. i.AOB^-4 BOC = i. BOC + 4. COD.
Take away from these equal sums the common angle BOC^ and we
have
4. AOB = 4COD.
(89. If equals be taken from equals, the remainders are equal.)
In the same way we may prove
4 BOC = 4DOA.
Exercises, i. Two angles are formed at a point on one
side of a line. Show that the lines which bisect these angles
contain a right angle.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. 21
Theorem IV.
109. If four lines go out from a point so as to make each
angle equal to the one not adjacent to it, the four lines will form
only two intersecting li7ies.
Hypothesis. Let OA, OB, OC, OD, be four lines, with the
common point O ; and let 4. AOB = ^ COD, and 4.BOC =
^ DOA.
Conclusions. AO and OC are in one line.
BO and OD are in one line.
Proof. ^ AOB + 4. BOC -f 4 COD + 4. DOA = a perigon,
(71. The sum of all the angles about a point in a plane is a perigon.)
By hypothesis, 4 AOB = 4 COD, and 4 BOC = 4 DOA,
.*. twice 4 AOB ■+■ twice 4 BOC = a perigon;
.-. 4 AOB + 4 BOC = a st. 4.
(107. Half a perigon is a straight angle.)
.'. AOand OC form one line.
(65. If two supplemental angles be added, their exterior arms will form one line.)
In the same way we may prove that 4 AOB and 4 DOA are sup-
plemental adjacent angles, and so that BO and OD form one Hne.
no. Corollary. The four bisectors of the four angles
formed by two lines intersecting, form a pair of lines perpen-
dicular to each other.
22 THE ELEMENTS OF GEOMETRY.
II. Angles about Two Points.
111. A line cutting across other lines is called a Transversal.
112. If in a plane two lines are cut in two distinct points
by a transversal, at each of the points four angles are deter-
mined.
Of these eight angles, four are between the two given
lines (namely, ^, J, «, ^), and are called Interior Angles ; the
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 y
or if two interior or two exterior angles on the same side of the
transversal are supplemental, then every angle is equal to its, cor-
respondijtg and to its alternate angle, aiid 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. 23
First Case. — Hypothesis, -^.a — 4. i.
Conclusions. 2<dJ = 2(^z = ^j = ^^.
%.a -^ 4.4 = 4.1 ■\- 4.d ^ 4.2 + 4.C = 4d -\- 4s = St. 4,
c d
Proof. 4a = 4c, 4h ^ 4d, 41 = 43, 42 = 44.
{108. If two lines intersect, the vertical angles are equal.)
Moreover, since, by hypothesis, 4^ — 4- ^i ^^^ir supplements are
equal, ox 4 ^ — 4- ^^
(98. All straight angles are equal.)
(8g. If equals be taken from equals, the remainders are equal.)
/. 4a = 41 = 4c = 43, 4b = 42 = 4d=4 4,
and St. ^ = 4 ^ + 4 d = 4 a -\- 4 4 = 4 i -\- 4 d = 4 2 + 4 c
(88. If equals be added to equals, the sums are equal.)
Second Case. If, instead of tsvo corresponding, we have given two
alternate angles equal, we substitute for one its vertical, which gives the
First Case again.
Third Case. — Hypothesis. 4 ^ + 4 4 = ^^- 4*
But ^ ^ + 2^ 7 = St. ^ ,
.-. 4a = 41,
which gives again the First Case.
Fourth Case. — Hypothesis. 2J^ 7 + 2j^ ^ = ^f. ^ .
But^ « + ^ ^= St. 4,
.-. 4a = 41,
which gives again the First Case.
24
THE ELEMENTS OF GEOMETRY.
III. Triangles.
114. An Equilateral Triangle is one in which the three sides
are equal.
115. An Isosceles Triangle is one which has two sides
equal.
116. A 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. When we speak of the angles of a triangle, we mean
the three interior angles.
119. A Right-angled Triangle has one of its angles a right
angle. The side opposite the right angle is called the Hypothe-
nuse.
120. An Obtuse-angled Triangle has one of its angles obtuse.
121. An Acute-angled Triangle has all three angles acute.
122. An Equiangular Triangle 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
angles are called Homologous Angles. The pair of sides oppo-
site homologous angles are called Homologous Sides.
RELATIONS OF LINES, ANGLES, AND TRIANGLES.
25
Theorem VI.
124. Two triangles are coftgruent if tivo sides and the in-
eluded angle in the one ai'e equal respectively to two sides ajzd
the included angle in the other.
M
I 3L
Hypothesis. ABC and LMN two triangles, with AB — LM,
BC = MN, 4.B = ^M.
Conclusion. The two triangles are congruent; or, using A for
"triangle," and = for "congruent,"
t.ABC^ t.LMN,
Proof. Apply the triangle LMN to A ABC in such a manner that
the vertex M shall rest on B, and the side ML on BA, and the point
N on the same side of BA as C.
Then, because the side ML equals BA, the point L will rest upon
A ; because ^ M = :^ B, the side MJV will rest upon the line BC;
because the side MN equals BC, the point JV will rest upon the
point C. Now, since the point L rests upon A, and the point IV rests
upon C, therefore the side LN coincides with the side A C.
(95. If two lines have two points in common, the two sects between those points
coincide.)
Therefore every part of one triangle will coincide with the corre-
sponding part of the other, and the two are congruent.
(94. Magnitudes which will coincide are congruent.)
125. In any pair of congruent triangles, the homologous
sides are equal.
26 THE ELEMENTS OF GEOMETRY.
Theorem VII.
126. In an isosceles triangle the angles opposite the equal
sides are equal.
Hypothesis. ABC a triangle, with AB = BC.
Conclusion. 4. A = :^C.
Proof. Imagine the triangle ABC to be taken up, turned over, and
put down in a reversed position ; and now designate the angular points
A' (read A prime), B\ C, to distinguish the triangle from its trace
ABC left behind.
Then, in the triangles ABC, C'B'A,
AB = C'B' and BC = B'A',
since, by hypothesis, AB = CB.
Also ^B = ^B',
.'. ^A = ^ C
(124. Triangles are congruent if two sides and the included angle are equal in each.)
But ^ C = ^ C,
.-. ^A = :^C
{87. Things equal to the same thing are equal to one another.)
127. Corollary. Every equilateral triangle is also equi-
angular.
Exercises. 2. The bisectors of vertical angles are in the
same line.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. 2/
Theorem VIII.
128. Two triangles are congrtient if two angles and the in-
cluded side in the one are equal respectively to two angles and
the i7icluded side iti the other.
Hypothesis. ABC and LMN two iriangles, with ^ A = ^ Z,
^C = ^JV, AC=LN.
Conclusion. The two triangles are congruent.
A ABC ^ A LMN.
Proof. Apply the triangle LMN to the triangle ABC so that the
point L shall rest upon A, and the side LN lie along the side A C, and
the point M lie on the same side of AC as B.
Then, because the side AC = ZN,
.*. point iV will rest upon point C.
Because ^ Z = ^ A,
.'. side ZAf will rest upon the line AB,
Because ^ JV = ^ C,
/. side JVM will rest upon the line CB.
Because the sides ZM and NM rest respectively upon the lines AB
and CB,
.*. the vertex M, resting upon both the lines AB and CB, must rest
upon the vertex B, the only point common to the tvvo lines.
Therefore the triangles coincide in all their parts, and are congruent
.,-]
28
THE ELEMENTS OF GEOMETRY.
Theorem IX.
129. Two triangles are congruent if the three sides of the
one are equal respectively to the three sides of the other.
Hypothesis. Triangles ABC and LMN, having AB = ZM,
BC = MJV, and CA = JVZ.
Conclusion, a ABC ^ a ZMJV.
Proof. Imagine A ABC to be applied to A ZMN in such a way
that ZN coincides with A C, and the vertex M falls on the side oi AC
opposite to the side on which B falls, and join BM.
Case I. When BM cuts the sect AC.
Then in A ABM, because, by hypothesis, AB = AM,
.: ^ ABM = ^ AMB.
(126. In an isosceles triangle the angles opposite the equal sides are equal.)
And for the same reason, in A BCM, because BC = CM,
.'. ^ CBM = ^ CMB.
Therefore ^ ABM + ^ CBM = ^ AMB + ^ CMB;
(88. If equals be added to equals, the sums are equal.)
that is, ^ABC^4. AMC,
:. A ABC ^ J^ ZMN.
{124. Triangles are congruent if two sides and the included angle are equal in each.
RELATIONS OF LINES, ANGLES, AND TRIANGLES. 29
Case II. When BM passes through one extremity of the
sect AC, as C.
Then, by the first step in Case I.,
^B = ^M',
and for the same reason as before,
A ABC ^ A LMN.
Case III. When BM falls beyond the extremity of the
sect AC.
Then in A ABMy because, by hypothesis, AB = AM,
/. ^ ABM = ^ AMB.
And in A BCM, because BC = CM,
/. ^ CMB = ^ CBM,
Therefore the remainders, ^ ABC = ^ AMC;
(89. If equals be taken from equals, the remainders are equal.)
and so, as in Case I., A ABC ^ A LMN,
30 THE ELEMENTS OF GEOMETRY.
CHAPTER IV.
PROBLEMS.
130. A Problem is a proposition in which something is
required to be done by a process of construction, which is
termed the Solution.
131. The solution of a problem in elementary geometry
consists, —
(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 I.
132. To describe an equilateral triangle upon a given sect.
,. (T.. .
^\
Given, the sect AB,
Required, to describe an equilateral triangle on AB.
PROBLEMS. 31
Construction. With center A and radius AB^ describe the circle
BCD.
With center B and radius BA^ describe O A CF
(using O for " circle ").
(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. A line may be drawn from any one point to any other point.)
Then will ABC be an equilateral triangle.
Proof. Because A is the center of O BCD,
.'. AB = AC, being radii of the same circle ;
and because B is the center of the O ACB,
.-. BA = BC, being radii of the same circle.
Therefore AC = AB = BC,
(87. Things equal to the same thing are equal to one another.)
and an equilateral triangle has been described on AB»
Problem II.
133. On a giveii line, to mark off a sect equal to a given sect.
^ J!>\
Given, AB, a lino ; a, a sect
Required, to mark off a sect on AB equal to a.
Construction. From any point (7 as a center, on AB, describe
the arc of a circle with a radius equal to a.
If D be the point in which the arc intersects AB, then CD 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 III.
134. To bisect 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 AB, On AB, on the side remote from (9, describe, by 132, an
equilateral triangle, ABC.
Join OC. The line OC will bisect the given angle AOD.
Proof. In the triangles CAO and CBO we have
OA = OB,
CA = CB by construction,
and the side CO common.
.-. A CAO ^ A CBO,
(129. Triangles with the three sides respectively equal are congruent.)
.-. ^ COA = ^ COB,
.'. CO is ihe Usector of ^ A OB.
Exercises, i. Describe an isosceles triangle having each
of the equal sides double the base.
2. Having given the hypothenuse and one of the sides of a
right-angled triangle, construct the triangle.
PROBLEMS, 33
Problem IV.
135. Through a given point on a given line, to draw a perpen-
dicular to this line.
^\ C i :b
Given, ihe line AB, and the point C in it
Required, to draw from C a perpendicular to AB.
Solution. By 1 34, bisect the st. 2^ A CB,
Exercises. 5. Solve 134 without an equilateral triangle.
6. Divide a given angle into four equal parts.
3. Find a point in a line at a given distance from a givea
point. When is the problem impossible }
4. Find a point in a line equally distant from two given
points without the line.
5. From a given point without a given line, draw a line
making with the given line an angle equal to one-half a right
angle.
6. y4 is a point without a given line, EC. Construct an
isosceles triangle, having A as vertex, whose base shall lie
along EC, and be equal to a given sect.
II. If a right-angle triangle have one acute angle double
the other, the hypothenuse is double one side.
34
THE ELEMENTS OF GEOMETRY.
Problem V.
136. To bisect a given sect.
A
# I *
/ • *«
/ I «
/ ' •
* ' \
c : 1
Given, the sect AB.
Required, to bisect it.
Construction. On AB describe an equilateral triangle ABCj by
132.
By 134, bisect the angle ACB by the line meeting AB in D, Then
AB shall be bisected in Z>.
Proof. In the triangles A CD and BCD
^ACD= ^BCD,
and AC = BC by construction, and CD is common.
.-. AD = BD,
(124. Triangles are congruent if two sides and the included angle are equal in each.)
AB is bisected at D.
137. Corollary I. The line drawn to bisect the angle at
the vertex of an isosceles triangle, also bisects the base, and is
perpendicular to it.
138. Corollary II. The line drawn from the vertex of an
isosceles triangle to bisect the base, is perpendicular to it, and
also bisects the vertical angle.
Exercises. 12. Construct a right-angled isosceles triangle
on a given sect as hypothenuse.
PROBLEMS. 35
Problem VI.
i39» From a point without a given line^ to drop a perpendicu-
lar upon the line.
?€
i^
Given, Me line AB, and the point C without it
Required, to drop from C a perpendicular upon AB.
Construction. Take any point, Z>, on the other side of AB from
C, and, by 102, from the center C, with radius CDy describe the arc
FDG, meeting AB at F and G.
By 136, bisect FG at H. Join CH,
C^ shall be ±^^
(using ± for the words "perpendicular to ").
Proof. Because in the triangles CHF and CHGy by construction,
CF = CG, and HF = HG, and CH is common,
.-. ^ CBF = ^ C^6J.
(129. Triangles with three sides respectively equal are congruent.)
.-. 4. FHC, being half the st. 4. FHG, is a rt. ^ ; and CH is
perpendicular to AB.
140. Corollary. The line drawn from the vertex of an
isosceles triangle perpendicular to the base, bisects it, and also
bisects the vertical angle.
Exercises. 13. Instead of bisecting FGy would it do to
bisect the angle FCG }
36 THE ELEMENTS OF GEOMETRY.
CHAPTER V.
INEQUALITIES.
141' The symbol > is called the Sign of Inequality y and
a> b means that a is greater than b\ so a <d means that a is
less than d.
Theorem X.
142. An exterior angle of a triangle is greater than either of
the two opposite interior angles.
G
Hypothesis. Let the side AB of the triangle ABC be produced
to D.
Conclusions. %. CBD > ^ BCA.
4.CBD> ^BAC.
Proof. By 136, bisect BC in H.
By 100 and 10 1, join AH^ and produce it to F', making, by 133,
HF = AH.
By 100, join BF,
INEQUALITIES. 37
Then, in the triangles AHC and FHB, by construction,
CH = HB, AH = HF, and 4. AHC = ^ FHB ;
(108. Vertical angles are equal.)
/. A AHC ^ A FHB,
(124. Triangles are congruent if two sides and the included angle are equal in each.)
.-. 4 HCA = 4 HBF.
Now, 4. CBD > 4 HBF,
(85. The whole is greater than its part.)
.-. 4 CBD > 4. HCA,
Similarly, if CB be produced to G, it may be shown that
4ABG>4 BAC,
.-. the vertical 4 CBD > 4 BAC.
Theorem XL
143. Any two angles of a triangle are together less than a
straight angle, jb
Hypothesis. Let ABC be any A.
Conclusion. 4 A 4- ^ ^ < st. ^ .
Proof. Produce CA to D.
Then 4 BAD >4B,
(142. An exterior angle of a triangle is greater than either opposite interior angle.)
.-. 4 BAD + 4 BAC>4 B -{- 4 BAC,
.', st.4>4B-\-4 A,
144. Corollary I. No triangle can have more than one
right angle or obtuse angle.
145. Corollary II. There can be only one perpendicular
from a point to a line.
38 THE ELEMENTS OF GEOMETRY,
Theorem XII.
146. If one side of a triangle be greater than a second^ the
angle opposite the first must be greater than the angle opposite^
the second. ^
Hypothesis, a ABC, with side a > side c.
Conclusion. ^ BA C> 4.C.
Proof. By 133, from a cut off BD = c.
By 100, join AD.
Then, because BD = c, :. ^ BDA = ^ BAD.
(126. In an isosceles triangle the angles opposite the equal sides are equal.)
And, by 142, the exterior ^ BDA of A CD A > the opposite inte-
rior ^ C, .-. also ^ BAD >^C.
Still more \s^BAC>:^C.
147. Corollary. If one side of a triangle be less than the
second, the angle opposite the first will be less than the angle
opposite the second.
For, if ^ < ^, :. b > a\ :. by 146, 4. B>4.A, .\ '4.A<:4.B.
148. Inverses. We have now proved,
By 146, \i a> b, .'. 4. A > 4. B \
Byi26, if« = ^, .-. ^A—^B'j
By 147, ii a < b, .*. :^ A < ^ B.
Therefore, by 33, the inverses are necessarily true, namely,
If :^A > ^B, .'. a > b;
If ^A = ^B, .-. a =: b;
If 4 A < 4.B, .-. a < b.
149. Corollary. An equiangular triangle is also equi-
lateral.
INEQUALITIES. 39
Theorem XIII.
150. The perpendicular is the least sect between a given
point and a given line.
Ji ^
Hypothesis, lei A he a given point, and BC a given line.
Construction. By 139, from A drop AD A.BC.
By 100, join A to Fy any point oi BC except D.
Conclusion. AD < AF.
Proof. Since, by construction, ^ ADF is rt. 4. ,
:. -4. ADF > i. AFD,
(143. Any two angles of a triangle are together less than a straight angle.)
.-. AF>AD.
(148. If angle ADF is greater than angle AFD, therefore side AF is greater than
side AD.)
151. Except the perpendicular, any sect from a point to a
line is called an Oblique.
Exercises. 14. Determine the least path from one point
to another, subject to the condition that it shall meet a given
straight line.
15. The four sides of a quadrilateral figure are together
greater than its two diagonals.
40
THE ELEMENTS OF GEOMETRY.
Theorem XIV.
152. Two obliques from a point, making equal sects from the
foot of the i)erpendiculary are equal, and make equal angles with
the line.
Hypothesis. AC X. BD.
BC = DC.
Conclusions. AB = AD.
4. ABC = 4. ADC.
Proof. AACB^t:^ACD.
(124. Triangles are congruent if two sides and the included angle are equal in each.)
153- Corollary. For every oblique, there can be drawn
one equal, and on the other side of the perpendicular.
Exercises. 16. ABC is a triangle whose angle A is
bisected by a line meeting BC at D. Prove AB greater than
BD, and A C greater than CD.
17. Prove that the sum of the sects from any point to the
three angles of a triangle is greater than one-half the perime-
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 obliques betwee^t a given point and line^
that which makes the greater sect from the foot of the perpen-
dicular is the greater.
Hypothesis. AC 1. BD,
CB > CF,
Conclusion. AF < AB.
Proof. By 153, take F on the same side of the perpendicular as
B 'j then, in A AFCy because, by hypothesis, ^ ACF is rt.,
/. ^ AFC < rt. ^,
(143. Any two angles of a triangle are together less than a straight angle.)
.-. ^ AFB > rt. ^,
since ^ AFC + ^ AFB = st. ^.
/. ^ ABF < i AFB,
(143. Any two angles of a triangle are together less than a straight angle.)
/. AF<AB.
(148. If angle A is less than angle By therefore a is less than b.)
^55' Corollary. No more than two of all the sects can
be equal.
Proof. For no two sects on the same side of the perpendicular
can be equal.
42 THE ELEMENTS OF GEOMETRY.
Theorem XVI.
156. Any two sides of a triangle are together greater than
the third side.
Jr/
D"'"
Hypothesis. Let ABC be a triangle.
Conclusions, a + b>c.
a -{- ob.
b -\- oa.
Proof. By loi, produce BC to D; making, by 133, CD = b.
By 100, join AD.
Then, because AC = CD, ■
.'. t CDA = :^ CAD,
Now, ^ BAD > 4. CAD,
(85. The whole is greater than its part.)
.*. also ^ BAD > 4. BDA.
By 148, .-. BD>BA,
But BD = a + b, and BA is c,
.*. a -{- b>c.
Similarly it may be shown that a -j- ob, and that b + oa,
157. Corollary. Any side of a triangle is greater than
the difference between the other two sides.
INEQUALITIES. 43
Theorem XVII.
158. If from the ends of one of the sides of a triangle two
sects be drawn to a point within the triangle^ these will be
together less than the other two sides of the triangle^ but will
contain a greater angle.
Hypothesis. D is a point within A ABC.
Conclusions. (I.) AB + BC>AD + DC.
(II.) 4. ADC >B.
Proof. (I.) Join AD and CD.
Produce CD to meet AB in F.
Then {CB + BF) > {CF), and \_DF + FA;\ > [DA-].
(156. In a triangle any two sides are together greater than the third.)
/. CB -h BA = (CB + BF) ^ FA> {CF) + FA
=^ CD + IDF + FA'\ >CD + IDA'].
(II.) Next, in A ^i^Z?
:^ADC>:^AFD.
(142. Exterior angle of a triangle is greater than opposite interior angle.)
And for the same reason, in A CFB
4AFD>4.B, /. i.ADC>4.B.
Exercises. 19. The perimeter of a triangle is greater than
the sum of sects from a point within to the vertices.
44 THE ELEMENTS OF GEOMETRY.
Theorem XVIII.
159. If two triangles have two sides of the one equal respec-
tively to two sides of the other y but the included angle of the
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.
Hypothesis. ABC and DEF are two fr/ang/es, in which
AB = DE,
BC = EF,
^ ABC > 4. DEF,
Conclusion. AC> DF.
Proof. Place the triangles so that EF shall coincide with BC, and
the point D fall on the same side of j5C as A.
By 134, bisect ^ ABD, and let G be the point in which the bisector
meets A C.
By 100, join DG.
Then, in the triangles ABG and BDG, by construction, ^ ABG =
-4. DBG'i by hypothesis, AB = BD; and BG is common.
.-. A ABG ^ A DBG,
(124. Two triangles are congruent if two sides and the included angle are equal in
each.)
.-. AG = DG,
.-. CA = CG i- GA = CG -{- GD.
INEQUALITIES.
45
But CG -\- GD> CD,
(156. Any two sides of a triangle are together greater than the third.)
.-. CA>FD,
160. Corollary. U c = f^ a = dy and i.B<^ E, then
i.E>'4.B\ and so, by 1 59,
e>by .'. b<e.
161. Inverses. We have now proved, when a •=. d^ and
By 159, \i4.B > 4.E,
By 124, if ^^ = ;^^,
By 160, if ^B < 4.E,
Therefore, by 33, Rule of Inversion,
3 > <?j
b=e;
b < e.
lib > e,
:. ^B > ^E]
If d = e,
.-. ^B = ^E;
If b < e,
.'. ^B < ^E,
Exercises. 20. In a ABC, AB < AC. D is the middle
point of BC. Prove the angle ADB acute.
46 THE ELEMENTS OF GEOMETRY.
Problem VI I.
162. To construct a triangle of which the sides shall be equal
to three given sects, but any two whatever of these sects must be
greater than the third.
\
^ '
\
»
:d\
\ :
a
Z-^"
3F
^ » '
'•• *m^m'
Tr
c -
Given, fhe three sects a, b, c, any two whatever greater than
the third.
Required, to make a triangle having its sides equal respectively to a,
b, c.
Construction. On a line DF, by 133, take a sect OG equal to one
of the given sects, as b. From O as a, center, with a radius equal to
one of the remaining sects, as ^, describe, by 102, a circle.
From (9 as a center, with a radius equal to the remaining sect a,
describe an arc intersecting the former circle at X. Join XO and XG,
KGO will be the triangle required.
Proof. By the construction and the equality of all radii of the same
circle, the three sides GK^ KO, and OG are equal respectively to a, b, c.
Limitation. It is necessary that any two of the sects
should be greater than the third, or, what amounts to the same,
the difference of any two sides less than the third. For, if
a and c were together less than b, the circles in the figure would
not meet ; and, if they were together equal to b, the point K
INEQUALITIES.
47
would be on OG, 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.
163. Corollary. If a tri-
angle be made of hinged rods,
though the hinges be entirely
free, the rods cannot turn upon
them ; the triangle is rigid.
Problem VIII.
164. At a given point in a given liney to make an angle equal
to a given angle.
Given, the point A, and the line AB and 4. C,
Required, at A, with arm AB, to niake an angle = ^C
Construction. By 100, join any two points in the arms of ^ C,
thus forming a A DCE, By 162, make a triangle whose three sides
shall be equal to the three sides oi DCE] making the sides equal to
CD and CE intersect at A.
.-. -2^ FAG ^ i. DCE,
(zag. Triangles with three sides respectively equal are congruent.)
48 THE ELEMENTS OF GEOMETRY.
CHAPTER VI.
PARALLELS.
Theorem XIX.
^^S* -^ t'^o lines cut by a transversal make alternate angles
egualy the lines are parallel.
/
7
This is the contranominal of 142, part of which may be
stated thus : If two lines which meet are cut by a transversal,
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 of these suppositions
makes also the alternate angles equal.
Exercises. 21. From a given point without a given line,
draw a line making an angle with the given line equal to a
given angle.
PARALLELS. 49
Problem IX.
167. Through agivenpomtf to draw a line parallel to a given
line.
/-?
Given, a line AB and a point P.
Required, to draw a line through P || AB
(using II for the words "parallel to").
Construction. In AB take any point, as C. By 100, join PC,
At P in the Une CP, by 164, make 4. CPD = 4. PCB.
PD is II AB.
Proof. By construction, the transversal PC makes alternate angles
equal ; 4. CPD = 4 PCB,
.', by 165, PD is II AB.
Exercises. 22. Draw, through a given point between two^
lines which are not parallel, a sect, which shall be terminated
by the given lines, and bisected at the given point.
23. If a line bisecting the exterior angle of a triangle be-
parallel to the base, show that the triangle is isosceles.
24. Sects from the middle point of the hypothenuse of a/,
right-angled triangle to the three vertices are equal.
25. Through two given points draw two lines, forming,,
with a given line, a triangle equiangular with a given triangle.
26. AC, BD, are equal sects, drawn from the extremities of.
the sect AB on opposite sides of it, such that the angles BAC„
ABD, are together equal to a straight angle. Prove that AB
bisects CD.
50 THE ELEMENTS OF GEOMETRY,
Theorem XX.
1 68. If a transversal cuts two parallels y the alternate angles
are equal.
/L
Hypothesis. AB H CD cuf at H and K by transversal FG.
Conclusion. 4. HKD = ^ KHB.
Proof. A line through H, making alternate angles equal, is paral-
lel to CD, by 165.
By our assumption, 99, two different Imes through H cannot both
be parallel to CD ;
.*• by 31, the line through H |I CD is identical with the line which
makes alternate angles equal.
But, by hypothesis, AB is parallel to CD through H,
.-. i. KHB = ^ HKD.
169. Corollary I. If a transversal cuts two parallels, it
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. Corollary II. If a line be perpendicular to one of
two parallels, it will be perpendicular to the other also.
171. CoNTRANOMiNAL OF i68. If alternate angles are un-
equal, the two lines meet.
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. CoNTRANOMiNAL OF 99. Lines in the same plane par-
allel to the same line cannot intersect, and so are parallel to
one another.
Theorem XXI.
173. Each exterior angle of a triafigle is equal to the sum of
the two interior opposite angles.
Hypothesis. ABC any a, wiih side AB produced to D,
Conclusion. ^ CBD = ^A -^ ^C.
Proof. From B, by 167, draw BF \\ AC. Then, by 168, ^ C
^ CBF, and by 169, :^ A = ^ DBF\
:. by adding, 4. A ^ t^C = 4. FBD + ^ CBF = 4. CBD.
Exercises. 2y. Each angle of an equilateral triangle is
two-thirds of a right angle. Hence show how to trisect a right
angle.
28. If any of the angles of an isosceles triangle be two-
thirds of a right angle, the triangle must be equilateral.
52 THE ELEMENTS OF GEOMETRY.
Theorem XXII.
174. The sum of the three interior angles of a triangle is
equal to a straight angle.
Hypothesis. ABC any A.
Conclusion. ^ CAB + 4. B -^r ^C=st. ^.
Proof. Produce a side, as CAy to D.
By 173, tB ^-tC = 4 DAB.
Add to both sides 4. BAC.
.-. 4 CAB + ^^ + ^C = ^ DAB + 4. BAC = st. ^.
175. Corollary. In any right-angled triangle the two
acute angles are complemental.
TRIANGLES. 53
CHAPTER VII.
triangles.
Theorem XXIII.
176. Two triangles are congruent if two angles and an oppO'
site side in the one are equal respectively to two angles and the
corresp07iding side in the other.
Hypothesis. ABC and DFG As, with
^A= tD,
^C = ^G,
AB = DF.
Conclusion. t^ABC^ t. DFG,
Proof. Byi74, 2^^-f-^^H-^C' = st. :^ = ^ D '\- -4. F -\-
^G,
By hypothesis, ^ A -[- :^C = 4. D -\- ^Gy
(89. If equals be taken from equals, the remainders are equal.)
.-. A ABC ^ A DFG,
(128. Triangles are congruent if two angles and the included side are equal in each.)
54 THE ELEMENTS OF GEOMETRY,
Theorem XXIV.
"^11' If two triangles have two sides of the one equal respec-
tively to two sides of the other^ and the angles opposite to 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 and FGH as, with
¥^ =^^,
AB = FG,
BC = GH,
^ABC > i.G.
Conclusion. ^ C -f ^ -^ = st. 4^ .
Proof. By 164, make the 4. ABD = :^Gy
.-. A ABD ^ A FGH,
(128. Triangles are congruent if two angles and the included side are equal in each.)
.-. ^ BDA ^ tH>
and
BD = GH,
TRIANGLES. 55
But, by hypothesis, BC =^ GH,
BC = BD,
:, i. BDC = 4.C.
(126. In an isosceles triangle the angles opposite equal sides are equal.)
But 4. BDA + t BDC = St. ^,
/. ^H ^ 4.C ^ St. ^.
178. Corollary I. If two triangles have two sides of the
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
angles opposite the other pair of equal sides is a right angle, or
if they are oblique but not supplemental, or if the side opposite
the given angle is not less than the other given side, the tri-
angles are congruent.
179. Corollary II. Two right-angled triangles are con-
gruent if the hypothenuse and one side of the one are equal
respectively to the hypothenuse and one side of the other.
On the Conditions of Congruence of Two Triangles.
180. A triangle 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) Two sides and the angle between them.
(128) Two angles and the side between them.
(129) The three sides.
(176) Two angles and the side opposite one of them.
In the only other case, two sides and the angle opposite one
56 THE ELEMENTS OF GEOMETRY.
of them, 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
ambiguous case.
Suppose that the side a and side r, and 4. C, are given. If
a> c, then, making the 2^ Cy and cutting off CB = a, taking B
as center, and describing an arc with radius equal to c, it may
cut CD in two points ; and the two unequal triangles ABC
and ABC will satisfy the required conditions.
In these the angles opposite the side BC are supplemental,
by 177.
Loci.
181. The aggregate of all points and only those points
which satisfy a given condition, is called the Loctis of points
satisfying that condition.
182. Hence, in order that an aggregate of points, L, may be
properly termed the locus of a point satisfying an assigned con-
dition, d it is necessary and sufficient to demonstrate the fol-
lowing pair of inverses : —
(i) If a point satisfies C, it is upon L.
(2) If a point is upon L, it satisfies C
We know from 18, that, instead of (i), we may prove its
contranominal : —
(3) If a point is not upon Z, it does not satisfy C
TRIANGLES.
57
Also, that, instead of (2), we may prove the obverse of
(I):-
(4) If a point does not satisfy C it is not upon L,
Theorem XXV.
183. The locus of the point to which sects from two give7t
points are equal is the perpendicular bisector of the sect joining
them.
T
1
\
\
,
,'
*\
i
;
C
:b
Hypothesis. AB a sect, C its mid-point
PA = PB,
Conclusion. PC J_ AB.
Proof. Join PC.
Then A ACP is ^ A BCP,
(129. Triangles with the three sides respectively equal are congruent.)
.-. i.ACP = 4.BCP,
CP is the perpendicular bisector of AB.
184. Inverse.
Hypothesis. P, any point on CP ± AB at its mid-point C.
Conclusion. PA = PB.
Proof, t^ ACP ^ aBCP.
(124. Two triangles are congruent if two sides and the included angle are equal in
each.)
58
THE ELEMENTS OF GEOMETRY,
Theorem XXVI.
185. The locus of points from which perpendiculars on two
given intersecting lines are equal, consists of the two bisectors of
the angles betwee7i the given lines.
Hypothesis. Given AB and CD intersecting at O, and P any
point such that PM 1. AB ^ PN 1. CD,
Conclusion. ^ POM = 4. PON.
Proof, a POM ^ A PON,
(179. Right triangles having hypothenuse and one side equal are congruent.)
Therefore all points from which perpendiculars on two intersecting
lines are equal, lie on the bisectors of the angles between them.
186. Inverse.
Hypothesis. P, any point on bisector.
Conclusion. PM = PN.
Proof, a POM ^ PON.
(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. Where it is required to find points satisfying two con-
ditions, if we leave out one condition, we may find a locus of
points satisfying the other condition.
TRIANGLES. 59
Thus, for each condition we may construct the correspond-
ing locus ; and, if these two loci have points in common, these
points, and these only, satisfy both conditions.
Theorem XXVII.
188. There is one^ and only one, point from which sects to
three given points not on a line are equal.
o ...-V ^
Hypothesis. Given A, B, and C, not in a fine.
Construction and Proof. By 183, every point to which sects from
A and B are equal, lies on Z>iy, the perpendicular bisector of AB.
Again : the locus of points to which sects from B and C are equal is
the perpendicular bisector J^J^\
Now, Z>Zy and FF' intersect, since, if they were parallel, AB, which
is perpendicular to one of them, would be perpendicular to the other also,
by 170, and ABC would be one line. Let them intersect in O. By
95, they cannot intersect again.
189. Thus, OB = OA,
and
OC = OA,
.-. OB = OC,
Therefore, by 183, O is on the perpendicular bisector of y^ C There-
fore
Corollary. The three perpendicular bisectors of the sides
of a triangle meet in a point from which sects to the three ver-
tices of the triangle are equal.
60 THE ELEMENTS OF GEOMETRY.
Problem X.
190. To Ji7td points from which perpendiculars on three given
lines which form a triangle are equal.
Given, a, b, r, ihres lines inierseciing in the three disiinci
points Af By C.
Solution. The locus of points from which perpendiculars on the
two lines a and b are equal consists of the two bisectors of the angles
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
line of one pair cuts each Hne of the second pair in one point ; so that
we get four points, O, O^, O2, O^, common to the two loci.
191. By 185, the four intersection points of the bisectors of the
angles between a and by and between b and Cy lie on the bisectors of
the angles between a and c. Therefore
Corollary. The six bisectors of the interior and exterior
angles of a triangle meet four times, by threes, in a point.
POLYGONS. 6 1
CHAPTER VIII.
POLYGONS.
I. Definitions.
192. 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
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 broken line, makes the Perim-
eter of the polygon.
62
THE ELEMENTS OF GEOMETRY.
196. The angles between the consecutive sides and towards
the enclosed surface are called the Interior Angles of the poly-
gon. Every polygon has as many interior angles as sides.
197. A polygon is said to be Convex when no one of its
interior angles is reflex.
198. The sects joining the vertices not consecutive are
called Diagonals of the polygon.
199. When the sides of a polygon are all equal to one
another, it is called Equilateral,
200. When the angles of a polygon are all equal to one
another, it is called Equiangular,
POL YGONS. 63
201. A polygon which is both equilateral and equiangular
is called Regular,
202. Two polygons are Mutually Equilateral if the sides of
the one are equal respectively to the sides of the other taken
m the same order.
203. Two polygons are Mutually Equiangular if the angles
of the 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.
64 THE ELEMENTS OF GEOMETRY.
' 206. 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
sides is a Heptagon ; one of eight, an Octagon ; of nine, a
Nonagon ; of ten, a Decagon; of twelve, a Dodecagon ; of fif-
teen, a Quindecagon.
207. The Surface of a polygon is that part of the plane
enclosed by its perimeter.
208. A Parallelogram is a quadrilateral whose opposite
sides are parallel.
/
209. A Trapezoid is a quadrilateral with two sides parallel.
II. 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 ? How
could a triangle be considered a trapezoid t
POL YGONS.
65
Theorem XXIX.
211. The sum of the interior angles of a polygon is two less
straight angles than it has sides.
Hypothesis. A polygon of n sides.
Conclusion. Sum of ^^s = (n — 2) st. 2^'s.
Proof. If we can draw all 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
(« — 2) triangles, whose angles make the interior angles of the polygon.
But, by 174, the sum of the angles in each triangle is a straight
angle,
.'. Sum of ^*s in polygon = (n — 2) st. ^'s.
212. Corollary I. From each vertex of a polygon of n
sides are (u — 3) diagonals.
213. Corollary II. The sum of the angles in a quadrilat-
eral is a perigon.
214
gon of 71 sides is
Corollary III. Each angle of an equiangular poly-
(n — 2) St. ^'s
66 THE ELEMENTS OF GEOMETRY,
Theorem XXX.
215. In a convex poly go7t^ the sum of the exterior angles, one
at each vertex, made by producing each side in order, is a perigon.
Hypothesis. A convex polygon of n sides, each in order pro-
duced at one end.
Conclusion. Sum of exterior 2(^ 's = perigon.
Proof. Every interior angle, as ^ A, and its adjacent exterior
angle, as :^ x, together = st. ^ .
.*. all interior 2^ 's + all exterior ^ 's = n St. ^ 's.
But, by 211, all interior ^'s = {n — 2) st. ^'s.
.*. sum of all exterior 2^'s = a perigon.
Exercises. 30. How many diagonals can be drawn in a
polygon of n sides }
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 tri-
angles.
34. In a right-angled triangle, the sect from the mid-point
of the hypothenuse to the right angle is half the hypothenuse.
POL YGONS. 67
III. Parallelograms.
Theorem XXXI.
216. If two opposite sides of a quadrilateral are equal and
parallel^ it is a parallelogram.
Hypothesis. ABCD a quadrilateral, with AB = and \\ CD,
Conclusion. AD \\ BC.
Proof. Join AC. Then ^BAC = ^ A CD.
(168. If a transversal cuts two parallels, the alternate angles are equal.)
By hypothesis, AB = CD, and ^ C is common,
/. ^ BCA = ^ CAD,
(124. Triangles are congruent if two sides and the included angle are equal in each.)
.-. BC I! AD.
(165. Lines making alternate angles equal are parallel.)
Exercises. 35. Find the number of elements required to
determine a parallelogram.
36. The four sects which connect the mid-points of the
consecutive sides of any quadrilateral, form a parallelogram.
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 of the triangle.
38. If BE bisects the angle ^ of a triangle ABC, and CE
bisects the exterior angle A CD, the angle E is equal to one-half
the angle A.
6S THE ELEMENTS OF GEOMETRY.
Theorem XXXII.
217. The opposite sides and angles of a parallelogram are
equal to one another^ and each diagonal bisects it.
Hypothesis. Let AC be a diagonal of /z? ABCD (using the sign
£y for the word " parallelogram ").
Conclusions. AB = CD.
BC = DA.
t DAB = ^ BCD.
^B =: 4.D.
A ABC ^ A CDA.
Proof. ^ BAC = ^ ACD, and
^ CAD= ■4.BCA)
(168. If a transversal cuts parallels, the alternate angles are equal.)
therefore, adding,
4. BAC + 4. CAD = 4. BAD = 4 ACD + 4 ACB = 4 BCD.
Again, the side A C included between the equal angles is common,
.-. aABC^ a CDA,
(128. Triangles are congruent if two angles and the included side are equal in each.)
.-. AB = CD, BC = DA, and 4 B = 4 D.
Exercises. 39. If one angle of a parallelogram be right, all
the angles are right.
40. If two parallelograms have one angle of the one equal
to one angle of the other, the parallelograms are mutually equi-
angular.
POLYGONS.
69
218. Corollary I. Any pair of parallels intercept equal
sects of parallel transversals.
219. Corollary II. If two lines be respectively parallel
to two other lines, any angle made by the first pair is equal or
supplemental to any angle made by the second pair.
220. Corollary III. If two angles have their arms respec-
tively perpendicular, they are equal or supplemental.
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.
70
THE ELEMENTS OF GEOMETRY.
221. Corollary IV. If one of the angles of a parallelo-
gram is a right angle, all its angles are right angles, and it is
called a Rectangle,
Corollary V. If two consecutive sides of a parallelo-
equal, all its sides are equal, and it is called a
Rhombus.
gram are
223. A Square is a rectangle having consecutive sides equal.
224. Both diagonals, AC SiVid 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. AB II CD.
2. BC II DA. C ^ ^777^
3. AB = CD.
BC = DA.
i. DAB = 4. BCD.
4. ABC = 4 CDA.
The bisection of AC hy BD.
The bisection of BD by AC.
The bisection of the njhy AC.
The bisection of the £17 by BD.
4.
5-
6.
7-
8.
9-
10.
POLYGONS. 71
These ten combined in pairs will give forty-five pairs ; with
each of these pairs it may be required to establish any of the
eight other properties, and thus three hundred and sixty ques-
tions respecting such quadrilaterals may be raised.
For example, from i and 2, 217 proves 3,
/. A ABE ^ A CDE,
(128. Two triangles are congruent if two angles and the included side are equal in
each.)
.*. the diagonals of a parallelogram bisect each other.
The inverse is to prove i and 2 from 7 and 8 :
If the diagonals of a quadrilateral bisect each other, the
figure is a parallelogram.
225. Since 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, there-
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 paral-
lelogram is a rectangle.
43. If the diagonals of a parallelogram cut at right angles,
it is a rhombus.
44. If the diagonals of a parallelogram bisect the angles, it
is a rhombus.
45. If the diagonals of a rectangle cut at right angles, it is
a square.
72 THE ELEMENTS OF GEOMETRY.
Theorem XXXIII.
226. The three perpendiculars froin the vertices of a triangle
to the opposite sides ineet in a point.
Hypothesis. Lei Djy, EE\ FF\ be the three perpendiculars
from the vertices D, E, F, io the opposite sides of A DEF.
Conclusion. Dlf , EE' , FF' , meet in a point O.
Proof. By 167, through D, E, F, draw AB, BC, CA \\ FE, DF,
DE.
Then the figures ADFE, DBFE, are parallelograms.
By 217,
.♦. AD = FE ^ DBy
:. Z> is mid-point of AB.
In same way, E and F are mid-points oi AC and BC.
But, since DIf , EE, FF', are respectively ± EF, FD, DEy
.'. they must he ±AB, AC, and BC,
(170. A line perpendicular to one of two parallels is perpendicular to the other.)
.'. by 189, they meet in a point.
Exercises. 46. The intersection of the sects joining the
mid-points of opposite sides of any quadrilateral is the mid-
point of the sect joining the mid-points of the diagonals.
POL YGONS.
73
Theorem XXXIV.
227. If three or more parallels intercept equal sects on one
transversal^ they intercept equal sects on every transversal.
Hypothesis. AB = BC = CD =, etc., are sects on the trans-
versa/ AC, intercepted by the parallels a, b, c, d, etc.
FGj GHy BKy etc., are corresponding sects on any other trans-
versal.
Conclusion. FG = GH = HK =, etc.
Proof. From F, G, and B, draw, by 167, FL, GM, and UN, all
parallel to AD ;
.-. they are all equal, because AB z=z BC = CD =, etc.
(2x7. Opposite sides of a parallelogram are equal.)
But ^ GFL = 4. HGM = 4. KHN, and
4. FGL = 4 GHM = 4 HKN,
(169. If a transversal cuts two parallels, the corresponding angles are equal.)
/. A FGL ^ A GHM ^ A HKN,
(176. Triangles are congruent if they have two angles and a corresponding side
equal in each.)
/. FG= GH =: HK.
228. Corollary. The intercepted part of each parallel will
differ from the neighboring intercepts by equal sects.
74
THE ELEMENTS OF GEOMETRY.
Theorem XXXV.
229. The line drawn through the mid-point of one of the non-
parallel sides of a trapezoid parallel to the parallel sides^ bisects
the remaining side.
Hypothesis. In the trapezoid ABCD, BF = FC, and AB \\ CD
WFG.
Conclusion. AG ^ GD.
Proof. By 227, if parallels intercept equal sects on any transversal,
they intercept equal sects on every transversal.
230. 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 II. If through the mid-point of one side
of a triangle a line be drawn parallel to a second side, it will
bisect the third side ; and, inversely, the sect joining the mid-
points of any two sides of a triangle is parallel to the third side
and equal to half of it.
232. A sect from any vertex of a triangle to the mid-point
of the opposite side is called a Medial of the triangle.
POL YGONS. 75
233. Three or more lines 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 a triangle are concurrent in a
trisection point of each.
Proof. Let two medials, AD and BE, meet in O, Take G the
mid-point of OA, and Hoi OB, Join GH, HD, DE, EG.
By 231, in A ABO, GH is equal and parallel to \AB \ so is DE
in A ABC',
.*. by 216, GHDE is a parallelogram.
But, by 224, the diagonals of a parallelogram bisect each other,
/. AG ^ GO ^ on, and BH ^ HO = OE.
So any medial cuts any other at its point of trisection remote from its
vertex,
/, the three are concurrent.
76
THE ELEMENTS OF GEOMETRY.
236. The intersection point of medials is called the Centroid
of the triangle. The intersection point of perpendiculars from
the vertices to the opposite sides is called the Orthocenter of the
triangle. The intersection point of perpendiculars erected at
the mid-points of the sides is called the Circumcenter of the
triangle.
IV. Equivalence.
237. Two plane figures are called equivalent if we can prove
that they must contain the same extent of surface^ even if we do
not show how to cut them into parts congruent in pairs.
238. The base of a figure is that one of its sides on which
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 from its
highest point to the line of its base.
So the altitude of a parallelogram is the perpendicular
dropped from any point of one side to the line of the opposite
side.
240. The words " altitude " and " base " are correlative ; thus,
a triangle may have three different altitudes.
POL YGONS.
77
Problem XI.
241. To describe a square upon a given sect.
Given, Me SGci AB.
Required, to describe a square on AB.
Construction. By 135, from A draw AC 1. AB.
By 133, in AC make AB> = AB.
By 167, through D draw DI^ || AB.
By 167, through B draw BF \\ AD.
AF will be the required square.
Proof. By construction, AF is a parallelogram,
AB = FD, and AD = BF,
4.F = ^Ay and ^B = ^ D.
(217. In a parallelogram the opposite sides and angles are equal.)
But, by construction, AB = AD,
.'. AF is equilateral.
Again, ^ A -^ :i^ B — ?X, ■^.
(169. If two parallel lines are cut by a transversal, the two interior angles are
supplemental.)
But, by construction, ^ A\s rt.,
.-. 4.A = 4.B,
and all four angles are rt.
242. Corollary.
it is a rectangle.
If a parallelogram have one angle right,
78
THE ELEMENTS OF GEOMETRY.
Theorem XXXVII.
243. In any right-angled triangle y the square on the hypothe-
nuse is equivalent to the sum of the squares of the other two
sides.
\3t
ya
Hypothesis, a ABC, right-angled at B.
Conclusion. Square on AB -\- square oTi BC = square on AC.
Proof. By 241, on hypothenuse A C, on the side toward the A ABC,
describe the square ADFC.
On the greater of the other two sides, as BC, by 133, lay off CG
= AB. Join FG.
Then, by construction, CA = FC, and AB = CG, and ^ CAB =
•4. FCG, since each is the complement of A CB \
.'. A ABC ^ A CGF.
Translate the A ABC upward, keeping point A on sect AD, and
point C on sect CF, until A is on D, and C on F; then call B' the
position of B.
Likewise translate CGF to the right until C is on A, and F on Z> j
then call G^ the position of G.
POL YGONS.
79
The resulting figure, AG'DB'FGBA^ will be the squares on the
other two sides, AB and BC.
For, since the sum of the angles at 2) = st. 2>^,
.*. G'D and DB' are in one line.
Produce GB to meet this line at H.
Then GF = FB' = BC, and t\. 4. B' = ^ GFB' = ^ FGH,
.'. GFB'H equals square on BC,
Again, G^'^ = AB, and rt. ^ G^' = ^ G'^^ = ^ ^^^y,
/. AG'HB is the square on ^^,
.-. sq. oiAC= sq. of ^^ + sq. of ^C
244. Corollary. Given any two squares placed side by
side, with bases AB 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.
In AC take AD = BC, and join D to the corners of the
squares opposite B. Two right-angled triangles are thus pro-
duced, with hypothenuses perpendicular to one another. Trans-
late each triangle along the hypothenuse of the other.
80 THE ELEMENTS OF GEOMETRY.
Theorem XXXVIII.
245. Two triangles are congruent if they are mutually equi-
angular^ and have corresponding altitudes equal.
Hypothesis. As ABC and BDF mutually equiangular, and alti-
tude BH = altitude GB.
Conclusion, a ABC ^ A BDF.
Proof. Rt. a BDG ^ rt. a BCH,
(176. Triangles are congruent when two angles and a side in one are equal to two
angles and a corresponding side in the other.)
BD = BC,
.: A ABC ^ A BDF.
(124. Triangles are congruent when two angles and the included side are equal in
each.)
Exercises. 47. If from the vertex of any triangle a per-
pendicular be drawn to the base, the difference of the squares
on the two sides of the triangle is equal to the difference of the
squares on the parts of the base.
48. Show how to find a square triple a given square.
49. Five times the square on the hypothenuse of a right-
angled triangle is equivalent to four times the sum of the
squares on the mcdials to the other two sides.
POL YGONS.
8l
Theorem XXXIX.
246. If two consecutive sides of one rectangle he respectively
equal to two consecutive sides of anothery the rectangles are con-
gruent.
X <? € S
Two quadrilaterals ABCD and FGHK, wHh all
Hypothesis.
^s rt.
AB = FG, BC = GH,
Conclusion. ABCD ^ FGHK.
Proof. AB \\ CD, and AD \\BC,
(166. K interior angles on same side of transversal are supplemental, lines are
parallel.)
/. a rectangle is a parallelogram,
/. AB = CD, and AD = BC,
(217. Opposite sides of a parallelogram are equal.)
In the same way, FG = BK, and GH = FK,
/. ABCD ^ FGHK,
(2x0. If two polygons be mutually equilateral and mutually equiangular, they are
congruent.)
247. Corollary. A rectangle is completely determined by
two consecutive sides ; so if two sects, a and ^, are given, we
c&
may speak of the rectangle of a and b, or we may call it the
rectangle ab. It can be constructed by the method given in
241 to describe a square.
82
THE ELEMENTS OF GEOMETRY.
Theorem XL.
248. A parallelogram is equivalent to the rectangle of its base
and altitude.
C jr 3
Hypothesis. ABCD any £=7, of which side AD is taken as base;
AF = altitude of ^.
Conclusion. /17ABCD = rt. /uA^FGD.
Proof. AF ^ DG, and AB = DC.
(217. Opposite sides of a parallelogram are equal.)
By construction, ^ G and 2^ F are rt.,
.-. aABF^^DCG.
(179. Right triangles are congruent if hypothenuse and one side are respectively
equal in each.)
From the trapezoid ABGD take away A DCG, and then is left
n? ABCD. From the same trapezoid take the equal A ABF^ and there
isleftthert. zi^^i^G^/?.
.-. ^ABCD = rt. ^ AFGD.
(8g. If equals be taken from equals, the remainders are equal.)
249. Corollary. All parallelograms having equal bases
and equal altitudes are equivalent, because they are all equiva-
lent to the same rectangle.
Exercises. 50. Equivalent parallelograms on the same
base and on the same side of it are between the same
parallels.
POL YGONS. 83
Exercises. 51. Prove 248 for the case when (7 and F coin-
cide.
52. If through the vertices of a triangle lines be drawn
parallel to the opposite sides, and produced until they meet,
the resulting figure will contain three equivalent parallelo-
grams.
53. On 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
is the greater which stands on the greater base. Prove also an
inverse of this.
56. Equivalent parallelograms situated between the same
parallels have equal bases.
57. Of parallelograms on equal bases, that is the greater
which has the greater altitude.
58. A trapezoid is equivalent to a rectangle whose base is
half the sum of the two parallel sides, and whose altitude is the
perpendicular between them.
59. The sect joining the mid-points of the non-parallel sides
of a trapezoid is half their sum.
60. If E and F are the mid-points of the opposite sides, AD,
BC, of a parallelogram ABCD, the lines BE, DF, 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 ?
84
THE ELEMENTS OF GEOMETRY.
Theorem XLI.
Alternative Proof.
250. All parallelogrmns having equal bases and equal alti-
tudes are equivalent.
2)> zrC
Hypothesis. T\no ^s with equal bases and equal aliiiudes.
Conclusion. They can be cut into parts congruent in pairs.
Construction. Place the parallelograms on opposite sides of their
coincident equal bases, AB. Produce a side, as FB, which when con-
tinued will enter the other parallelogram. If it cuts out of the parallel-
ogram at H before reaching the side CD opposite AB, then will the
other cutting side, as CB, when produced, also leave the ^ ABFG
before reaching the side FG opposite AB ; that is, the point, K, where
CB cuts the line A G will be on the sect A G.
For, through H and K draw BL and KM \\ AB,
Then, in As ABB and ABK
^1 = ^1, ^2 = ^2,
(i68. If a transversal cuts two parallels, the alternate angles are equal.)
and side AB is common ;
.-. A ABB ^ A ABK,
POL YGONS. 85
But altitude of A AHB < altitude of £jABCD,
/. altitude of A ABK < altitude of £y ABFG, and K is on sect A G.
Now, A ABH ^ A BHL, and A ABK ^ A BKM,
(217. The diagonal of a parallelogram makes two congruent triangles.)
Taking away these four congruent triangles, we have left two paral-
lelograms, HLCD and KMFG, with equal bases, and^ equal but dimin-
ished altitudes. Treat these in the same way as the parallelograms first
given ; and so continue until a produced side, as FRQ,2irvd so the other,
CRN, also, reaches the side opposite the base before leaving the paral-
lelogram.
Then, as before, the As FRN and QRC are mutually equiangular;
but now we know their corresponding altitudes are equal.
.'. by 241, they are congruent,
.-. FN =. CQ,
.-. GN = DQ,
Also, QR ^ RF = PG, and NR ^ RC ^ PD,
4.G = ^ FRP = 4. RQD,
4.D =^4. CRP = 4. RNG,
4. DPR = 4. PRN,
4GPR = 4.PRQ',
and therefore the remaining trapezoids, PRQD ^ PRNG.
(210. If two polygons be mutually equilateral and mutually equiangular, they
are congruent.)
251. Corollary. Since a rectangle is a parallelogram,
therefore a parallelogram is equivalent to the rectangle of its
base and altitude.
Exercises. 65. How do you know that HLCD and KMFG
have equal altitudes "i
66. How do you know, that, if Q is on the sect CD, N is
on the sect FG ?
S6 THE ELEMENTS OF GEOMETRY.
Theorem XLII.
252. A triangle is equivalent to half the rectangle of its base
and altitude.
Hypothesis, a ABC, with base AC and altiiude BD.
Conclusion, a AB C — half the rectangle oi AC and BD.
Proof. Through A draw AF \\ CB, and through B draw BF \\ CA,
meeting AF in F;
.'. L^ACB ^ ^BFA.
(217. The diagonal of a parallelogram bisects it.)
But £jAFBC = rectangle oi AC and BD,
(248. A parallelogram is equivalent to the rectangle of its base and altitude.)
.-. t.ABC=^\ ^AFBC = J rectangle oi AC and BD.
253. Corollary I. All triangles on the same base having
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.
255. Corollary III. If a parallelogram and a triangle be
upon the same base and between the same parallels, the paral-
lelogram is double the triangle.
Exercises. 6j. Equivalent triangles on equal bases have
equal altitudes.
POL YGONS. Zj
Theorem XLIII.
256. If through any point on the diagonal of a parallelogram
two lines be drawn parallel to the sideSy the two parallelograms y
one on each side of the diagonal y will be equivalejit.
Hypothesis. P any point on diagonal BD of /=; ABCD\ FG
and HK lines ih rough P || AB and BC respectively, and meeting
the four sides in the points F, G, Hy K,
Conclusion. ^AKPG=^PFCH.
Proof, a ABD = a BCD.
A KBP = A BFP,
A GPD = A PBD.
(217. The diagonal of a parallelogram bisects it.)
From A ABD take away A KBP and A GPD, and we have lisft
C7 AKPG. From the equal A BCD take away the equal As BFP and
PHDy and we have left the £=7 PFCH,
.-. ^AKPG=^PFCH,
(89. If equals be taken from equals, the remainders will be equal.)
257. In figures like the preceding, parallelograms like KBFP
and GPHD are called parallelograms about the diagonal BD ;
while Z3^s AKPG and PFCH are called complements of par-
allelograms about the diagonal BD.
Exercises. 68. When are the complements of the paral-
lelograms about a diagonal of a parallelogram congruent ?
88 THE ELEMENTS OF GEOMETRY.
V. Problems.
Problem XII.
258. To describe a parallelogram equivalent to a given tri-
angle, and having an angle equal to a given angle.
M JU
Given, a ABC and ^ G.
Required, to describe a parallelogram — A ABC, while having an
angle = 2^ G.
Construction. Bisect ACmD.
At D, by 164, make ^ ADF = 4. G,
By 167, through A draw AH \\ DF,
By 167, through B draw BFH \\ CA.
AHFD will be the parallelogram required.
Proof. Join DB. Then A ABD = aBCD,
(254. Triangles having their vertices in same point, and for bases equal sects of
the same line, are equivalent.)
/. A ABC is double A AB£>.
But /I7AHFD is also double A ABD,
(255. If a parallelogram and a triangle be upon the same base and between the
same parallels, the parallelogram is double the triangle.)
/. npAHFD = A ABC,
and, by construction, ^ ADF = ^ G,
POLYGONS.
89
Problem XIII.
259. On a given sect as base, to describe a parallelogram
equivalent to a given triangle^ and having an angle equal to a
given angle,
X ^^ Q /V
Given, Mo seci AB, a CDF, 4. G.
Required, to describe on AB a parallelogram = A CDF, while having
an angle = ^ G.
Construction. By 258, make ^ BHKL — A CDF, and having
^ HBL = 7^ G, and place it so that BH is on line AB produced.
Produce KL. Draw AM \\ BL. Join MB.
4. HKM + i- KMA = St. ^,
(169. If a transversal cuts two parallels, it makes the two interior angles supple-
mental.)
/. ^HKM ^ 4KMB<%\..4.,
Therefore KH and MB meet if produced through H and B.
(171. If a transversal cuts two lines, and the interior angles are not supplemental,
the lines meet.)
Let them meet in Q. Through Q draw QN || HA, and produce
LB and MA to meet QN'm P and N.
:. ^ NPBA = C7 BHLK,
(256. Complements of parallelograms about the diagonal are equivalent.)
/. £yNPBA = A CFD, and 4. ABP = 4 G.
260. Corollary. Thus we can describe on a given base a
rectangle equivalent to a given triangle.
90 THE ELEMENTS OF GEOMETRY.
Problem XIV.
261. To describe a triangle equivalent to a given polygon.
J)
Given, a polygon ABCDFG,
Required, to constntct an equivalent triangle.
Construction. Join the ends of any pair of adjacent sides, as AB
and BC, by the sect CA.
Through the intermediate vertex, B, draw a hne H CA^ meeting GA
produced in H. Join HC.
Polygon ABCDFGA = polygon BCDFGB,
and we have obtained an equivalent polygon having fewer sides.
Proof. /\ABC= a ABC.
(253. Triangles having the same base and equal altitudes are equivalent.)
Add to each the polygon A CDFGA,
.'. ABCDFGA = BCDFGB.
In the same way the number of sides may be still further diminished
by one until reduced to three.
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.
263. Corollary II. Thus we can describe on a given
base a rectangle equivalent to a given polygon.
264. Remark. To compare the surfaces of different poly-
gons, 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
of the surfaces.
POL YGONS.
91
VI. Axial and Central Symmetry.
265. If two figures coincide, every point A in the one coin-
cides with a point A' in the other. These points are said to
correspond.
Hence to every point in one of two congruent figures there
corresponds one, and only one, point in the other ; those points
being called " corresponding " which coincide if one of the two
figures is 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 con-
gruent figures are themselves congruent.
Symmetry with Regard to an Axis.
266. If we start with two figures in the position of coinci-
dence, and take in the common plane any line or, we may turn
the plane of one figure about this line x until its plane, after
half a revolution, coincides again with the plane of the other
figure.
The two figures themselves will then have distinct positions
in the same plane ; but they will have this property, that they
THE ELEMENTS OF GEOMETRY.
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 we take in the common plane of two coincident
figures any point Xy we may turn the one figure about this
point so that its plane slides over the plane of the other figure
without ever separating from it.
Let this turning be continued until one line to Jf, and
therefore the whole moving figure, has been turned through a
straight angle about X.
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 sym-
metrical with regard to the point X as center of symmetry.
268. Any 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
that every line drawn through it cuts the figure in two points
symmetrical with respect to this center.
POL YGONS.
93
Theorem XLIV.
269. If a figure has two axes of symmetry perpendiciUar to
each othery then their intersection is a center of symmetry.
:
;
^^.
_\;>1
.'""'" :
-V-l^
-
>\
- ^
'^i
J
For, if X and y be two axes at right angles, then to a point A will
correspond a point A' with regard to x as axis.
To these will correspond points A^ and A/ with regard to j as axis.
These points A^ and A/, will correspond to each other with regard to x.
To see this, let us first fold over along y ; then A falls on A,y and A'
on A/.
If we now, without folding back, fold over along x. A, and with it
Ai, will fall on A\ which coincides \vith A^.
At the same time OA and OA^ coincide, so that the angles A Ox
and AiOx' are equal, where x' denotes the continuation of ^ beyond O.
It follows, that AOAl are in a line, and that the sect AA^ is bisected at
O, or (P is a center of symmetry for AAlj and similarly for Aj_ and A' ,
BOOK 11.
RECTANGLES.
270. A Contimious Aggregate is an assemblage in which two
adjacent parts have the same boundary.
271. A Discrete or Discontinuous Aggregate is one in which
two adjacent parts have different boundaries.
A pile of cannon balls is a discrete aggregate. We 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.
273. There is 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.
This proposition is involved in the meaning of the phrase
*• distinct things."
Any one and any other of them make two. If they are
attached to two of my fingers in a certain order, they can also
be attached to the same fingers in the other order. Thus, one
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
96 THE ELEMENTS OF GEOMETRY.
274. By generalizing the use of the fingers in counting,
man 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,
and so will only correspond exactly to discrete aggregates.
275. In a row of shot, we can find between any two, only a
finite number of others, and sometimes none at all.
Just so in regard to any two numbers. A row of six shot
can be divided into two equal parts ; but the half, which is
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
parts. So a line is not a discrete aggregate of points. It is
something totally different in kind from the natural series of
numbers.
277. The science of numbers is founded on the hypothesis
of the distinctness of things. The science of space is founded
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 num-
bers : we will now show how the laws for the addition and
multiplication of these discrete aggregates are applicable to
sects, which are continuous aggregates.
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. g/
It depends entirely on the interchangeability of any pair
of the units of numeration.
281. The sum of two sects is the sect obtained by placing
them on the same line so as not to overlap, with one end point
in common.
Thus, the sum of the sects a and b means the sect AC^
which can be divided into two parts,
AB = a, and BC = b,
282. The commutative law holds for the summation of sects.
a ■\- b = b -^ a.
M
AC= CA')
for 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
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. .
98
THE ELEMENTS OF GEOMETRY.
283. The sum of two rectangles is the hexagon formed by
superimposing two sides, and bringing the bases into the same
line.
Thus, if two adjacent sides of one are a and b, and of
the other c and d^ the sum of the rectangles ab and cd is
ABCDFGA.
284. The commutative law holds for the addition of rect-
angles ; that is, the sum is independent of the order of summa-
tion.
ab •\- cd ^ cd -\- ab\
for ABCDFGA turned over may be superimposed upon
AB'CD'FCA\ and will coincide with it.
Since, by 263, we can describe on a given base a rectangle
equivalent to a 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 num-
bers 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 -{■ {y ■\- z) = X ^ y -\- z.
286. The associative law holds for the summation of sects.
a + {3+c) = a + d-\-c = AD.
J£ 2
<x
ty
C
I 1
C 2)
o.
287. The associative law holds for the summation of rect-
angles.
ad + {bf + eg) = ad + bf -^ eg.
f
9
JS.
lOO
THE ELEMENTS OF GEOMETRY.
THE COMMUTATIVE LAW FOR MULTIPLICATION.
288. The product of numbers remains unaltered if the
factors be interchanged.
xy = yx.
289. The commutative law holds for the rectangle of two |
sects. I
a.
or
If a and b are any two sects, rectangle ab = rectangle ba,
ab = ba,
for rectangle ab may be so applied to rectangle ba as to coin-
cide with it.
THE DISTRIBUTIVE LAW.
290. To multiply a sum of numbers by a number, we may
multiply each term of the sum, and add the products thus
obtained.
x{^y ■\- z) — xy ■\- xz.
291. The distributive law holds when for numbers and prod-
ucts we substitute sects and rectangles.
a{b -{■ c) — ab -\- ac 'y
^ (^tuuu^
RECTANGLES.
lOI
for 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
other, the sum makes a rectangle whose base is ^ + ^ and
whose altitude is a ; that is, the rectangle a{p + c).
In the same way, by adding three rectangli.s . <?£; the sanje ,
altitude, we get ,•,""''* ' ' "•>'' »
a{b -\- c ■\- a)— ab -^ ac ^ adfj\ f,, \ >'"{'' *'* ;'.;
We may state this in words as follows :
If there be any two sects one of which is divided into any
number of partSy the rectaitgle contained by the two sects is
equivalent to the recta^tgles contained 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 con-
tained by the whole and each of the parts are together equivalent
to the square on the whole sect.
102
THE ELEMENTS OF GEOMETRY.
293. If r = ^, then
a{b -^ c) — a{b -\- a) — ab ■\- aa ^=- ab -\- a^.
amC
Therefore
If a sect he divided into any two parts^ 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.
294. The rectangle of two equal sects is a square, and
{a + by is only a condensed way of writing {a + b){a -\- b).
But, by the distributive law,
{a + b){a + b) = {a -\- b)a -\- {a + b)b.
By the commutative law,
(a + b)a = a{a + b).
By the distributive law,
a{a -\- b) = aa -\- ab = a^ -\- ab.
In the same way,
^ g
(a + b)b = ab -\- b^,
{a + by = a^ + ab -^ {ab + b^).
RECTANGLES.
103
By the associative law,
a^ -ir ab -^ {ab + <^) = ^' + {ab + ab) + b^ = a^ -^ 2ab + b',
,', {a + by = a" -\- 2ab + ^.
Therefore
If a sect be divided into a7iy two parts, the square on the
whole sect is equivalent to the squares on the two parts y together
with twice the rectangle contained by the two parts.
295. Corollary. The square on a sect is four times the
square on half the sect.
a
:-»■ «.
c z. J) a, :b
I
296. By the distributive law, and 294,
a{a -{- b -\- b) -\- b^ = a"" -h ab -^ ab + b^ = (a ■}- b)'.
From this, if we bisect the sect AB at C, and divide it into
two unequal parts at D, and for BD put a, and for DC put b,
we get the theorem :
If a sect is divided into two equal parts, and also into two
unequal parts, the rectangle contained by the uneqtial parts,
together with the square oji the line between the points of section,
is equivalent to the square on half the sect.
297. If CD = b, and AC = a + b, then their sum AD = a
•}- b -\- b, and their difference is a ; therefore the rectangle con-
tained by their sum and difference equals a(a + 2b), which, by
296, is the difference between (a + by and b^.
104-
THE ELEMENTS OF GEOMETRY.
Therefore
The rectangle contained by the sum and difference of any two
sects is equivalent to the difference between the squares on those
sects.
298. If we bisect the sect AB at C, and produce it to Z?,
and for BD put a, and for BC put 3, then the above equation,
a{a ■\- b '\- b) -\- b"^ =i {a •\- by, gives us the theorem :
If a sect be bisected^ and.produced to any pointy the rectangle
contained by the whole line thus produced and the part of it pro-
duced, together with the square on half the sect bisected, is equiv-
alent to the square on the line which is made up of the half and
the part produced.
299. By 294,
{a + by -{- a^ = a^ + 2ab + ^^ + cC,
By the associative law,
a" ^ 2ab -^ b" -^ a^ =^ 20" + 2ab + ^^
By the distributive law,
2a^ 4- 2ab -{- b^ = 2a{a -^ b) -\- b*.
By the commutative law,
2a(a -\- b) -{- b'' = 2{a + b")a + ^,
/. {a -\- by -^ a^ = 2(a -{- b)a + b^.
Therefore
If a sect be divided into any two parts, the squares on the
whole line and on one of the parts are equivalent to twice the
RECTANGLES.
•105
rectangle co7ttained by the whole and that party together with
the square on the other part.
Z- «
300. By the commutative and distributive laws, and 294,
4(« + b)a + b^ = 4a^ + 4ab + ^^ = (2a + by.
By the associative and commutative laws,
(2a -{-by=^(a + a + by = (la + ^] + «)%
• .-. 4{a 4- b)a 4- ^^ = {[a 4- ^] + ^y.
a 4t ^
^
Therefore
7/" a sect be divided into any two parts, four times the rect-
angle contained by the whole sect and one of the parts, together
a
a
I
I06 THE ELEMENTS OF GEOMETRY.
with the square on the other part ^ is equivalent to the square on
the line which is made up of the whole sect and the first part.
301. By the associative and commutative laws, and 294,
{\_a -f /5] + ay + ^^2 = (<5 4- ^ay + ^'
= /^ 4- /i^ab + 4^^ + ^* = 20" + A,ab + 2<^ + 2«^
By the distributive law, and 294,
20" 4- Adb + 2b'- 4- 2a^ — 2{a^ -\- 2ab + b^) ■\- 20" = 2{a -\- by + 2<a!%
.-. {[a 4- ^] 4- ay -\- b^ — 2{a 4- by + 20",
, ^""^^ » " . ^ .
Therefore
If a sect be divided into two equal and also two unequal
partSy the squares on the two unequal parts are together double
the squares on half the sect and on the line between the points
of section.
302. By 294, and the associative and distributive laws,
(dz + ^)2 + ^ = «^ 4- 2ab + b^ + b""
= J + ^IV 2ab + 2bA = 2ffl + 2 (^ + bj.
Therefore
If a sect be bisected and produced to any pointy 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 made up of the half and the part produced.
303. The projection of a point on a line is the foot of the
perpendicular from the point to the line.
(TU,
V--JL- — -^M"
•^-^^,., ir:x.
a
c,
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' is the projection of the sect AB on the
line c.
Theorem I.
305. In an obtuse-angled triangle y the square on the side oppo-
site the obtuse angle is greater than the sum of the squares on
the other two sides by twice the rectangle co7itained by either of
those sides and the projection of the other side upon it.
Hypothesis, a ABC, wiih ^ CAB obtuse.
Conclusion. ^^ = ^ + ^ + 2bf
Proof. By 294,
(b + J'Y = 3^ + 2hj + j\
Adding h^ to both sides,
{b + JY J^ h- = b^-\- 2bj + y- + h\
But
{b + j'Y -\- h^ = a", and j^ -\- h^ z= c",
(243. In a right triangle, the square of the hypothenuse equals the sum of the
squares of the other sides.)
.-. ^2 _ ^2 _|_ 2^- ^ ^.
i-
io8
THE ELEMENTS OF GEOMETRY.
Theorem II.
306. In any triangle^ the square on a side opposite any acute
angle is less than the sum of the squares on the other two sides
by twice the rectangle contained by either of those sides and the
projection of the other side upon it.
Hypothesis, a ABC, with ^ C acute.
Conclusion, c^ -{- 2bJ = a^ -{- b^.
Proof. By 295,
b'' + j^ = 2b/ + Alf,
Adding k^ to both sides,
b"^ + J^ + y^" = 2b/ + A& + /i\
But
y^ 4. ^2 ^ «2^ and Alf + h^ = c^,
(243. In a right triangle, the square of the hypothenuse equals the sum of the
squares of the other sides.)
.*. /^^ + «2 = 2bj + c^,
307. Having now proved that in a triangle,
By243, if ;^^ = rt. 2^, .'. a' = b- -{- c- ;
By 305, if ^A > rt. ^, .'. a^ > b^ •{- c^ ;
By 306, if ^A < rt. ^, /. a^ < b^ + c^.
Therefore, by 33, Rule of Inversion,
If a"" = b^ -{- f% .-. ^ ^ == rt. ^ j
If a' > b^ + ^, .-. ^A> rt. ^ ;
If a^ < b^ + ^, /. ^A< rt. i.
RECTANGLES. 109
Theorem III,
308. In any triangle, if a medial be drawn from the vertex
to base^ the sum 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.
Hypothesis, a ABC, with medial BM = i.
Conclusion, a"^ + c^ =. 2{^by + 21^.
Proof. — Case I. If ^ BMA = ^ BMC, then
a^ = {\by + /% and c^ = {^by + /^
(243. In a right triangle, the square of the hypothenuse equals the sum of the
squares of the other sides.)
Case II. If ^ BMA does not equal ^ BMC, one of them must
be the greater. Call the greater BMC.
Then, in the obtuse-angled triangle BMC, by 305,
a-^ {^dy + i^+ 2{\bj).
In A BMA, by 306,
^ + 2(|^y) = {iby + i\
Adding,
«- + ^ 4- 2{\bj) = 2{\by + 2/« + 2{\bj),
,\ a^ + c" = 2{iby 4- 2i^.
309. Corollary. The difference of the squares on the
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 IV.
310. 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
diagonals. ^ -q
Hypothesis. EF is the sect joining fhe mid-points of the diag-
onals AC and BD of the quadrilateral ABCD.
Conclusion. ~A& + 'BC^ + C^' -f 'DA'
= ^C" + BL^ -\-4£F\
Proof. Draw BE and DE.
By 308, _ ,.r\2
AB" + BC' = 2\—-] + 2^^,
and
C&-h^'= 2{^ + 2DE'.
Adding,
AB" -^ BC^ -\- CD" -\- DA" =^ ^(^T + ^^^ + ^^^'
But, by 308,
BE'+ DE'^: 4^y + 2KF^,
= AC"" + B& + 4^^^-
RECTANGLES.
Ill
311. Corollary. The sum of the squares on the four
sides of a parallelogram is equivalent to the sum of the squares
on the diagonals, because the diagonals of a parallelogram
bisect each other.
Problem I.
312. To square atty polygon.
. H
r^-
Given, any polygon N.
Required, to describe a square equivalent to N.
Construction. Describe, by 263, the rectangle ABCD = N.
Then, if AB — BC^ the required square is ABCD.
If AB be not equal to BC, produce BA^ and cut off AF = AD,
Bisect BF in G, and, with center G and radius GB, describe FLB.
Produce DA to meet the circle in H.
The square on AH shall be equal to N.
Proof. Join GH. Then, by 296, because the sect BF is divided
equally in G and unequally in A^
rectangle BA,AF + AC^ ^ GF" ^ GH"" = AH"" + AC^,
by 243 ;
AH^ = rectangle BA,AF = rectangle BA,AD = N.
112
THE ELEMENTS OF GEOMETRY.
Problem II.
313. To divide a given 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,
Jfr ,X
Given, ihe sect AB.
Required, to find a point C such that rectangle AB,BC = AC*.
Construction. On AB^ by 241, describe the square ABDF,
By 136, bisect AF in G. Join BG,
Produce FA, and make GH = GB.
On AH describe the square AHKC.
Then AB,BC = ZC^
Proof. Produce KC to Z. Then, by 298, because FA is bisected
in G and produced to H, rectangle FH,BA + 'A& = 'GH^ = 'GB"
= GA' + AB\
.-. FB,HA = AB",
But, since HK = HA,
.-. FH,HK = AB",
Take from each the common part AL,
.', HC== CD;
that is,
AC^ = CB,BD = CB,BA,
RECTANGLES. 1 13
Problem III.
314. To describe an isosceles triangle having each of the
angles at the base double the ajigle at the vertex.
Construction. Take any sect, AB, and, by 313, divide it in Cso
that rectangle AB,AC = ~^^. With center C and radius CB, describe
a circle.
With same radius, but center A, describe a circle intersecting the
preceding circle in D. Join AD, BD, CD.
ABD will be the triangle required.
Proof. By construction, CB = CD = DA,
By 134, bisect ^ ADC by DF. By 137, DF \% Jl AC, and bisects
AC.
Then 'AB' -f Z^ = 'BD' + 2AB,AF.
(306. The square on a side opposite an acute angle is less than the squares on the
other two sides by twice the rectangle of either and the projection of the other
on it.)
But ^C = 2AFy .'. AB,AC == AB,2AF - 2AB,AF\
.'. by our initial construction, BC^ == 2AB,AF,
.-. AB" + AD" = Bif + BC\
But, by construction, AD ^ BC, .'. AB" = BD', .-. AB = BD,
.'. ^ ADB = ^ DAB = ^ ACD = ^ B -^ ^ BDC = 2 ^ B.
(173. The exterior angle of a triangle equals the sum of the two opposite interior
angles.)
BOOK III.
THE CIRCLE.
I. 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. As the motion of a sect does not enlarge or diminish
it, all radii are equal.
319. Since the moving sect, after revolving through a peri-
gon, returns to its original position, therefore the moving end
point describes a closed curve.
This divides the plane into two surfaces, one of which is
swept over by the moving sect. This finite plane surface is
called the surface of the circle.
Any part of the circle is called an Arc,
SS5
Il6 THE ELEMENTS OF GEOMETRY.
Theorem I.
320. The sect to a point fivm the ce^iter of a circle is less
tha?t, equal to, or greater than, the radius, according as the point
is within, on, or without the circle.
Proof. If a point is on the circle, the sect drawn to it from the
center is a radius, for it is one of the positions of the describing sect.
Any point, Q, within the circle lies on some radius, OQR,
/. OQ < OR.
If ^S* is without the circle, then the sect OS contains a radius OR,
:. OS > OR.
321. By 33, Rule of Inversion, a point is within, on, or with-
out the circle according as its sect from the center is less than,
equal to, or greater than, the radius.
322. A Secant is a line which passes through two points on
the circle.
THE CIRCLE. 11/
Theqrem II.
323. A secant can meet the circle in only two points.
Proof. By definition, all sects joining the center to points on the
circle are equal, but from a point to a line there can be only two equal
sects.
(155. No more than two equal sects can be drawn from a point to a line.)
324. A Chord is the part of a secant between the two points
where it intersects the circle.
325. A Segment of a circle is the figure made by a chord
and one of the two arcs into which the chord divides the circle.
326. When two arcs together make
an entire circle, each is said to be the
Expleinejit of the other.
327. When two explemental arcs are
equal, each is a Semicircle.
328. When two explemental arcs are
unequal, the lesser is called the Minor
Arc, and the greater is called the Major
Arc.
329. A segment is called a Major or Minor Segment ac-
cording as its arc is a major or minor arc.
Il8 THE ELEMENTS OF GEOMETRY.
Theorem III.
330. A circle has only one center.
Hypothesis. Lei F, G, and H be points on a O.
Conclusion. O FGH has only one center.
Proof. Join FG and 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 F and G is in
the perpendicular bisector of FG, and any center of a circle through
G and H is in the perpendicular bisector of GH.
(183. The locus of the point to which sects from two given points are equal is the
perpendicular bisector of the sect joining them.)
But these two perpendicular bisectors can intersect in only one
point,
/. O FGH has only one center.
331. Corollary I. The perpendicular bisector of any
chord passes through the center.
332. Corollary II. To find the center of any given cir-
cle, or of any given arc of a circle, draw two non-parallel
chords and their perpendicular bisectors. The center is the
point where these bisectors intersect.
333. A Diameter is a chord through the center.
334. A diameter is equal to two radii : so all diameters are
bisected by the center of the circle, and are equal.
THE CIRCLE. II9
Theorem IV.
335. Circles of equal radii are congruent.
Hypothesis. Two circles of which C and O are the centers,
and radius CD = radius OP.
Conclusion. The circles are congruent.
Proof. Apply one circle to the other so that the center O shall
coincide with center C, and sect OF fall upon line CD. Then, because
OF = CD, the point F will coincide with the point D. Then every
particular point in the one circle must coincide with some point in the
other circle, because of the equality of radii.
(321. A point is on the circle when its sect from the center is equal to the radius.)
/. C ^ O 6>.
336. Corollary. After being applied, as above, the second
circle may be turned about its center ; and still it will coincide
with the first, though the point P no longer falls upon D.
Hence, considering one circle as the trace of the other, —
A circle can be made to slide along itself by being turned
about 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.
120 THE ELEMENTS OF GEOMETRY.
337. Circles which have the same center are called Con-
centric.
Theorem V.
338. Different concentric circles cannot have a point in com-
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. Two different circles
with a point in common are not concentric.
340. Second Contranominal of 338. Two concentric cir-
cles with a point in common coincide.
341. The 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 Sym-
metry, and the fundamental property that the circle slides
along itself when turned about its center, and so coincides
with itself after turning about the center through any angle.
The circle is the only closed curve which will slide upon its
trace.
THE CIRCLE. 121
Theorem VI.
342. The perpendicular from the center of a circle to a secant
bisects the chord ; and, if a line thivngh the center bisect a chord
not passing through the center^ it cuts, it at right angles.
Proof. For any chord, there is only one perpendicular front the
center, only one line through the mid-point and center, only one per-
pendicular bisector; and these, by 331, are identical.
(331. The perpendicular bisector of any chord passes through the center.)
343. Every diafneter 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.
Inverse. Every line which is an axis of symmetry of a
circle contains a diameter.
For, if 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.
(330. A circle has only one center.)
Each circle has therefore, besides its center of symmetry^
an infinite number of axes of symmetry.
122 THE ELEMENTS OF GEOMETRY.
Theorem VII.
344. Every chord lies wholly within the circle.
Hypothesis. Lei A and B be any two points in ABC.
Conclusion. Every point on chord AB between A and B is within
the O ABC.
Proof. Take any point, B>, in chord AB.
By 332, find O, the center of the circle.
. Join OA, 0Z>, OB.
Then OB makes a greater sect than OD from the foot of the per-
pendicular from O on the line AB,
(342. The perpendicular from center on line AB bisects chord AB.)
.-. OB > OD,
(r54. The oblique which makes the greater sect from the foot of the perpendicular
is the greater.)
/. D is within the circle.
.<320. A point is within the circle if its sect from the center is less than the radius.)
345' Corollary. If 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
center to the line, each equal to the radius ; that is, the line
will pass through two points on the circle.
Thus, again, the circle is a closed curve.
THE CIRCLE. 1 23
Theorem VIII.
346. In a circle, 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. AB and CD do not mutually bisect each other.
Proof. If one of them pass through the center, it is not bisected
by the other, which does not pass through the center.
If neither pass through the center, find the center (9, and join OF,
If F is the bisection point of one of the chords, as AB, then
4. OFB = rt. 4.,
(342. If a line through the center bisect a chord not passing through the center, it
cuts it at right angles.)
.*. "4- OFD is oblique,
.-. OF does not bisect CD.
Exercises. 69. What is the locus of mid points of parallel
chords }
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.
124
THE ELEMENTS OF GEOMETRY.
Theorem IX.
347. If from any point not the center^ sects he drawn to the
different points of a circle, the greatest is that which meets
the circle after passing through the center ; the least is part of the
same line.
Hypothesis. From any point A, sects are drawn to a Q DBHK,
whose center is C.
First Conclusion. A CB > AD.
Proof. Join CD.
Then, because CB = CD,
.: AB = AC + CB = AC -j- CD > AD.
(156. Any two sides of a triangle are greater than the third.)
Second Conclusion. AH < AK.
Proof. When A is within the circle,
BC = KC <KA + AC, by 156.
Taking away A C from both sides, .*. AH < AK.
When A is on the circle, AH is a point.
When A is without the circle,
AC = AH -{- HC <AK ■\- KC, by 156.
Taking away HC = KC, .\ AH < AK.
348. Corollary. The diameter of a circle is greater than
any other chord.
THE CIRCLE. 1 25
Theorem X.
349* If from any point three sects drawn to a circle are equal,
that point is the center.
Hypothesis. From ihe point O to Q ABC, lei OA = OB = OC.
Conclusion. O is center of O ABC.
Proof. Join AB, and BC.
O is on the perpendicular bisector of chord AB, and also on that
of chord BC,
(183. The locus of a point to which sects from two given points are equal is the
perpendicular bisector of the sect joining them.)
.-. O is center of O ABC.
(332. The center is the intersection of perpendicular bisectors of two non-parallel
chords.)
350. CoNTRANOMiNAL OF 349. From any point not the
center, there cannot be drawn more than two equal sects to a
circle.
351. Corollary I. If two circles have three points in
common, they coincide.
Because from the center of one circle three equal sects can be drawn
to points on the other circle,
/. they are concentric, and they have a point m common,
/. they coincide.
(340. Two concentric circles with a point in common coincide.)
126 THE ELEMENTS OF GEOMETRY.
352. Corollary II. Through three points not more than
one circle can pass.
353. Corollary III. Two different circles cannot meet
one another in more than two points.
354. A circle is circumscribed about a polygon when it
passes through all the vertices of the polygon.
Then the polygon is said to be inscribed in the circle.
Problem I.
355' Through any three points not in the same line^ to de-
scribe a circle.
yo
^-r"'/ \2>
^>x.
c
Given, ihree points, A, B, and C, not in t/ie same fine.
Required, to describe a circle which shall pass through all of
them.
Construction. By 188, find the point 6>, such that
OA = OB = OC,
356. Corollary. To 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. 1 27
Theorem XI.
358. Perpendiculars from the center on equal chords are
equal ; andy on unequal chords ^ that on the greater is the
lesser.
C
Hypothesis. From center O,
OH, OK, OL, J. chords AB = CD> FG,
Conclusion. OH = OK < OL.
Proof. Draw the radii OA, OB, OC, OD, OF.
By hypothesis, AB = CD,
,\ A OAB ^ A OCjD,
(129. Triangles with three sides respectively equal are congruent.)
.*. the altitude OH = corresponding altitude OK.
Again, because CZ> > FG,
.'. CK > FL,
But 'CK' + KD* = ~0C' = OF" = FL' + T&.
But CX' > 'FL\ _
.-. KO^KLCf",
/. OK < OL,
359* By 33, Rule of Inversion,
Chords having equal perpendiculars from the center are
equal ; and, of chords having unequal perpendiculars, the one
with the lesser is the greater.
128
THE ELEMENTS OF GEOMETRY.
II. Angles at the Center.
360. The explemental angles at the center of a circle, whose
arms are the same radii, are said to be subtended by^ or to stand
upon^ the explemental arcs opposite them intercepted by the
radii, the reflex angle upon the major arc.
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.
364. 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.
j<5t
THE CIRCLE. 1 29
Theorem XII.
365. In the same or equal circles ^ equal arcs are congruent^
subtend equal angles at the center^ determine equal sectors^ and
are stibtended by equal chords.
Hypothesis. Radius OA — radius CF, and arc AB = arc FG.
Conclusions. I. ^ AOB = 4. FCG.
II. Sector AOB = sector FCG.
III. Chord AB = chord FG.
Proof. Place the sector A OB over the sector FCG so that the
center O shall fall on C, and the radius OA on the line CF. Then,
because OA = CF^
.*. point A falls on point F.
Again, because the radii are equal, every point of the arc AB will
fall on some part of the circle FG.
But arc AB = arc FG,
.'. point B falls on G,
.'. chord AB = chord FG,
i.AOB ^ 4 FCG,
and sector AOB = sector FCG.
366. In the same or equal circles the stim of two minor arcs
is the arc obtained by placing them on the same circle so as
not to overlap, with one end point in common.
I30 THE ELEMENTS OF GEOMETRY,
Theorem XIII.
367. A sum of two arcs of the same or equal circles subtends
an angle at the center equal to the sum of the angles which each
arc subtends separately.
o
Proof. Placing the arcs with two end points in common at B, join
B and the other end points, A and C, to the center O,
Then the angles are in such a position, that, by definition 61,
^ AOB + -4. BOC = i. AOC.
But 4 AOC is subtended by arc AC, which is the sum of arc AB
subtending ^ AOB, and arc BC subtending ^ BOC.
368. Corollary. Two sectors of the same or equal circles
may be so placed as to form a sector whose arc is the sum of
their arcs, whose angle is the sum of their angles, and whose
surface is the sum of their surfaces.
Theorem XIV.
369. In the same or equal circles, of two uneqtial arcs, ths
greater subtends the greater angle at the center, and determines
the greater sector.
Proof. If the first arc is greater than the second, it is equal to the
second plus a third arc; and so, by 367, the angle which the first sub-
tends is greater than the angle which the second subtends by the angle
which the third arc subtends at the center.
And like is true of the sectors.
370. From 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.
372. Again, 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.
Theorem XV.
373' I''^ t^e same or equal circles y of two unequal minor arcSy
the greater is subtended by the greater chord ; of two unequal
major arcSy the greater is subtended by the lesser chord.
J9.
D
Hypothesis. Minor arc AB > arc CD,
Conclusion. Chord AB > chord C£>.
Proof. 4. AOB > ^ COD,
(369. The greater arc subtends the greater angle at the center.)
.*. in As A OB and COD, side AB > side CD.
(159. Two triangles with two sides equal, but the included angle greater in the first,
have the third side greater in the first.)
Secondly, because the minor arc with its major arc together make
up the entire circle, therefore to a greater major arc will correspond a
lesser minor arc, and therefore a lesser chord.
132
THE ELEMENTS OF GEOMETRY.
374. 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
subtends the greater minor arc and the lesser major arc.
Theorem XVI.
375. An angle at the center of a circle is double the inscribed
angle standing up07t the same arc.
Fig. I.
Fig. 2.
Hypothesis. Lei AB be any arc, O the center, C any point
on the circle not on arc AB.
Conclusion. ^ A OB = 2^ A CB.
Proof. Join CO, and produce it to Z>.
Because, being radii, OA = OC,
.-. ^ OCA = ^ OAC,
(126. The angles at the base of an isosceles triangle are equal.)
.-. ^ AOD = 4. OCA 4- 4. OAC = 2^ OCA.
(173. The exterior angle of a triangle equals the sum of the interior opposite
angles.)
Similarly, ^ DOB = 2^ OCB.
Hence (in Fig. i) the sum or (in Fig. 2) the difference of the
angles AOD, DOB, is double the sum or difference of OCA and
OCB\ that is, ^ AOB = 2^ ACB.
THE CIRCLE.
133
III. 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.
For each of the angles ACBy ADB is half of the angle
subtended at the center by the arc ARB,
378. Corollary II. If a circle is divided into two seg-
ments by a chord, any pair of angles, one in each segment, will
be supplemental.
For they are halves of the explemental angles at the center
standing on the same explemental arcs.
134
THE ELEMENTS OF GEOMETRY.
379. Corollary III. The opposite angles of every quad-
rilateral inscribed in a circle are supplemental.
For they stand on explemental arcs, and so are halves of
explemental angles at the center.
Theorem XVII.
-380. From a point on the side toward the segment ^ its chord
subtends an angle less than^ equal tOy 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 any segment, ABC; P any point
without the segment on the same side of AC as B ; Q any point
within the segment.
Conclusions. Since, by 377, we know all angles inscribed in the
segment are equal, we have only to prove
I. ^ AFC < 4. ABC.
II. 4AQC> 4 ABC,
THE CIRCLE. 1 35
Proof. I. Let R be the point where PC meets the circle, and join
RA, making A APR) then 4. APC <4.ARC,
(142. An exterior angle of a triangle is greater than either interior opposite angle.)
II. For the same reason, if ^^ is produced to meet the circumfer-
'ence at S, and SC joined, making A CSQ, 4- ^Q.^ > 4- CSA.
381. By 33, Rule of Inversion,
On 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,
or greater than, an angle in the segment.
Theorem XVIII.
382. If two opposite angles of a quadrilateral are supple-
mental, a circle passing through any three of its vertices will
contaifz the fourth.
Hypothesis. ABCD is a quadrilateral, with ^ A -\- ^ C = st. ^.
Conclusion. The four vertices lie on the circle determined by any
three of them.
Proof. By 355, pass a circle through the three points B, C, D.
Take any point, F, on the arc DFB of the segment, on the same
side of DB as A. Join FB,FD. Then^ 7^ + ^ C = st. ^.
{379. The opposite angles of an inscribed quadrilateral are supplemental.)
From hypothesis, .*. ^ A = :4 F, .*. ^ is on the arc DFB.
(381. On 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.)
136 THE ELEMENTS OF GEOMETRY.
Theorem XIX.
383. The angle in a segment is greater than^ equal to^ or less
than, a right angle, accordijtg as the arc of the segment is less
than, equal to, or greater than, a semicircle.
Hypothesis. Lei AD be a diameter of a circle whose center
is O. Take B and C points on the same circle.
Conclusions. I. ^ ACD = rt. 2^, being half the straight angle
A on.
(375. An angle at the center is double the inscribed angle on the same arc.)
II. .-. ^ADC<xt.4..
(143. Any two angles of a triangle are together less than a straight angle.)
III. .-. 4.ABC>x\..4.,
since
4.ADC -\- }^ABC = st. 4..
(379. Opposite angles of an inscribed quadrilateral are supplemental.)
384. By 33, Rule of Inversion,
A segment is less than, equal to, or greater than, a semi-
circle, according as the angle in it is greater than, equal to, or
less than, a right angle.
THE CIRCLE. 1 37
Theorem XX.
385. If two chords intersect within a circle y an angle formed
and its vertical are each equal to half the angle at the center
standing on the sum of the arcs they intercept.
Hypothesis. Let the chords AC,BD intersect at F within the
circle.
Conclusion. ^ BFC = half ^ at center standing on (arc BC +
arc DA) .
Proof. Join CB>.
4. BFC = %. BDC + 4. DC A,
(173. The exterior angle of a triangle is equal to the sum of the opposite interior
angles.)
But 2 4. BDC = ^ at center on BC, and 2 4 DCA = 4 at cen-
ter on DA ;
(375. An angle at the center is double the inscribed angle upon the same arc.)
.-. 2 4 BFC — 4a.t center on arc (BC + DA).
(367. A sum of two arcs subtends an angle at the center equal to the sum of the
angles subtended by the arcs.)
Exercises. 71. The end points of two equal chords of a
circle are the vertices of a symmetrical trapezoid.
72. Every trapezoid inscribed in a circle is symmetrical.
138 THE ELEMENTS OF GEOMETRY.
Theorem XXI.
386. An angle formed by two secants is half the angle at
center standing on the difference of the arcs they intercept.
Hypothesis. Lei two lines from F cut the circle whose center
is O, in the points A, B, C, and D.
Conclusion. ^ ^ = | 2j^ at (9 on difference between arc AB and
arc CD.
Proof. Join^C
^ CAD = 4.F^- 4.C.
Doubling both sides, :^ at 6> on arc CD =2^i^-f^at^0n
arc AB ;
.*. twice ^ F = difference of 2j^ s at O on arcs CD and AB.
IV. Tangents.
387. A line which will meet the circle in one point only
is said to be a Tangent to the circle.
388. The point at which a tangent touches the circle is
called the Point of Contact.
THE CIRCLE. 1 39
Theorem XXII.
389. Of lines passiiig through the e7id of any radius, the
perpendicular is a tangent to the circley and every other line is a
secant.
Hypothesis. A radius OP perpendicular fo PB, oblique to PC.
Conclusions. I. PB is a tangent at P.
II. PC is a secant.
Proof. (I.) The sect from O to any point on PB, except P,
is > OP,
(150. The perpendicular is the least sect between a point and a line.)
.*. every point of PB except P is outside the circle.
{321. A point is without a circle if its sect from the center is greater than the
radius.)
Proof. (II.) A sect perpendicular to PC is less than the oblique
OP,
.*. a point of PC is within the circle ;
/. PC is a secant.
390. Corollary I. One and only one tangent can be
drawn to a circle at a given point on the circle.
391. Corollary II. To draw a tangent to a circle at a
point on the circle, draw the perpendicular to the radius at the
point.
140
THE ELEMENTS OF GEOMETRY.
392. Corollary III. The radius to the point of contact
of any tangent is perpendicular to the tangent.
393* Corollary IV. The perpendicular to a tangent from
the point of tangency passes through the center of the circle.
394. Corollary V. The perpendicular drawn from the
center to the tangent passes through the point of contact.
On the Three Relative Positions of a Line and a
Circle.
395' Corollary VI. A line will be a secant, a tangent, or
not meet the circle, according as its perpendicular from the
center is less than, equal to, or greater than, the radius.
396. 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 angle formed by a tangent and a chord from the
point of contact is half the angle at the center standing on the
intercepted arc. y
Htpothesis, AB is tangeni at C, and CD is a chord of O
wUh center at O.
THE CIRCLE. 141
Conclusions. ^DCB = ^^DOC,
4.DCA = |explementZ>(9C
Proof. At C erect chord CF Jl AB.
CF is a diameter of the circle.
(393. The perpendicular to a tangent from the point of contact passes through the
center of the circle.)
Join OD. Then
rt. ^ OCA = l^st. ^at O,
also
^ OCD = i^ FOB,
(375. The angle at the center is double the inscribed angle on the same arc.)
Therefore, adding
^ £>CA = ^reflex ^ Z>OC,
therefore the supplement of ^ VGA, which is ^ DCB, is half the exple-
ment of reflex ^ DOC, which is 4. DOC,
398. Inverse. If 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.
Proof. There is but one line which will make this angle ; and we
already know, from 397, that a tangent makes it.
Exercises. 73. The chord which joins the points of con-
tact of parallel tangents to a circle is a diameter.
74. How may 397 be considered as a special case of 375 1
75. A parallelogram inscribed in a circle must be a rect-
angle.
y6. If a series of circles touch a given line at a given point,
where will their centers all lie }
'jj. The angle of two tangents is double that of the chord
of contact and the diameter through either point of contact.
142 THE ELEMENTS OF GEOMETRY.
Theorem XXIV.
399. The angle formed by a tangent and a secant is half the
angle at the center standing on the difference of the intercepted
arcs.
Hypothesis. Of iwo lines from Z>, one cuts at B and C the O
whose center is O, the other is tangent to the same O at A.
Conclusion. 2 ^ D — difference between ^ AOC and ^ AOB.
Proof. Join AB.
^ ABC = ^D + ^ DAB,
(173. The exterior angle of a triangle is equal to the two interior opposite angles.)
.-. 4. AOC = 24.D + :^ AOB,
(375. An angle at the center is double the inscribed angle on the same arc.)
and
(397. An angle formed by a tangent and chord is half the angle at the center on
the intercepted arc.)
.-. twice ^ Dis the difference between ^ AOC and ^ AOB.
400. Corollary. The angle formed by two tangents is
half the angle at the center standing on the difference of the
intercepted arcs.
THE CIRCLE. I43
Problem II.
401. From a given point without a circle to draw a tangent
to the circle.
Given, a O ly/M center O, and a point P outside it
Required, to draw through P a tangent to the circle.
Construction. Join OP. Bisect OP in C.
With center C and radius CP describe a circle cutting the given
circle in F and G.
Join PF and PG.
These are tangents to O FGB.
Proof. Join OF.
^ OFF is a rt. ^,
(383. The angle in a semicircle is a right angle.)
.-. PF is tangent at i^ to O BFG.
(389. A line perpendicular to a radius at its extremity is a tangent to the circle.)
402. Corollary. Two 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.
144
THE ELEMENTS OF GEOMETRY.
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 common chord must
pass through the centers of the two circles.
(331. The perpendicular bisector of any chord passes through the center.)
404. Corollary. 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 circles have no second point in common.
405. Two circles 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.
406. By 343,
Two circles^ not concentric^ have always one, and only one^
common axis of symmetry ; namely, their line of centers.
For this is the only line which contains a diameter of
each.
THE CIRCLE. 1 45
Theorem XXVI.
Obverse of 404.
407* Jf t'^0 circles have one common point not on the line
through their centers^ they have also another common point.
Hypothesis. O with center O and O with center C, having a
common point B not on OC.
Conclusion. They have another common point.
Proof. Join OC, and from B drop a line perpendicular to OC 2X
D, and prolong it, making DF = BD.
F is the second common point.
For A ODB ^ A OFD, and A CBD ^ A CFD-,
(124. Triangles having two sides and the included angle equal in each are con-
gruent.)
.-. OF = OB, and CF =^ CB -,
.'. F is on both Os.
{321. A point is on the circle if its distance from the center is equal to the radius.)
408. CONTRANOMINAL OF 40/. If tWO circIcS tOUCh OHC
another, the line through their centers passes through the point
of contact.
409. Corollary. Two circles which touch one another
have a common tangent at their point of contact ; namely, the
perpendicular through that point to the line joining their
centers.
146
THE ELEMENTS OF GEOMETRY,
410. Calling the sect joining the centers of two circles their
center-sect c, and calling their radii r^ and rj, we have, in
regard to the relative positions of two circles, —
I. If ^>ri + ^2, therefore the Os are wholly exterior.
2. If ^ = /'j + r^y therefore the Os touch externally.
3. \i c <r^-\- ^2, but c > the difference of radii, therefore
the Os cut each other.
THE CIRCLE. 1 47
4. If ^ = the difference of radii, therefore the Os touch
internally.
5. If ^ < the difference of radii, therefore one is wholly
interior to the other.
411. By 33, Rule of Inversion, the five inverses to the
above are true.
Exercises. 78. How must a line through one of the com-
mon points of two 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
sect.
80. If any two lines be drawn through the point of contact
of two circles, the lines joining their second intersections with
each circle will be parallel.
148
THE ELEMENTS OF GEOMETRY,
VI. Problems.
Problem III.
412. To bisect a given arc.
Given, the arc BD.
Required, to bisect it.
Construction. Join BD, and bisect the sect BD
m F; 2X F erect a perpendicular cutting the arc in C.
C is the mid point of the arc.
Proof. Join BC, CD.
aBCF^ aDFC,
(124. Triangles having two sides and the included angle in each equal are con-
gruent.)
.-, chord BC = chord CD,
a.TcBC = arc CD.
(374. In the same circle, equal chords subtend equal minor arcs.)
413. A polygon is said to be circumscribed about a circle
when all its sides are tangents to the circle.
The circle is then said to be inscribed in the polygon.
414. A circle which touches one side of a triangle and the
other two sides produced is called an Escribed Circle.
THE CIRCLE.
149
Problem IV.
415. To describe a circle touching three given lines which are
not all parallel^ and do not all pass through the same point.
Given, three lines interseciing in A^ B, and C,
Required, to describe a circle touching them.
Construction. Draw the bisectors of the angles at A and C.
These four bisectors will intersect in four points, (9, (9i, O^, Oy
A 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. By 186, every point in the bisector of an angle is equally
distant from its arms ;
Therefore, since O is on the bisector of ^ A, the perpendicular
from O on AB equals the perpendicular from O on AC, which also
equals the perpendicular from O on CB, since O is also on the bisector
of ^ a
150 THE ELEMENTS OF GEOMETRY.
416. The four tangents common to two circles occur in two
pairs intersecting on the common axis of symmetry.
Problem V.
417. In a given circle to inscribe a triangle equiangular to a
given triangle.
Given, a and A ABC.
Required, to describe in the (d a t^ equiangular to A ABC.
Construction. Draw a tangent GH touching the circle at the
point D.
Make 4. HDK = 2^: C, and 4. GDL = ^ A.
K and L being on the circumference, join KL.
DKL is the required triangle.
Proof. ^ KDL = ^ B.
(174. The three angles of a triangle are equal to a straight angle.)
4K = 7^A, and ^ L = }^ C.
(375' An inscribed angle is half the angle at the center on the intercepted arc.)
(397. An angle formed by a tangent and a chord is half the angle at the center
on the intercepted arc.)
BOOK IV.
REGULAR POLYGONS.
I. Partition of a Perigon.
Problem I.
418. To bisect a perigon.
Solution. To bisect the perigon at the point (9, 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 peri-
gon into 8, 16, 32, 64, etc., equal parts.
152 THE ELEMENTS OF GEOMETRY.
Problem II.
420. To trisect a perigon.
Solution. To trisect the perigon at the point O, to O draw any
line BO; on BO produced take a sect 0C\ on OC construct, by 132,
an equilateral triangle CDO.
^ DOB is one-third of a perigon.
For ^ DOC i?, one-third of a st. :^,
(174. The three angles of a triangle are equal to a straight angle.)
.*. ^ DOB is two-thirds of a st. 2(^.
421. Corollary. Since we can bisect any angle, so we
may cut the perigon into 6, 12, 24, 48, etc., equal parts.
422. Remark. To trisect a7ty given angle is a problem
beyond the power of strict 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.
REGULAR POLYGONS: 1 53
Problem III.
423. To cut a perigon mto five equal parts.
Solution. By 314, describe an isosceles triangle -^4-5C having
^A = 4.C^24.B,
Then ^ ^ is two-fifths of a st. ^ .
(174. The three angles of any triangle are equal to a straight angle.)
Therefore, to get a fifth of a perigon at a point O, construct, by 164,
4. GOB = ^A.
424. Corollary. Since we can bisect any angle, we may-
cut a perigon into 10, 20, 40, 80, etc., equal parts.
Problem IV.
425. To cut a perigon into fifteen equal parts,
'J3
Solution. At the perigon point (9, by 420, construct the ^ AOC
= one-third of a perigon.
By 423, make 2^ AOB = one-fifth of a perigon.
Then of such parts, as a perigon contains fifteen, -i^ AOC contains
five, and ^ AOB contains three, therefore ^ BOC contains two.
So bisecting ^ BOC giwts one-fifteenth of a perigon.
154
THE ELEMENTS OF GEOMETRY,
426. Corollary. Hence a perigon may be divided into 30,
60, 120, etc., equal parts.
II. Regular Polygons and Circles.
Problem V.
427. To mscribe i?t a circle a regular polygon having a given
number of sides. c
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 the circle in Ay
By 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 OA coincides
with the trace of OB, therefore, because the angles are all equal, OB
will coincide with the trace of OC, and 6>C with the trace of OD, etc. ;
then AB will coincide with the trace of BC, and BC with the trace of
CD, etc. ;
.-. AB = BC == CD = etc.,
.•4 the polygon is equilateral.
Moreover, since then ABC will coincide with the trace of BDCy
.-. t ^BC = 4. BCD = etc.,
.*. the polygon is equiangular.
Therefore ABCD, etc., is a regular polygon, and it is inscribed in
the given circle.
REGULAR POLYGONS. 1 55
428. Remark. From the time of Euclid, about 300 B.C.,
no advance was made in the inscription of regular polygons
until Gauss, in 1796, 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
or more different prime numbers of that form, or else the
product of a power of 2 by one or more different prime num-
bers of that form.
In other words, it is necessary that n should contain no odd
divisor not of the form 2"' + i, nor contain the same divisor of
that form more than once.
Below 300, the following 38 are the only possible values of
/^: 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, 24, 30, 32, 34, 40, 48,
51, 60, 64, 6Z, 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 diame-
ter.
^2, Prove that each diagonal is parallel to a side of the
regular pentagon.
83. 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
equivalent to one-sixth of a regular hexagon described on the
same sect.
85. If a triangle is equilateral, show that the radius of
the circumscribed circle is double that of the inscribed ; and the
radius of an escribed, triple.
%6. The end points of a sect slide on two lines at right
angles : find the locus of its mid-point.
156
THE ELEMENTS OF GEOMETRY.
Problem VI.
429. To circumscribe about a given circle a regular polygon
having a given number of sides.
JW C X'
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 angles, and extend their arms to meet the circle in A^
Bj C, D, etc. Draw perpendiculars to these arms at A, B^ C, D, etc.
These will be tangents.
Call their points of intersection K, L, M, etc.
Then shall KLM, etc., be a regular polygon.
For if the figure be turned about its center O until OA coincides
with the trace of OB, then, because the angles are all equal, OB will
coincide with the trace of OC, and 6>Cwith the trace of OD, 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 ;
also
.-. KL = LM = etc.,
^K = :^L = :^M = etc. ;
therefore the polygon is regular, and it is circumscribed about the given
circle.
REGULAR POLYGONS. 1 5/
430. Corollary. Hence we can circumscribe about a cir-
cle regular polygons of 3, 4, 5, 6, 8, 10, 12, 15, 16, 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.
Construction. Bisect -4. EAB and 4- ABC by lines intersecting in
O. With center O and radius OA describe a circle.
This shall be the required circle.
Proof. Join OC. Then
A OBC'^ A OBAy
(124. Triangles having two sides and the included angle in each equal are con-
gruent.)
.*. ^ OCB = 4 OAB = half one ^ of the regular polygon,
.-. 4 B CI? is bisected.
Similarly prove each ^ of the polygon bisected,
.-. OA = OB = oc = on = OE,
(148. In a triangle, sides opposite equal angles are equal.)
therefore a circle with radius OA passes through B^ C, D, Ey and is
circumscribed about the given polygon.
158 THE ELEMENTS OF GEOMETRY,
Problem VIII.
432. To inscribe a circle in a given regular polygon.
T
c
Given, a regular polygon, ABODE,
Required, to inscribe a circle in it.
Construction. Bisect 4. EAB and ^ ABC by lines intersecting in
O. From O drop OP perpendicular to AB.
The O with center O and radius OP shall be the O required.
Proof. Join 00, OD, OE, and draw 0Q1.B0, OR A. CD, OS
± DEy and OT J. EA, Then
A OBC^ A OBA,
(124. Triangles having two sides and the included angle in each equal are con-
gruent.)
.*. ^ OCB = 2^ OAB = half one ^ of the regular polygon,
.•; ^ BCD is bisected.
Similarly prove each ^ of the polygon bisected.
Again,
A OBP ^ A OBQ,
(176. Triangles having two angles and a corresponding side in each equal are con-
gruent.)
/. OP = OQ,
REGULAR POLYGONS. 1 59
Similarly,
OQ =^ OR = OS = OT,
therefore a circle with radius OF will touch AB, BC, CD, DE, EA,
at points P, Q, R, S, T;
.'. it is inscribed in the given polygon.
433* Corollary I. The inscribed and circumscribed cir-
cles of a regular polygon are concentric.
434. Corollary II. The bisectors of the angles of a reg-
ular polygon all meet in a point which is the center both of the
circumscribed and inscribed circles, and is called the center
of the regular polygon.
435' Corollary III. The 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
is called its apothem.
437. The side of a regular hexagon inscribed in a circle is
equal to the radius. For the sects from the center to the ends
of a side make an isosceles triangle, one of whose angles is
one-third a straight angle ; therefore it is equilateral.
III. Least Perimeter in Equivalent Figures. — Greatest
Surface in Isoperimetric Figures.
438. Any two figures are called Isoperimetric when their
perimeters are equal.
l60 THE ELEMENTS OF GEOMETRY.
Theorem I.
439' Of all equivalent triangles having the same base^ that
which is isosceles has the least perimeter.
Hypothesis. Lei ABC be an isosceles iriangle, and ABC any
equivalent iriangle having ihe same base.
Conclusion. AB + AC<A'B + A'C,
Proof. A A' \\ BC.
(253. Inverse. Equivalent triangles on the same base, and on the same side of it,
are between the same parallels.)
Draw CND ± AA\ meeting BA produced in D. Join A'D,
4.NAC = ^ACB,
(168. If a transversal cuts two parallels, the alternate angles are equal.)
•4.ACB = ^ABC,
(126. In an isosceles triangle the angles opposite the equal sides are equal.)
^ ABC = 4. DAN,
(169. If a transversal cuts two parallels, the corresponding angles are equal.)
.-. i^ACN^ ^ADN,
(128. Triangles having two angles and the included side equal in each are con-
gruent.)
/. AN is the perpendicular bisector of CD,
.-. AD = AC, and A'D = A' C.
(183. The locus of the point to which sects from two given points are equal, is the
perpendicular bisector of the sect joining them.)
REGULAR POLYGONS. l6l
But
BD < A'B + A'D,
(156. Any two sides of a triangle are together greater than the third.)
.-. AB ^ AC<A'B ■\- AC,
440. Corollary. Of all equivalent triangles, that which
is equilateral has the least perimeter.
For the triangle having the least perimeter enclosing a
given surface must be isosceles whichever side is taken as the
base.
Theorem II.
441. Of all isoperimetric triangles having the same hase^
that which is isosceles has the greatest surface.
MAS
Hypothesis. Lei ABC be an isosceles friangle ; and lei A'BC
standing on ihe same base BC, have an equal perimeier; thai is,
A'B + A'C = AB + AC.
Conclusion, a ABC > a A'BC.
Proof. The vertex A' must fall between BC and the parallel AN\
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
greater.
Therefore the altitude of A ABC is greater than the altitude of
A A'BC, and hence also its surface.
442. Corollary. Of all isoperimetric triangles, that which
is equilateral is the greatest.
For the greatest triangle having a given perimeter must be
isosceles whichever side is taken as the base.
1 62
THE ELEMENTS OF GEOMETRY.
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
greatest surface.
Hypothesis. Lei ABC, A'BCy be two triangles having the sides
AB, BC, respectively equal to A'B, BC\ and let ^ ABC be right
Conclusion, a ABC > A A'BC.
Proof. Taking BC sls the common base, the altitude AB of
A ABC is greater than the altitude A'B> of A A'BC.
Theorem 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, among all the figures of the same perimeter, there must
be one greatest figure, or several equivalent greatest figures of different
forms.
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, joining two of the points in its perimeter, lying without
the figure ; then if the re-entrant portion A CB be revolved about the
line AB into the position AC'B, the figure AC'BNA has the same
perimeter as the first figure, but a greater surface.
Now let AFBCA be a figure of greatest surface formed with a
given perimeter ; then, taking any point A in its perimeter, and drawing
AB to bisect the perimeter, it also bisects the surface. For if the sur-
face of one of the parts, as AFB, were greater than that of the other
part, A CB, then if the part AFB were revolved upon the line AB into
the position AF'B, the surface of the figure AF'BFA would be greater
than that of the figure AFBCA, and yet would have the same perim-
eter.
Now the angles AFB and AF'B must be right angles, else the tri-
angles AFB and AF'B could be increased, by 443, without varying
the chords AF,FB, AF',F'B, and then (the segments AGF, FEB,
AG'F' , F'E'B, still standing on these chords) the whole figure would
have increased without changing its perimeter.
But F is any point in the curve AFB \ therefore this curve is a
semicircle.
(384. The arc of a segment is a semicircle if the angle in it is right.)
Therefore the whole figure is a circle.
164
THE ELEMENTS OF GEOMETRY.
Theorem V.
445* Of all equivalent plane figures^ the circle has the least
peri7neter.
Hypothesis. Lei C be a circle, and A any other figure having
the same surface as C.
Conclusion. The perimeter of C is less than that of A.
Proof. Suppose B a circle with the same perimeter as the figure
A ; then, by 444, A <B :. C <B.
But, of two circles, that which has the less surface has the less
perimeter ;
.-. perimeter of C < perimeter of B, or of A.
Theorem VI.
446. Of all the polygons constructed with the same given
sides y that is the greatest which can be inscribed in a circle.
Hypothesis. Lei P be a polygon consirucied wiih ihe sides
a, b, c, d, e, and inscribed in a circle S, and lei P' be any oiher
REGULAR POLYGONS. 1 65
polygon constructed with the same sides, and not inscriptible in a
circle.
Conclusion. F > P'.
Proof. Upon the sides a^ b, c, etc., of the polygon P' construct
circular segments equal to those standing on the corresponding sides of
P. The whole figure S' thus formed has the same perimeter as the
circle Sy 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 sami nttmber
of sides y the regular polygon is the greatest.
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
surface by substituting for the triangle AB'C the isoperimetric isosceles
triangle ABC.
II. The greatest polygon constructed with the same number of
equal sides must, by 446, be inscriptible in a circle. Therefore it is a
regular polygon.
Exercises, ^y. Of all triangles that can be inscribed in a
given triangle, that v^hose vertices are the feet of the altitudes
of the original triangle has the least perimeter.
i66
THE ELEMENTS OF GEOMETRY.
Theorem VIII.
448. Of all eqiiivalefit polygons having the same number of
sides ^ the regular polyg07i has the least perimeter.
Hypothesis. Let P be a regular polygon, and M any equiva-
lent irregular polygon having the same number of sides as P.
Conclusion. The perimeter of P is less than that of M.
Proof. Let A" be a regular polygon having the same perimeter
and the same number of sides as M \ then, by 447,
M <N, or P <N.
But, of two regular polygons having the same number of sides, that
which has the less surface has the less perimeter ; therefore the perim-
eter of P is less than that of N or of M.
Theorem IX.
449. If a regular polygon be constructed with a given perim-
eter, its surface will be the greater^ the greater the number of its
sides.
REGULAR POLYGONS.
167
Proof. Let P be the regular polygon of three sides, and Q the
regular polygon of four sides, constructed with the same given perim-
eter.
In any side AB of P take any arbitrary point D \ the polygon P
may be regarded as an irregular polygon of four sides, in which the
sides AD J DB, make a straight angle with each other; then, by 447,
the irregular polygon P of four sides is less than the regular isoperi-
metric polygon Q of four sides.
In the same manner it follows that Q is less than the regular isoper-
imetric polygon of five sides, and so on.
Theorem X.
450. Of equivalent regular polygons^ the perimeter will be
the less J the greater the number of sides.
p
R
Hypothesis. Let P and Q be equivalent regular polygons, and
let Q have the greater number of sides.
Conclusion. The perimeter of P will be greater than that of Q.
Proof. Let i? be a regular polygon having the same perimeter as
Q and the same number of sides as P; then, by 449,
Q>P, or P>P;
therefore the perimeter of P is greater than that of P or of Q,
BOOK V.
RATIO AND PROPORTION.
Multiples.
451. Notation. In Book V., capital letters denote magni-
tudes.
Magnitudes which are or may be of different kinds are
denoted by letters taken from different alphabets.
.^ + ^ is abbreviated into 2A.
A + A + A = sA.
The small Italic letters w, «, /, ^, denote whole numbers.
mA means A taken m times ;
nA means A taken n times j
.*. mA + nA = {m + n)A.
nA taken m times is mnA.
452. A greater magnitude is said to be a Multiple of a lesser
magnitude when the greater is 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
I/O THE ELEMENTS OF GEOMETRY.
453. A lesser magnitude is a Submultiple^ or Aliquot Party
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 of, or
exactly contains, a third magnitude, they are said to be Com-
mensurable.
455. If there is no magnitude which each of two given mag-
nitudes will contain an exact number of times, they are called
Incommensurable.
456. Remark. It is important the student should know,
that of two magnitudes of the same kind taken at hazard, or
one being given, and the other deduced by a geometrical con-
struction, 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 I.
457. To find the greatest common submultiple or greatest
common divisor of two given magnitudeSy if any exists.
I 1 1 ^ 1 1 1 ^ J£JJ
Let AB and CD be the two magnitudes.
From AB, the greater, cut off as many parts as possible, each equal
to CZ>y the less. If there be a remainder PB, set it off in like manner
as often as possible upon CZ>. Should there be a second remainder
HD, set it off in like manner upon the first remainder, and so on.
RATIO AND proportion: 171
The process will terminate only if a remainder 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 common divisor.
For suppose HD the last remainder.
Then HD is an aliquot part of FB, and so of CH, and therefore
of 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.
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 Gy 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. Sim-
ilarly, CD and the first remainder FB being divisible by G,
the second remainder HD must be so, and in the same way the
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.
459. CoNTRANOMiNAL OF 457. The above process applied
to incommensurable magnitudes is interminable.
460. Obverse of 457. If the above process be intermin-
able, the magnitudes are incommensurable.
On this depends the demonstration, given in 461, of a
remarkable theorem proved in the tenth book of Euclid's
" Elements."
1/2
THE ELEMENTS OF GEOMETRY.
Theorem I.
461. The side and diagonal of a square are incommensurable.
F ,
Hypothesis. Lei ABCD be a square; AB, a side; AC, a
diagonal.
Conclusion. Then will AB and AChQ incommensurable.
Proof. AC> AB, but < 2AB,
Therefore a first remainder ^C is obtained by setting off on AC di
part AE = AB.
Erect EF perpendicular to ^C and meeting CD in F. Join AF.
A ADF ^ A AEF,
(179. Right triangles are congruent when the hypothenuse and one side are equal
respectively in each.)
.\ DF = FE.
Again, rt. A CEF is isosceles, because one of the complemental
angles, ECF, is half a rt. ^,
.;. CE = EF = FD.
Hence a common divisor oi EC and DC would be also a common
divisor oi EC and EC.
But EC and EC are again the side and diagonal of a square, there-
fore the process is interminable.
RATIO AND PROPORTION. 1 73
462. Another demonstration that the side and diagonal of
any sqiiare are incommensurable.
If you suppose them commensurable, let P 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 VI represent the number of times P is contained in the side 6",
and n the number of times P is contained in the diagonal D ; so that
S = mP, and D = nP, where m and n are whole numbers.
Then, by 243, the square on D = 2 sq. on S, and
therefore n^ is an even number,
therefore n is an even number, since the square of an odd number is
odd j therefore n may be represented by 2^,
.*. 2^^ = m^,
therefore 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 z, and so on forever.
This is impossible ; for even if m and n were high powers of 2, they
would by constant division reduce to i, an odd number.
Scales of Multiples.
463. By taking magnitudes each equal to A, one, and two,
three, four, etc., of them together we obtain a set of magni-
tudes depending upon A, and all known when A is known;'
namely. A, 2^4, $A, 4A, sA, 6A, . . . and so on, each being
obtained by putting A to the preceding one.
This we shall call the scale of multiples of ^.
464. If in be a whole number, 7nA and mBare called equi-
multiples of A and P, or the same multiples of ^and B.
1/4 THE ELEMENTS OF GEOMETRY.
465. We assume: As A is greater than, equal to, or less
than By so is mA greater than, equal to, or less than mB
respectively.
466. By 33, Rule of Inversion,
As mA is greater than, equal to, or less than mB^ so is A
greater than, equal to, or less than B respectively.
Theorem II.
467. Commensurable magnitudes have also a common mul-
tiple.
If A and B are commensurable magnitudes, there is some multiple
of A which is also a multiple of B.
Proof. Let C be a common divisor of A and B.
The scale of multiples of C is
C, 2C, 3C ....
Now, by hypothesis, one of the multiples of this scale, suppose pC,
is equal to A, and one, suppose qC, is equal to B.
Hence, by 465, the multiple pqC is equal to qAy and the same
multiple is equal to pB \ therefore,
qA = pB.
468. Inverse of 467. Magnitudes which have a common
multiple are commensurable.
Proof. If pA = qB, then — will go p times into B, and q times
into A.
469. Any whole number or fraction is commensurable with
every whole number and fraction, being each divisible by unity
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 frac-
tions.
RATIO AND PROPORTION. 1 75
470. Multiplication by a whole number or fraction is dis-
tributive,
m{A ■\- B -\- . . , ) =^ mA -^ vtB -^ , . .
471. Multiplication being commutative for whole numbers
or fractions,
.*. m{nA) = mnA — nmA = n{mA).
472. Magnitudes which are of the same kind can, being
multiplied, exceed each the other.
Scale of Relation.
473. The Scale of Relation of two magnitudes of the same
kind is a list 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 its diagonal
D, their scale of relation will commence thus, —
S, D, 2S, 2D, zS, 46", 3A 5^,4A 65, 76*, 5 A ZS, 6D, gS, jD,
loS, iiS, SZ>, 12S, gD, 135, 14S, loDj 156", iiX>, 166*, 12I?, 176",
186", i^D, igS, i^D, 20S, 21S, !$£>, etc.
141^, looD, 142S, 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
1/6 THE ELEMENTS OF GEOMETRY.
kind, however small A may be, we may, by continuing the scale
of multiples of A sufficiently far, at length obtain a multiple
of A greater than B.
So we are justified in saying,
I. We can always take mA greater than B or pB.
II. We can always take nA such that it is greater than/^,
but not greater than qB, provided that A is less than B, and /
than q.
476. The scale of relation of two magnitudes will be changed
if one is altered in size ever so little ; for some multiple of the
altered magnitude can be found which will exceed, or fall short
of, the same multiple of it before alteration, by more than the
other original magnitude, and consequently the interdistribu-
tion of the multiples of the two original magnitudes will differ
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 is
known, and also one of the original magnitudes, the other is of
fixed size, even though we may not yet be in a condition to
find it.
477' The order in which the multiples of A lie among the
multiples of B, when all 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 Antecedent, 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, triple, quadruple, quintuple, sextuple, are spe-
cial cases of a more subtile relation which exists between
every two magnitudes of the same kind.
RATIO AND proportion: 1 7/
Ratio is the relation of one magnitude to another with
respect to quantuplicity.
479. If the multiples of A and B interlie in the same order
as the multiples of C and D, then the ratio ^ to ^ is the same
as the ratio C to Dy and the four magnitudes are said to be
Proportional^ or to form a Proportion.
This is written A : B \: C : D, and read " ^ to ^ is the
same as ^'to />."
The second pair of magnitudes may be of a different kind
from the first pair.
A and D are called the Extrejues, B and C the Means ^ and
D 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
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.
482. Thus, the ratio oi A \.o B is the same as the ratio of C
to D when mA is greater than, equal to, or less than nBy
according as mC is greater than, equal to, or less than nBy
whatever whole numbers m and n may be.
483. Three magnitudes (Ay 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 A and
By and B the Mean Proportional between A and C.
178 THE ELEMENTS OF GEOMETRY.
Theorem III.
484. Ratios which are the same as the same ratio are the
same as 07te ajiother.
Hypothesis. A \ B w C \ D^'axA A \ B w X -. V.
Conclusion. C \ D w X \ V.
Proof. By the inverse of 479, the multiples of C and D have the
same interorder as those of A and B ; and, the same being true of
the multiples of X and V, therefore the multiples of X and V have the
same interorder as those of C and Z).
485. Remark. Thus ratios come under the statement,
"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 the inverse of a complete definition is true,
therefore if a pair of numbers / and ^ can be found such that
pA is greater than, equal to, or less than gB when pC is not
respectively greater than, equal to, or less than gDy then the
ratio oi A to B is unequal to the ratio of C to D.
487. The first is to be to the second in a greater (or less)
ratio than the third to the fourth when a multiple of the first
takes a more (or less) advanced position among the multiples
of the second than the same multiple 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
is greater than nB, while 7nC is not greater than nD ; or such
that mA is equal to nB, while mCis less than nD.
Theorem IV.
489. Equal magnitudes have the same ratio to the same
magnitude^ and the same has the same ratio to equal magni-
tudes.
RATIO AND proportion: 1 79
For if A = B,
.*. mA = mB.
So if mA > nC,
.*. mB > nC;
and if equal, equal ; if less, less ;
/. A : C :: B : a
And also if ^T > mA,
.'. pT> mB;
and if equal, equal ; if less, less ;
/. T : A :: T ', B,
Theorem V.
490. Of two uneqtial magnitudes, the greater has a greater
ratio to any other magnitude than the less has ; and the same
magnitude has a greater ratio to the less, of two other magni-
tudes, tlian it has to the greater,
\i A > B,m can be found such that mB is less than mA by a
greater magnitude than C.
Hence, if mA — nC, or if mA is between wCand {n -f- 1)6', mB
will be less than nC \ therefore, by 488,
A '. C > B : C,
Also, since nC > mB, while « C is not > mA,
/. C : B > C : A,
491. From 489 and 490, by 33, Rule of Inversion,
If ^ : C :: ^ : C, or if T : A :: T : B,
:. A = B.
li A :C>B:C, or a T :A<T :B,
.-. A > B.
If A : C<B :C,oriiT:A>T:B,
/. A < B.
l80 THE ELEMENTS OF GEOMETRY,
Theorem VI.
492. If any number of magnitudes he proportionals y as one
of the antecedents is to its consequent^ so will all the antecedents
taken together be to all the consequents.
Let
A ', B ',', C ', D w E '. F,
then
A \ B '.'. A -ir C ^ E ', B -\- D -ir F,
For as mA >, =, or < nB^ so, by inverse of 482, is »?C >, =, or
< nD \ and so also is mE >, =, or < nF;
.', so also is mA + mC -{• mE >, =, or < nB + nD + nF,
and therefore so is m{A -\- C + E) >, —, or < n{B + D ■}- F);
whence
A : B :: A + C + E : B -\- Z> -i- F.
Theorem VII.
493- The ratio of the equimultiples of two magnitudes is the
same as the ratio of the magnitudes themselves,
mA \ mB w A \ B,
For as pA >, =, or < qB, so is mpA >, =, or < mqB ',
but mpA = pmA, and mqB = qmBy
.*. as pA >, =, or < qB,
so is pmA >, =, or < qmB, whatever be the values of p and q;
.-. A : B :: mA : mB.
RATIO AND proportion: i8i
Theorem VIII.
494. If four magnitudes be proportionals y then^ if the first be
greater than the third, the second will be greater than the fourth;
and if equal, equal ; and if less, less.
Given, A \ B \\ C \ D.
\i A > C,
/. A '. B > C : B;
therefore, from our hypothesis,
C I r> > C '. B;
therefore, by 491,
If ^ = C,
If ^ < C,
B > D.
,'. A : B :i C : B,
,\ C : V :: C : B,
.-. B =^n,
/. A : B < C : By
/. C '. D < C ; B,
/. B < D,
Theorem IX.
495' If four magnitudes of the same kind be proportionals,
they will also be proportionals when taken alternately.
Let A \ B w C : /?, the four magnitudes being of the same kind,
then alternately,
A '. C :. B \ D.
For, by 493,
mA\ mB \\ A \ B '.: C . D \\ nC ', nD,
.*. mA : mB \\ nC \ nD \
therefore, by 494, mA >, =, or < «C, as mB >, =, or < nD', and, this
being true for all values of m and «,
.-. A \ C ','. B \ D,
1 82 THE ELEMENTS OF GEOMETRY,
Theorem X.
496. 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
consequent.
li A : B :: C '. D,
then
A ^ B '. B ','. C + D : D,
Proof. Whatever multiple of ^ + ^ we choose to examine, take
the same multiple of A, say 1 7^, and let it lie between some two mul-
tiples of Bf say 23^ and 24^ ; then, by hypothesis, 1 7 C lies between
23Z) and 24D.
Add I "jB to all the first, and i "jD to all the second ; then 1 7 (^ +
B) lies between 40^ and 41^, and i7(C + D) between 40Z) and
412? ; and in the same manner for any other multiples.
BOOK VI.
RATIO APPLIED.
I. Fundamental Geometric Proportions.
Theorem I.
497* J^f i'^o ^i^^s ^^^ ^^^ by three parallel lines ^ the intercepts
on the one are to 07ie aiiother in the same ratio as the correspond-
ing intercepts on the other.
Hypothesis. Lei the three parallels AA', BB\ CC, cut two
other lines in A^ B, C, and A', B\ C , respectively.
183
1 84
THE ELEMENTS OF GEOMETRY.
Conclusion. AB : BC :: A'B' : B' C ,
Proof. On the line ABC, by laying off m sects = AB, take BM
= mAB, and, in the same way, BN = n.BC, taking J/ and N on the
same side of B. From J/ and N draw lines 1| AA', cutting A'B'C
in J/' and iV^'.
.-. B'M' =^ m,A'B\ and B'N'^n.B'C.
(227. If three or more parallels intercept equal sects on one transversal, they
intercept equal sects on every transversal. )
But whatever be the numbers m and n, as BM (or m . AB) is >,
= , or <BN {orn.BC),
so is B'M' (or w . A'B') respectively >, =, or < B'N" (or n . ^'C') ;
.-. AB : BC :: A'B' : ^'6".
498. Remark. Observe that the reasoning holds good,
whether B is between A and C, or beyond A^ or beyond C,
499. Corollary I. If the points A and ^4' coincide, the
figure ACC will be a triangle ; therefore a line parallel to one
side of a triangle divides the other two sides proportionally.
500. Corollary II. If two lines are cut by four parallel
lines, the intercepts on the one are to one another in the same
ratio as the corresponding intercepts on the other.
501. If a sect AB is produced, and the line cut at a point P
outside the sect AB, the sect AB is said to be divided exter-
nally at jP, and AP and BP are called External Segments of
AB,
RATIO APPLIED. 1 85
In distinction, if the point P is on the sect AB^ it is said
to be divided ifiternally.
Theorem II.
502. A given sect can be divided internally into two seg-
ments having the same ratio as any two given sectSy and also
externally unless the ratio be 07te of equality ; aftdj in each case,
there is only o?te such point of division.
I jr G
:b
-N M
\
^^4
\\
w
\
%
^. 1
\
\
\
\
\
\l
\
JC^
Given, ihe sect AB.
On a line from A making any angle with AB, take A C and CD
equal to the two given sects. Join BD.
Draw CF \\ DB and meeting AB in F. By 497, AB is divided
internally at F in the given ratio.
If it could be divided internally at G in the same ratio, BH being
drawn || CG to meet AD in H,
A G would be to GB as ^ C to CH, and therefore not as ^ C to CD.
(476. The scale of relation of two magnitudes will be changed if one is altered in
size ever so little.)
Hence F is the only point which divides AB internally in the given
ratio.
i86
THE ELEMENTS OF GEOMETRY,
Again, if CD be taken so that A and D are on the same side of C,
the like construction will determine the external point of division.
In this case the construction will fail if CD = AC, for D would
coincide with A.
As above, we may also prove that there can be only one point of
external division in the given ratio.
C\
503. Inverse of 499. A line which divides two sides of a
triangle proportionally is parallel to the third.
For a parallel from one of the points would divide the
second side in the same ratio, but there is only one point of
division of a given sect in a given ratio.
Theorem III.
504. Rectangles of equal altitude are^ to one another in the
same ratio as their bases.
: 1
B M
j)r
Let AC, BC, be two rectangles having the common side OC, and
their bases OA, OB, on the same side of OC.
In the line OAB take OM = m .OA, and ON = n.OB, and
complete the rectangles MC and NC.
RATIO APPLIED. 1 87
Then MC = m.AC, and JVC = n.BC; and as OM is >, =, or
< ON, so is MC respectively >, =, or < NC ;
.*. rectangle ^C : rectangle ^C :: base OA : base OB.
505. Corollary. Parallelograms or triangles of equal alti-
tude are to one another as their bases.
Theorem IV.
506. In the same circle, or in equal circles, angles at the cen-
ter and sectors are to one another as the arcs on which they stand.
Let O and C be the centers of two equal circles ; AB, KL, any
two arcs in them.
Take an arc AM = m . AB ;
then the angle or the sector between OA and OM equals m . A OB.
(365. In equal circles, equal arcs subtend equal angles at the center.)
Also take an arc KN = n . KL ; then the angle or sector between
CK and CN equals n . KCL.
But as AOM>, =, or < KCN, so respectively is arc AM >, =, or
< arc 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 :: aicAB : aic KL.
1 88 THE ELEMENTS OF GEOMETRY,
II. Similar Figures.
507. Similar figures are those of which the angles taken in
the same order are equals and the sides between the equal angles
proportional
The figure ABCD is similar to A'B'C'jy \i 4. A ^ 4. A\ 4. B ^
4. B\ etc., and also
AB : A'B' :: BC : B' C x\ CD \ CU : : etc.
Theorem V.
508. Mutually equiangular triangles are similar.
Hypothesis, t^s ABC, FGH, having ^.s at A, B, and C, = ^.s
at F, G, and H.
Conclusion. AB \ FG w BC \ GH .. CA -. HF,
Proof. Apply A FGH to A ABC so that the point G coincides
with B, and GH falls on BC
Then, because ^ 6^ = ^ ^,
.-. GF faUs on BA.
RATIO APPLIED. 1 89
Now, since 2< BF'H' = ^ A by hypothesis,
.-. J^'H' II AC,
(166. If corresponding angles are equal, the lines are parallel.)
therefore, by 498, AB : BF' -.: CB : BH\
In the same way, by applying the A FGH so that the ^ s at ZT and
C coincide, we may prove that BC \ GH : : CA : HF,
509. Corollary. A triangle is similar to any triangle cut
off by a line parallel to one of its sides.
Theorem VI.
510. Triangles having their sides taken in order proportional
are similar.
T J£
Hypothesis. AB : FG :: BC -. GH -,: CA : HF,
Conclusion. :^ C = ^ H, and -4. A — -i^ F.
Proof. On BA take BF" = GF, and draw FH' \\ CA ;
.-. AB : F'B '.'. BC '. BH' : : CA : HF'.
(508. Mutually equiangular triangles are similar.)
Since FG = F'B, :, BC -. GH :: BC -. BH\
(484. Ratios equal to the same ratio are equal.)
therefore, by 491, GH = BH'.
In the same way HF = H'F',
.-. L.FGH^ t.F'BH'.
But F'BH' is similar to ABC.
1 90 THE ELEMENTS OF GEOMETRY.
Theorem VII.
511. Two triangles having one angle of the one equal to one
angle of the othevy and the sides about these angles proportional,
are similar.
Hypothesis. ^ B = ^ G, and AB : BC .. FG : GH,
Conclusion, a AB C ~ a FGH (using ^ for the word " similar ") .
Proof. In BA take BF' = GF, and draw F'L \\ AC,
.'. A F'BL - A ABC,
(508. Equiangular triangles are similar.)
.-. AB : BC :: F'B : BL.
But F'B — FG by construction ; therefore, from our hypothesis,
AB ', BC '.'. F'B : GH,
/. FB X BL i: F'B : GH,
therefore, by 491,
BL = GH,
.-. A F'BL ^ A FGH,
(124. Triangles having two sides and the included angle respectively equal are
congruent.)
.-. A FGH - A ABC.
RATIO APPLIED. IQI
Theorem VIII.
512. If two triangles have two sides of the 07ie proportional
to two sides of the other^ and an angle in each opposite one cor-
responding pair of these sides equals the angles opposite the other
pair are either equal or supplemental.
The angles included by the proportional sides are either
equal or unequal.
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. ABC and FGH as with AB \ BC w FG \ GH,
^A==^F, ^B>^G,
Conclusion. ^ C -f- ^ ^ = st. ^ .
Proof. Make ^ ABV = ^ 6^ ;
.♦. A ABD - A FGH;
(508. Equiangular triangles are similar.)
.-. AB : BD :: FG : GH;
therefore, from our hypothesis,
AB : BV :: AB : BC]
therefore, by 491, BD = BC,
.*. ^ C = ^ BJDC.
But ^ BDC + ^ BDA = st. ^,
.-. ^ C + ^ Zr = St. ^.
192 THE ELEMENTS OF GEOMETRY.
513. Corollary. If two triangles have two sides of the
one proportional to two sides of the other, and an angle in each
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
side, the triangles are similar.
514. In similar figures, sides between equal angles are called
Homologous^ or corresponding. The ratio of a side of one
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 vAv^n each side of the one is parallel
to the corresponding side of the other.
Theorem IX.
515. If two unequal similar figures are similarly placed^ all
lines joining a vertex of one to the corresponding vertex of the
other are concurrent.
Hypothesis. ABCD and A'B'C'iy two similar figures simi-
larly placed.
Conclusion. The lines AA\ BB\ CC, etc., meet in a point P.
Proof. Since AB and A'B' are unequal,
.-. AA' and BB' are not |1 .
Call their point of intersection F, Then
AF : A'F '.: BF '. B'F :: AB : A'B,
(508. Equiangular triangles are similar.)
RATIO APPLIED. 1 93
In the same way, if BB' and CC meet in Q, then
BQ : B'Q :: BC : B' C\
But, by hypothesis,
AB : A'B' '.', BC '. B' C\
.-. BP : B'P :: BQ : B' Q,
.'. the point Q coincides with P.
(502. A line can be cut only at a single point into external segments having a given
ratio.)
516. 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 cor-
responding vertices would so divide them.
517. The point of concurrence of the lines joining the equal
angles of two similar and similarly placed figures is called the
Center of Similitude of the two figures.
518. Corollary. The sects from the center of similitude
along any line to the points where it meets corresponding sides
of the similar figures are in the ratio of those sides.
Exercises, %'j. Construct a polygon similar to a given
polygon, the ratio of similitude of the two polygons being
given.
194
THE ELEMENTS OF GEOMETRY.
Theorem X.
519. y4 perpendicular from the right angle to the hypothenuse
divides a right-angled triangle into two other triangles similar
to the whole and to one another.
Hypothesis, a ABC right-angled at C.
CD ± AB.
Conclusion, t. ACD ^ t. ABC -^ t. CBD.
Proof. ^ CAD = ^BAC, and rt. ^ ADC = rt. ^ ACB,
.-. ^ACD = ^ABC,
(174. The three angles of any triangle are equal to a straight angle.)
.-. aACD ~ hABC.
In the same way we may prove A CBD ~ A ABC;
.-. aACD ~ A CBD.
520. Corollary I. Each side of the right triangle is a
mean proportional between the hypothenuse and its adjacent
segment.
For since A A CD ~ A ABC,
.-. AB : AC :: AC : AD.
521. Corollary II. The perpendicular is a mean propor-
tional between the segments of the hypothenuse.
522. Corollary III. Since, by 383, lines from any point
in a circle to the ends of a diameter form a right angle, there-
fore, 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.
Theorem XL
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.
Hypothesis. ABC any A. BD ihe bisector of 4. at B.
Conclusion. AB : BC .'. AD \ DC.
Proof. Draw AF \\ BD.
Then, of the two angles at B given equal by hypothesis, one equals
the corresponding interior angle at F, and the other the corresponding
alternate angle at A,
(168. A line cutting two parallels makes alternate angles equal, and (169) corre-
sponding angles equal.)
.-. AB = BF.
(126. If two angles of a triangle are equal, the sides opposite are equal.)
But BF : BC :-. AD -. DC,
(499. If a line be parallel to a side of a triangle, it cuts the other sides propor-
tionally.)
/. AB :BC::ADi DC,
196 THE ELEMENTS OF GEOMETRY.
524. Inverse. Since, by 502, a sect can be cut at only one
point internally in a given ratio, and at only one point exter-
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 har-
monically.
526. Corollary. The bisectors of an interior and exterior
angle at one vertex of a triangle divide the opposite side har-
monically.
Theorem XII.
527. If a secty AB, is divided harmonically at the points P
and Q, the sect PQ will be divided harmonically at the points
A and B.
I ! , ,
3L ^ 7» a
Hypothesis. THq sect AB divided internally at F, and exter-
nally at Q, so that
AP '. BP '.'. AQ : BQ',
therefore, by inversion,
BP '. AP '.'. BQ '. AQ;
therefore, by alternation,
BP : BQ '.: AP : AQ.
528. The points A, B, and P, Q, of which each pair divide
harmonically the sect terminated by the other pair, are called
four Harmonic Points.
RATIO APPLIED,
197
III. Rectangles and Polygons.
Theorem XIII.
529. If four sects are proportionaly the rectangle contained by
the extremes is equivalent to the rectangle contained by the
means.
ac
ad
^
Let the four sects a^ b, c, ^/, be proportional.
Then rectangle ad — be.
Proof. On a and on b construct rectangles with altitude = c.
On c and on d construct rectangles of altitude a.
Then
a : b :: ac : bCy and c : d :-, ac -, ad,
(504. Rectangles of equal altitudes are to each other as their bases.)
But, by hypothesis,
a '. b '.: c ', d,
therefore, by 491,
.*. ac : be : : ae : ad;
be = ad.
198
THE ELEMENTS OF GEOMETRY.
530. Inverse. If two rectangles are equivalent, the sides
of the one will form the extremes, and the sides of the other
the means, of a proportion.
^■c
a^.
Hypothesis. Reciangle ad = be.
Conclusion, a \ b w c \ d.
Proof. Since ad = be, therefore, by 489, ac : be
but ae : be :: a : bj and ae : ad '.'. e : d,
.'. a : b we : d.
ae : ad\
531. Corollary. If three sects are proportional, the rect-
angle of the extremes is equivalent to the square on the mean.
Theorem XIV.
532' If two chords intersect either within or without the
circle, the rectangle contained by the segments of the ofie is equiv-
alent to the rectangle contained by the segments of the other.
RATIO APPLIED.
199
Hypothesis. Let the chords AB and CD intersect in P.
Conclusion. Rectangle AP . PB = rectangle CP . PD.
Proof. ^PAC == ^ PDB,
(376. Angles in the same segment of a circle are equal.)
and
^ APC = ^ BPD,
.-. A APC - A BPD,
(508. Equiangular triangles are similar.)
/. AP : CP ::PI> : PB,
.-. AP.PB = CP.PD,
533. Corollary. Let the point P be without the circle,
and suppose DCP to revolve about P until C and D coincide ;
then the secant DCP becomes a tangent, and the rectangle
CP . PD becomes the square on PC. Therefore, if from a point
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.
200 THE ELEMENTS OF GEOMETRY.
Theorem XV.
534. The rectangle of two sides of a triangle is equivalent to
the rectangle of two sects drawn from that vertex so as to make
equal angles with the two sides ^ and produced^ one to the base,
the other to the circle circumscribing the triangle.
Hypothesis. ^ ABE = ^ CBD.
Conclusion. Rectangle AB .BC — DB . BE,
Proof. Join AE. Then
4-C^i.E,
(376. Angles in the same segment of a circle are equal.)
^ CBD = 4. ABE, by hypothesis,
.-. A CBD ~ A ABEy
.-. AB : BE :: DB '. BC,
.-. AB.BC ^ DB.BE.
535- Corollary I. If BD and BE coincide, they bisect
the angle B ; therefore rectangle AB . BC ^ DB . BE =
DB{BD + DE) = BD^ + BD . DE = BD^ + CD . DA (by
532). Therefore, when 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. Corollary II. If BD be a perpendicular, BE is a
diameter, for angle BAE is then right ; therefore in any triangle
-2> 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.
Exercises. %Z. Prove that the inverse of 535 does not
hold when AB — BC,
89. Prove 535 when it is an exterior angle which is bi-
sected.
202 THE ELEMENTS OF GEOMETRY,
Theorem XVI.
537. The rectangle of the diagonals of a quadrilateral in-
scribed in a circle is equivalent to the sum of ihe two rectangles
of its opposite sides.
Hypothesis. ABCD an inscribed quadrilaieral.
Conclusion. Rectangle A C . BD =^ AB . CD ■\- BC . DA,
Proof. By 164, make ^ DAF = ^ BAG.
To each add ^i^^C.
Then, in A s ^ CZ? and ABF,
^DAC = ^BAF',
also
4.ACD = ^ABD,
(376. Angles inscribed in the same segment are equal.)
.-. AACD'^ ^ABF,
(508. Equiangular triangles are similar.)
/. AC : AB :: CD i BF;
therefore, by 529,
AC .BF= AB. CD,
Again, by construction, ^ DAF = ^ BAC,
RATIO APPLIED.
203
Moreover, ^ ADF = ^ ACB,
(376. Angles inscribed in the same segment are equal.)
(508. Equiangular triangles are similar.)
.-. AC : AD :: CB : DF'y
therefore, by 529,
AC ,FD = AD.BC,
.-. AB .CD + BC ,DA = AC .BF+ AC .FD
= AC{BF + FD) = AC . ^Z>.
Problem I.
538. To alter a given sect in a given ratio.
^
3
/ k
^.^ .
*
Given, Me raf/b as ihat of sect a to sect b^ and given the
sect AB.
Required, to find a sect to which AB shall have the same ratio as
a to b.
Construction. Make any angle C,
On one arm cut off CD = AB. On the other arm cut off CF = a,
and FG = b.
Join DFy and through G draw GH \\ DF,
.-. AB : DH '.: a : b.
204 ^-^^ ELEMENTS OF GEOMETRY.
539. This is the same as finding a fourth proportional to
three given sects.
To find a third proportional to a and by make AB = ^ in
the above construction.
540. Every alteration of a magnitude is an alteration in
some ratio.
Two or more alterations are jointly equivalent to some one
alteration, and then this single alteration which produces the
joint effect of two is said to be compounded of those two.
The composition of the ratios of a\,Q b and (7 to Z> is per-
formed by assuming F, altering it into G, so that F : G w a : by
then altering G into 77, so that G : H '.: C : D.
The joint effect turns F into H) and the ratio oi F to H is
the ratio compounded of the two ratios, a : b and C : D.
541. A ratio arising from the composition of two equal
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.
Hypothesis. In /=? ACy ^ BCD = ^ HCF of £y CG.
Conclusion. /^ACi^yCG^ ratio compounded oi DC : CF,
and^C : CH,
RATIO APPLIED. 205
Proof. Place the /u^ so that HC and CB are in one line ; then,
by log, DC and CF are in one line. Complete the £j BF.
Then
CD AC : C7BF .i DC : CF,
and
£yBF '. n7CG \\ BC '. CH,
(505. Parallelograms of equal altitude are as their bases.)
.'. c7 AC has to z:7 CG the ratio compounded oi DC : CF and
BC : CH,
543. Corollary I. Triangles which have one angle of the
one equal or supplemental 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.
544. Corollary II. Since all rectangles are equiangular
parallelograms, therefore the ratio compounded of two ratios
between sects is the same as the ratio of the rectangle con-
tained by the antecedents to the rectangle of the consequents.
If the ratio compounded of a : a\ and d : d\ be written
— . — , this corollary proves ^ . — = -^ ; and the composition
ad a ab
of ratios obeys the same laws as the multiplication of fractions.
Thus ^ . ^ =^, and so the duplicate ratio of two sects is the
00 b^
same as the ratio of the squares on those sects.
It will be seen hereafter that the special case of 542, when
the parallelograms are rectangles, is made the foundation of all
mensuration of surfaces.
Exercises. 90. If mutually equiangular parallelograms are
equivalent, so are rectangles with the same sides.
91. Equivalent parallelograms having the same sides as
equivalent rectangles are mutually equiangular.
206 THE ELEMENTS OF GEOMETRY.
Theorem XVIII.
545. Similar triangles are to one another as the squares on
their corresponding sides.
Let the similar triangles ABC^ AHK, be placed so as to have the
sides AB, AC, along the corresponding sides AH, AK, and therefore
BC II HK. Join CH,
Since, by 505, triangles of equal altitude are as their bases,
.-. A ABC : A AHC w AB \ AH,
and A AHC : A AHK :: AC : AK = AB : AH, by hypothesis ;
AB AB AB^
A ABC : A AHK =
AH AH AH^
Exercises. 92. The ratio of the surfaces of two similar
triangles is the square of the ratio of similitude of the tri-
angles.
93. If the bisector of an angle of a triangle also bisect a
side, the triangle is isosceles.
94. In every quadrilateral which cannot be inscribed in a
circle, the rectangle contained by the diagonals is less than the
sum of the two rectangles contained by the opposite sides.
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, ^ -g-
Hypothesis. ABCD and A'B'Clf two similar polygons, of
which BC and B' C are corresponding sides.
Conclusion. ABCD : A'B'C'iy : : AB^ : A'B'\
Proof. By 516, the polygons may be divided into similar triangles.
By 545, any pair of corresponding triangles are as the squares on
corresponding sides,
t^ABD BJy /\BCD BC^
AA'B'iy B'D^ AB'Cjy B'C'^
therefore, by 496,
A ABD + A BCD BC
A A'B'D^ + A B'C'D^ B'C'^
In the same way for a third pair of similar triangles, etc.
208 THE ELEMENTS OF GEOMETRY.
547. Corollary. If three sects form a proportion, a poly-
gon on the first is to a similar polygon similarly described on
the second as the first sect is to the third.
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.
7)
Proof. The angles of two regular polygons of the same number
of sides are all 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,
therefore, by 496,
AB
BC
r
" A'B'~
B'C
■?'
AB + BC
r
CD
A'B' 4- B'C
r'
c'jy'
AB + BC -{- CD
r
A'B' ^B'C ^ C'jy r'
RATIO APPLIED. 209
Problem II.
549. To find a mean proportional between two giveji sects.
3)
Let AB, BC, be the two given sects. Place AB, BC, in the same
line, and on AC describe the semicircle ADC. From B draw BD
perpendicular to ^C
BD is the mean proportional.
(383. An angle inscribed in a semicircle is a right angle.)
(521. If from the right angle a perpendicular be drawn to the hypothenuse, it will
be a mean proportional between the segments of the hypothenuse.)
Exercises. 97. If the given sects were ^C and ^C placed
as in the above figure, how would you find a mean proportional
between them 1
98. Half the sum of two sects is greater than the mean
proportional between them.
99. The rectangle contained by two sects is a mean propor-
tional between their squares.
100. The sum of perpendiculars drawn from any point
within an equilateral triangle to the three sides equals its
altitude.
loi. The bisector of an angle of a triangle divides the
triangle into two others, which are proportional to the sides
of the bisected angle.
102. Lines which trisect a side of a triangle do not trisect
the opposite angle.
103. In the above figure, if AF ± AD meet DB produced at
F, then a ABD — a FOB.
2IO
THE ELEMENTS OF GEOMETRY,
Problem III.
550. On a given sect to describe a polygon similar to a given
polygon.
Let ABCDE be the given polygon, and A'B' the given sect.
^OYCiAD, AC,
4. A'B'C = 4. ABC,
^B'A'C = 4BAC, ■
.-. tB'C'A' = 4BCA,
Make
and
Then make
A A'B'C
h.ABC,
and
^.A'C'jy = 4ACD,
4. C'A'jy = 4 CAD,
:, 4 CD A' = 4 CDA,
t^AC'iy ~ A^CA etc.
Therefore, from the first pair of similar triangles,
A'B' : B'C :: AB : BC,
and
B'C : CA' :: BC : CA.
From the second pair of similar triangles,
CA' : C'ly :: CA : CD,
,', B'C : CD' :: BC : CD,
and so on.
Thus all the ^^s in one polygon are = the corresponding 2<s in the
other, and the sides about the corresponding ^s are proportional.
.*. the polygons are similar.
RATIO APPLIED.
211
Theorem XXI.
551. In a right-angled triangle, any polygon upon the hypoth-
enuse is equivalent to the sum of the similar and similarly
described polygons on the other two sides.
Draw CD ± AB.
Therefore, by 520,
AB : AC :: AC : AD',
therefore, by 547,
AB '. AD :: Q : S.
Similarly,
AB : BD :: Q : R,
Therefore, by 496,
AB : AD -{- DB :: Q : R -{■ S,
.', 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.).
212 THE ELEMENTS OF GEOMETRY.
Problem IV.
552. To describe a polygon equivalent to one^ and -similar to
anothery given polygon.
JJC
A.
Let D be the one, and ABC the other, given polygon.
By 262, on ^C describe any ^7 CE = ABC, and on AB describe
/:7 AM = Z>, and having ^ AEM = ^ CAE.
By 549, between AC and AF find a mean proportional GH.
^y 55°> o^ ^^ construct the figure KGH similar and similarly
described to the figure ABC.
KGH is the figure required.
It may be proved, as in 542, that ^C and AF are in one line, and
also LE and EM-,
therefore, by 505,
AC '. AF '.'. n7 CE '. n7 AM : : ABC : £>,
But
AC : AF :: ABC : KGH,
(547. If three sects form a proportion, a polygon on the first is to a similar polygon
on the second as the first sect is to the third.)
.-. ABC \ D .'. ABC : KGH-,
therefore, by 491,
KGH = D,
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
two points of a line are in a plane, the whole line lies in that
plane.
Two other assumptions will now be made : —
555. Any number of planes may be passed through any line.
z
2
^
556. A plane may be revolved on any line lying in it.
ax3
214 THE ELEMENTS OF GEOMETRY.
Theorem I.
557. Through two intersecting lines y one plane y and only one^
passes.
a
Hypothesis. Two lines AB, BC, meeting in B.
Conclusion. One plane, and only one, passes through them.
Proof. By 555, let any plane FG be passed through AB, and, by
556, be revolved around on AB as an axis until it meets any point C
of the line BC,
The line BC then has two points in the plane FG'j and therefore,
by 554, the whole line BC'i^m this plane.
AlsOy any plane containing AB and BC must coincide with FG.
For let Q be any point in a plane containing AB and BC.
Draw QMN in this plane to cut AB, BC, inM and JV. Then,
since M and JV are points in the plane FF, therefore, by 554, Q is a
point in the plane FF.
Similarly, any point in a plane containing AB, BC, must lie in FF;
therefore any plane containing AB, BC, must coincide with FF.
558- Corollary I. Two lines which intersect lie in one
plane, and a plane is completely determined by the condition
that it passes through two intersecting lines.
559* Corollary II. Any number of lines, each of which
intersects all the others at different points, lie in the same
plane ; but a line may pass through the intersection of two
others without being in their plane.
OF PLANES AND LINES. 21 5
560. Corollary III. A line y and a point without that line,
determine a plane,
s
Proof. Suppose AB the line, and C the point without AB.
Draw the hne CD to any point D in AB. Then one plane con-
tains AB and CD, therefore one plane contains AB and C.
Again, any plane containing AB must contain D ; therefore, any
plane containing AB and C must contain CD also.
But there is only one plane that can contain AB and CD.
Therefore there is only one plane that can contain AB and C.
Hence the plane is completely determined.
561. Corollary IV. Three points not in the same line
determine a plane.
57
For let A, B, C be three such points. Draw the line AB.
Then a plane which contains A, B, and C must contain AB
and C\ and a plane which contains AB and C must contain A,
B, and C. Now, AB and C are contained by one plane, and one
only ; .therefore A, B, and C are contained by one plane, and
one only. Hence the plane is completely determined.
2l6 THE ELEMENTS OF GEOMETRY.
562. Two parallel lines determine a plane.
For, by the definition of parallel lines, the two lines are in
the same plane ; and, as only one plane can be drawn to con-
tain 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 another^ their common section must
be a straight line,
JL
Hypothesis. Lei AB and CD be two intersecting planes.
Conclusion. Their common section is a straight line.
Proof. Let M and N be two points common to both planes.
Draw the straight line MN. Therefore, by 554, since M and N are in
both planes, the straight line MN lies in both planes.
And no point out of this line can be in both planes ; because then
two planes would each contain the same line and the same point without
it, which, by 560, is impossible.
Hence every point in the common section of the planes lies in the
straight line MN.
OF PLANES AND LINES. 21/
564. CoNTRANOMiNAL OF 554. A line which does not lie
altogether in a plane may have no point, and cannot have more
than one point, in common with the plane.
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 to two planes lie on a line, and this line can
have only one point in common with the third plane.
565. All planes are congruent ; hence properties proved for
one plane hold for all. A plane will slide upon its trace.
Principle of Duality.
566. When any figure is given, we may construct a reciprocal
figure by taking planes instead of points, and points instead of
planes, but lines where we had lines.
The figure reciprocal to four points which do not lie in a
plane will consist of four planes which do not meet in a point.
From any theorem we may infer a reciprocal theorem.
Two points determine a line.
Three points which are not in
a line determine a plane.
A line and a point without it
determine a plane.
Two lines in a plane determine
a point.
Two planes determine a line.
Three planes which do not pass
through a line determine a point.
A line and a plane not through
it determine a point.
Two lines through a point de-
termine a plane.
There is also a more special principle of duality, which, in
the plane, takes points and lines as reciprocal elements ; for
they have this fundamental property in common, that two ele-
ments 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 corre-
sponding lines is bisected by the
axis.
The sect between tvvo corre-
sponding points is bisected by the
center.
2l8
THE ELEMENTS OF GEOMETRY.
Theorem III.
567* If ^ ^^^^ ^^ perpendicular to two lines lying in a plane y
it will be perpendicjilar to every other li7te lying in the plane and
passing through its foot.
Hypothesis. Let the line EF be perpendicular to each of the
lines AB, CD, at E, the point of their intersection.
Conclusion. EF _L GH, any other line lying in the plane ABCD,
and passing through E.
Proof. Take AE = BE, and CE = DE.
Join AD and BC, and let G and H be the points in which the
joining lines intersect the third line GH.
Take any point F in EF. Join FA, FB, EC, FD, EG, FH.
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 = HE, and AG = BH.
OF PLANES AND LINES. 219
Then
A AEF ^ A BEF,
(124. Triangles having two sides and the included angle equal are congruent.)
AF = BF.
In the same way
CF = DF,
Then
t.ADF^ t.BCF,
(129. Triangles having three sides in each respectively equal are congruent.)
/. 4. DAF = 4. CBF,
Then
A AFG ^ A BFH,
(124. Triangles having two sides and the included angle equal are congruent.)
FG = FH,
Then
A FEG ^ A FEH,
(129. Triangles having three sides in each respectively equal are congruent.)
/. 4 FEG = i. FEB,
FE A. GH.
As GH is any line whatever lying in the plane ABCDy and passing
through E,
/. EF J_ every such line.
568. A line meeting a plane so as to be perpendicular to
every line lying in the plane and passing through the point of
intersection, is said to be perpendicular to the plane. Then
also the plane is said to be perpendicular to the line.
569. Corollary. At a given point in a plane, only one
perpendicular to the plane can be erected ; and, from a point
without a plane, only one perpendicular can be drawn to the
plane.
220
THE ELEMENTS OF GEOMETRY.
570. Inverse of 567. All lines perpendicular to another
line at the same point lie in the same plane.
Hypothesis. AB any line ± BD and BE, BC any other
line _L AB.
Conclusion. ^C is in the plane BDE,
Proof. For if not, let the plane passing through AB, BC, cut the
plane BDE in the line BF. Then AB, BC, and BE are all in
one plane ; and because AB J_ BD and BE, therefore, by 567,
AB J. BF,
But, by hypothesis, AB 1.BC;
therefore, in the plane ABE we have two lines BC, BF, both ± AB at
B, which, by 105, is impossible ;
.-. BC lies in the plane DBE.
571. Corollary. If a right angle be turned round one of
its arms as an axis, the other arm will generate a plane ; and
when this second arm has described a perigon, it will have
passed through every point of this plane.
572. Through any point D without a given line AB to pass
a plane perpendicular to AB. In the plane determined by
the line AB and the point D, draw DB J_ AB ; then revolve the
rt. 4. ABD about AB.
OF PLANES AND LINES.
221
Theorem IV.
573. Lines perpendicular to the same plane are parallel to
each other.
Hypothesis. AB, CD JL plane MN at the points B, D.
Conclusion. AB \\ CD.
Proof. Join BD, and draw DE i. BD in the plane MN,
Make DE = AB, and join BE, AD, AE. Then
A ABD ^ A EDB,
(124. Triangles having two sides and the included angle equal are congruent.)
/. AD = BE,
Then
A ABE ^ A EDA,
(129. Triangles having three sides in each respectively equal are congruent.)
.-. 4. ABE = 4. EDA,
But, by hypothesis, ABE is a rt. 4,
:. EDA is a rt. ^,
.'. AD, BD, CD, are all in one plane.
(570. All lines perpendicular to another at the same point lie in the same plane.)
But AB is in this plane, since the points A and B are in it,
.'. AB and CD He in the same plane ;
and they are both X BD,
,'. AB II CD.
(166. If two interior angles are supplemental, the lines are parallel.)
222
THE ELEMENTS OF GEOMETRY.
574. Inverse of 573. If one of two parallels is perpen-
dicular to a plane, the other is also perpendicular to that plane.
7
Hypothesis. AB \\ CD, CD ± plane MN.
Conclusion. AB _L plane MN.
Proof. For if AB^ meeting the plane MN at B^ is not ± it, let
BF \>Q l.iX.'y
therefore, by 573, BF \\ CD, and, by hypothesis, BA \\ CD,
But this is impossible,
(99. Two intersecting lines cannot both be parallel to the same line.)
.-. ^^ ± plane JfiV.
575. Corollary I. If one plane be perpendicular to one of
two intersecting lines, and a second plane perpendicular to the
second, their intersection is
perpendicular to the plane of
the two lines. -
For their intersection DF
is perpendicular to a line
through D parallel to ABy
and also perpendicular to a
line through D parallel to
BC.
(574. If, of two parallels, one be perpendicular to a plane, the other is also.)
576. Corollary II. Two lines, each parallel to the same
line, 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 I.
577. To draw a line perpendicular to a given plane from a
given point without it.
GrvTEN, ihe plane BH and the point A without it.
Required, to draw from A a line ± plane BH.
Construction, In the plane draw any line BC, and, by 139, from
A drsLW AD ±BC.
In the plane, by 135, draw Z>jF ± BC; and from A, draw, by 139,
AF ± DF. AF will be the required perpendicular to the plane.
Proof. Through F draw GH \\ BC.
Then, since BC 1. the plane ADFy
(567. A line perpendicular to any two lines of a plane at their intersection is
perpendicular to the plane.)
.\ GH ± the plane ADF,
(574. If two lines are parallel, and one is perpendicular to a plane, the other is
also.)
/. GH ± the line AF of the plane ADF,
:. AF1.GH.
But, by construction, AF _L DF,
,', AF J_ the plane passing through GH, DF.
224
THE ELEMENTS OF GEOMETRY.
Problem II.
578. To erect a perpendicular to a given plane from a given
point in the plane.
1
Let A be the given point.
From any point B, without the plane, draw, by 577, -5 C ± the
plane, and from A, by 167, draw AD \\ BC,
/. AD ± the plane.
{574. If one of two parallels is 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 is a line.
Theorem V.
581. A line makes with its own projectiojt upon a plane a less
angle than with any other line ijt the plane.
Hypothesis. Lq^ BA meet the plane MN at B, and let BA'
be its project/on upon the plane MN^ and BC any other line drawn
through B in the plane MN.
Conclusion. ^ ABA' < ^ ABC.
Proof. Take BC ^ BA. Join A C and A' C.
Then, in As ABA' and ABC,
AB = AB,
BA' = BC,
AA' < AC,
(150. The perpendicular is the least sect from a point to a line.)
.-. ^ ABA' <^ABC.
(161. In two triangles, ii a = a', 6 = y, c Kc*, therefore C< C.)
226 THE ELEMENTS OF GEOMETRY,
582. The angle between a line and its projection on a plane
is called the Inclinatioji of the line to the plane.
583. Parallel planes are such as never meet, how far soever
they may be produced.
584. A line is parallel to a plane when they never meet, how
far soever they may be produced.
Theorem VI.
585. Planes to 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 perpen-
dicular. Then from this point X we would have two perpendiculars to
the same line, which is impossible,
(145. There can be only one perpendicular from a point to a line.)
.*. the planes cannot intersect, and
.*. are parallel.
586. If two lines are parallel, every plane through one of
them, except the plane of the parallels, is parallel to the other.
Let AB and GH be the parallels, and DEF any plane
through GH\ then the line AB and the plane DEF are par-
allel. For the plane of the parallels ABHG intersects the
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 can-
not 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
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.
227
Theorem VII.
587- If ci pair of ijttersecting lines be parallel to another
pair, bnt not in the same plane with them, the plane of the first
pair is parallel to the plane of the secojtd pair.
Hypothesis. AB \\ DE, and BC \\ EF.
Conclusion. Plane ABC II plane DEF.
Proof. From B draw BH J_ plane DEF, meeting it in H,
Through B draw GB \\ DE, and BK \\ EF,
.-. GB II AB, and ' BK \\ BC,
(576. Two lines each parallel to the same line are parallel to each other.)
/. -2^ ABB + ^ BBG = St. ^.
But because BB was drawn J_ plane DEF,
/. ^BBG^xt.^.
So '^,
and, in same way,
:^ABB= rt. ^;
^ CBB = rt. ^ ;
.-. BB A_rg\^nQABC,
.♦. plane ABC \\ plane DEF.
(585. Planes to which the same line is perpendicular are parallel.)
228
THE ELEMENTS OF GEOMETRY.
Theorem VIII.
588. If 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.
Hypothesis. AB \\ DE, and BC II EF.
Conclusion. 2^ ABC = or supplemental to ^ DEF,
Proof. Join BE.
Since AB \\ DE,
.'. AB, BE, and ED, are in one plane.
In this plane, from A draw a line I| BE.
It must meet the line DE. Call the intersection point D.
.'. ABED is a ,C7 ;
.-. AB = DE, and AD = BE.
In same way,
BC = EF, and CF = BE,
.-. AD = CF.
But since, by construction, AD \\ BE, and CF \\ BE, therefore,
hysi6, AD II CF,
/. ACFD isa,/=7,
(216. If any two opposite sides of a quadrilateral are equal and parallel, it is a
parallelogram.)
OF PLANES AND LINES. 229
.-. AC = DF,
,\ A ABC ^ A £>£F,
(129. Triangles with three sides of the one equal to three of the other are con-
gruent.)
.-. ^ ABC = ^ JDjEK
Theorem IX.
589. If two parallel planes be cut by another plane y their
common sections with it are parallel.
Call the parallel planes A and B, and the third plane X.
Then the lines of intersection are in one plane, since they both lie
in the plane X.
Again, because one of these lines is in the plane A^ and the other
in the parallel plane B, they can never meet.
Therefore the two lines are in one plane, and can never meet ; that
is, they aire parallel.
590. Corollary. Parallel sects included between two par-
allel planes are equal.
230
THE ELEMENTS OF GEOME^TRY.
Theorem X.
591. Parallel lines intersecting the same plane are equally
inclined to it.
Hypothesis.
Conclusion.
AB II CD,
BB' ± plane ACF,
niy A. plane ACF.
^ BAB' = V Dcjy,
Proof. ^ ABB' = or supplemental to ^ CDL/,
(588. If 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.)
But 2< ABB' cannot be supplemental to ^ CDL/ , since each is acute,
.-. -4. ABB' = 4. CDjy-,
and therefore their complements are equal.
592. Two lines not in the same plane are regarded as mak-
ing with one another the angles included by two intersecting
lines drawn parallel respectively to them.
We know, from 588, that these angles will always be the
same, whatever the position of the point of intersection.
OF PLANES AND LINES, 23 1
Theorem XL
593* V ^^^ //«^J be cut by three parallel planes, the corre-
spondi7ig sects are proporiio7ial.
Let the lines AB, CD, be cut by the || planes MN, FQ, RSy in the
points Aj E, By and C, F, D.
Conclusion. AE : EB :: CF : FD.
Join AD, cutting the plane FQ in G,
Join A C, BD, EG, EG. Then
EG II BD.
(589. If parallel planes be cut by a third plane, their common sections with it are
parallel.)
In the same way, AC \\ GF\
,\ AE : EB :: AG : GD,
and
AG '. GD :: CF ', FD',
(499. A line parallel to the base of a triangle divides the sides proportionally.)
/. AE \ EB \\ CF '. FD.
232 THE ELEMENTS OF GEOMETRY.
Theorem XII.
594. Two lines not in tJu same plane have one, and only oite^
common j)erpendicular.
3L T
Hypothesis. AB and CD, two lines not in tha same plane, and
therefore neither parallel nor intersecting each other.
Conclusion. There is one line, and no more, perpendicular to both
AB and CD.
Proof. Through one of the lines, as CD, pass a plane, and let it
revolve on CD as axis until it is parallel to AB. Call this plane MN.
Let A'B' be the projection of AB on the plane MN, and let O be the
point in which this projection intersects CD. Then O, like every point
in the projection, is the foot of a perpendicular to the plane from some
point of AB. Call this point P. Then, since FO is perpendicular to
the plane MN, it is perpendicular to CD and A'B'. But A'B' is par-
allel to AB, because, being in a plane parallel to AB, it can never meet
AB ; and, being the projection of AB, it is, by 580, in the same plane
with AB.
.'. -since FO _L A'B', it is also ± AB,
(170. A line perpendicular to one of two parallels is perpendicular to the other
also.)
,\ OF ± both AB and CD.
If there could be any other common perpendicular, call it F'Q.
OF PLANES AND LINES,
233
Through Q draw in the plane MN, QR \\ AB.
Since FQ ± AB, .'. FQ J_ QR-,
but, by hypothesis, F Q 1. CD,
:. i^'Q ± plane J/W;
/. ^ is a point in A'B' , the projection of AB,
.'. Q is O, the only point common to A'B' and CD,
.-. jP' ^ coincides with /'C?.
(569. At a given point in a given plane, only one perpendicular to the plane can be
erected.)
Theorem XIII.
595. The smallest sect between two lines not i7t the same
plane is their common perpeitdictdar.
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. The perpendicular is the smallest sect from a point to a line.)
596. Remark. If two planes intersect, and two intersecting
lines are drawn, one in each plane, these lines may, for the same
two planes, make an acute, right, or obtuse angle, according to
their relation to the line of intersection of the planes.
234
THE ELEMENTS OF GEOMETRY.
Theorem XIV.
597. The smallest sect from a point to a plane is the perpen-
dicidar.
Hypothesis. AD ± plane MN. B any other point of MN.
Conclusion. AD < AB.
Proof. Join AB, BD. In A ABDyX ^^^ is a rt. ^,
.-. AD < AB.
(150. The perpendicular is the least sect from a point to a line.)
Theorem XV.
598. Equal obliques from a point to a plane m,eet the pla7te
in a circle whose center is the foot of the perpendicular from the
point to the plane.
OF PLANES AND LINES.
235
Hypothesis. AB, AC, equal sects from A to the plane MN.
AD ± plane MN,
Conclusion. DB = DC.
Proof. Rt. A ABD ^ rt. A A CD.
(179. If two right triangles have the hypothenuse and one side respectively equal,
they are congruent.)
599. Corollary I. To draw a perpendicular from a point
to a plane, draw any oblique from the point to the plane ;
revolve this sect about the point, tracing a circle on the plane ;
find the center of this circle, and join it to the point.
600. Corollary II. If 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.
Theorem XVI.
601. The locus of all points from which the two sects drawn
to two fixed poiftts are equal, is the pla7ie bisecting at right angles
the sect joining the two given points.
X
For the line from any such point to the mid point of the joining
sect is perpendicular to that sect ; therefore all such lines form a plane
perpendicular to that sect at its mid point.
(570. All lines perpendicular to another line at the same point lie in the same plane.)
2^6
THE ELEMENTS OF GEOMETRY.
Theorem XVII.
602. The locus of all points from which the two perpendicu-
lars onto the same sides of two fixed planes are equal, is a plane
determined by one such point and the intersection line of the tzvo
given planes.
Hypothesis. Lei AB and CD be the two given planes, and K
one such point.
Conclusion. Perpendiculars dropped on to AB and CD from any
point in the plane determined by X and the intersection line BC are
equal.
Proof. Take P any point in the plane XCB. Draw PH ± plane
AB, and BP _L plane CB>. Call G the point where the plane FBI/
cuts the line BC. Join FG, BG, GH.
Draw KC || BG. Because the perpendiculars from K are given
equal, therefore KC is equally inclined to the two planes. But BG has
the same inclination to each as KC ;
(591. Parallels intersecting the same plane are equally inclined to it.)
/. ^ BGF = ^ BGH,
/. A BGF ^ A BGH,
BF = BH,
OF PLANES AND LINES. 237
Theorem XVIII.
603 • If three lines not in the same plane meet at one pointy
any two of the angles formed are together greater than the third.
Proof. The theorem requires proof only when the third angle
considered is greater than each of the others. In the plane of the
greatest 4. BAC, make 4. BAF = 4 BAD. Make AF = AD.
Through F draw a line BFC cutting AB in B^ and AC m C. Join
DB and DC.
A BAD ^ A BAF,
(124. Triangles having two sides and the included angle respectively equal are
congruent.)
.-. BD = BF.
But from A BCD we have BD + DC> BC.
{156. Any two sides of a triangle are together greater than the third.)
And, taking away the equals BD and BF,
DC > FC,
.-. in As CAD and CAF, we have 4 CAD > 4 CAR
(161. If two triangles have two sides respectively equal, but the third side greater
in the first, its opposite angle is greater in the first.)
Adding the equal 4 s BAD and BAF gives
4 BAD + 4 CAD > 4 BAC.
238 THR ELEMENTS OF GEOMETRY.
Theokem XIX.
604. If the vertices of a convex polygon be joined to a point
not in its plane ^ the snin of the vertical angles of the triangles so
made is less than a perigon.
s
Proof. The sum of the angles of the triangles which have the
common vertex S is equal to the sum of the angles of the same number
of triangles having their vertices 2X O in the plane of the polygon.
But
%. SAB + ^ SAE > ^ BAE,
%. SBA 4- ^ SBC > t ABC, etc.
(603. If three Hnes not in a plane meet at a point, any two of the angles formed
are together greater than the third.)
Hence, 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 sum of the angles at ^ is less than the sum of the angles at O, that
is, less than a perigon.
BOOK VIII.
TRI-DIMENSIONAL SPHERICS.
605. If one end point of a sect is fixed, the locus of the
other end point is a Sphere.
606. The fixed end point is called the Center of the sphere.
607. The moving sect in 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
240 THE ELEMENTS OF GEOMETRY.
6io. 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,
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-
out, the sphere, according as the sect to it from the center is
less than, equal to, or greater than, the radius.
Theorem I.
613. The common section of a sphere and a plane is a ciirle.
Take any sphere with center O.
Let A, B, C, etc., be points common to the sphere and a plane,
and OD the perpendicular from O to the plane. Then
OA = OB = OC =^ etc.,
being radii of the sphere,
.*. ABC, etc., is a circle with center D.
(598. 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. 24 1
614. Corollary. The line through the center of any circle
of a sphere, perpendicular to its plane, passes through the cen-
ter of the sphere.
615. A Great Circle of 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,
since each has for its 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.
618. Corollary. Since the perpendicular passes through
the center of the sphere, the two poles of any circle are oppo-
site points, and the diameter between them is called the Axis
of that circle.
Theorem II.
619. Every great 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 its former position, but inverted,
the great circle will coincide with its own trace, and the two hemispheres
will coincide.
242
THE ELEMENTS OF GEOMETRY.
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 circle.
621. If any number of great circles pass through a point,
they will also pass through the opposite point.
622. Through any two points in a sphere, not the extremi-
ties of a diameter, one, and only one, great circle can be passed ;
for the two given points and the center of the sphere determine
TRI-DIMENSIONAL SPHERICS. 243
its plane. Through opposite points, an indefinite number of
great circles can be passed.
623. Through any three points in a sphere, a plane can be
passed, and but one ; therefore three points in a sphere deter-
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 hypoth-
enuse, one side of a right-angled triangle decreases as the other
increases. ^««.^
625. A Zone is a portion of a sphere included between two
parallel planes. The circles made by the parallel planes are
the Bases of the zone.
626. A line or plane is tangent to a sphere when it has one
point, and only one, in common with the sphere.
627. Two spheres are tangent to each other when they have
one point, and only one, in common.
Exercises. 105. If 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
intercepted chords is constant. Also the sum of the squares
of the six segments of these chords is constant.
106. If three radii of a sphere, perpendicular to each other,
are projected upon any plane, the sum of the squares of the
three projections is equal to twice the square of the radius of
the sphere.
244 ^-^^ ELEMENTS OF GEOMETRY.
Theorem III.
628. A plane perpendicular to a radius of a sphere at its
extremity is tangent to the sphere.
For, by 597, this radius, being perpendicular to the plane, is the
smallest sect from the center to the plane ; therefore every point of
the plane is without the sphere except the foot of this radius.
629. Corollary. Every line perpendicular to a radius at
its extremity is tangent to the sphere.
630. Inverse of 628. Every plane or line tangent to the
sphere is perpendicular to the radius drawn to the point of
contact. 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
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 of points whose sect from point A is ^,
and from point B \s b.
TRI-DIMENSIONAL SPHERICS.
245
Theorem IV.
631. If two spheres cut one another ^ their miersection is a cir-
cle whose plane is perpendicnlar to the line Joining the centers of
the spheres^ and whose center is in that line.
Hypothesis. Let C and O be the centers of the spheres, A
and B any two points in their intersection.
Conclusion. A and B are on a circle having its center on the line
OC, and its plane perpendicular to that line.
Proof. Join CA, CB, OA, OB. Then
A CAO^ A CBO,
because they have OC common, CA = CB, and OA = OBj radii of
the same sphere.
Since these As are =, therefore perpendiculars from A and B upon
OC are equal, and meet OC at the same point, Z>.
Then AB> and DB are in a plane ± OC ; and, being equal sects,
their extremities A and B axe in a circle having its center at Z>.
632. Corollary. By moving 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
center ; therefore, if two spheres are tangent, either internally
or externally, their centers and point of contact lie in the same
line.
246 THE ELEMENTS OF GEOMETRY.
Theorem V.
633. Through any four pohits not in the same plane, one
spJiere, and only one, can be passed.
Let A^ B, C, D, be the four points. Join them by any three sects,
as AB, BC, CD. Bisect each at right angles by a plane.
The plane bisecting BC has the line EH'm. common with the plane
bisecting AB, and has the line FO in common with the plane bisecting
CD. Moreover, EH i. plane ABC, and FO ± plane BCD,
(575. If one plane be perpendicular to one of two intersecting lines, and a second
plane perpendicular to the second, their intersection is perpendicular to the
plane of the two lines.)
.-. EH LEG, and FO JL FG,
,'. EH and FO meet, since, by hypothesis, ^ at 6^ > o and < st. ^.
Call their point of meeting O. O shall be the center of the sphere
containing A, B, C, and D,
TRI-DIMENSIONAL SPHERICS.
247
For O is in three planes bisecting at right angles the sects AB, BC,
CD,
(601. The locus of all points from which the two sects drawn to two fixed points
are equal, is the plane bisecting at right angles the sect joining the two given
points.)
634. A Tetrahedron is a solid bounded by four triangular
plane surfaces called its faces. The sides of the triangles are
called its edges.
635. Corollary I. A sphere may be circumscribed about
any tetrahedron.
636. Corollary II. The lines perpendicular to the faces
of a tetrahedron through their circumcenters, intersect at a
common point.
637. Corollary III. The six planes which bisect at right
angles the six edges of a tetrahedron all pass through a com-
mon point.
248 THE ELEMENTS OF GEOMETRY.
Problem I.
638. To inscribe a sphere in a given tetrahedron.
Construction. Through any edge and any point from which per-
pendiculars to its two faces are equal, pass a plane.
Do the same with two other edges in the same face as the first, and
let the three planes so determined intersect at O. O shall be the center
of the required sphere.
(602. The locus of all points from which the perpendiculars on the same sides of
two planes are equal, is a plane determined by one such point and the intersec-
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 solid bounded by a sphere is called a Globe.
641. A globe-segment is a portion of a globe included be-
tween two parallel planes.
The sections of the globe made by the parallel planes are
the bases of the segment.
Any sect perpendicular to, and terminated by, the bases, is
the altitude of the segment.
642. A figure with two points fixed can still be moved by
revolving it 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 Revohction, If the axis of
revolution is any line passing through the center, a sphere
slides upon its trace. This is because every section of a sphere
by a plane is a circle.
Any figure has central symmetry if it has a center which
bisects all sects through it terminated by the surface. The
sphere has central symmetry, 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 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
another. 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
sphere : on the same sphere similar figures cannot exist.
643. Two points are symmetrical with respect to a plane
when this plane bisects at right angles the sect joining them.
Two 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 figures symmetrical with respect to that plane. The
sphere is symmetrical with respect to every plane through its
center. Any two spheres are symmetrical with respect to
every plane through their line of centers.
Every such plane cuts the spheres in two great circles ; and
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.
250
THE ELEMENTS OF GEOMETRY,
Theorem VI.
644. If a sphere he tangent to the parallel plafzes containing
opposite edges of a tetrahedron^ and sections made in the globe
and tetrahedron by one plane parallel to these are equivalenty
sections made by any parallel plaiie are equivalent.
K
K
\^. ij Jo
L\ /^ ^^
Hypothesis. Let KJ be the sect J_ the edges EF and GH in
the II tangent planes.
Then Kf = DT, the diameter.
Let sections made by plane ± Kf at R and ± DT at I, where
KR = DI, be equivalent; that is, ^ MO = O PQ.
Draw any parallel plane ACBLSN,
Conclusion, ^l^ LN — O AB.
Proof. Since A LEU ^ h. MEW, and A LHV ^ A MHZ,
:. MW \ LU w EM -. EL :: fR : fS;
MZ '. LV -.'. HM '. HL '.: KR : KS;
(593. If lines be cut by three parallel planes, the corresponding sects are propor-
tional.)
.-. WM. MZ : C/L.LF :: fR .RK : fS . SK ;
:, ^MO : ^ LN :: fR .RK : fS .SK (i)
{542. Mutually equiangular parallelograms have to one another the ratio which is
compounded of the ratios of their sides.)
TRI-DIMENSIONAL SPHERICS.
251
But
OPQ \ Q)AB '.'. PP '. AC- :: TI . ID : TC . CD, (2)
{546. 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, it
will be a mean proportional between the segments of the diameter.)
By hypothesis and construction, in proportions (i) and (2), the
first, third, and fourth terms are respectively equal,
Theorem VII.
645. The sects Joining its pole to points on any circle of a
sphere are equal.
Proof. (600. If 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.)
646. Corollary. Since chord PA equals chord PB, there-
fore the arc subtended by chord PA in the great circle PA
equals the arc subtended by chord PB in the great circle PB.
Hence the great-circle-arcs joining a pole to points on its
circle are equal.
252
THE ELEMENTS OF GEOMETRY.
So, if an arc of a great circle be revolved in a sphere about
one of its extremities, its other extremity will describe a circle
of the sphere.
647. One-fourth a great circle is called a Quadrant.
648. The great-circle-arc joining any point in a great circle
with its pole is a quadrant.
p
^
o\
"'^N
/ 'N
\3j
:/
::^
649. If a point P be a quadrant from two points, A^ B, which
are not opposite, it is the pole of the great circle through A, B ;
for each of the angles POA, FOB, is right, and therefore PO
is perpendicular to the plane OAB.
650. The angle between two intersecting curves is the angle
between their tangents, at the point of intersection.
When the curves are arcs of great circles of the same
sphere, the angle is called a Spherical Angle.
TRI-DIMENSIONAL SPHERICS, 253
651. If from the vertices, A and F, of any two angles in a
sphere, as poles, great circles, BC and GH^ be described, the
angles will be to one another in the ratio of the arcs of these
circles intercepted between their sides (produced if necessary).
For the angles A and F are equal respectively to the angles
BOC and GOH.
(591. Parallels intersecting the same plane are equally inclined to it.)
But BOC and GOH are angles at the centers of equal cir-
cles, and 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 is perpendicular to that circle.
FG is ± GH. For, by hypothesis, FG is a quadrant ; there-
fore the great circle described with G as pole passes through F,
and so the arc intercepted on it between GF and GH is, by
648, also a quadrant. But, by 651, the angle at G is to this
quadrant as a straight angle is to a half-line.
Inversely, any great-circle-arc perpendicular to a great cir-
cle will pass through its pole.
For if we use G as pole when the angle 'at (7 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.
254
THE ELEMENTS OF GEOMETRY.
Theorem VIII.
653. The smallest line in a sphere^ between two points, is the
great-circle-arc not greater than a semicircle^ which joins them.
Hypothesis. AB is a greaf-c/rc/s-arc, not greater than a
semicircle, joining any two points A and B on a sphere.
First, let the points A and B be joined by the broken line A CB,
which consists of the two great- circle -arcs A C and CB.
Conclusion. AC + CB > AB.
Proof. Join O, the center of the sphere, with A, By and C.
^ AOC -f ^ COB > ^ AOB.
(603. If three lines not in the same plane meet at one point, any two of the angles
formed are together greater than the third.)
But the corresponding arcs are in the same ratio as these angles,
.-. AC -\- CB>AB.
Second, let F be any point whatever on the great-circle-arc AB.
The smallest line on the sphere from A to B must pass through B.
For by revolving the great-circle-arcs AB and BB about A and B
as poles, describe circles.
These circles touch at B, and lie wholly without each other ; for let
F be any other point in the circle whose pole is B, and join FA, FB
by great-circle-arcs, then, by our First,
FA ^ FB> AB,
.*. FA > FA, and F lies without the circle whose pole is A.
TRI-DIMENSIONAL SPHERICS. 255
Now let ADEB be any line on the sphere from AXo B not passing
through /*, and therefore cutting the two circles in different points, one
in D, the other in E. A portion of the line ADEB, namely, DE, lies
between the two circles. Hence if the portion AD be revolved about
A until it takes the position AGP, and the portion BE be revolved
about B into the position BHP, the line AGPHB will be less than
ADEB. Hence the smallest line from A to B passes through P, that
is, through any or every point in AB ; consequently it must be the arc
AB itself.
654. Corollary. A sect is the smallest line in a plane
between two points.
BOOK IX.
TWO-DIMENSIONAL SPHERICS.
INTRODUCTION. .
655. Book IX. will develop the Geometry of the Sphere,
from theorems and problems almost identical with those whose
assumption gave us Plane Geometry. In Book VIII., these
have been demonstrated by considering the sphere as contained
in ordinary tri-dimensional space. But, if we really confine our-
selves to the sphere itself, 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
characteristic line, the great circle ; just as the assumptions in
our first book were the most fundamental properties of the
plane and its characteristic line.
So now we will call a great circle simply the spherical line ;
and, whenever in this book the word line is used, it means
spherical line. Sect now means a part of a line less than a
half-line.
257
258 THE ELEMENTS OF GEOMETRY.
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
change. Suppose a portion of this line is such, that, when
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
the trace of its first end point. Call such a portion a half-
line, and any lesser portion a sect. Suppose, that, while the
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
other change ; that is, figures are independent of their place on
the sphere.
658. Through any two points in the sphere can be passed a
line congruent with the generating line of the sphere. In
Book IX., the word line will always mean such a line, and sect
will mean a portion of it less than half.
659. Two sects cannot meet twice on the sphere ; that is, if
two sects have two points in common, the two sects coincide
between those two points.
660. If two lines have a common sect, they coincide
throughout. Therefore through two points, not end points of
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. 259
662. A piece of the sphere from along one side of a line
will fit either side of any other portion of the line.
DEFINITIOXS.
663. If one end point of a sect is kept fixed, the other end
point moving in the sphere describes what is called an arc,
while the sect describes at the fixed point what is called a
spherical angle. The angle and arc are greater as the amount
of turning in the sphere is greater.
664. When a sect has turned sufficiently to fall again into
the same line, but on the other side of the fixed point or ver-
tex, the angle described is called a straight angle, and the arc
described is called a semicircle.
665. Half a straight angle is called a right angle.
666. The whole angle about a point in a sphere is called a
perigon : the whole arc is called a circle.
The point is called the pole of the circle, and the equal
sects are called its spherical radii.
ASSUMPTIONS.
667. A circle can be described from any pole, with any sect
as spherical radius.
668. All straight angles are equal.
26o
THE ELEMENTS OF GEOMETRY.
669. Corollary I. All perigons are equal.
670. Corollary II. If one extremity of a sect is in a line,
the two angles on the same side of the line as the sect are
together a straight angle.
671. Corollary III. Defining adjacent angles as two
angles having a common vertex, a common arm, and not over-
lapping, it follows, that, if two adjacent angles together equal
a straight angle, their two exterior arms fall into the same
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, 26 1
675. Corollary II. Every line bisects the sphere, — cuts
it into hemispheres.
Reduced Properties of the Sphere.
Theorem I.
676. All lilies in a sphere ijttersect, and bisect each other at
their points of intersection.
Let BB' and CC be any t\vo lines.
Since, 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
sect AB coincides with some point in the sect AB' y 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 vertical, are,
by 672, equal, AC A' will lie along AC A'. Therefore the second point
of intersection of AB and A C must coincide with the second point of
intersection of AB' and A C, and ABA' be a half-line, as also A CA' .
262
THE ELEMENTS OF GEOMETRY.
677. A spherical figure such as ABA'CA^ which is contained
by two half -lines, is called a Lune.
Theorem II.
678. The angle contained by the sides of a hme at one of their
points of intersection equals the angle contained at the other
For, if the two angles are not equal, one must be the greater. Sup-
pose :^ A> :^ 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
A CA' . Then, since we have supposed -4- A> ^ A' , the half-Hne A' CA
would start from the trace of A between the trace of AC A' and the
trace of ABA' . Starting between them, it could meet neither again
TWO-DIMENSIONAL SPHERICS.
263
until it reached the trace of A' ; and so we would have the surface of the
lune less than its trace, which is contrary to our first assumption (657).
.*. the two angles cannot be unequal.
DEFINITIONS.
679. The Supplement of a Sect is the sect bywhich it differs
from a half-line.
680. One-quarter of a line is called a Quadrant.
618. A Spherical Polygoii is a closed figure in the sphere
formed by sects.
682. A Spherical Triangle is a convex spherical polygon of
three sides.
683. Symmetrical Spherical Polygons are those in which the
sides and angles of the one are respectively equal to those of
264 THE ELEMENTS OF GEOMETRY,
the other, but arranged in the reverse order. If one end of a
sect were pivoted v^ithin one polygon, and one end of another
sect pivoted within the symmetrical polygon, and the two sects
revolved so as to pass over the equal parts at the same time,
one sect would move clockwise, while the other moved counter-
clockwise.
684. 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.
Any two figures are symmetrical with respect to an axis
when every point of one has its symmetrical point on the
other.
Theorem III.
685. The perimeter of a convex spherical polygon wholly con-
tained within a second spherical polygon is less than the perimeter
of the second.
Let ABFGA be a convex spherical polygon wholly contained in
ABCDEA,
TWO-DIMENSIONAL SPHERICS. 26$
Produce the sides A G and GF to meet the containing perimeter at
K and L respectively.
By 66 1, a sect is the smallest path between its end points,
.-. BF<BCLF, and LG<LKG, and KA<KDEA;
/. BFGA < BCLGA < BCKA < BCDEA,
Theorem IV.
686. The stint of the sides of a convex spherical polygon is
less than a line.
Let ABCDEA be the polygon, and let its sides BA and ^C be
produced to meet again at M. Since the polygon is convex, the
^BCD <?,t. :^, and also ^ BAB < st. ^ ; therefore ABCDEA lies
wholly within ABCMA ; therefore, by 685, the perimeter oi ABCDEA
< 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
plane through a straight angle. This holds good when for
"plane" we substitute "sphere."
If two figures have axial symmetry in a plane, they can be
made to coincide by folding the plane over along the axis, but
not by any sliding in their plane. That is, we must use the
266 THE ELEMENTS OF GEOMETRY.
third dimension of space, and then their congruence depends
on the property of the plane that its two sides are indistin-
guishable, so that any piece will fit its trace after being turned
over. This procedure, folding along a line, can have no place
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 V.
688. Two spherical tria7tgles having two sides and the in-
cluded angle of one equal respectively to two sides and the
included angle of the other, are either congruent or symmetrical.
(i) If the parts given equal are arranged in the same order, as in
DEF and ABC, then the triangle DEF can be moved in the sphere
until it coincides with ABC.
(2) If the parts given equal are arranged in reverse order, as in
DEF and A' B' C\ by making a triangle symmetrical to A'B'C, as
ABC, we get the equal parts arranged in the same order as in DEF,
which proves DEF congruent to any triangle symmetrical to A'B' C\
TWO-DIMENSIONAL SPHERICS.
267
Theorem VI.
689. If two spherical triangles have two sides of the one
equal respectively to two sides of the other, but the ijicluded angle
of tJie 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.
Hypothesis. AB = DE, and AC ^ EF, bui 4. BAC> 4. EDF,
Conclusion. BC > EF.
Proof. When AB coincides with DE, if the points C and F are
on the same side of the line AB, then, since ^ BAC > ^ EDF, the
side DF will lie between BA and A C, and DF must stop either within
the triangle ABC, on BC, or after cutting BC.
When F is within the triangle, by 685, ^C + CA > EF + FD ;
but^C= DF,
.-. BC>EF,
When F lies on BC, then EF is but a part of BC. In case DF
cuts BC, call their intersection point G. Then BG + GF > EF, and
GC -^ GD> AC,
.-. BC ^ FD >AC ^ EF',
hwi FD = AC,
.'. BC>EF.
If, when one pair of equal sides coincide, the other pair lie on oppo-
site sides of 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
side of the second triangle, and therefore the third side of the first
greater than the third side of the second.
268 THE ELEMENTS OF GEOMETRY.
690. From 6'^% and 689, by 33, Rule of Inversion, if two
spherical triangles have two sides of the one equal respectively
to two sides of the other, the included angle of the first is
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.
691. 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.
Problem I.
692. To bisect a given spherical angle.
Let BA C be the given 2^ .
In its arms take equal sects AB and A C each less than a quadrant.
Join BC. With B as pole, and any sect greater than half BC, but
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 EDF at D within the lune BAC. Join DB,
DC, DA. i9^ shall bisect the angle ^^C.
For the 'as ABD, A CD, having AD common and AB = AC, and
BD = DC (spherical radii of equal circles), are, by 691, symmetrical,
.'. ^ BAD = ^ CAD.
TWO-DIMENSIONAL SPHERICS.
269
693. Corollary I. To bisect the re-entrant angle BAC,
produce the bisector of the angle BAC.
694. Corollary II. To erect a perpendicular to a given
line from a given point in the line, bisect the straight angle at
that point.
Problem II.
695. To bisect a given sect.
With A and B, the extremities of the given sect, as poles, and equal
spherical radii greater than half AB, but less than a quadrant, describe
arcs intersecting at C.
Join AC and BC; and, by 692, bisect the angle ACB, and produce
the bisector to meet AB at Z>.
B> is the mid point of the sect AB.
For, by 688, a" A CD is symmetrical to 'aBCJD,
696. Corollary. The angles at the base of an isosceles
spherical triangle are equal.
270
THE ELEMENTS OF GEOMETRY.
Problem III.
697. To draw a perpendicular to a given line from a given
point in the sphere not in the line.
Given, the line AB, and point C.
Take a point D on the other side of the line AB, and with C as
pole, and CD as spherical radius, describe an arc cutting AB in F
and G. Bisect the sect FG at B, and join CB. CB shall be ± AB.
For, by 691, 'as GCB and FCB are symmetrical.
Theorem VII.
698. If two lines be drawn in a sphere, at right angles to a
given line, they will intersect at a point from which all sects
drawn to the given line are equal.
Let AB and CB, drawn at right angles to AC, intersect at B, and
meet A C again at A' and C respectively.
TWO-DIMENSIONAL SPHERICS. 27 1
Then 4. BA'C = ^ BAC, and 4. EC A' = 4 BCA' -,
(678. The angles contained by the sides of a lune, at their two points of intersection,
are equal.)
moreover,
AC = A'C\
for they have the common supplement A C . Hence, keeping A and C
on the line AC, slide ABC until AC comes into coincidence with A' C .
Then, the angles at A, C, A\ C, being all right, AB will lie along A'B,
and CB along C'B, and hence the figures AB C and A'B C coincide.
.-. each of the half-Unes ABA' and CB C is bisected at B.
In like manner, any other line drawn at right angles to ^ C passes
through B, the mid point of ABA' .
Hence every sect from AC to B is a. quadrant.
699. Corollary I. A line is a circle whose spherical
radius is a quadrant.
700. Corollary II. A point which is a quadrant from
two points in a line, and not in the line, is its pole.
701. Corollary III. Any sect from the pole of a line to
the line is perpendicular to it.
702. Corollary IV. Equal angles at the poles of lines
intercept equal sects on those lines.
703. Corollary V. If K be the pole, and FG a sect of
any other line, the angles and semilunes ABC and FKG are to
one another as ^ 6^ to FG.
2/2
THE ELEMENTS OF GEOMETRY.
704. We see, from 698, that a spherical triangle may have
two or even three right angles.
If a spherical triangle ABC has two right angles, B and C^
it is called a bi-rectaiigular triangle. By 698, the vertex A is
the pole of BC, and therefore AB and AC are quadrants.
705. The Polar of a given spherical triangle is a spherical
triangle, the poles of whose sides are respectively the vertices
of the given triangle, and its vertices each on the same side of
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.
Hypothesis. Lei A'B'C be the polar of ABC,
Conclusion. Then ABC is the polar of A'B'C.
TWO-DIMENSIONAL SPHERICS, 2/3
Proof. Join A'B and A' C,
Since B is the pole of A' C , therefore BA' is a quadrant ; and since
C is the pole of A'B\ therefore CA' is a quadrant ;
.*. by 700, A' is the pole of BC.
In like manner, B' is the pole of A C, and C of AB,
Moreover, A and A' are on the same side oi B'C'y B and B' on the
same side oi A'C, and C and C on the same side oi A'B\
.-. ^.5C is the polar of A'B'C,
Theorem IX.
707. In a pair of polar triangles, any angle of either inter-
ceptSy on the side of the other which lies opposite to it, a sect which
is the supplement of that side.
Let ABC and A'B^C be two polar triangles.
Produce A'B' and B' C to meet BC at Z> and ^ respectively.
Since B is the pole of B' C , therefore BE is a quadrant ; and since C
is the pole of A'B' , therefore CD is a quadrant ; therefore BE -f CD
= half-line, but BE -{- CD = BC -{- DE. Therefore DE, the sect
of BC which A' intercepts, is the supplement of BC.
Exercises. 109. Any lune is to a tri-rectangular triangle
as its angle is to half a right angle.
274
THE ELEMENTS OF GEOMETRY.
Theorem X.
708. If two angles of a spherical triangle be equals the sides
which subtend them are equal.
Hypothesis. In ^ ABC let 4. A = ^ C,
Conclusion. BC = AB.
Proof. For draw A'B'C, the polar of ABC.
Now, on B' C and A'B' the equal 7^^ A and C intercept equal sects.
Therefore B' C and A'B'y being, by 707, the supplements of these
equal sects, are equal, .*. "4- A' — -^ C,
(696. The angles at the base of an isosceles spherical triangle are equal.)
.'. the supplements oi BC and AB are equal,
/. BC ^ AB,
Theorem XI.
709. If one angle of a spherical triangle be greater than a
second^ the side opposite the first must be greater than the side
opposite the second.
In'^ABC, 4 C>4.A. Make ^ ACD = ^^;
/. by 708, AD = CD,
TWO-DIMENSIONAL SPHERICS. 2y$
But, by 66i, CD -^ DB> BC,
.-. AD -^ DB>BC.
710. From 708 and 709, by 33, Rule of Inversion,
If one side of a spherical triangle be greater than a second,
the angle opposite the first must be greater than the angle
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 congrue7it or symmetricaL
For the first triangle can be moved in the sphere into coincidence
with the second, or with a triangle made symmetrical to the second.
Theorem XIII.
712. Two spherical triangles having three angles of the one
equal respectively to three angles of the other^ are either congruetit
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.)
2/6 THE ELEMENTS OF GEOMETRY.
Hence these polars, being, by 691, congruent or symmetrical, are
respectively equiangular, and therefore the original spherical triangles
are respectively equilateral.
Theorem XIV.
713. An exterior angle of a spherical triangle is greater than^
eqnal to, or less tkaUy either of the interior opposite angles, accord-
ing as the medial from the other interior opposite angle is less
than, equal to, or greater than, a quadrant.
Let ACD be an exterior angle of the a" ABC. By 695, bisect AC
at E. Join BE, and produce to E, making EE = BE. Join EC»
A ABE ^ A CEE,
(688. Spherical triangles having two sides and the included angle equal are congru-
ent or symmetrical.)
.'. ^ BAE = ^ ECE.
If, now, the medial BE be a quadrant, BEE is a half-line, and, by
676, E lies onBD;
.'. 4.DCE coincides with ^ ECE,
.\ ^ DCE = t BAE.
If the medial BE be less than a quadrant, BEE is less than a half-
line, and E lies between A C and CD ;
.♦. ^ DCA > ^ ECE,
.-. ^ DCA > ^ BAC.
And if BE be greater than a quadrant, BEE is greater than a half-
hne, and E lies between CD and A C produced ;
.-. ^ DCA < -4. ECE,
/. ^ DCA < 4 BAC.
TWO-DIMENSIONAL SPHERICS. 2'J'J
Thus, according as BE is greater than, equal to, or less than, a
quadrant, the exterior ^ A CD is less than, equal to, or greater than, the
interior opposite 2(1 BA C,
714. From 713, by 33, Rule of Inversion,
According as the exterior angle ACD is greater than, equal
to, or less than, the interior opposite angle BAC, the medial
B£ is less than, equal to, or greater than, a quadrant.
Theorem XV.
715. Two spherical triangles having the a^tgles at the base of
the 07ie eqical to the a^tgles at the base of the other^ and the sides
opposite one pair of equal angles eqical, are either coJigricent or
symmetrical, provided that in neither triangle is the third
angular point a quadrant from any point iii that half of its
base not adjaceftt to 07ie of the sides equal by hypothesis.
First, if the parts given equal lie clockwise in the two spherical
triangles, as in ABC and DEF, where we suppose ^ B = }^ E,
4. C = 4. F,2ir\dAB == DE, make DE coincide with AB ; then EF
will lie along BC, and DF must coincide with AC. For if it could
take any other position, as AG, it would make 2^^ AGC with exterior
^ AGB = interior opposite ^ C, and therefore, by 714, with medial
AH a quadrant, which is contrary to our hypothesis.
Second, if the equal parts lie in one spherical triangle clockwise,
in the other counter-clockwise, as in 'KA'B'C and ^ DEF, then
A B>EF ^^ ABC, which is symmetrical to a A' B' C,
2/8 THE ELEMENTS OF GEOMETRY.
Theorem XVI.
716. If a sect drawn in the sphere from a point perpendictilar
to a line be less than a qnadranty it is the smallest which can be
drawn in the sphere from the point to the line ; of others^ that
which is nearer the perpendicular is less than that which is more
remote ; also to every sect drawn on one side of the perpendicular
there can be drawn one^ and only one^ sect equal on the other side.
Let A be the point, BD the Hne, AB J. BD^ and AD nearer than
AE to AB.
Produce AB to C, making BC ^ AB. Join CD, CE,
Then AD = CD.
(688. Spherical triangles having two sides and the included angle equal are con-
gruent or symmetrical.)
But, by 661,
AD -j- CD or iAD > AC or 2AB,
.-. AD > AB.
Also, by 685,
AE + CE or 2AE > AD -f CD or 2 AD.
Again, make BE — BD. Join AE.
As before, AE = AD, and we have already shown that no two
sects on the same side of the perpendicular can be equal.
TWO-DIMENSIONAL SPHERICS. 279
717. Corollary. The greatest sect that can be drawn from
A to BD is the supplement of AB.
Theorem XVII.
718. If two spherical triangles have two sides of the one
equal to two sides of the other^ and the angles opposite one pair
of equal sides equals the angles opposite the other pair a7'e either
equal or supplemental.
First, given the equal parts AB = DE, AC = DF, and ^ B =
^ E arranged clockwise in the two spherical triangles.
\i -2^ A =. }^ D, the spherical triangles are congruent.
\i ^ A is not = ^ Z>, one must be the greater.
Suppose -4. A> -i^ D. Make DE coincide with AB. Then, since
^B = :4E, side EF will he along BC; since ^ A> ^ D, side VF
will lie between AB and A C, as at A G.
Now, the two angles at G are supplemental ; but one is -^ F and
the other = ^ C, because "a CFG is isosceles.
Second, if the parts given equal lie clockwise in one spherical tri-
angle and counter-clockwise in the other, as in "a A'B' C and 'a DEF,
then, taking the "a ABC symmetrical to A'B' C, the above proof shows
^ F equal or supplemental to ^ C, that is, to ^ C\
280 THE ELEMENTS OF GEOMETRY.
Theorem XVIII.
719. Symmetrical isosceles spherical triangles are congruent.
For, since four sides and four angles are equal, the distinction
between clockwise and counter-clockwise is 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.
For every point in this perpendicular bisector, and no point out of
it, 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.
722. Any angle made with a side of a spherical triangle by
joining its extremity to the circwnc enter ^ equals half the angle-
sum less the opposite aiigle of the triangle.
FoT^A-}-^B-{-^C=2^ OCA + 2 ^ OCB ± 2 ^ OAB,
^A 4- ^B 4- ^ C
t OCA = ^ \ U OCB ± ^ OAB)
^A -\- ^B + 4.C
^ ^B.
282 THE ELEMENTS OF GEOMETRY.
723. Corollary. 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 = lune ABHCA.
For DF — BC, having the common supplement CD ) and FA —
CH, having the common supplement HF'^ and AD = BH^ having the
common supplement HD ;
.-. aADF^aBCFT,
.-. A ABC + A ADF = A ABC -f a BCH = lune ABHCA.
725. The Spherical Excess of a spherical triangle is the
excess of the sum of its angles over a straight angle. In gen-
eral, the spherical excess of a spherical polygon is the excess
of the sum of its angles over as many straight angles as it has
sides, less two.
TWO-DIMENSIONAL SPHERICS. 283
Theorem XXII.
726. A spherical triangle is equivalent to a lime whose angle
is half the t7 iattgles spherical excess.
Proof. Produce the sides of the "a ABC until they meet again,
two and two, at Dy F^ and H. The a" ABC now forms part of three
lunes, whose angles are A, Bj and C respectively.
But, by 724, lune with 2^ ^ = a ABC + a ADF.
Therefore the lunes whose angles are Ay By and C, are together
equal to a hemisphere plus twice 'a ABC.
But a hemisphere is a lune whose angle is a straight angle,
.-. 2^ ABC = lune whose ^ is (^ + ^ + C - st. ^)
= lune whose ^ is e,
727. Corollary I. The sum of the angles of a spherical
triangle is greater than a straight angle and less than 3 straight
angles.
728. Corollary II. Every angle of a spherical triangle is
greater than ^e.
T2.g, 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
as their spherical excesses, since, by 703, lunes are as their
angles.
284 THE ELEMENTS OF GEOMETRY.
731. Corollary V. To construct a lune equivalent to any
spherical polygon, add its angles, subtract {71 — 2) straight
angles, halve the remainder, and produce the arms of a half
until they meet again.
Theorem XXIII.
732. If the line be completed of which the base of a given
spherical triangle is a sect^ and the other two sides of the triangle
be produced to meet this line, and a circle be passed through these
tzvo points of intersectio7i and the vertex of the triangle, the locns
of the vertices of all triangles equivalent to the give?z triajtgle^
and on the same base with it, and on the same side of that base,
is the arc of this circle terminated by the intersection points and
cojttaining the vertex.
C
Let ABC be the given spherical triangle, AC its base. Produce
AB and CB to meet AC produced in D and E respectively. By 719,
pass a circle through B,D,E. Let F be any point in the arc EBD.
Join AP and CP. AP produced passes through the opposite point
D, and CP through E, forming with DE the a PDE. Join F, its
circumcenter, with P, D, and E, Since, by 678, the two angles of a
lune are equal,
.*. 2^: PAC = St. ^ - :^ PDE,
^ PCA = St. ^ - 2< PED,
4.APC ^ 4.DPE',
TWO-DIMENSIONAL SPHERICS. 285
/. 4.{PAC ■\- PCA 4- APC) = 2St. ^s - ^ {PDE + PED - DPE)
= 2 St. ^ s -- 2 ^ EZ>E = a constant.
(722. Any angle made with a side of a spherical triangle by joining its extremity
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 when it
meets the circle, but will not cut it.
Theorem XXIV.
734. T/ie line drawn at right angles to the spherical radius of
a circle at its extremity touches the circle.
Let BD be perpendicular to the spherical radius AB, Join A with
any point C in BD.
By 7 1 6, ^ C > AB, therefore C is without the circle.
And no other line through B, as BF, can be tangent.
For draw AE _L BF. By 716, AB > AE,
,'. E is within the circle.
286
THE ELEMENTS OF GEOMETRY.
Theorem XXV.
735. In a sphere the stun of one pair of opposite angles of a
quadfilateral inscribed in a circle equals the sum of the other
pair.
Join E^ the circumcenter, with A, By C, D, the vertices of the
inscribed quadrilateral.
By 696, ^ ABC = 4. BAE + ^ BCE, and 4. ADC = 4. DAE
+ 4 DCE,
/. 4. ABC + 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.
p
TWO-DIMENSIONAL SPHERICS.
287
Theorem XXVII.
737. EquQ,l spherical chords cut equal circles into the same
two arcs.
Theorem XXVIII.
738. In equal circles^ angles at the corresponding poles have
the same ratio as their arcs.
Theorem XXIX.
739' Four pairs of equal circles ca7t be drawn to touch three
non-concurrent lines in a sphere.
BOOK X.
POLYHEDRONS.
740. A Polyhedron is a solid bounded by planes.
741. The bounding planes, by their intersections, determine
the Faces of the polyhedron, which are polygons.
742. The Edges of a polyhedron are the sects in which its
faces meet.
290 THE ELEMENTS OF GEOMETRY.
743. The Summits of a polyhedron are the points in which
its 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 is called the Apex^ and the face
not passing through the apex is taken as the Base.
747. The faces and edges which meet at the apex are called
Lateral Faces and Edges.
748. Two polygons are said to be parallel when each side of
the one is parallel to a corresponding side of the other.
749. A Prism is a polyhedron two of whose faces are con-
gruent parallel polygons, A»d the other faces are parallelo-
grams.
750. The Bases of a prism are the congruent parallel poly-
gons.
751. The Lateral Faces of a prism are all except its bases.
752. The Lateral Edges are the intersections of the lateral
faces.
753. A Right Section of a prism is a section by a plane per-
pendicular to its lateral edges.
POL YHEDRONS.
291
754. The Altitude of a Prism is any sect perpendicular to
both bases.
755. The Altitude of a Pyramid is the perpendicular from
its vertex to the plane of its base.
756. A Right Prism is one whose lateral edges are perpen-
dicular to its bases.
\
757. Prisms not right are oblique.
758. A Parallelopiped is a prism whose bases are parallelo-
grams.
759. A Quader is a parallelopiped whose six faces are rect-
angles.
760. A Citbe is a quader whose six faces are squares.
292
THE ELEMENTS OF GEOMETRY.
Theorem I.
761. All the summits of any polyhedron may be joined by
one closed line breaking only in themy and lying wholly on the
surface.
For, starting from one face, ABC . . . , each side belongs also to a
neighboring polygon.
Therefore, to join A and B, we may omit AB, and use the remain-
der of the perimeter of the neighboring polygon a. In the same way,
to join B and C, we may omit BC, and use the remainder of the
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 in the same way from C around the neighboring
polygon c.
When this procedure has taken in all summits in faces having an
edge in common with ABC . . . , we may, by proceeding from the
closed broken line so obtained, in the same way take in the summits
on the next series of contiguous faces, etc.
So continue until the single closed broken Hne goes once, and only
once, through every summit.
POL YHEDRONS.
593
Theorem II.
762. Cutting by diagonals the faces not triangles into tri-
angleSy the whole surface of aiiy polyhedron contains four less
triattgles than double its finmber of summits.
For, joining all the summits by a single closed broken line, this cuts
the surface into two bent polygons, each of which contains S — 2 tri-
angles, where S is the number of summits.
763. Corollary. The sum of all the angles in the faces
of any polyhedron is as many perigons as the polyhedron has
summits, less two.
764. Remark. Theorem II. is called Descartes' Theorem,
and is really the fundamental theorem on polyhedrons, though
this place has long been held by Theorem III., called Euler's
Theorem, which follows from it with remarkable elegance.
294
THE ELEMENTS OF GEOMETRY.
Theorem III.
765. The number of faces and summits iri any polyhedron^
taken together, exceeds by two the member of its edges.
Case I. If all the faces are triangles. Then, by 762,
7^ = 2(6'- 2).
But also
2E = zF,
for each edge belongs to two faces, and so we get a triangle for every
time 3 is contained in 2E.
By adding, we have 2E = 2F + 2{S — 2); that is,
F -\- S = E -^ 2,
Case II. 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.
295
equality or inequality relation oi F -\- S \.o E -{■ 2 ; for such replace-
ment only adds the same number to F as to E^ and changes one face
to a summit. But, after all polygonal faces have been so replaced,
F ■\- S =■ E -\- 2y \yy Case I. Therefore always the relation was
equality.
Theorem IV.
766. Qtiaders having Congruent bases are to each other as
their altitudes.
1
jm~- - - —
*' \
1
y
7
at
A /
A 1
-•r
hH
,>-.-
7
Hypothesis. Let a and c^ be the aliiiudes of two quaders, Q
and Q, having congruent bases B.
Conclusion. Q : Q w a \ a'.
Proof. Of a take any multiple, ma ; then the quader on base B
with altitude ma is mQ.
In the same way, take equimultiples na' , nQ.
According as m Q is greater than, equal to, or less than, n Q, we
have ma greater than, equal to, or less than na' \ therefore, by defini-
tion,
Q \ Q \'. a \ a\
Exercises, no. In no polyhedron can triangles and three-
faced summits both be absent ; together are present at least
eight. Not all the faces nor all the summits have more than
five sides.
III. There is no seven-edged polyhedron.
296
THE ELEMENTS OF GEOMETRY.
Theorem V.
767. Quaders having equal altitudes are to each other as their
bases.
"/HA A
A A
a
1
I
1
1
/
f
Jo-
JP
V
A /
i
Hypothesis. Let the rectangles he and h'<f be the bases of two
quaders, Q and Q, of altitude a.
Conclusion. Q \ Q w be \ b'(f ,
Proof. Make P a third quader of altitude a and base hc\
Now, considering the rectangles ab as the bases of Q and P, by 766,
Q ', P '.', c : c';
considering the rectangles «/ as bases of P and ^, by 766,
P '. Q '.'. b : b'.
Therefore, compounding the ratios,
Q \ q '.'. be ', b'<f.
POL YHEDRONS.
297
Theorem VI.
768. Tivo qtiaders are to each other ift the ratio compounded
of the ratios of their bases and altitudes.
y\
7
L
/
7o
A
•
7
f
If
F-
A /
/
^.
Hypothesis. Lei Q and Q be two quaders, of aliifude a and a\
and base be and b'c', respectively.
Conclusion. Q : Q w abc \ a'b'cf.
Proof. Make P a third quader of altitude a' and base be.
Then, by 766,
Q : P '.: a -.a'-,
and, by 767,
P ', q '.'. be '. b'd.
Therefore, compounding,
Q '. Q '.'.abc '. a'b'(f.
298
THE ELEMENTS OF GEOMETRY.
Theorem VII.
769. Any parallelopiped is equivalent to a quader of equiva-
lent base and equal altitude.
For, supposing AB an oblique parallelopiped on an oblique base,
prolong the four edges parallel to AB^ take sect CD = AB, and draw-
ee J_ C£>, and DF \\ CE. Through CE and /?i^pass parallel planes.
Now the solids DFB and CEA are congruent, having all their angles
and edges respectively equal. Taking each in turn from the whole solid
DFAj leaves parallelopiped AB = CD.
In the line CD take GH = CD, and through G and H pass planes
perpendicular to GH.
The solids DFH and CEG are congruent, therefore parallelopiped
CD = GH.
Now prolong the four edges not parallel to GH, and take LM —
HK, and through L and M pass planes perpendicular to LM.
As before, parallelopiped GH = LM\ but LM is a quader of
equivalent base and equal altitude to parallelopiped AB.
POL YHEDRONS.
299
Theorem VIII.
770. A plane passed through two diagonally opposite edges of
a parallelopiped divides it into two equivalent triangidar prisms.
fa>\
V
/!x' >
B
>. ./'.'T--^^^
<^/Jr^^
JT
5^s^-r
n
-k
^
. ' ^
^
'• ^«
^*'
2
^^ ^ •
7 /-"^ >^
K^
^
If the lateral faces are all rectangles, the two prisms are congruent ;
if not, the prisms are still equivalent.
For draw planes perpendicular to AA' at the points A and
A'. Then the prism ABC A'B'C is equivalent to the right prism
AEM A'E'M', because the pyramid AEBCM'\% congruent to the pyr-
amid A'E'B'C'M'. In the same way, ADC A'D'C is equivalent to
ALM A'UM'. But AEM A'E'M' and ALM A'L'M' are congruent,
/. ABC A'B'C and ADC A' U C are equivalent.
771. Corollary. Any triangular prism is half a parallelo-
piped of twice its base but equal altitude.
300
THE ELEMENTS OF GEOMETRY.
Theorem IX.
772. If a pyramid be ctit by a plane parallel to its base, the
section is to the base as the square of tJie perpe7idicular on it,
from the vertex, is to the square of the altitude of the pyramid.
The section and base are similar, since corresponding diagonals
cut them into triangles similar in pairs because having all their sides
respectively proportional.
For
A'C : AC :: VC : VC : : VO" : VO,
and
B'C : BC :: VC : VC,
.'. AC : AC :: B' C : BC;
and in the same way,
B'C : BC :: A'B' : AB,
.'. t.A'B'C'^t.ABC,Q\c.,
section : base :-. A' C"" -. AC^ w VO^ : VC
773. Corollary. In pyramids of equivalent bases and
equal altitudes, two sections having equal perpendiculars from
the vertices are equivalent.
POL YHEDRONS.
301
Theorem X.
774. Tetrahedra {triangular pyramids) having equivalent
bases and equal altitudes are equivale7it.
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
302 THE ELEMENTS OF GEOMETRY.
as we please by taking n sufficiently great ; but it is always greater than
the constant difference between the tetrahedra, and so that constant
difference must be nought.
Theorem XI.
775. A triangular pyramid is one-third of a triangular prison
of the same base and altitude.
Let E ABC be a triangular pyramid. Through one edge of the
base, as A C, pass a plane parallel to the opposite lateral edge, EB, and
through the vertex E pass a plane parallel to the base. The prism
ABC DEF has the same base and altitude as the given pyramid. The
plane AFE cuts the part added, into two triangular pyramids, each
equivalent to the given pyramid; for E ABC and A DEF have the
same altitude as the prism, and its bottom and top respectively as bases ;
while E AFC and E AFD have the same altitude and equal bases.
BOOK XL
MENSURATION, OR METRICAL GEOMETRY.
CHAPTER I.
THE METRIC SYSTEM. — LENGTH, AREA.
776. In practical science, every quantity is expressed by a
phrase consisting of two components, — one a number, the
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. The Meastirement of a magnitude consists in finding
this number.
778. Measurement, then, is 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
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 tem-
perature of melting ice.
780. This material meter is the ultimate standard univer-
sally chosen, because of the advantages of the metric system
303
304
THE ELEMENTS OF GEOMETRY.
of subsidiary units connected with it, which uses only decimal
multiples and sub-multiples, being thus in harmony with the
decimal nature of the notation of arithmetic.
781. The metric system designates multiples by prefixes
derived from the Greek numerals, and sub-multiples by pi^e-
fixes from the Latin numerals.
Prefix.
Meaning as used.
Derivation.
Abbreviation.
myria-
ten thousand
fJiVpLOL
kilo-
one thousand
XlXlol
k-
hecto-
one hundred
iKarov
h-
deka-
ten
SeKa
da-
deci-
one-tenth
decem
d-
centi-
one-hundredth
centum
c-
milli-
one-thousandth
mille
m-
The abbreviation for meter is m. ; hence km. for kilometer, mm. for millimeter.
782. The adoption of the meter gives the world one stand-
ard sect as fundamental unit.
783. The Levgth of any sect is its ratio to the meter ex-
pressed approximately in numbers.
10
mm
I decimeter = 10 centimeters = 100 millimeters.
Men of science often express their measurements in terms
of a subsidiary unit, the Centimeter.
The length of a sect referred to the centimeter as unit is
one hundred times as great as referred to the meter.
MENSURATION-, OR METRICAL GEOMETRY.
305
784. An accessible sect may be practically measured by the
direct application of a known sect, such as the edge of a ruler
suitably divided.
But because of incommensurability, any description of a
sect in terms of the standard sect 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, ^■'^).
In science the Square Centimeter (^'^•^) is adopted as the
primary unit of surface.
788. To find the area of a rectangle.
3 f
Rule. Multiply the base by the altitude.
Formula. R = ab.
Proof. — 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
306 THE ELEMENTS OF GEOMETRY.
numbers. In the rectangle ABCD, divide AD into ^, and AB
into b, equal parts. Through the points of division draw lines
parallel to the sides of the rectangle. These lines divide the
rectangle into a number of squares, each of which equals the
assumed unit of surface. In the bottom row, there are b such
squares ; and, since there are a rows, we have b squares re-
peated a times, which gives, in all, ab squares.
Note. The composition of ratios includes numerical multiplication
as a particular case.
But ordinary multiplication is also an independent growth from
addition.
In this latter point of view, the multiplier indicates the number of
additions or repetitions, while the multiplicand indicates the thing
added or repeated. This is not a mutual operation, and the product
is always in terms of the unit of the multiplicand. The multiplicand
may be any aggregate ; the multiplier is an aggregate of repetitions.
To repeat a thing does not change it in kind, so the result is an aggre-
gate of the same 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 com-
mutative.
This is the only sort of multiplication needed in mensuration ; for
all ratios are supposed to be expressed exactly or approximately in num-
bers, and in our rules it is only of these numbers that we speak. Thus,
when the rule says, " Multiply the base by the altitude," it means, Mul-
tiply 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 is a tmmber, which we prove to be the area of the rectangle ;
that is, its numerical measure in terms of the superficial unit. This is
the meaning to be assigned whenever in mensuration we speak of the
product of one sect by another.
General Proof. If L represent the unit for length, and
5 the unit of surface, and ^ ab the rectangle, the length of
MENSURATION, OR METRICAL GEOMETRY.
307
whose base is b and the length of whose altitude is a^ then, by
542,
C7 cib aL bL
But the first member of this equation is the area of the rect-
angle, which number we may represent by R ; and the second
member is equal to the product of the numbers a and b ;
.'. R = ab.
clZ
Example i. Find the area of a ribbon I*"- long and i^i"- wdde.
Answer, ydit'"** = loC^"^-^.
Since a square is a rectangle having its length and breadth
equal, therefore
789. To find the area of a square.
Rule. Take the second power of the mimber denoting the
length of its side.
Note. This is why the product of a number into itself is called
the square of that number.
790. Given, the area of a square, to find the length of a side.
Rule. Extract the square root of the number deftoting the
arecL.
308 THE ELEMENTS OF GEOMETRY,
791. Metric Units of Surface.
I hectar (^^•) = i sq. hectometer = loooo sq. meters.
I dekar = = 1000 sq. meters.
I ar (a-) = I sq. dekameter = 100 sq. meters.
I deciar = = 10 sq. meters.
I centiar = i sq. meter = i sq. meter.
I milliar = = .1 sq. meter.
I sq. decimeter = .01 sq. meter.
I sq. centimeter = .0001 sq. meter.
I sq. millimeter = .000001 sq. meter.
Example 2. How many square centimeters in 10 millimeters square?
Answer, {iQ^'^-y = loo™"^-^ = i^m.^,
792. Remark. 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 " (s''"^-)^ means a square
whose sides are each 5 kilometers long, so that the figure con-
tains 25^"^•^
Example 3. A square is looo'"-''. Find its side.
Answer, yiooo"^- = 31.623™-
793. Because the sum of the squares on the two sides
of a right-angled triangle is the square of the hypothenuse,
therefore, also, >
Given, ihe hypothenuse and one side, to find the other side.
Rule. Multiply their sum by their difference^ and extract
the square root.
Formula, c^ — a^ =:i {c -\' a) {c — a) = ^^
MENSURATION, OR METRICAL GEOMETRY. 309
From this it follows, that, in an acute-angled triangle, if we
are given two sides and the projection of one on the other, or
two sides and an altitude, we can find the third side.
Exercises. 112. What must 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.
310
THE ELEMENTS OF GEOMETRY.
CHAPTER II.
RATIO OF ANY CIRCLE TO ITS DIAMETER.
Problem I.
794. Given^ the perimeters of a regular inscribed and a sim^
ilar circiLfHscribed polygon^ to compute the perimeters of the
regular inscribed and circumscribed polygons of double the tium"
ber of sides.
a JO
Take AB a side of the given inscribed polygon, and CD a side of
the similar circumscribed polygon, tangent to the arc AB at its mid-
point E.
Join AE, and at A and B draw the tangents AF and BG; then
AE is a side of the regular inscribed polygon of double the number of
sides, and FG is a side of the circumscribed polygon of double the
number of sides.
Denote the perimeters of the given inscribed and circumscribed
polygons by p and q respectively, and the required perimeters of the
inscribed and circumscribed polygons of double the number of sides
by/' and / respectively.
MENSURATION, OR METRICAL GEOMETRY. 311
Since <9C is the radius of the circle circumscribed about the poly-
gon whose perimeter is q,
.-. q \ p w OC \ OE.
(548. The perimeters of two similar regular polygons are as the radii of their cir-
cumscribed circles.)
But, since OF bisects the 4- COE,
.'. OC : OE :: CE : FE,
(523. The bisector of an angle of a triangle divides the opposite side in the ratio of
the other two sides of the triangle.)
/. q \ p \\ CF \ FE,
whence, by composition,
p ■\- q '. 2p \\ CF '\- FE \ 2FE ^ CE \ EG \\ q \ ^,
since FG is a side of the polygon whose perimeter is /, and is con-
tained as many times in / as CE is contained in q,
.-. p -{- q : 2p '.'. q '. q'.
If, now, the letters be taken to represent lengths in terms of the unit
sect Z, this proportion is
pL -f- qL : 2pL w qL \ q'L,
which gives the number
^ = 7+i- <'>
Again, right A AEH ~ EFN, since acute ^ EAH = FEN,
,\ AH '. AE :: EN : EF,
.'. p :/ ::/ : /,
since AH and AE are contained the same number of times in / and/'
respectively, and EN and EF are contained twice that number of
times in /' and / respectively. If, now, the letters be taken to repre-
sent length in terms of the same unit sect Z, this proportion is
pL : /L : : /Z ; /Z,
which gives the number , , — -,
Therefore from the given lengths p and q we compute q' by equa-
tion (i), and then with / and / we compute /' by equation (2).
312
THE ELEMENTS OF GEOMETRY.
Theorem I.
795* The length of a circle whose diameter is unity is
3.141592+.
The length of the perimeter of the circumscribed square is 4. A
side of the inscribed square is -|V2, therefore its perimeter is 2V 2.
Now, putting p = 2V2 and ^ = 4 in 794, we find, for the perime-
ters of the circumscribed and inscribed octagons,
2/^
/ =
= 3-3137085,
/ = V'//= 3.0614675.
Then, taking these as given quantities, we put / = 3.0614675, and
^ = 3-3137085, and find by the same formulae, for the polygons of
sixteen sides, / = 3.1825979, and / = 3.1214452.
Continuing the process, the results will be found as in the following
table : —
Number of
Sides.
Perimeter
of Circumscribed
Polygon.
Perimeter
of Inscribed
Polygon.
4
4.0000000
2.8284271
8
3-3137085
3.0614675
16
32
3.1825979
3-I517249
3.1214452
3-1365485
64
3.1441184
3.1403312
128
3.1422236
3-I412773
256
3.1417504
3-1415138
512
1024
3.I41632I
3.1416025
3-I415729
3-I415877
2048
3-I415951
3.1415914
4096
3-1415933
3-I415923
8192
3.1415928
3.1415926
MENSURATION, OR METRICAL GEOMETRY. 313
But since, by 654, a chord is shorter than its arc, therefore the
circle is longer 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
perimeter ; therefore the circle whose diameter is unity is longer than
3.1415926 and not longer than 3.1415928.
796. A Variable is a quantity which may have successively
an indefinite number of different values.
797* If ^ variable which changes its value according to
some law can be made to approach some fixed value as nearly as
we please y but can never become equal to it, the constant is called
the Limit of the variable.
Example 4. The limit of the fraction -, as x increases indefitiitely,
X
is zero ; for, by taking x sufficiently great, we can make - less than any
X
assigned quantity, but we can never make it zero.
PRINCIPLE OF LIMITS.
798. Ify while tending toward their respective limits, two
variables are always in the same ratio, their limits will have
that ratio.
I 1 i hH-H M— H H^
^ ' y C' x' C Z' y Z 2
Let the sects AC and AL represent the limits of any two
variable magnitudes which are always in the same ratio, and
let Ax, Ay, represent two corresponding values of the variables
themselves ; then
Ax -.Ay '.'.AC : AL.
For if not, then
Ax : Ay :: AC : to some sect > or < AL,
314 THE ELEMENTS OF GEOMETRY,
Suppose, in the first place, that
Ax \ Ay \\ AC \ AL',
where AL' is less than AL.
By hypothesis, the variable Ay continually approaches AL,
and may be made to differ from it by less than any given quan-
tity.
Let Ax and Ay, then, continue to increase, always remaining
in the same ratio, until Ay differs from AL by less than the
quantity L'L ; or, in other words, until the point y passes
the point Z', and reaches some point, as j/', between V 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,
Ax : Ay :: Ax' : A/,
But, by hypothesis.
Ax : Ay :: AC : AL\
.-. Ao^ : A/ :: AC : AL',
But
Ax'<AC, .-. Ay<AL\
which is absurd. Hence the supposition that Ax : Ay : :
AC : AL'y or to any quantity less than AL, is absurd.
Suppose, then, in the second place, that
Ax : Ay :: AC : AL",
where AL" > AL. Now, there is some sect, as AC, less than
AC, which is to AL 2iS AC is to AL".
Substituting this ratio for that oi AC to AL'\ we have
Ax '.Ay :: AC : AL,
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 pro-
portion can be neither greater nor less than ALy it must equal
AL ; that is,
Ax \ Ay w AC \ AL.
799. Corollary. If two variables are always equal, their
limits are equal.
Theorem II.
800. Ally two circles are to each other as their radii.
Proof. By 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.
The inscribed regular polygons remaining similar to each other when
the number of sides is doubled, their perimeters continue to have the
same ratio. Assuming the circle to be the limit toward which the
perimeter of the inscribed polygon increases, by 798, Principle of
Limits, the circles have the same ratio as their radii.
801. Since c : c'
2r',
c_
2r
2r'
that is, the ratio of any one circle to its diameter is the same
as the ratio of any other circle to its diameter.
3l6 THE ELEMENTS OF GEOMETRY.
This constant ratio is denoted by the Greek letter tt. But,
by 795, this ratio for the circle with unit diameter, and there-
fore for every circle, is
TT = 3.I4I592-I-.
802. For any circle.
Formula, c = 2r7r.
CIRCULAR MEASURE OF AN ANGLE.
803. When its vertex is at the center of the circle, by 506,
any ^ _ its intercepted arc _ arc
St. ^ semicircle rTr'
any ^ _ arc
TT
So, if we adopt as unit angle the radian^ or that part of a peri-
st ^
gon denoted by -^^, that is, the angle subtended at the center
Ih
of every circle by an arc equal to its radius, and hence named
a radian, then
The number which expresses any angle in radianSy also
expresses its intercepted arc in terms of the raditis.
If ti denote the number of radians in any angle, and / the
length of its intercepted arc, then
/
u — -,
r
MENSURATION, OR METRICAL GEOMETRY, 317
The fraction arc divided by radius^ or Uy is called the circu-
lar 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
as the including angle contains the radian. In this sense an
angle at the center is measured by the arc intercepted between
its sides.
3i8
THE ELEMENTS OF GEOMETRY.
CHAPTER III.
MEASUREMENT OF SURFACES.
805. By 248, any parallelogram is equivalent to the rect-
angle of its base and altitude ; therefore,
ziy
To find the area of any parallelogram.
Rule. Multiply the base by the altitude.
Formula, /u =. ab.
806. Corollary. The area of a parallelogram divided by
the base gives the altitude.
807. By 252, any triangle is equivalent to one-half the rect-
angle of its base and altitude ; therefore,
Given, one side and the perpendicular upon it from the opposite
vertex, to find the area of a triangle.
Rule. Take half the product of the base into the altitude.
Formula, a = '^ib.
MENSURATION, OR METRICAL GEOMETRY.
319
808. Given, the three sides, to find the area of a triangle.
Rule. From half the sum of the three sides subtract each
side separately ; multiply together the half-sitm and the three
remainders. The square root of this product is the area.
Formula, t. :=i \s {s — a) {s — b) {s — c).
Proof. Calling j the projection of c on b, by 306,
a' =, b^ j^ c" — 2bj\
. . l^ -^ c^ - a""
2b
Calling the altitude //, this gives
h^ =, c^ - j^ = c^ -
{h" -\- c' - a^y
4b^
/. 4h^b^ = 4^V — (/r* + ^ - a^Yy
2/ib = V^4^V2 - (b^ ■\- c' - a'Y,
2hb
2hb^
sl{2bc -h b^ -\- c^ -
- a'){2bc - b'
— c^
+ «^),
= Vc^
-{-b -\- c){b + c
- a){a -\- b —
c){^a
- b ■\-
0,
^ /(^ + ^ + (-^ + ^
-a)(a + b -
c){a
-b^
c)
.,.. a -\- b -\- c ' b At c
Writing s — \^ gives ^
.-. \hb = \ls{s - a){s - b){s - c).
But, by 807, Ihb = A.
320 THE ELEMENTS OF GEOMETRY.
809. To find the area of a regular polygon.
Rule. Take half the prodtict of its perimeter by the radius
of the inscribed circle.
rp
Formula. TV = — .
2
Proof. Sects from the center to the vertices divide the
polygon into congruent isosceles triangles whose altitude is
the radius of the inscribed circle, and the sum of whose bases
is the perimeter of the polygon.
810. To find ihe area of a circle.
Rule. Multiply its squared radius by -k.
Formula. O == rV.
If a regular polygon be circumscribed about the circle, its
area TV, by 809, is \rp.
MENSURATION, OR METRICAL GEOMETRY, 32 1
If, now, the number of sides of the regular polygon be con-
tinually doubled, the perimeter / decreases toward c as limit,
and N toward circle. But the variables N and / are always in
the constant ratio \r\ therefore, by 798, Principle of Limits,
their limits are in the same ratio,
.-. O = \rc.
But, by 802, c = 2r7r,
811. To find the area of a sector.
Rule. Multiply the length of the arc by half the radius.
Formula. 5 = J/r = \ur''.
Proof. By 506, 5
\ :
/ :
c :
\\ u \
27r,
s
=
0/
c
=
t
27r
s
=
c
=
r'^irU
27r
s
=
¥r
=
lur^.
812. An Amnihis is the figure included between two con-
centric circles. Its height is the difference between the radii.
322
THE ELEMENTS OF GEOMETRY.
813. To find the area of a sector of an annu/us.
Rule. Multiply the sum of the bounding arcs by half the
difference of their radii.
Formula. 5 . ^ . = J (^3 — r,) (/, + 4) = \h (/, + Q.
Proof. The annular sector is the difference between the
two sectors \rj^ and \rj^.
But /i and 4 are arcs subtending the same angle ; therefore,
by 804,
^ = 4
.-. -1^24 - -1^14 = Wi + i^2^2 - i^i4 — 1^,4
= K^. -^0(^ + 4).
814. To find the lateral area of a prism.
^
MENSURATION^ OR METRICAL GEOMETRY,
323
Rule. Multiply a lateral edge by the perimeter of a right
section.
Formula. P =z Ip.
Proof. The lateral 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
lateral surface of the prism.
815. A Cylindric Surface is generated by a line so moving
that every two of its positions are parallel.
816. The generatrix in any position is called an Elemejtt of
the surface.
817. A Cylinder is a solid bounded by a cylindric surface
and two parallel planes.
324
THE ELEMENTS OF GEOMETRY.
8i8. The Axis of a circular cylinder is the sect joining the
centers of its bases.
.--r-^
819. A Truncated Cylinder is the portion between the base
and a non-parallel section.
820. To find the lateral area of a right circular cylinder.
Rule. Multiply its length by its circle.
Formula. C = cl.
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 I. The curved surface of a truncated
circular cylinder is the product of the circle of the cylinder by
the intercepted axis.
^vK-
For, by symmetry, substituting an oblique for the right sec-
tion 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
is equivalent to a circle whose radius is a mean proportional
between the altitude of the cylinder and the diameter of its
base.
115. A plane through two elements of a right circular
cylinder cuts its base in a chord which subtends at the center
an angle whose circular measure is u \ find the ratio of the
curved surfaces of the two parts of the cylinder.
326 THE ELEMENTS OF GEOMETRY.
824. A Conical Surface is generated by a straight line
moving 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.
826. A right circular cone may be generated by revolving a
right triangle about one pependicular. All the elements are
equal, and each is called the Slant Height of the cone.
MENSURATION, OR METRICAL GEOMETRY.
327
827. The Frustum of a pyramid or cone is the portion
included between the base and a parallel section.
828. To find ihe lateral area of a right circular cone.
Rule. Multiply the circle of its base by half the sla7it
height.
Formula. K = \ch = -n-rh.
Proof. If the curved surface be slit along a slant height,
and spread out flat, it becomes a sector of a circle, with slant
height as radius, and the circle of cone's base as arc ; therefore,
by 811, its area is ^ch.
829. Corollary. Since, by 810, the base O = ^cr, there-
fore 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.
328 THE ELEMENTS OF GEOMETRY.
830. To find ihe lateral area of the frustum of a right circular
cone.
Rule. Midtiply the slajit height of the frustum by half the
sum of the circles of its bases.
Formula. F ^=^ \h{c^-\' c^ = irh {r^ + r^.
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).
831. The axis of a right circular cone, or cone of revolution,
is the line through its vertex and the center of its base.
Exercises. 116. Given the two sides of a right-angled
triangle. Find the area of the surface described when the
triangle revolves about its hypothenuse.
Hint. Calling a and b the given altitude and base, and x the per-
pendicular from the right angle to the hypothenuse, by 828, the
area of the surface described by a is itxa, and by b is irxb.
Also a \ X w sla" -^ 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 r^^ find the radius of the circle in
which the two cones cut.
MENSURATIONy OR METRICAL GEOMETRY.
329
Theorem III.
832. The lateral area of a friistmn of a cone of revolution is
the product of the projection of the frnstiims slant height on the
axis by twice tr times a perpendicular erected at the mid-point of
this slant height^ and terminated by the axis.
Proof. By 830, the lateral area of the frustum whose slant height
is PR and axis MN is
F = tt{PM -{- RN)FR,
But if Q is the mid-point of PR, then PM + RN = 2QO,
.'. P = 27r X PR X QO.
But A PRL ~ A QCO, since the three sides of one are drawn per-
pendicular to the sides of the other ;
.-. PR X QO = PL X QC,
/. F = 2it{LP X QC) = 27r{MN X CQ).
Exercises. 118. Reckon the mantel from the two radii
when the inclination of a slant height to one base is 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.
330
THE ELEMENTS OF GEOMETRY.
833. To find the area of a sphere.
Rule. Multiply four times its squared radius by tt.
Formula. H = 4r*7r.
Proof. In a circle inscribe a regular polygon of an even
number of sides. Then a diameter through one vertex passes
through the opposite vertex, halving the polygon symmetrically.
Let PR be one of its sides ; draw PM, RNy perpendicular
to the diameter BD. From the center C the perpendicular
CQ bisects PR. Drop the perpendiculars PL^ QO.
Now, if the whole figure revolve about BD as axis, the
semicircle will generate a sphere, while each side of the in-
scribed polygon, as PR, will generate the curved surface of the
frustum of a cone. By 832, this
F = 2ir{MN X CQ);
and the sum of all the frustums, that is, the surface of the solid
generated by the revolving semi-polygon, equals 2itCQ into the
sum of the projections, BM, MN, NC, etc., which sum is BD,
/. Sum of surfaces of frustums = ^-kCQ X BD.
mensuration; or metrical geometry. 331
As we continually double the number of sides of the inscribed
polygon, its semi-perimeter approaches the semicircle as limit,
and its surface of revolution approaches the sphere as limit,
while CQ, its apothem, approaches r, the radius of the sphere,
as limit. Representing the sum of the surfaces of the frus-
tums by 2/s and BD by 2r, we have
CQ ^
That is, the variable sum is to the variable CQ in the constant
ratio 4r7r; therefore, by 798, Principle of Limits, their limits
have the same ratio,
H
.-. — = 4r7r,
r
,'. H = 4r27r.
834. A Calot is a zone of only one base.
835. The last proof gives also the following rule : —
To find the area of a zone.
Rule. Multiply the altitude of the segment by twice ir times
the radius of the sphere.
Formula. Z = 2ar'K.
332
THE ELEMENTS OF GEOMETRY.
CHAPTER IV.
SPACE ANGLES.
836. A Plane Angle is the divergence of two straight lines
which meet in a point.
837. A Space Angle is the spread of two or more planes
which meet in a point.
838. Symmetrical Space Angles are those which cut out
symmetrical spherical polygons on a sphere, when their vertices
are placed at its center.
839. A Steregon is the whole amount of space angle round
about a point in space.
MENSURATION, OR METRICAL GEOMETRY.
333
840. A Steradian is the angle subtended at the center by
that part of every sphere equal to the square of its radius.
841. The space angle made by only two planes corresponds
to the lune intercepted on any sphere whose center is in the
common section of the two planes.
842. A Spherical Pyramid is a portion of a globe bounded
by a spherical polygon and the planes of the sides of the poly-
gon. The center of the sphere is the apex of the pyramid ;
the spherical polygon is its base.
843. Just as plane angles at the center of a circle are pro-
portional to their intercepted arcs, and also sectors, so space
angles at the center of a sphere are proportional to their inter-
cepted spherical polygons y and also spherical pyramids.
334 "^^^ ELEMENTS OF GEOMETRY.
Example. Find the ratio of the space angles of two right cones
of altitude a^ and a^, but 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
is
2Trh{h — a,) _ h — Uj^
2i:h{h — a^ h — a^
the ratio of the calot altitudes.
For the equilateral and right-angled cones this becomes
2 - V^ ^
2 - V2*
844. To construct a space angle of two faces equivalent to
any polyhedral angle, only involves constructing a lune equiva-
lent to a spherical polygon, as in 731.
845. To find the area of a lune.
Rule. Multiply its angle in radians by twice its squared
radius.
Formula. L = 2r^u.
Proof. By 703, lunes are as their angles,
/, a lune is to a hemisphere as its angle is to a straight angle,
Z _ u
2r^ir TV
.', L = 2r^u,
MENSURATION, OR METRICAL GEOMETRY. 335
846. Corollary I. A lime measures twice as many stera-
dians as its angle contains radians.
847. Corollary II. If two-faced space angles are equal,
their lune angles are equal ; so a dihedral angle may be meas-
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 poly-
gon, 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
infer an analogous property of space angles.
For example, the following properties of trihedral angles
have been proved in our treatment of spherical triangles : —
I. Trihedral angles are either congruent or symmetrical
which have the following parts equal : —
336 THE ELEMENTS OF GEOMETRY.
(i) Two face angles and the included dihedral angle.
(2) Two dihedral angles and the included face angle.
(3) Three face angles.
(4) 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
hypothesis.
(6) Two face angles and the dihedral angle opposite one of
them, provided the other pair of opposite dihedral angles are
not supplemental.
II. 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
angle subtended by the other.
III. Symmetrical trihedral angles are equivalent.
IV. Space angles having the same number and sum of dihe-
dral angles are equivalent.
V. The face angles of a convex space angle are together
less than a perigon.
849. To find ihe area of a spherical iriangle.
Rule. Multiply its spheHcal excess in radians by its squared
radius.
Formula, a" = er^.
Proof. By 711, a triangle is equal to a lune whose angle
is \e ; therefore, by 845, 'a = r'^e.
850. Corollary I. A spherical triangle measures as many
steradians as its e contains radians.
851. Corollary II. By 714, to find the area of a spherical
polygon, multiply its spherical excess in radians by its squared
radius.
MENSURATION, OR METRICAL GEOMETRY.
337
CHAPTER V.
THE MEASUREMENT OF VOLUMES.
852. The Volume of a solid is its ratio to an assumed unit.
853. The Unit for Measicremejit of Volume is a cube whose
edge is the unit for length.
Cubic Centimeter.
854. Metric Units for Volume.
Solid.
1 kilostere
I hectostere
I dekastere
I stere
I decistere
I centistere
Liquid. Cubic.
= I cubic dekameter =
I kiloliter = i cubic meter =
I hectoliter =
I dekaliter =
I liter = I cubic decimeter =
I deciliter =
I centiliter =
I cubic centimeter =
1000 cubic meters.
100 cubic meters.
10 cubic meters.
I cubic meter.
.1 cubic meter.
.01 cubic meter.
.001 cubic meter.
.0001 cubic meter.
.ocxx)i cubic meter.
.ocxxxDi cubic meter.
I cubic millimeter = .000000001 cubic meter.
The authorized abbreviation for cubic is the index ^ as in icm.3 foj. j cubic centi-
meter ; that for stere is s., and for liter 1..
338
THE ELEMENTS OF GEOMETRY,
855. To find the volume of a quader.
Rule. Multiply together its lengthy breadth, and thickness.
Proof. By "jG^, two quaclers have the ratio compounded
of the ratios of their bases and altitudes.
856. Corollary. The volume of any cube is the third
power of the length of an edge. This is why the third power
of a number is called its cube.
857. To find the volume of any parallelepiped .
Rule. Multiply its altitude by the area of its base.
Proof. By 769, any parallelopiped is equivalent to a quader
of equivalent base and equal altitude.
858. To find the volume of any prism.
Rule. Multiply its altitude by its base.
Formula. V . P = aB.
Proof. By 771, any three-sided prism is half a parallelo-
piped, with base twice the prism's base and the same altitude.
Thus the rule is true for triangular prisms, and consequently
for all prisms ; since, by passing planes through one lateral
edge, and all the other lateral edges excepting the two adja-
cent to this one, we can divide any prism into triangular prisms
of the same altitude, whose triangular bases together make the
polygonal base.
MENSURATION, OR METRICAL GEOMETRY.
339
859. To find the volume of any cylinder.
Rule. Multiply its altitude by its base.
Proof. The cylinder is the limit of an inscribed prism
when the number of sides of the prism is indefinitely increased,
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 its altitude by its base.
Formula. Y = \aB.
340 THE ELEMENTS OF GEOMETRY.
Proof. By 775, any triangular pyramid is one-third of a
triangular prism of the same base and altitude.
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, we can divide any pyramid into triangular pyramids
of the same 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. K = ^ar^-n-.
Proof. The base of a cone is the limit 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 ratio,
and V.K z=z ^aB.
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.
341
862. Corollary. The volume of 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.
F= fTT^A.
PRISMATOID.
863. A Prismatoid is a polyhedron whose bases are any two
polygons in parallel planes, and whose lateral faces are triangles
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.
342
THE ELEMENTS OF GEOMETRY.
864. A number of different prismatoids thus pertain to the
same two bases.
865. If two basal edges which form with the same lateral
edge two sides of two adjoining faces are parallel, then these
two triangular faces fall in the same plane, and together form a
trapezoid.
MENSURATION, OR METRICAL GEOMETRY. 343
866. A Prismoid is a prismatoid whose bases have the same
number of sides, and every corresponding pair parallel.
867. A frustum of a pyramid is a prismoid whose two bases
are similar.
868. Corollary. Every three-sided prismoid is the frus-
tum of a pyramid.
869. If both bases of a prismatoid become sects, it is a
tetrahedron.
870. A Wedge is a prismatoid whose lower base is a rect-
angle, and upper base a sect parallel to a basal edge.
871. The altitude of a prismatoid is any sect perpendicular
to both bases.
872. A Cross-Sectio7t of a prismatoid is a section made by a
plane perpendicular to the altitude.
344 THE ELEMENTS OF GEOMETRY.
873. To find the volume of any prismaioid.
Rule. Multiply one-fourth its altitude by the sum of one
base and three times a cross-section at two-thirds the altitude
from that base.
Formula. D — -{B -{- z T).
4
Proof. Any 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
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
sect in the plane of each base of the prismatoid, as AC 3.nd OG.
Therefore, if the formula holds good for tetrahedra in these
three positions, it holds for the prismatoid, their sum.
In (i), call 7"i the section at two-thirds the altitude from
the base B^ ; then T^ is \a from the apex. Therefore, by 772,
4 4 3 3
which, by 860, equals Y, the volume of the tetrahedron.
MENSURATION, OR METRICAL GEOMETRY.
345
In (2), B^ = o, being a point, and T^ is ^a from the apex ;
/. T, : B, :: (|«)^ : a\ /. T, = |^„
... A = ^(^, + 37;) = ^(o + ^^.) = -aB, = K
4 4 3 3
In (3), let KLMN be the section T^.
Join CK, CN, OM, ON. Now
A ANK : A ^(?6> : : AN^ : ^6^^ : : (^ay : a^ : : 1 : g.
A GNM: A (9-^C : : GN^ : 6^^^ : : (|^)^ : tz^ : : 4 : 9.
But the whole tetrahedron D^ and the pyramid C ANK may
be considered as having their bases in the same plane, AGO,
and the same altitude, a perpendicular from C;
C ANK : Z>3 : : A ANK : A AGO : : i : 9,
.-. C ANK = 4Z>3.
346 THE ELEMENTS OF GEOMETRY.
In same way,
O GNM \ D^ : : A GNM : A GAC : : 4 : 9,
/. OGNM=%D^.
C ANK + O GNM = %D^,
.'. C KLMN + O KLMN = %D^,
But, by 860,
C KLMN -f O KLMN = \ . \aT, 4- i . 1^7; = \aT,,
.-. 4_z)3 = i^r, /. i^3 = ^3r, = ^{B, 4- 37;),
9 3 4 4
since here
B^ = o.
874. Corollary I. Since, in the frustum of a pyramid, B
and T are similar.
4 \ ^i' /
where w^ and ^e'^ are corresponding sides of B and T.
875. Corollary II. For the frustum of a cone of revolu-
tion,
where rj is the radius of T.
V,F = -7rr,^
4
MENSURATION, OR METRICAL GEOMETRY. 247
876. To find the volume of a globe.
Rule. Multiply the cube of its radius by |-7r.
Formula. G = ^-kt^.
Proof. By 644, a tetrahedron on edge, and a globe with
the tetrahedron's altitude for diameter, have all their corre-
sponding cross-sections equivalent if any one pair are equiva-
lent.
Hence the volumes may be proved equivalent, as in 774,
and
G = \aT,
But a = 2r, and, by 522, T = fr . f r . tt,
.-. dP = I . 2r . |r . f r . TT = f 7rr3.
877. Corollary I. 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.
879. If 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
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.
881. The part of a line which could be generated by a tra-
cing-point, starting from a given point on that line, and moving
on the line without ever turning back, is called a Ray from
that given point as origin.
348 THE ELEMENTS OF GEOMETRY.
882. Two rays from the same origin are said to have the
same direction if they coincide ; otherwise, they are said to
have different directions.
883. Two rays which have no point but the origin in com-
mon, and fall into the same line, are said to have opposite direc-
tions.
884. Two rays lying on parallel lines have parallel-same
directions if they are on the same side of the line joining their
origins. .
885. Two rays lying on parallel lines have parallel-opposite
directions if they are on opposite sides of the line joining their
origins.
886. A sect is said to have the same direction as the ray of
which it is a part.
887. A sect is definitely fixed if we know its initial point,
its parallel-same direction, and its length.
888. The operation by which a sect could be traced if we
knew its initial point, that is, the operation of carrying a
tracing-point in a certain parallel-same direction until it
passes over a given number of units for length, is called a
Vector.
889. The position of B relative to A is indicated by the
length and parallel-same direction of the sect AB drawn from
A to B. If you start from A, and travel, in the direction indi-
cated 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 ^^ is completely
defined by the parallel-same direction and length of the trans-
ferrence. All vectors which are of the same magnitude and
parallel-same direction, and only those, are regarded as equal.
Thus AB is not equal to BA.
MENSURATION^ OR METRICAL GEOMETRY. 349
PRINCIPLE OF DUALITY.
JOIN OF POINTS AND OF LINES.
891. The line joining two points is called the _/(?/;/ of the
Two Points. The point common to two intersecting lines is
called the Join of the Two Lines.
892. Pencil of Lines. A fixed point, A^ may be joined to
all other points in space.
We get thus all the lines which can be drawn through the
point A. The aggregate of all these lines is called a Pencil of
Lines. The fixed point is called the Base of the pencil. Any
one 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 line passes through the point.
893. In most cases, we can, when one figure is given, con-
struct 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 corre-
sponding theorem concerning the second figure. Figures or
theorems related in this manner are called Reciprocal Figures
or Recipi'ocal Theorems.
894. Small letters denote lines, and the join of two elements
is denoted by writing the letters indicating the elements, to-
gether.
Thus the join of the points A and B is the line AB, while
ab denotes the join, or point of intersection, of the lines a and b.
ROW OF points, pencil OF LINES.
895. A line contains an infinite number of points, called a
Row of Points, of which the line is the Base,
A row is all points in a line.
350
THE ELEMENTS OF GEOMETRY.
The reciprocal figure is all lines in a pointy or all lines pass-
ing through the point.
A Flat Pe7icil is the aggregate of all lines in a plane which
pass through a given point. In plane geometry, by a pencil we
mean a flat pencil.
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 posi-
tion, B.
896'. A line turning about a
point describes a pencil.
897'. An angle is a part of a
pencil described by a line turning
from one position, a, to another
position, b.
Thus 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
fixed point of the cell, pivot the other extremity to a fixed
point. Then pivot the first end to one of the free angles of
the rhombus. The opposite vertex of the rhombus will now
describe a straight line, however the linkage be pushed or
moved.
Proof. By the bar FD, the point D is constrained to move
on the circle ADR ; therefore -^ ADR^ being the angle in a
semicircle, is always right.
If, now, E moves on EM ± AMy
A ADR ~ A AME,
.-. DA : AR : : AM : AE,
/. DA , AE = RA , AM.
Therefore, if AE . AD is constant, E moves on the straight
line EAf. But because BDCE is a rhombus, and AB =: AC,
.*. D and iV are always on the variable sect AE.
352 THE ELEMENTS OF GEOMETRY.
Always
AB^ = AN^ + NB% and ~BE' = EN^ + NB\
:. AB^ - ~BE' =■ AN' - JV£^
= {AN + iV^^) (^iV^ - NE) = ^^ . AD.
900. Corollary. The efficacy of our cell depends on its
power to keep, however it be deformed, the product of two
varying sects a constant.
Therefore a Hart four-bar cell, constructed as in the accom-
panying figure, may be substituted for the Peancellier six-bar
cell, since AE . AD equals a constant.
Also, for the Hart cell may be substituted the quadruplane,
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 —— : -— — is written {ABCD).
CB DB
Distinguishing the "step" AB from BA, as of opposite
** sense," and taking the points in the two groups of two in a
definite order, to write out a cross-ratio, make first the two
MENSURATION, OR METRICAL GEOMETRY. 353
bars, and put crossing these the letters of the first group of
A A
two, thus — - : — - ; then fill up, crosswise, the first by the first
B B
letter of the second group of two, the second by the second.
If we take the points in a different order, the value of the
cross-ratio may change.
We can do this in twenty-four different ways by forming all
permutations of the letters.
But of these twenty-four cross-ratios, groups of four are
equal, so that there are really only six different ones.
We have the following rules : —
I. If in a cross-ratio the two groups be interchanged^ its value
remains unaltered.
{ABCD) = {CDAB).
II. If in a cross-ratio the tzvo points belonging to one of the
two groups be interchanged^ the cross-ratio cha?iges to its recip-
rocal.
I
(ABCD)
{ABDC)
1. and II. are proved by writing out their values.
III. Fjvm II. it follozuSj that, if we inteixhange the elements
in each pair^ the cross-ratio remains unaltered.
{ABCD) = {BADC).
IV. If in a cross-ratio the two middle letters be interchanged^
the cross-ratio changes into its complement.
{ABCD) = I - {ACBD),
This is proved by taking the step-equation for any four col-
linear points,
BC .AD ^ CA . BD + AB . CD = o,
and dividing it by CB . AD.
354
THE ELEMENTS OF GEOMETRY.
Theorem IV.
go2. If any four concurrent lines are cut by a transversal,
any cross-ratio of the four points of intersection is constant for
all positions of the transversal.
Hypothesis. Let A, B, C, D, be the intersection points for f,
the transversa/ in one posit/on, and let A\ B', C , jy, be the cor-
responding intersection points for t\ the transversal in another
position.
Conclusion. {ABCD) = {A'B'Cjy); that is,
AC , AD ^ A[C_ . A'D'
CB ' DB C'B' ' jyB''
Proof. Through the points A and B on t draw parallels to /',
which cut the concurrent lines in C2, D2, B^, and A„ C^, D^.
AACC^-^ £^BCC„
AC ^ AC^
CB C,B'
and
and
A ADD:, ~ A BDD,,
AD
DB
AD,
D,B'
where account is taken of " sense.'
MENSURATION, OR METRICAL GEOMETRY. 355
Hence
CB * Z>B C,B ' D,B AD^ ' Z>,b''
but
ii^ = f^' and C^^C:^^
AD, A'ly D,B jyB"
. AC , AD ^ A'C . CB' ^ y^^C^ . A'jy
" CB' DB AD ' DB' CB' '' D'B''
EXERCISES.
BOOK I.
1 20. Show how to 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 congru-
ent.
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.
1 24. Draw a perpendicular to a sect at one extremity.
125. Draw three figures to show that an exterior angle of a triangle
may be greater than, equal to, or less than, the interior adjacent angle.
126. Any two exterior angles of a triangle are together greater than
a straight angle.
127. The perpendicular from any vertex of an acute-angled triangle
on the opposite side falls within the triangle.
128. The perpendicular from either of the acute angles of an obtuse-
angled triangle on the opposite side falls outside the triangle.
129. The semi-perimeter of a triangle is greater than any one side^
and less than any two sides.
3S7
358 THE ELEMENTS OF GEOMETRY.
130. The perimeter of a quadrilateral is greater than the sum, and
less than twice the sum, of the diagonals.
131. If a triangle and a quadrilateral stand on the same base, and
the one figure fall within the other, that which has the greater surface
shall have the greater perimeter.
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.
133. If any sect joining two parallels be bisected, this point will
bisect any other sect drawn through it and terminated by the parallels.
134. Through a given point draw a line such that the part inter-
cepted between two given parallels may equal a given 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 quadri-
lateral, and the included angle equal to either of the angles at which the
diagonals intersect.
137. Describe a square, having given a diagonal.
138. ABC is a right-angled triangle; BCED is the square on the
hypothenuse; ACKH and ABFG are the squares on the other sides.
Find the center of the square ABFG (which may be done by drawing
the two diagonals), and through it draw two lines, one parallel to BC,
and the other perpendicular to BC. This divides the square ABFG
into four congruent quadrilaterals. Through each mid-point of the sides
of the square BCED draw a parallel to AB or A C. If each be extended
until it meets the second of the other pair, they will cut the square
BCED into a square and four quadrilaterals congruent to ACKH and
the four quadrilaterals in ABFG.
139. The orthocenter, the centroid, and the circumcenter of a tri-
angle are collinear, and the sect between the first two is double of the
sect between the last two.
140. The perpendicular from the circumcenter to any side of a
triangle is half the sect from the opposite vertex to the ortho-
center.
141. Sects drawn from a given point to a given circle are bisected ;
find the locus of their mid-points.
EXERCISES. 359
142. The intersection of the Hnes joining the mid-points of opposite
sides of a quadrilateral is the mid-point of the sect joining the mid-points
of the diagonals.
143. A parallelogram has central symmetry.
SYMMETRY.
144. No triangle can have a center of symmetry, and every axis of
symmetry is a medial.
145. Of two sides of a triangle, that is the greater which is cut by
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 ^s-ill be acute, right, or obtuse, according
as the medial through its vertex is greater than, equal to, or less than,
half the opposite side.
148. If a quadrilateral has axial symmetry, the number of vertices
not on the axis must be even ; if none, it is a symmetrical trapezoid ; if
two, it is a kite.
149. A kite has the following seven properties ; from each prove all
the others by proving that a quadrilateral possessing it is a kite.
(i) One diagonal, the axis, is the perpendicular bisector of tlie
other, which will be called the transverse axis.
(2) The axis bisects the angles at the vertices which it joins.
(3) The angles at the end-points of the transverse axis are equal, and
equally divided by the 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) The transverse axis divides the kite into two triangles, each of
which is symmetrical.
(7) The Hnes joining the mid-points of opposite sides meet on the
axis, and are equally inclined to 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.
360 THE ELEMENTS OF GEOMETRY.
(i) Two opposite sides are parallel, and have a common perpendicu-
lar bisector.
(2) The other two opposite sides are equal, and equally inclined to
either of the other sides.
(3) Each angle is equal to one, and supplemental to the other, of
its two adjacent angles.
(4) The diagonals are equal, and divide each other equally.
(5) One median line bisects the angle between those sides produced
which it does not bisect, and likewise 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 tri-
angle 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
parts, the rectangle contained by the two sects is equivalent to the sum
of the rectangles contained by all the parts of the one, taken separately,
with all the parts of the other.
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 sect.
157. The rectangle contained by two internal segments of a sect
grows less as the point of section moves from the mid-point.
158. The sum of the squares on the two segments of a sect is least
when they are equal.
159. If the 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
parallelogram is constant.
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. Prove
that the bisector of their angle contains the center of the circle.
167. Describe a circle of given radius to pass through a given point
and have its center in a given line.
168. Equal chords in a circle are all tangent to a concentric circle.
169. Two concentric circles intercept equal sects on any common
secant.
1 70. If two equal chords intersect either within or without a circle,
the segments of the one equal the segments of the other.
171. Divide a circle into two segments such that the angle in the
one shall be seven times the angle in the other.
172. ABC and ABC zx^ two triangles such that AC = AC ',
prove the circle through A, B, C, equal to that through A, B, C,
362 THE ELEMENTS OF GEOMETRY.
1 73. Pass a circle through four given points. When is the problem
possible ?
174. \i AC and BD be two equal arcs in a circle ABCD, prove
chord AB parallel to chord CD.
175. Circles described on any two sides of a triangle as diameters
intersect on the third side, or the third side produced.
1 76. If one circle be described on the radius of another as diameter,
any chord of the larger from the point of contact is bisected by the
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 equi-
lateral triangle.
181. If two chords intersect at right angles, the sum of the squares
on their four segments is equivalent to the square on the diameter.
182. The opposite sides of an inscribed quadrilateral are produced
to meet ; prove that the bisectors of the two angles thus formed are at
right angles.
183. The feet of the perpendiculars drawn from any point in its
circumscribed circle to the sides of a triangle are collinear.
184. From a point P outside a circle, two secants, PAB, PDC, are
drawn to the circle ABCD ; AC, BD, are joined, and intersect at O.
Prove that O lies on the chord of contact of the tangents drawn from P
to the circle. 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 ;
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 comers 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.
190. Draw a line which would bisect the angle between two hnes
which are not parallel, but which cannot be produced to meet.
191. The angle between the altitude of a triangle and the line
through vertex and circumcenter equals half the difference of the angles
at the base.
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 circum-
center.
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 radius.
197. The intersections of the diagonals of a regular pentagon are
the vertices of another regular pentagon ; so also the intersections of
the alternate sides.
198. The sum 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 alti-
tudes, and bisects the sects between the orthocenter and vertices.
Hint. Let ABC be the triangle ; H^ A', Z, the mid-points of its sides ; Xy
V, Z, the feet of its altitudes ; O its orthocenter ; C/, F, JV, the mid-points
of AO, BO, CO.
LHWU is a rectangle ; so is HKUV\ and the angles at X^ Y, Z, are right.
364 THE ELEMENTS OF GEOMETRY.
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 equi-
angular if the number of sides be odd.
203. Given the vertical angle of a triangle and the sum of the
sides containing it ; find the locus of the circumcenter.
204. Given the vertical angle and base of a triangle; find the
locus of
(i) The center of the inscribed circle.
(2) The centers of the escribed circles.
(3) The orthocenter.
(4) The centroid.
(5) The mediocenter.
205. Of all rectangles inscribable in a circle, show 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 rect-
angle 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 extrem-
ities are concyclic.
208. Does the magnitude of the third proportional to two given
sects depend on the order in which the sects are taken ?
209. Through a given point inside a circle draw a chord so that it
shall be divided 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.
211. Find two sects from any two of the six following data: their
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 given point a line making equal angles with
two given Hnes.
216. From a given point place three given sects so that their
extremities may 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
parts shall be half the square on the whole sect.
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.
221. Through two given points on opposite sides of a line draw
lines to 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
mid-points of its sides.
223. If a quadrilateral be described about a circle, the angles sub-
tended at the center by two opposite sides are supplemental.
224. Two circles touch at A and have a common tangent BC.
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.
366 THE ELEMENTS OF GEOMETRY.
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.
232. The sect of a common tangent between its points of contact
is a mean proportional between the diameters of two tangent circles.
233. Any regular polygon inscribed in a circle is a mean propor-
tional between the inscribed and circumscribed regular polygons of half
the 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 com-
pletely, and five hundred and twenty-four exercises and problems, of
which also more than twenty are solved completely, and many others
have hints appended.
si
RETURN CIRCULATION DEPARTMENT
TO^- r/lnip'i^^rnry. 198 Main Stocks
LOAN PERIOD 1
HOME USE
Ll I BOOKS MAY BE RECALLED AFTER 7 DAYS.
J^nirond Recharges may ^ made 4 days prior ,o me due date.
Books may be Renewed by calling 642-3405.
DUE AS STAMPED BELOW
FORM NO. DD6
^^fJ^T^i^^^lSS^S
U. C. BERKELEY LIBRARIES
CO
SfiS
bEDl