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ELEMENTS
v r ,- ./' "
G E O M E T R Y:
<;. \. v ENTWORTH, A. If.,
raomsoK UK MA I It
^ OF THE
BOSTON :
PUBLISHED BY GINN AND HEATH.
L881.
Copyright, 1877.
By GINN AND HEATH.
I'kks.s of Rockwell and Churchill,
39 Arch St., Boston.
PREFACE
Most persons do not possess, and do not easily acquire, the
powei of abstraction requisite for apprehending the Geometri-
cal conceptions, and for keeping in mind the successive steps
of a continuous argument Bence, with a very large proportion
of beginners in (leometry^it depends mainly upon the form in
which the subject i- ] vrhetner they pursue the study
with indifferen o say aversion, or with Increasing i nt
and pleaeura
In oompiling the preeenl treatise* this lad has 1 n kept con-
stantly in view. All unnecessary discussions and Bcholia have
been avoided h methods hare been adopted aa experi-
comhined with repeated trials,
have shown to be most readily comprehended. No attempt has
been made to render more intelligible the simple notions of
position, magnitude, and direction, which every child derives
lVoin obsi : hut it is believed that these notions have
nd defined with mathematical precision.
A few symbols, which stand tor words and not for operations,
have been used, hut these are of so great utility in giving style
and v to the demonstrations that no apology seems
for their introduction.
(Ircat pains have been taken to make the page attractive.
The Inures are large and distinct, and are placed in the middle
of the page, so that they fall directly under the eye in imme-
diate connection with the corresponding text. The given lines
PKEFACE.
of the figures are full lines, the lines employed as aids in the
demonstrations are short-dotted, and the resulting lines are long-
dotted.
In each proposition a concise statement of what is given is
printed in one kind of type, of what is required in another, and
the demonstration in still another. The reason for each step
is indicated in small type between that step and the one follow-
ing, thus preventing the necessity of interrupting the process of
the argument by referring to a previous section. The number
of the section, however, on which the reason depends is placed
at the side of the page. The constituent parts of the propo-
sitions are carefully marked. Moreover, each distinct assertion in
the demonstrations, and each particular direction in the construc-
tions of the figures, begins a new line ; and in no case is it neces-
sary to turn the page in reading a demonstration.
This arrangement presents obvious advantages. The pupil
perceives at once what is given and what is required, readily
refers to the figure at every step, becomes perfectly familiar with
the language of Geometry, acquires facility in simple and accu-
rate expression, rapidly learns to reason, and lays a foundation
for the complete establishing of the science.
A few propositions have been given that might properly bo
considered as corollaries. The reason for this is the great diffi-
culty of convincing the average student that any importance
should be attached to a corollary. Original exercises, however,
have been given, not too numerous or too difficult to discourage
the beginner, but well adapted to afford an effectual test of the
degree in which he is mastering the subjects of his reading.
Some of these exercises have been placed in the early part of
the work in order that the student may discover, at the outset,
that to commit to memory a number of theorems and to repro-
duce them in an examination is a useless and pernicious labor;
but to learn their uses and applications, and to acquire a readi-
ness in exemplifying their utility, is to derive the full benefit
of that mathematical training which looks not so much to the
1' KM FACE.
M to the 'dudplvu of the mental fac-
It only remains to express my sense of obligation to Dr.
I). F. \Vi:u> fox valuable assistance, and to the University
for the irith which the book has been printed;
and also to give assurance tliat any suggestions relating to the
work will be thankfully received.
G. A. WENTWOKTH.
Phillips Kxkter Acadi
January, 1878.
NOTE TO THIRD EDITION
In this edition I have ei i more rF
but not less simple, treatment dlela, Ratio, and
Limits. The ehangea are not sufficient to prevent the Bimultar
ia use of the old and new editiona in the data : still the]
very important, and have been made after the most careful and
prolonged conai
I have to express my thanks for valuable suggestions received
in my correspondents; and a special nt is due
from me to Professor C. EL Judson, of Furman [Jnivei
Greenville, South Carolina, to whom I am indebted for assist-
aiice in effecting many improvemente in thia edition.
TO THE TEACHER.
When the pupil Lb reading eacli Book for the Brat time, it will be
well to let him write hia proofa <>n the blackboard in hia own Ian-
re being taken that hia language^ be the sunniest possible,
that the arrangement of work be vertical (without side work), and
that the figures be accurately constructed.
This method will furnish a valuable exercise as a language lesson,
will cultivate the habit of neat and orderly arrangement of work,
and will all«>w a brief interval for deliberating on each Btep,
After a Book has been read in thia way the pupil should review
Book, and should be required to draw the figures free-hand. He
PKEFACE.
should state and prove the propositions orally, using a pointer to
indicate on the figure every line and angle named. He should be
encouraged, in reviewing each Book, to do the original exercises ; to
state the converse of propositions ; to determine from the statement,
if possible, whether the converse be true or false, and if the converse
be true to demonstrate it ; and also to give well-considered answers
to question's which may be asked him on many propositions.
The Teacher is strongly advised to illustrate, geometrically and
arithmetically, the principles of limits. Thus a rectangle with a
constant base b, and a variable altitude x, will afford an obvious
illustration of the axiomatic truth contained in [4], page 88. If x
increase and approach the altitude a as a limit, the area of the rec-
tangle increases and approaches the area of tKe rectangle a b as a
limit ; if, however, x decrease and approach zero as a limit, the area
of the rectangle decreases and approaches zero for a limit. An arith-
metical illustration of this truth would be given by multiplying a
constant into the approximate values of any repetend. If, for exam-
ple, we take the constant 60 and the repetend .3333, etc., the approxi-
mate values of the repetend will be T 8 ^-, fifc, i% s $j, ^Atf* etc -> anc l
these values multiplied by 60 give the series 18, 19.8, 19.98, 19.998,
etc., which evidently approach 20 as a limit ; but the product of 60
into J (the limit of the repetend .333, etc.) is also 20.
Again, if Ave multiply 60 into the different values of the decreasing
series, ^, ^-, ^^ sooocp etc -> which approaches zero as a limit,
we shall get the decreasing series, 2, £, -gV> jhfr etc - > ail( l tn * s series
evidently approaches zero as a limit.
In this way the pupil may easily be led to a complete comprehen-
sion of the whole subject of limits.
The Teacher is likewise advised to give frequent written examina-
tions. These should not be too difficult, and sufficient time should
be allowed for accurately constructing the figures, for choosing the
best language, and for determining the best arrangement.
The time necessary for the reading of examination-books will be
diminished by more than one-half, if the use of the symbols employed
in this book be permitted.
G. A. W.
Phillips Exeter Academy,
January, 1879.
CONTENTS.
PLANE GEOMETRY.
BOOK I. Eta -tii.ini-.ai: l Paob
bmtOPUOT Om T Remarks 3
00110*1 4
Strak.iii Links . . • G
J'i. am. An«;i.ks 7
Angular M acmitde D
IMPOSITION 10
Math km a i km. Tnjn 11
Axioms and Postulates IS
Symbols a 1 :j
i« -ULAR AM) OM IS 14
Pabaiul lam 24
Triangles 37
i.s 58
POLYQOOT al 68
BOOK II. Circles.
Dkkinhk»ns 73
SiKAK.nr Lam and Circles 75
Measurement 86
Theory of Limits 87
SUPPLKMI MARY PROPOSITIONS 100
i ructions 103
BOOK III. Proportional Lines and Similar Polygons.
Theory of Proportion 128
Proportional Lines 139
Similar Polygons 143
era . . . • 164
Mil CONTENTS.
BOOK IV. Comparison and Measurement of the Sur-
faces of Polygons.
Comparison and Measurement of Polygons . . .174
Constructions 194
BOOK V. Begular Polygons and Circles.
Regular Polygons and Circles 210
Constructions 224
ISOPERIMETRICAL POLYGONS. SUPPLEMENTARY . . . 237
Symmetry. Supplementary 245
ELEMENTS OF GEOMETRY.
OP THE
JWIVERSITT]
BOOK I.
RECTILINEAR FIGURES.
Introductory 1 1 km arks.
A BOUOB Mock of marble, under the stone-cutter's hammer,
may 1 I assume regularity of form.
H a block be eat in the shape repre- , /
I in this diagram,
It will have six flat faces.
Each face of the block is called a Sur- /'
face. I *'
It these surfaces be made smooth by pol-
ishing, m that, vrhen a sti ge is applied t<> any cue of
them, the straight-edge in every pari will touch the surface, the
« s are called Plane Surfaces.
The sharp edge in which any two of these surface
called a I
Ihe place aft which any three of these lines meet is called a
.
I t now the block be removed, we may think of the place
pied by the block as being of precisely the same shape and
lock itself; also, as having surfaces or boundaries
whicl bom surrounding space. AVe may likewise
think of these surfaces as having lines for their boundaries or
limits ; and of these lines as having points for their extremities
or limits.
A Solid, as the term is used in Geometry, is a limited por-
tion of space.
r we acquire a clear notion of surfaces as boundaries of
solids, we can easily conceive of surfaces apart from solids, and
GEOMETRY. BOOK I.
suppose them of unlimited extent. Likewise we can conceive of
lines apart from surfaces, and suppose them of unlimited length;
of, points apart from lines as having position, but no extent
Definitions.
1. Def. Space or Extension has three Dimensions, called
Length, Breadth, and Thickness.
2. Def. A Point has position without extension.
3. Def. A Line has only one of the dimensions of exten-
sion, namely, length.
The lines which we draw are only imperfect representations
of the true lines of Geometry.
A line may be conceived as traced or generated by a point in
motion.
4. Def. A Surface has only two of the dimensions of ex-
tension, length and breadth.
A surface may be conceived as generated by a line in motion.
5. Def. A Solid has the three dimensions of extension,
length, breadth, and thickness. Hence a solid extends in all direc-
tions.
A solid may be conceived as generated by a surface in motion.
Thus, in the diagram, let the upright
surface A B CD move to the right to
the position E F II K. The points
A, B, C, and D will generate the lines
AE, BF, CK, and D II respectively. C K
And the lines A B, B D, DC, and A C will generate the sur-
faces A F, B II, D K, and A K respectively. And the surface
ABC D will generate the solid A H.
The relative situation of the two points A and H involves
three, and only three, independent elements. To pass from A t< I //
it is necessary to move East (if we suppose the direction A E to
I
I
J//
DEFINITIONS.
be i i distance equal to A E, North a distance equal to
/•//', and down a distance equal to F II.
These three dimensions we designate for convenience length,
Lth, and thickness.
6. Tin- limits (extremitiee) of lines are points.
The limits (houndariee) of surfaces an tinea*
The limits (boundaries) of solids are surfaces.
7. I >i:i . Extension is also called Magmtt
When reference is had to extent, lines, . and solids are
called ma
8. Def. A Straight line is a line which has
the same direction throughout its who)
!». I>rr. X Curved line which changes
its directi int.
10. I'm A Broken line is a series of con-
i:t lines.
When the word line is used a straight lino is meant, and
when the woid eui I line is meant.
11. 1 >i i . A Flam St in which,
if any two points be taken, the straight line joining these points
will lie wholly iii the surface.
1 -. l»i:i. A «* is a surface no part of which
is plana
13. Figure or form depends upon the relative position of
points. Thus, the figure oi form of a line (straight or curved)
depends upon the relative position of points in that line; the
figure or form of a Burface depends upon the relative position of
points in that surface.
When reference is had to form or shape, lines, surfaces, and
solids are .-all |
G GEOMETRY. BOOK I.
14. Def. A Plane Figure is a figure, all points of which
are in the same plane.
15. Def. Geometry is the science which treats of position,
magnitude, and form.
Points, lines, surfaces, and solids, with their relations, are
the geometrical conceptions, and constitute the subject-matter of
Geometry.
16. Plane Geometry treats of plane figures.
Plane figures are either rectilinear, curvilinear, or mixtilinear.
Plane figures formed by straight lines are called rectilinear
figures ; those formed by curved lines are called curvilinear fig-
ures ; and those formed by straight and curved lines are called
mixtilinear figures.
17. Def. Figures which have the same form are called
Similar Figures. Figures which have the same extent are called
Equivalent Figures. Figures which have the same form ami
1 are called Equal Figures.
On Straight Lines.
18. If the direction of a straight line and a point in the
line be known, the position of the line is known; that is, a
straight line is determined in position if its direction and one of
its points be known.
Hence, all straight lines which pass through the same point in
the same direction coincide.
Between two points one, and but one, straight line can be
drawn ; that is, a straight line is determined in position if turn of
its points be known.
Of all lines between two points, the shortest is the straight
line; and the straight line is called the distance between the
two points.
]»K1 INITIONS.
The point from which a line isjdrawn is called its origin.
19. II' B line, as C B, A f B j be produced through C,
the portions C B and C A may be regarded as different lines
having opposite directions from the point C\
Hence, even line, as A B, t f, has two opposite
directions, namely from A toward B, which is expressed by say-
ing line A B, and from B toward A, which is expressed by
line A' A.
20. If a straight line change its magnitude, it must become
!• or shorter. Thus by prolonging A B to C, - f ?,
- A 11 + B C ; and conversely, /;('=.!('-;( B.
If a line iia-rease so tliat it is prolonged by its own magnitude
several times in succession, the line is muk suit-
ing line is called a oi the given line. Thus, if A B =
B( -CD, etc., A f \ £_£, then AC 'I A /;, AD =
:\A /;.
It must also be possible to divide B il line into an
ued number of equal parte. For, assumed thai the *th
part of a given line were not attainable, then the double, triple,
quadruple, of the *th part would imt be attainable. Among
these multiples, however, we should reach the nth multiple of
this ath part, that is, the line itself Hence, the line itself would
not be attainable ; which contradicts the hypothesis that we have
ven line before us.
Therefore, U i.<, and the latter as the angle E H F.
hKFIXITIONS.
D
The magnitude of an angle depends wholly upon the extent
i of its sides, and not upon their
length. Thus if the sides of the angle BAG,
namely, A B and A G, be prolonged, their
extent of opening will not be altered, and the
of the angle, consequently, will not be
chan
24. Dkf. Adjacent Anglet are an
bavin, mon vertex and a common
side between them. Tims the angles
G D E and G D F are adjacent angles. E
25. Def. A Right Angle is an angle included between two
lit lines which meet each other SO that the two adjacent
angles Formed by producing one of the tin
through ' Thus it' the
straight line A B meal theatraighl line G D
so tint ;1 ut angles ABC and ABB
are equal to one another, each of these an-
gles is call t angle.
26. DlP. Perpen
which make a righl angle with each other.
27. Def. An A< is an an
BA G.
ie is an an
iter than a right angle; as the angle
p /; r.
29. n gles, i n
distinction from right angles, are called ob-
B
'D
C
-D
rlet; and intersecting. lines which are not perpendicular
to each other are called line*, A
30. Def. The Complejnent of an angle is
the difference between a right angle and the
n angle. Tims A B D is the complement
of the angle D BO\ also I) BC is the com-
plement pf the angle A B D.
10
GEOMETRY. — BOOK I.
31. Def. The Supplement of an angle
is the difference between two right angles
and the given angle. Thus A CD is the
supplement of the angle D C B; also D C B
is the supplement of the angle AC D.
32. Def. Vertical Angles are angles
which have the same vertex, and their
sides extending in opposite directions.
Thus the angles AOD and COB are
vertical angles, as also the angles A C
and DOB.
D
-D
^ On Angular Magnitude.
A'-
B'
A
33. Let the lines B B' and A A' be in B
the same plane, and let B B' be perpen-
dicular to A A' at the point 0.
Suppose the straight line C to move
in this plane from coincidence with A,
about the point as a pivot, to the po-
sition C ; then the line C describes or
generates the angle A C.
The amount of rotation of the line, from the position A to
the position C, is the Angular Magnitude A C.
If the rotating line move from the position A to the po-
sition B, perpendicular to A, it generates a right angle ; to
the position A 1 it generates two right angles ; to the position
B f , as indicated by the dotted line, it generates three right
angles ; and if it continue its rotation to the position A,
whence it started, it generates four right angles.
Hence the whole angular magnitude about a point in a plane
is equal to four right angles, and the angular magnitude about
a point on one side of a straight line drawn through that point
is equal to two right angles.
DEFINITIONS.
11
c
^
\
/
/ B
\
D
7
K
A
A
A
£»
/
If
N>
/>
11
A
Fig
1.
2.
:;i. ice the an. pritude about the point is
bet increased nor diminished by the number of lines which
radiate from thai poinl . i of <>!l tl
in a plane, as A B + B (7 + C D, etc, in Fig. 1, is equal
; an« I the sum of all U
on one side of a straight lint dra ih that
A <> /; + /; o C + CO I>. ' to two right
Henoe twn tdjacenJ angles, OCA and OCB, jj
formed by two itraighl linesj of which one is
produced from the pew in both di-
rects :
l)e called
Ox mi Method of Superposition'.
of the equality of two geometrical magnitudes
IS that they coincide point for point,
Thus, two equal, if they can be so placed
thai tli*- points at fcheii tities coincide. Two angles are
equal, if they can laced that their vertices coincide in
tion and their sides in direction.
In applying this test of equality, we assume that a line may
be moved from one place t<> another without altering its length;
may he taken up, turned over, and put down,
without altering the difference in direction of its sides.
12
GEOMETRY. BOOK I.
D
This method enables us to com-
pare unequal magnitudes of the
same kind. Suppose we have two
angles, ABC and A' B' C Let
the side B C be placed on the side
B' C, so that the vertex B shall fall on B', then if the side B A
fall on B l A r , the angle ABC equals the angle A' B' C ; if the
side B A fall between B' C and B' A' in the direction B' D, the
angle ABC is less than A' B' 6" ; but if the side B A fall in the
direction B'E, the angle A B C is greater than A' B f C.
This method of superposition en- g C
ables us to add magnitudes of the
same kind. Thus, if we have two c
straight lines AB and CD, by A
placing the point C on B, and keeping C D in the same direc-
tion with A By we shall have one continuous straight line A D
equal to the sum of the lines A B
and C D.
in : if we have the angles
A 11 and D E F, by placing
the vertex B <>n K and the side
BC in the direction of ED, the
angle AB C will take the position
A ED, and the angles D E F and
J B C will together equal the an-
gle A EF.
-D
-B
Mathematical Terms.
36. Def. A Demonstration is a course of reasoning by which
the truth or falsity of a particular statement is logically established.
37. Def. A Theorem is a truth to be demonstrated.
38. Def. A Construction is a graphical representation of
a geometrical conception.
39. Def. A Problem is a construction to be effected, or a
question to be investigated.
DEFINITIONS. 13
40. Def. An Axiom is a truth which is admitted without
demonstration.
41. DsF. A Postulate is a problem which is admitted to
I*.- possible.
42. DSP. A /V< 7 is i it her a theorem or a problem.
43. Ih.i. A Corollary is a truth easily deduced from the
proposition to which it is attached.
1 1. Def. A Scholium is b remark upon some particular fea-
ture of a proposition.
45. Ihr. An Hypothesis is a supposition made in the
enunciation <>f a proposition, ox in the course of a demonstration.
Axiom.
1. Things which are equal to the same tiling arc equal to each
oti
•_'. When equals arc added to equals the sums are equal.
3. When equals arc taken I are equal.
1. When equals are added to iiiicquals I arc unequal
5. When equals are taken bom anequala the remainders
■iual.
G. Things which arc double tl thing, or equal things,
re equal to each oth<
7. Thin a which are halves of the same thing, or of equal
tliih ch other.
8. The w r than any of its parts.
'.». The wlml .1 to all its parts taken together.
17.
ated —
1. That a straight line can be drawn from any one point to any-
other point.
l\ Thai a straight line can be produced to any distance, or can
be terminated at any point.
3. That the circumference of a circle can be described about any
i any distance from that centre.
14
GEOMETRY. BOOK I.
48. Symbols and Abbreviations.
.*. therefore.
= is (or are) equal to.
Z angle.
A angles.
A triangle.
A triangles.
II parallel,
O parallelogram
HJ parallelograms.
_L perpendicular.
Jl perpendiculars.
rt. Z right an
it angles.
> is (or are) greater than.
< is (or are) less than.
rt. A right triangle.
rt. A right triangli
O circle.
© circles.
+ increased by.
— diminished by.
X multiplied by.
-4- divided by.
Post, postulate.
Def. detinition.
Ax. axiom.
Hyp. hypothesis.
Cor. corollary.
Q. E. D. quod erat demonstran-
dum.
Q. E. F. quod erat faciendum.
Adj. adjacent.
K\t.-int, exterior-interior.
Alt. -int. alternate-interior.
[den. identical.
( 'mis. construct ion.
Sup. supplementary.
Sup. adj. supplcinentary-adja-
Ex. exercise.
111. illustration.
PERPENDICULAR AND OBLIQUE LINES. 15
•x Perpendicular axd #blique Lixes.
ri:tr#srn#x I. The#kkm.
49. When one ttoraigkt line crosses another straight line
He * til.
o
D
I'
Let line OP cross A B at C.
We are to prove Z OCB = Z A CP.
Z C A + Z CB = 2 rt. A, §34
10 sup. -adj. A).
Z OCA + Z A CP = 2 rt. A, § 34
i.A).
. . <>< ' A + ZOC B = ZOCA + Z AC P. Ax. 1.
Tab of these equals the common ZOC A.
Tn ZOCB = ZACP.
In like mam, v prove
Z ACO = Z PCB.
Q. E. D.
50. COROLLARY. If two straight lines cut one another, the
four angles which they make at the point of intersection are
fchei equal to four right angles.
16 GEOMETRY. — BOOK I.
Proposition II. Theorem.
51. When the sum of two adjacent angles is equal to two
right angles, their exterior sides form one and the same
straight line.
Let the adjacent angles Z OCA + Z C B = 2 it. A.
We are to prove A C and C B in the same straight line.
Suppose C F to be in the same straight line with A C.
Thru Z OCA + Z OCF= 2 it. A. §34
[being sap. -adj. A ).
But Z OCA + Z OCB = 2 rt. A. Hyp.
.-.Z OCA + Z OC F=Z OCA + Z OCB. Ax. 1.
Take away from each of these equals the common Z C A .
Then Z OCF=Z OCB.
.'. C B and C F coincide, and cannot form two lines as rep-
resented in the figure.
. * . A C and C B are in the same straight line.
Q. E. D.
PBRPENDICULAB AND OBLIQUE LINES.
17
Proposition III. Theorem.
52. A perpentlh'iihir measure* the shortest distance from
a /jo i uf to a tiraigkt 1%
i
Let A B be the given straight line, C the given point
and CO the perpendicular.
\Y> em to prove C O < drown from CtoA />,
as C
luce CO to E, making OE= CO.
Drav, A'/:
On A B U a: (."' /' until it comes into the
plane rfO A'/'.
The line C will take the direction of E y
■■<• Z COF= ZEOF, eachbeing a rt. Z).
The point will fall upon the poinl
(*i?we C = OE by cons.).
.'. line CF"*lu*FB,
(Afl hi the same points).
(70+ 0E=2 CO.
C0+ OE d is the shortest distance between two points).
Substitute 2 C for CO + OF,
an.l 2 CFfor C F+ F E ; then wo have
2 C0< 2(7>.
Q E. D.
and
But
§ 18
Cons.
§ 18
18
GEOMETRY. BOOK I.
Proposition IV. Theorem.
53. Two oblique lines drawn from a point in a perpen-
dicular, cutting off equal distances from the foot of the per-
pendicular, are equal.
Let F C be the perpendicular, and C A and C two
oblique lines cutting off equal distances from F.
We are to prone C A = C 0.
Fold over C FA, on C F as an axis, until it comes into the
plane ofCFO.
FA will take the direction of FO,
(since ZCF A — ZCFO, each being art. Z).
Point A will fall upon point 0,
(FA = FO, by hyp.).
.'.line C A = line CO,
(their extremities being the same points).
§ 18
Q. E O.
PKKI'KNDK IIAR AND OBLIQUE LINES. 19
Proposition V. Theokkm.
54. The sum of two lines drawn from a point to the ex-
ities of a straight lii iter than tie sum of two
oihrr I . but included by them.
Let C A and Q /J be two lines drawn from the point
to the extremities of the straight line A ft Let O A
and B be two lines similarly drawn, but included
byCA and ft
i prove CA + CB>OA + OB.
Pru 1 E.
Tl AG+CB>AO+0£, §18
and /■> BO. § 18
Add . and we b
l+CE+BE+0E>0A + 0E+0B.
Substitute for CE + BE its equal C B,
and take an side of the inequality.
We have C-4 + CB>0'A + OB.
Q. E. D.
20
GEOMETRY. -
BOOK I.
Proposition VI. Theorem.
55. Of two oblique lines drawn from the same point in a
perpendicular, cutting off unequal dista?ices from the foot of
the perpendicular, the more remote is the greater.
C
Let C F be perpendicular to A B, and C K and C II two
oblique lines cutting off unequal distances from F.
1 1 V are to prove C II > C K.
Produce CF to E f making F E= C F.
Draw JTJTand K1L
CII = // h\ and C K = KE, § 53
(two oblique lines drawn from the same point in a J_, cutting off equal dis-
tances from the foot of the ±, are equal).
But CII+ II E >CK+ K E, § 54
(7 J he sum of two oblique lines drawn from a point to tlie extremities of a,
straight, line is greater than the sum of two other lines similarly drawn,
but included by them) ;
.-. 2 CII> 2 CK)
.'. CII>CK.
Q. E. D.
56. Corollary. Only two equal straight lines can be drawn
from a point to a straight line ; and of two unequal lines, the
greater cuts off the greater distance from the foot of the perpen-
dicular.
II ERF KXDICULAR AND OBLIQUE LINES. 21
Pro position VII. Theorem.
57. Two equal oblique lines, drawn from tie same point
in a perpendicular, cut off equal distances from the foot of
fie perpendicular.
Let C F be the perpendicular, and C E and C K be two
equal oblique lines drawn from the point C.
Wi m to i rove /'A'.
Fold over C FA on F as an axis, until it comes into the
plan.- of C /-'A'.
The line FB will take the direction /'A',
(Z CFE= Z C F A. i art. Z).
Then the point E mud fall upon the point K \
otherwise one of these oblique lines must be more remote from
the _L,
and .*. greater than the other; which is contrary to the
hypothesis. § 55
.\ F /;== FK.
Q. E. D.
22
GEOMETRY. — BOOK I.
Proposition VIII. Theorem.
58. If at the middle point of a straight line a per pen-
lot be erected,
I. Any point in the perpendicular is at equal dis fauces
from the extremities of the straight line.
II. Any point without the perpendicular is at unequal
distance* from the extremities of the straight line.
Let P 11 be a perpendicular erected at the middle oi
the straight line A B, O any point in PR, and C any
point without 7V.'.
I. Draw OA and OB.
We are to prove A — B.
Since PA = PB,
OA = OB, §53
•hliqnc lines drawn from the same point in a _L, cutting off equal dis-
tances from the foot of the _L, are equal).
II. BmwCA and C B.
We are to prove C A and C B unequal.
One of these lines, as CA, will intersect fche _L.
From D, the point of intersection, draw 1) 11.
I'KKPKNDh ll.AU AND OBLIQUE LINES. 23
DB = DA, § 53
[twt oh'i'i' " _L, cutting off equal dis-
fout of tJie J_, are equal).
CB< CD+ DB, § 18
(a sf is tlie shortest distance between two points).
Substitute f<»r D B its equal DA, then
CB < CD+ DA.
But CD + DA = CA, Ax. 9.
..CBf all points equallj from the extreini-
ties of tha! imr.
60. Scholium. Si fcermine the position of
a straight line, two points equally distant from the extremities
of a straighl Line det rmine the perpendicular at the middle
1>- »int of that line.
Ex, 1. If an an hat is its complement?
L\ If an and*' he a right its supplement ?
3. If an angle he £ of a right angle, what is its complement?
1. If an angle be £ of a right anjgle, what is its supplement?
5. Show that tlie hisectors of ,xwo vertical angles form one
and the same straight line.
w that the two straighl lines which bisect the two
d angles are perpendicular to each other.
24
GEOMETRY.
-BOOK I.
Proposition IX. Theorem.
61. At a point in a straight line only one perpendicular
to that line can be drawn ; and from a point without a
straight line only one perpendicular to that line can be drawn.
AE
/>*
Fig. 1.
D
Let B A (fig. 1) be perpendicular to C D at the point B.
We are to prove B A the only perpendicular to C D at the
pot Hi B.
If it be possible, let B E be another line J_ to C D at B.
Then Z EBD is art. Z. § 20
Bat Z AB1) is a rt. Z. § 26
.\Z EBD = Z ABD. Ax. 1.
That 18, a part is equal to the whole ; which is impossible.
In like manner it may be shown that no other line but B A
is_L to CD at B.
Let AB (fig. 2) be perpendicular to C D from the point A.
We are to prcm A B the only _L to C D from the point A.
If it be possible, let A E be another line drawn from A _L
to CD.
Conceive Z A E B to be moved to the right until the ver-
tex E falls on B, the side E B continuing in the line CD.
Then the line E A will take the position B F.
Now if A E be _1_ to C D, B F is _L to C D, and there will
be two J* to C D at the point B ; which is impossible.
In like manner, it may be shown that no other linn hut
A B is JL to CD from A. Q E D
62. Corollary. Two lines in the s-dinc plane perpendicular
to the same straight line have the same direction; otherwise
they would meet (§ 22^), ami we should have two perpendicular
liius drawn from their point of meeting to the same line ; which
is impossible.
*v
PARALLEL LINES]
UlTIVSKS^
I >\ Parallel Lines.
Parallel Lines are straight lines which lie in the same
plane and have the Bame direction, or opposite directions.
rile! lines lie in the Bame direction, when they are on
iame Bide of the straight line joining their origins.
Parallel lines lie in opposite directions, when they are on
oppo of the straight line joining their origins.
64, Tu "Ot meet. § 21
-////' plane perpendicular to a given
line I I art therefore parallel,
\ a given point only om J >< drawn par-
allel to a g § 18
line EF cu1 two other straight lines AB
and ODj it makes with those lines eight angles, to which par-
ticular nai: iven.
The angles I. l. 6, 7 are called Interior angles.
Tli 2, 3, 5, 8 are called Exterior angles.
The pairs of angles 1 and 7, 4 and 6 are called Alternate-
The pairs of angles 2 and 8, 3 and 5 are called Alternate-
The pairs of angles 1 and 5, 2 and o, 4 and 8, 3 and 7 are
• all r angles.
26
GEOMETRY. — BOOK I.
Proposition X. Theorem.
67. If a straight line be perpendicular to one of two
parallel lines, it is perpendicular to the other,
H
-B
M
-N
K
Let A B and E F be two parallel lines, and let UK be
perpendicular to A B.
\Y> are t<>J,rove II K J_ to E F.
Through C draw MN JL to UK.
Then MN is I! to A B. § or,
(Two lines in the same plane ± to a given line are parallel).
But EF is II ioAB, llvp.
.'. E F coincides with M N. § 66
vugh Hi< aanu point only one line can he drawn II to a given line),
.\EFis±to // K,
that ia // K is JL to E F.
Q. E. D.
PARALLEL LINES. 27
Proposition XL Theorem.
68. If two parallel ttraight lines be cut by a third
straight line the alternate-interior angles are equal.
A B
Let EF and OH be two parallel straight lines cut by
the line BC.
We are to prove Z B = Z C.
Through 0, the middle point of B C, draw A Z)Xto G II.
Then A D is likewise _L to E § 67
(a ti ± to one of two lis is ± to the other),
that is, CD and B A are both ± to A D.
Apply tigure COD to figure £04 so that 02) shall fall
on
Thru 6' will fall on OB,
'' D = Z. B A, bei, A);
and point G will tall upon 2?, *
>ce C = B by construction).
Then _L 02> will coincide with _i_ AM. § 01
vm _L to tlmt line can be drawn),
.*. Z CD coincides with Z BA, and is equal to it.
Q. E. D.
s. SOLIUM. By i; r$e of a proposition is meant a
proposition which has the hypothesis of the first as conclusion
and the conclusion of the first as hypothesis. The converse of
a truth is not neceuarilg true. Thus, parallel lines never meet ;
r ?/*e^ are parallel, is not true unless
the lines lie in the same plane.
v.. The << >n verse of many propositions will be omitted,
but their statement and demonstration shomel be required as an
import rise for the student.
28 GEOMETRY. BOOK I.
Proposition XII. Theorem.
69. Conversely : When two straight lines are cut by a
third straight line, if the alternate-interior angles he equal,
the two straight lines are parallel.
Let E F cut the straight lines A B and C I) in the points
II and K, and let the Z A II K = Z II KB.
We are to }~>rove A B II to C D.
Through the point II draw MN II to CD;
thm Z MIIK=Z I/KIJ, §68
(being alt. -int. A ).
Bui Z AIIK = Z II K I), Hyp.
.'.Z M II K = Z A UK. Ax. 1.
.*. the lines M N and A B coincide.
But J/iVis II to CD; Cons.
.". A B, which coincides with M N, is II to C I>.
Q. E. D.
PABALLEL I.IXES. 29
Proposition XIII. Theorem.
70. If two parallel lines be cut by a third straight line,
ike ' Interior angle* are equal.
E
Let A B and CD be two parallel lines cut by the
straight line E /', in the points II and K.
H v / EU H = Z II KD.
Z EJIIi = ZAII K\ §49
A).
But ZAIIK=Z1IA />. §68
-.-////. A).
.-.Z E II Jl = Z H KD. Ax. 1
In like manner we may prove
ZEIIA=ZIIKC.
Q. E. D.
71. CoBOLLARY. The alternate-exterior angles, E II B and
( ' K I\ and also A II E and D K E, are equal.
30 GEOMETRY. — BOOK I.
Proposition XIV. Theorem.
72. Conversely : When two straight lines are cut by a
third straight line, if the exterior-interior angles be equal,
these two straight lines are parallel.
E
Let E F cut the straight lines AB and CD in the
points II and K f and let the Z K II B = Z UK/).
Wt are to prove A B II to C D .
Through the point II draw the straight Line M X II to CD.
Then Z KIIN=Z II K IK § 7C
ng r.rt.-'nil. A).
But Z KIIB = Z UK D. Hyp
.-. Z EHB - Z EHN. Ax. 1
.*. the lines MX and A B coincide.
Bui MX is II to CD, Cons
.'. A B, which coincides with M JX, is II to CD.
Q. E. D.
PABALLEL LINES. 31
Proposition XV. Theorem.
73. If two parallel lines be cut by a third straight line,
He turn of (//>' too inferior angles on the same side of the
>f H ne is equal (<> two fight
Let All and C D be two parallel lines cut by the
straight line EF in the points II and K
We a m Z B UK + Z UK I) = two rt. A.
Z EUB + Z BIIK=2 it. A, § 34
A).
Bui Z Ell B = Z UK lh § 70
'■"j c.rt.-int. A).
gtitate Z II K I) for Z #7/7? in the first equality;
then Z B II K + Z IIKO = 2 rt. A
Q. E. D.
32 GEOMETRY. BOOK I.
Proposition XYI. Theorem.
74. Conversely : When two straight lines are cut o// a
third straight line, if the two interior angles on the same side
of the secant line be together equal to tivo right angles, then
the two straight lines are parallel.
E
Let EF cut the straight lines AB and CD in the
points II and K, and let the Z B II K + Z II K D
equal two right angles.
W( "re to prove AB II to C D.
Through fche point II draw UN II to CD.
Then Z NIIK + Z II K D = 2 rt. A, § 73
(h ing two interior A on f/ir some side of the secant line).
But Z B1IK+ Z II Kl) = 2 rt. A. I [yp.
.'.ZX1IK+ Z II KI) = ZB1IK+ ZII K I). Ax. 1.
Take away from eacli of these equals the common Z II K IK
then Z NHK= Z B II K.
.'. the lines A B and M N coincide
But MNvt II to CD; Cons.
.*. A B, which coincides with M N, is II to CD.
Q. E D.
PARALLEL LINES.
33
Proposition XVII. Theorem.
7."). Two straight lines which are parallel to a third
Uraighi line are parallel to each other.
A-
C-
B
-D
F
Let A B and C D be parallel to E F.
We are toprem A B II to CD.
Draw //A'_L to EF.
Since CD and EFm 1, ffJTis J. to CD, § 67
gki /hie be J_ to one of two fb, it is ± to the other also).
Since A B and BF*te II, UK is also ±toAB, § 67
.'.Z IIOB = Z II PD,
{each being art. Z).
.'. A B is II to CD, § 72
we cut by a third straight line, if tlie cxt.-int. A
qnalf the two lines are II ).
Q. E. D.
34
GEOMETRY. •
■ BOOK I.
Proposition XVIII. Theorem.
76. Two parallel lines are everywhere equally distant
from each other.
i: M II
A 1 1 ; B
Let A B and CD be two parallel lines, and from any
two points in A B, as E and II, let EF and UK
be drawn perpendicular to A I>.
We ore t<> prove E F ■= // K.
Now EF and // K are i. to C h, § G7
{>' live _L t<> n„, of two \\s is _L to Ho other also).
Let M he the middle point of EH.
Draw MP± to A B.
On MP as an axis, fold over the portion of the figure on
the right of MP until it eoraee into the plane of the figure on
the left.
M B will fall on MA,
{for Z P M 11= ZP M /•:, i ach being a rt Z ) ;
the point // will fall on E,
(forMH= ME, by hyp.) ;
If K will fall on EF,
{for Z M JIK= Z M E F, each being a rt. Z) ;
and the point K will fall on E E, or E F produced.
Also, PD will fall on PC,
(Z MPK= Z MPF, earlt being a rt. Z) ;
and the point K will fall on P C.
Since the point K fills in both the lines EFmd PC,
it must fall at their point of intersection E.
.'./IE=EF, §18
(their 'xtremities being the same points).
Q. E. D.
PARALLEL LINES.
35
Proposition XIX. Theorem.
77. Two mi i } lee whose sides are parallel, two and two,
Hud lie in ike name direction^ or opposite directions, from their
■t'S, are equal,
A D
D>
Pig. l.
Let A 11 and E (Fig. 1) have their sides BA and ED,
and BO and BF respectively, parallel and lying
in the same direction from their vertices.
(Pi an to pro* the A B = A B,
ay) two ridea which an.* not II until they
intersect, as at // ;
A n = ZDHC, § 70
(beii A),
and AE = AD1I<\ §70
.-.z n = z /:. Ax. l
Let A B and K' (Fig. 2) have B' A' and E' /)', and B' C
and E' F respectively, parallel and lying in oppo-
site directions from their vertices.
PPi on / Z B' = Z E'.
Produce (if necessary) two sides which are not II until they
intersect, as at II'.
Th AB t = AE'irC' f §70
(being txL-inL A),
and A E' = A E'lI'C, § 08
(Jbeing alt. -int. A) ;
.-. A B' = A E', Ax. 1.
Q. E. D.
oG GEOMETRY. BOOK I.
Proposition XX. Theorem.
78. If tiro angle* have two sides parallel and lying in
the same direction from their vertices, while the other two
sides arc parallel and He in opposite directions, then the two
a, i pies are supplements of each other.
c
Let A EC and 1) E F be two angles having BC and E I)
parallel and lying in the same direction from their
vertices, while EF and B A are parallel and lie in
opposite directions.
W\ on to prove Z ABC and Z D E F supplements of each
Produce (if necessary ) two sides which ate not II until they
Intel //.
Z A BC = Z BUD, § 70
{being aot.-int. A).
z i) ef=z BB&t § 68
(h, fag , f 11. -int. A).
Bat Z n If I) and Z 11 If E are supplements of each other, § 3 t
(being sup. -adj. A ).
.-. Z A BO and Z I) EF, the equals of Z BUD and
Z B ll /;. are supplements <>f<-;trii other.
Q. E. D.
TRIANGLES. 37
On Triangles.
79. Def. A Triangle is a plane figure bounded by three
straight lines.
A triangle has six parts, three sides and three angles.
80. When lli.- rix parts of one triangle are equal to the six
parte of another h ich to each, the triangles are said to
>< l. I > i i . An Tmscdes triangle is one of which two sides
mal.
I >i:i . An Equilateral triangle is one of which the three
sides are equal
I'll. Tin- Base of a triangle is the side on which the
triangle is supposed to stand.
In an isosceles triangle, the side which is not one of the
equal sides is considered the base.
38
-UOOK I.
87. Def. A : angle is one which has one of the
_ ht angle.
88. DBF. The side opposite the right angle is called the
<>tenu*e.
89. \h.\ . An Obtuse triangle is one which has one of the
I an olAu.se angle.
90. 1 h.v. An Acute triangle is one which has all the angle*
CQUIANOULAK.
91. Ih.v. A f/ular triangle is one which lias all
Def. In anv to the hase is
le, and its vertex is called I
i -in the vertex to the haw-, or the bote produced.
Def. Tl
eluded between a side and an adjacen / CB h.
Def. The two angle* of a Mangle which are opp
. are called the tw<
I sad C.
TRIANGLES. 39
96. Any side of a triangle is less than the sum of the
other two sides.
e a straight line is the shortest distance between two '
points,
A C < A B + /;
97. Any side of a triangle is greater than the difference
of the other two sides.
In the inequality A C < A B + BC,
take away A B from each side of the inequal
Then -AB1. (Job. 6, In an equiangular triangle, each angle is one
third of two right an w<_> thirds of one right angle.
PROl XXII. TlII.oKKM.
105. / i r'miKj It is equal fo Ih e 8 u ui
'>/ thr //<■< tUtf/feS.
a
Let BC If be an exterior angle of the triangle ABC.
II, em to j'mve Z BCIf=Z A + Z B.
Z BC 11+ Z A C B = 2 it. A, § 34
A).
Z A + Z B + ZACB = 2 rt. A, § 98
(three AofaA = two rt. A ).
.'.ZBCH+, ^ZA + ZB + ZACB. Ax. 1.
Take away from each of these equals the common ZA CB\
then Z BC 11 = Z A + Z B.
Q. E. D.
42
GEOMETRY.
BOOK I.
Proposition XXIII. Theorem.
106. Two triangles are equal in all resjjects when two
sides and the included angle of the one are equal respectively
to two sides and the included auyle of the other.
B At
In the triangles ABC and A' £' C, let AB = A'B',
A C=A'C',A A=Z. A'.
We are to prove A A B C = A A' B' C.
Tab- 1 up the A A BC and place it upon the A J'/j'C so
that A B shall coincide with A' B'.
Then A C will take the direction of A 1 C,
(for /. A = A A', by /.),
the point C will fall upon tin' point O,
(forAC=A'C' y by hyp.) ;
.'. CB = C B',
(their extremities being the same, points).
•\ the two A coincide, and are equal in all respects.
§ 18
Q. E. D.
TRIANGLES. 43
Proposition XXIV. Theorem.
107. Tiro triangle* are equal in all respects when a side
and two adjacent angles of the one are equal respectively to a
side and two adjacent angles of ike other.
c a
n ,v
In the triangles ABC and A' B' C, let A B = A' B' y
Z A=Z A', A B = Z. B'.
We are to prove A A BC = A A' B' O.
Take up A A BC and place it upon A A' B' C, so that
A B shall • oiacide with At l>'.
A C will take the direction of A 1 C,
(/or Z A = Z A', by hyp.) ;
the poinl C. the extremity of A 6', will fall upon 4' C" or
A' C prodaced
B C will take the direction oiB'C,
(forZB = ZB f , by hyp.);
the poinl C. the extremity of BC, will fall upon B' C or
B 1 C prodn
.'. the point C, falling upon both the lines A' C and B 1 C,
must fall apon a point common to the two lines, namely, C.
.'. the two A coincide, and are equal in all respects.
Q. E. D.
44
GEOMETRY. BOOK I.
Proposition XXV. Theorem.
108. Tiro triangles are equal when the three sides of the
qne are equal respectively to the three sides of the other,
B B>
In the triangles A B C and A' B' C, let A B = A' B',
AC = A'C, BC = B' C
in on to prove A A BC = A A' B' C.
Place A A' B' C in the position A B' C, haying its greatest
side A' C in coincidence with its equal AC, and its vertex at
ll 1 , opposite B.
Draw B B' intersecting A C at //.
Snu-At AB = A /:\ Hyp.
point A is at equA] distances bom B and B'.
Since J? C = /? 7 C, Hyp.
point 6' La at equal distances from B and B 1 .
.'. A C is J_ to J5 J5' at its middle point, § 60
(two points at equal distant i from the extremities of a straight line deter'
" the J. at the middle of that line).
Now [f A A B' C be folded over on .4 (7 as an axis until it
comes into the plane of A ABC,
II H' will tall on EB }
{for Z A H B = Z A J I /;', each h ing >< rt, Z),
and point B 1 will fall on B,
{for H B' = H B).
.'. the two A coincide, and are equal in all respects.
Q. E. D.
TRIAXGLES.
45
Proposition XXVI. Theorem.
L09. Two right triangles are equal when a side and the
hypotenuse of the one are equal respectively to a side and the
hypotenuse of the other.
In the right triangles ABC and A' B' C, let A B = A' 5',
and AC = A'C.
We are to prove A A BC = A A' B' C.
Tftkfl up the A A BC and place it upon A A' B' C, so that
A B will oomdde with A' B'.
Then BC will fell upon n'C,
(forZABC=ZA'B'C \g a rt. Z),
and point C will fall upon C ;
otherwise the equal oblique lines A (7- and A' C would cut
off unequal distances from the foot of the JL, which is im-
possible, § 57
((>/•<> (qua J sfrom a point in a X cut off equal distances from the
foot of the _L ).
.*. the. two A coincide, and are equal in all respects.
Q. E. D.
46
GEOMETRY. BOOK I.
Proposition XXVII. Theorem.
110. Two right triangles are equal when the hypotenuse
and an acute angle of the one are equal respectively to the
hypotenuse and an acute angle of the other.
In the right triangles ABC and A' B' C, let AC = A' C,
and Z A= Z A'.
We are to prove A A B C = A A' B' ,
Hyp.
Hyp.
then Z C = Z C, § 101
100 rt. A have an acute Z of the one equal to an acute Z of the other,
then the other acute A are equal).
.'.AABC = AA'B'C, § 107
(two & are equal when a side and two adj. A of the one are equal
respectively to a side and two adj. A of the other),
Q. E. D.
111. Corollary. Two right triangles are equal when a
side and an aeute angle of the one are equal respectively to an
homologous side and acute angle of the other.
TRIANGLES. 47
Proposition XXVIII. Theorem.
112. In nit uoteete* triangle the angles opposite the
c
I B
Let ABC be an isosceles triangle, having the sides
A C and C B equal.
We are to prove Z A = Z B.
From C draw the straight line C E so as to bisect the
ZACB.
In the A it CA'an.l BOB,
AC=BC, Hyp.
C E=C E f Iden.
ZACE=ZBCE; Cons.
..AACE = ABCE, §106
- sides and the included Z. of the one are equal
respectively to two sides and the included Z. of the other).
r.Z A=Z B,
(being homologous A of equal A ).
Q. E. D.
If the equal sides of an isosceles triangle be produced,
show that the angles formed with the base by the sides produced
'[Ual.
48
GEOMETRY. — BOOK I.
Proposition XXIX. Theorem.
113. A straight line which bisects the angle at the vertex
of an /'.v.y -r/r.v triangle divides the triangle into tiro equal
triangles, is perpendi'-ular to the base, and bisert.s the base.
C
Let the line C E bisect the Z AC B of the isosceles
A ACB.
Wi em to prove I. AACE = ABCE;
II. line CE±toAB;
III. A E = BE.
I. In thoA ACE ai»\ BCE,
AC=BC, Hyp.
CE=C/.\ Iden.
Z ACE = Z BCE. Cons.
.-.A AC E = A BCE, § 10G
ag two sides and / , f ual respectively to two aides
and the included A of the otJier).
Also, II. Z CEA = Z CEB,
{being homologous A of equal &).
.*. CEis± to AB,
(a straight line meeting another, making tlie adjacent A equal, is ± to
that line).
Also, III. AE= E /;,
(being homologous sides of equal &).
Q. E. D.
TRIANGLES. 49
Proposition XXX. Theorem.
114. If Udo angle* of a triangle be equal, the sides op-
lr the equal angles are equal, and the triangle is isosceles.
B D
In the triangle ABC, let the Z B = Z C.
Wi ve A B = A C.
Draw A D ± to B C.
In the it. A A DBund A l><\
A I> - A D t Iden.
ZB = ZC,
.'. rt. A A 1) B = rt. A A DC, § 111
>g a side and an acute /L oj !>>al respectively to a side and an
Z. of the "I
:.AB = AC,
(being homologous sides of equal &).
Q. E. D.
Ex. Show that an equiangular triangle is also equilateral.
50 GEOMETRY. BOOK I.
Proposition XXXI. Theorem.
115. If two triangles have two sides of the one equal
respectively to two sides of the other, 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.
C
\\l
E *1e
In the A ABC and ABE, let A B = A B, B C = B E ;
but Z ABO Z ABE.
We are to prove AO A E.
Place the A so that A B of the one shall coincide with A B
<>f the other.
Draw BFaotBto bisect Z EBC.
Draw EF.
In the a BBFaad OBF
i: B = BC, Hyp.
BF=BF, Men.
Z EBF=Z CBF, Cons.
.'. the A EBFund C B F are equal, § 10G
{hnrng two sides and the included £ of one equal respectively to two sides
and the included Z. of the other).
.\EF=FC,
(being homologous tides of equal A).
Now A F + F 3 > A B f § 96
(the sum of two sides of a A is greater than the third side).
Substitute for F E its equal FC. Then
AF+ FOAE; or,
A C > A E.
Q. E. D.
TRIANGLES. 51
Proposition XXXII. Theorem.
116. Conversely: If two sides of a triangle be equal
respectively to two sides of another, but the third side of the
first triangle be greater than tie third side of the second, then
tlir angle opposite tie third side of the first triangle is greater
tin m tie angle opposite the third side of the second.
In the &ABCandA'B'C' J let A B = A' B', AC — A' C ;
but HO B' C.
We are to prove Z A > Z A'.
If Z A=Z A',
then would AABC = AA f B'C, §106
(having two sides and the included Z oftlie one equal respectively to two sides
and ( I £of tlie other),
and BC^B'C,
(being homologous sides of equal & ).
And if A < A',
thru would BC Z A 1 .
Q. E. D
52 GEOMETRY. — BOOK I.
Proposition XXXIII. Theorem.
117. Of two sides of a triangle] that is the greater
which is opposife the greater angle.
In the triangle ABC let angle AC B be greater than
angle /:.
Wi we A B > A C.
Draw CEbq aa bo make Z B C E = Z B.
Then EC = EB, §114
A ).
Now AE+ EOAC, §96
(/A- wm of two tides of a A /*■ greater than the third side).
t i t ute for i£ C its equal .# A Then
ii ^ + E B> AC, or
AB> A C
Q. E. D.
Ex. ABC and ABB are two triangles on the same base
^1 j5, and on the same side of it, the vertex of each triangle
being without the other. If A C equal A D, show that B C
cannot equal B D.
TRIANGLES. 53
Proposition XXXIY. Theorem.
lis. Of two angle* of a triangle, that is the greater
which is opposite the greater side.
B
C
In the triangle A BC let A B be greater than I C.
PPi tm to prove Z A C B > Z /;.
Tak ,ual to A C;
Draw EC.
Z AEC = Z ACE, § 112
i
But Z A EC> Z /;, § 105
(an exterior A of a A is gr Z. ),
an. I Z AC B > Z ACE.
Substitute foi Z A C E its equal Z A EC, then
Z ACB> Z A EC.
h mure is Z AC B> Z B.
Q. E. D.
If the angles ABC and ACB, at the base of an
be bisected by the straight lines B D, CD,
Bhow that D BC will be an isosceles triangle.
54 GEOMETRY. BOOK I.
Proposition XXXY. Theorem.
119. The three bisectors of the three angles of a triangle
m •trt in a point.
Let the two bisectors of the angles A and C meet
at 0, and B be drawn.
xm B I he Z B.
Draw the J8 OK, () />, and OH.
CP,
OC=OC, Iden,
Z OCK = Z OCP, (lis.
.'.A OCK = A OCP, § 110
M '■ i ofeq/kal A).
In the it, A Oil P and O J //,
0A*=OA t Lien.
-ZOAP = ZOAH, , Cona.
/.A 6>d P = AOA //, § 110
w /A-' KypoUnxm and cm acute Z of the one eq ""I respectively to the
hypotenuse and an acute Z of the 0th
.'. OP=0/t.
(being homologous tides "/equal A ).
But we have already shown P = O A\
.'.OH=()k\ Ax. I
Now in rt. A II B and KB
TRIANGLES. 55
Oil =0 K, a.ndOB = OB,
.'.AOHB = A OKB, § 109
of the one equal respectively to the hypote-
nuse and a side of the other),
.-. Z o/;// = Z OBK,
(being homologous A of equal A ).
Q. E. D.
Proposition XXXVI. Theorem.
120. The i peudicular* erected at the middle
point* qf the thn yf a triangle meet in a point.
A
F
Let I) D , / / , /'/', be three perpendiculars erected
at I), E, F, the middle points of A B, A C, and BO.
We are to pro* iat, as 0.
The two J* I) D' and BE 1 meet, otherwise they would be
parallel, and A B and A £7, being _l! to these lines from the same
point A, would he in the same straight line;
but this is impossible, since they are sides of a A.
Let be the point at which they meet.
Then, since is in J) D', which is J_ to A B at its middle
point, it is equally distant from A and B. § 59
Also, since is in* E E' , _L to A at its middle point, it is
equally distant from A and 0. /
.*. is equally distant from B and C ;
.'. is in F F' _L to BO at its middle point, . § 59
(the locus o) \ly distant from the extremities of a straight line
is the ± erected at the middle of that line).
Q. E. D.
56
GEOMETRY. BOOK I.
Proposition XXXYII. Theorem.
1:21. The three perpendiculars from the vertices of a tri-
angle to the opposite sides meet in a point.
B
In the triangle ABC, let B P, A II. OK, be the per-
pendiculars from the vertices to the opposite
sides.
We are to prove they meet in some point, as 0.
Through the vertices A, B, C, draw
A' B II- to BC,
A' C II to A C,
B' C II to A B.
In the A A B A' and A B C, we have
A B = A B,
ZABA' = Z BAC,
(being alternate interior A ),
Z BAA' = Z ABC.
Iden.
§ 68
§ 68
.*.A ABA' = A ABC, § 107
Hide and two adj. A of tlie one equal respectively to a side mm
two adj. A of the other).
.'.A'B = AC\
(being homologous sides of egical A ).
TIM 57
In the A CBC mid A B(
BC = BC, Idea
ZCBC' = ZBCA, §08
(being alternate interior A ).
ZBCC' = ZCBA §68
.-.A CBL" = AABC, §107
nrj a side o> A of the one equal respectively to a M
adj. A of the other).
BC' = AC,
{being homologous sides of equal A ).
But we have already shown A' B = A C,
A'B = BC, Ax. 1.
/.Bjb\ We point of A'C.
BPia±toAC, Hyp.
\s±toAU § 67
(a straight line which is ± to one of two \\s is ± (
middle point of A' C \
.-. BP is A. to A'C at its middle point
In like manner we may prove tl
A H i> _L t<» A' W at its middle point,
and C K A. to B 1 C at its middle point.
.-. 11 l\ A //. and CK are Js erect- middle points
. . A B'C.
.'. these J§ meet in a point. § 120
points of the sides of a A /// >int).
Q. E. D.
58
GEOMETRY. BOOK I.
On Quadrilaterals.
122. Def. A Quadrilateral is a plane figure bounded by
four straight lines.
123. Def. A Trapezium is a quadrilateral which has no
two sides parallel.
124. Def. A Trapezoid is a quadrilateral which has two
sides parallel.
125. Def. A Parallelogram is a quadrilateral which has
its opposite sides parallel.
TRAPEZIUM.
TRAPEZOID.
PARALLELOGRAM.
126. Def. A Rectangle is a parallelogram which has its
angles right angles.
127. Def. A Square is a parallelogram which has its
angles right angles, and its sides equal.
128. Def. A Rhombus is a parallelogram which has its
sides equal, but its angles oblique angles.
129. Def. A Rhomboid is a parallelogram which has its
angles oblique angles.
The figure marked parallelogram is also a rhomboid.
RECTANQLE.
QUADRILATERALS. 59
130. I'i.f. The side upon which a parallelogram stands,
tnd the opposite side, are called its lower and upper bases; and
(he parallel sides of a trapezoid are called its bases.
LSI. Def. The Altitude of a parallelogram or trapezoid is
thr perpendicular distance between its baa
132. Def. The Biaxial of a
quadrilateral La a straight line joining
any two opposite vertices.
PROPOSITI- \ WWII I. Thiohem.
LS3. / ^a par all the figure
into ftco eq
B C
A I
Let ABC K be a parallelogram, and A C its diagonal.
Wi are to prove A A B C = A A E < .
In the A ABC and A EC
AC = AC, Iden.
Z ACB = Z CA /:. §68
■inf. A).
ZCAB = ZACE, §68
.*. AABC = AAEC, § 107
" adj. A of the o vely to a wide end two
adj. A of the other).
Q. E. D.
60 GEOMETRY. BOOK I.
Proposition XXXIX. Theorem.
134. In a parallelogram the opposite sides are equal,
and the opposite angles are equal.
A E
Let the figure ABC E be a parallelogram.
We are to prove B C = A E, and AB = EC,
also, ZB = Z E,andZ BAE = Z BCE.
Draw A C
A ABC = AAEC, § 133
(the diagonal of a O divides the figure into two equal A ).
.'.BC = AE,
and AB = CE,
(being homologous sides of equal & ).
Z B = Z E,
(being Jiomologous A of equal A ).
Z BAC = Z ACE,
and ZEAC = ZACB,
(being Jiomologous A of equal &)•
Add these last two equalities, and we have
ZBAC + ZEAC = ZACE+ZACB;
or, ZBAE = ZBCE.
Q. E. D.
135. Corollary. Parallel lines comprehended between par-
allel lines are equal.
QUADRILATERALS. 61
Proposition XL. Theorem.
136. If u quadrilateral have two sides equal and par-
allel, then the otki ' parallel, and the
//// n re is a parallelogra m .
B C
Let the figure ABCE be a quadrilateral, having the
side A E equal and parallel to BC.
We are to prove A B equal and II to E (
Draw A C.
In the A ABC and A EC
BC = A A. Hyp.
A C = AC, [den.
Z BCA =Z CAE,
{being alt. -int. A).
..AABC = AACE, § 106
' sides and the included /Lof tlie < to two sides
and the included Z. of the M
.\AB = EC,
lologoits sides of equal A ).
Also, ZBAC = ZACE,
(being homologous A of equal & ) ;
.'.A Bis II to EC, §69
/ lines are cut by a third straight line, if the alt.-int. A be
equal the lines are pumUrl).
.'. the figure ABCEis&CJ, § 125
(the opposite sides being parallel).
Q. E D.
62 GEOMETRY. BOOK I.
Proposition XLI. Theorem.
137. If in a quadrilateral the opposite sides be equal, the
figure is a parallelogram.
A
Let the figure A B C E be a quadrilateral having
BC = AE and AB = EC.
We are to prove figure A B C E a O.
Draw A C.
InthzA ABC and A EC
BC = AE, Hyp.
AB = CE, Hyp.
A C = A C, Iden.
.'. A ABC = AAEC, § 108
(having three sides of the one equal respectively to three sides of the other).
.'./. ACB = Z CAE y
and ZBAC = ZACE,
(being homologous A of equal & ).
.'.BCis II toAE,
and A Bis II to EC, §69
(when two straight lines lying in the same plane are cut by a third straight
line, if the alt. -int. A be equal, tlie lines are parallel).
.'. the figure ABC E is a, EJ, § 125
(having its opposite sides parallel).
Q. E. D.
QUADEILA1 BBAL8.
63
Proposition XLII. Thsobkm.
138. The diagon from bisect each other.
B C
Let the figure A B C E be a parallelogram, and let
the diagonals A C and BE cut each other at 0.
We are to prove AO = OC, and B 0= ft
In the A A OB and BOC
AE=BC,
§ 134
pposite sides of a E3\
ZOAE=ZOCB,
§68
(being alt. -int. A ),
Z OEA=Z OBC; § 68
.-. AAOE = A BOC, § 107
ng a side and two adj. A of the <> spectively to a side and two
adj. A of the other).
..AO = OC,
and BO = OE.
(being homologous sides of equal & ).
Q. E. D.
64 GEOMETRY. BOOK I.
Proposition XLIII. Theorem.
139. The diagonals of a rhombus bisect each other at
right angles.
A E
Let the figure A B C E he a rhombus, having the
diagonals A C and BE bisecting each other at 0.
We are to prove Z AO E and Z A B rt. A.
In the A A E and A B,
AE = AB, §128
(being sides of a rhombus) ;
OE=OB, §138
(the diagonals of a CD bisect each otlier) ;
AO = AO, Iden.
.'. A AOE = A AOB, § 108
(having three sides of the one equal respectively to three sides of the other) ;
.'.ZAOE = ZAOB,
(being homologous A of equal A ) ;
.\ Z A E and Z A B are rt. A. § 25
When one straight line meets another straight line so as to make the adj. A
equal, each Z is a rt. Z).
Q. E. D.
(»IAI>RILATERAI>. 65
Proposition XLIV. Thboj
1 10. Two paralltb i two iifcs and the in-
cluded angle ofth % and the
angle of the other, are equal in all resp*
B CD' a
A'
In the parallelograms A B C D and A' B 1 O D', let
AB = A' B', AD = A'D', and Z. A = Z A'.
We are to prove that the HJ are eq
Apply O A BCD to O A'B'C'D', so that A D will fall
<>ii and coinejdi I D'.
Tli.:. A /I will fall 0]
(/orZA = Z A' t I
and B will fid! on B',
B = A' B\ \
Now, BC and B' O are both II V ad an- drawn
through point B'\
.-. tli- line* BCvbA W <" coincide, § 66
ed
In like manner I> and />' C are II to A' 11' and an drawn
through tin- point
.-. DC and /^ ' " coincide : § 66
.-. the point ('falls on I)' C , oi /> ' ' '•' produced ;
.'. (7 falls on both AB'C;
.*. 6' must fall on a point common to both, namely, C.
.'. the two UJ coincide, and are equal in all respects.
Q. E. D.
Ml. CoBOLLABT. Two red vring the same base and
aUitudi ,-iu'it ; fox they may be applied to each other and
will coincide.
66 GEOMETRY. BOOK I.
Proposition XLV. Theorem.
142. The straight line which connects the middle points
of the non-parallel sides of a trapezoid is parallel to the par-
allel sides, and is equal to half (heir sum.
C U
Let SO be the straight line joining the middle points
of the non-parallel sides of the trapezoid ABCE.
We are to prove SO II to A E and B C ;
also SO = h(A £ + HC).
Through the point draw F II II to A B,
and produce B C to meet F II at //.
In the A FOEbxA C II
OE=OC, Cons.
ZOEF=ZOCII, §68
( fc iiirj alt. -int. A ),
ZFOE = ZCOH, §49
(being vertical A ).
/.A FOE = A CO//. § 107
{having a side and two adj. A of the one equal respectively to a side and two
adj. A off/<<' other).
QUADRILATERALS. 67
.\FM=CH %
and 0F=01I,
\g homologous sides of equal A ).
Now F II - A H § 135
(II lines conqrrelumded between II lines arc equal) ;
..FO—AS. Ax. 7.
•\ the figure A FO Sis a O, § 136
(having two opposite sides equal
. . .S'Ois II to ^1 /■'. § 125
tig opposite sides of a O).
SO is also II to BC,
{a st i I to one of two II lines is II to the other also).
Now
SO = A I\
(being opposite sides of a CJ) t
§125
and
so - /;//.
§ 125
But
AF=AE-FE,
and
BH = BC+ C II.
Substitute for A F and BE their equals, ^i? — FE&ni
BC+ Cll.
and add, observing that C // = F E\
then 2S0 = AE+BC.
.'. SO = i(AE+ BC).
Q. E. D.
68
GEOMETRY. BOOK I.
On Polygons in General.
143. Def. A Polygon is a plane figure bounded by straight
lines.
144. Def. The bounding lines are the sides of the polygon,
and their sum, s&AB + BC+CD, etc., is the Perimeter of
the polygon.
The angles which the adjacent sides make with each other
are the angles of the polygon.
145. Def. A Diagonal of a polygon is a line joining the
vertices of two angles not adjacent.
B B'
C A
D F'
An Equilateral polygon is one which has all its
An Equiangular polygon is one which has all
E
146. Def.
sides equal.
147. Def.
its angles equal.
148. Def. A Convex polygon is one of which no side,
when produced, will enter the surface bounded by the perimeter.
149. Def. Each angle of such a polygon is called a Salient
angle, and is less than two right angles.
150. Def. A Concave polygon is one of which two or more
sides, when produced, will enter the surface bounded by the
perimeter.
151. Def. The angle FD E is called a Re-entrant angle.
AVhen the term polygon is used, a convex polygon is meant.
The number offsides of a polygon is evidently equal to the
number of its angles.
By drawing diagonals from any vertex of a polygon, the fig-
ure may be divided into as many triangles as it has sides less two.
POLYGONS. 69
152. I'ii. Two polygons are Equal, when they can be
divided by diagonals into the same number of triangles, equal
each to each, and similarly placed ; for the polygons can be
applied to each other, and the corresponding triangles will evi-
dently coincide. Therefore the polygons will coincide, and be
equal in all respects.
153. Def. Two polygons are Mutually 1. lar } if the
angles of the one be equal to the angles of the other, each to
each, when taken in the same order; as the polygons ABODE F,
and A 1 B 1 C D' E' F % in which Z A - Z A 1 , Z B = Z B',
ZC = ZC, etc.
L5 1. 1 >i i . The equal angles in mutually equiangular poly-
gons are called Homologous angles; and the sides which lie
between equal angles are called Homologous sides.
155. Def. Two polygons are Mute //, if the
sides of tie tl to the sides of the other, each to each,
when taken in the same order.
Fig. 1. Pig. 3. :. 4.
Two polygons may be mutually equiangular without being
mutually equflatei I and 2.
And, t*f two polygons may be
mutually equilateral without being mutually equiangular \ as
S and 4.
If two polygons be mutually equilateral and equiangular,
they are equal, for they may be applied the one to the othei
as to coincide.
156. Def. A polygon of three sides is a Trigon or Tri-
angle ; one of f«»ur sides is a Tetragon or Quadrilateral ; one of
fagon ; one of six sides is a Hexagon ; one of
ii sides is a Heptagon; one of eight sides is an Octagon; one
of fcen agon ; one of twelve sides is a Dodecagon.
70
GEOMETRY. BOOK I.
Proposition XLYI. Theorem.
157. The sum of the interior angles of a polygon is
equal to two right angles, taken as many times less two as
the figure has sides.
A B
Let the figure ABC DEF be a polygon having n sides.
We are to prove
A A + Z B + A C, etc., = 2 rt. A (n - 2).
From the vertex A draw the diagonals A C, A D, and A E.
The sum of the A of the A = the sum of the angles of the
polygon.
Now there arc (« — 2) A,
and the sum of the A of each A = 2 rt. A.
§98
.'. the sum of the A of the A, that is, the sum <>f the A of
the polygon = 2 rt. A (n — 2).
Q. E. D.
158. Corollary. The sum of the angles of a quadrilateral
equals two right angles taken (4 — 2) times, i. e. equals 4 right
angles; and if the angles be all equal, each angle is a right
angle. In general, each angle of an equiangular polygon of n
9 ( v — 9\
is equal to ^-A_ J. right angles.
POLYGONS. 71
PBOPOemOH XLVIL Theorem.
159. Tl i of a polygon, nufde by produ-
cing each < if lis -svVA'.y in succession, arc together equal to four
rig/tt angles.
Let the figure ABODE be a polygon, having its sides
produced in succession.
•IPJ // to prove the sum oj A.
Denote the int. A of the polygon by A,B,t\ i>. E \
ami the ext . 1 ■. (" the equal
s of the triangle.
12. If from the diagonal 11 1) of a square A BCD, BE be
• >fT equal to BC, and E F }>•■ drawn perpendicular to BD y
show that D E is equal to F F, and also to FC.
1 3. Show that the three lines drawn from the vert ices of a
triangle to the middle points of the opposite sides meet in a
point
BOOK II.
CIRCLES.
PiaiM'I ,
160. I > i : i . A ' a plane figure bounded by a curved
line, all the points of which are equally distanl from a point
within called ill'* &
1''. 1. I > i i . 1 nee of a circle is the line which
bounds tin- qu
162. of a circle is any straight line drawn
tram the centric t o the circumference, as -1. 1
163. 1 1 (meter of a cii< line paw-
be centre and havh ircum-
.
By :li definition of a circle, all its radii are equal. He i
all i; . since tin- dian I to twice
i liua
M .1/
16 I. 1 »i i . An Arc of a circle is any portion of the circum-
is an arc equal to one
half the circumference, as A M l>. Fig. 2.
166. Dl >rd of a circle is any straight line having
:ivniiti i ; i . In different cir . similar sectors,
Mid J > ~" it R'* ~~ B* '
Q. E. D.
— =
Then
». Corollary. Similar arcs, being like parts of their re-
ive circumferences, are to each other as their radii; similar
like parts of their respective circles, are to each
/res on their radii.
2*22 GEOMETRY. BOOK Y.
Proposition XII. Theorem.
386. Similar segments are to each other as the squares
on their radii. C
O A
P , p
Let A C and A' C be the radii of the two similar seg-
ments ABP and A' B' P'.
w . ABP AC 1
\\ e are to prove = .
1 A'B h P' jj~(jP-
The sectors ACB and A' C B' are similar, § 383
(having the A at the centre, C and (7, equal).
In the A ACB and A'C'B'
/_C = ZC, § 383
(being corresponding A of similar sectors).
AC = CB, § 1G3
A'C' = C'B'; § 163
.-. the A A C 11 and A' C B' are similar, § 284
ing an Z of the one equal to an Z of tlie other, and the including sides
proportional).
Now sector ACB _ AC 2 § 385
sector A'C'B' £Tj}?
i ilar sectors are to each other as the squares on their radii) ;
and AACB ,-£* §342
A A'C'B' JTq/*'
(similar & are to each other as the squares on their homologous sides).
jj sector ACB- A ACB ■ ATO 2
Hence
sector A' C B' - A A' C B' ^HJ' 2
or,
segment A B P J~C & 271
segment A' B' P' A 7 !!' 2
■■■■ quantities be increased or diminished by like parts of each, the results
will be in Uie same ratio as the quantities themseh
Q. E. D.
EXERCISES. 223
Exercises.
1. Show that an equilateral polygon circumscribed about a
circle is regular if the number of its sides be odd.
2. Show that an equiangular polygon inscribed in a circle is
ir if the number of its sides be odd.
■ . Show that any equiangular polygon circumscribed about a
circle is regular.
vl 4. Show that the side of a circumscribed equilateral triangle
LB double the side of an inscribed equilateral triangle.
5. Show that the area of a regular inscribed hexagon is
ilis of that of the regular circumscribed hexagon.
Show that the area of a regular inscribed hexagon is a
• nil 1m! areas of the inscribed and cir-
cumscribed equilateral brian
7. Show that the area of a regular inscribed octagon is equal
to that of a rectangle whose adjacent sides are equal to the
ibed an' I circumscribed squares.
8. Show that the area <>t' a regular inscribed dodecagon is
te square on the radius.
9. Given the di rff a circle 50; find the area of the
area of a sector of 80° of this circle.
1«>. Three equal circles touch each other externally and thus
inclose one acre of ground ; find the radius in rods of each of
these circles. '
11. Show that in two circles of different radii, angles at the
centres subtended by arcs of equal length are to each other in-
versely as the radii.
1 2. Show that the square on the side of a regular inscribed
pentagon, minus the Bquare on the side of a regular inscribed
decagon, is equal to the square on the radius.
224 GEOMETRY. BOOK V.
On Constructions.
Proposition XIII. Problem.
387. To inscribe a tegular jx>/t/gon of any number of
tides in a given circle*
"Let Q be the given circle, and n the number of sides
of the polygon.
It is required to inscribe in Q, a regular polygon having n
Divide the circumference of the O into n equal arcs.
Join the extremities of these .arcs.
Then we have the polygon required.
For the polygon is equilateral, § 181
{in the same O equal arcs are subtended by equal chords) ;
and the polygon is also regular, § 364
{an equilateral polygon inscribed in a O is regular).
Q. E. F.
CONSTRUCTIONS.
225
Proposition XIV. Problem.
388. To inscribe in a given circle a regular polygon
which has double the number of sides of a given inscribed
tegular polygon.
if
Let ABC D be the given inscribed polygon.
It is required to in$cribe a regular polygon having double the
N u mher of sides of A BC D.
Bisect the arcs A B, BC, etc.
v A /:. KB, BF, etc.,
The polygon A E B FC, etc., is the polygon required.
the chords A B, BC, etc., are equal, § 363
(being sides of polygon).
.'. the arcs A B, BC, etc., are equal, § 182
( M the same O equal chords subtend equal arcs).
Hence the halves of these arcs are equal,
or, AE, E B, BF, FC, etc., are equal ;
.'. the polygon A EB F, etc., is equilateral.
The polygon is also regular, § 364
(an equilateral polygon inscribed in a O is regular) ;
and has double the number of sides of the given regular
polygon.
Q. E. F.
226
GEOMETRY. BOOK V.
Proposition XV. Problem.
389. To inscribe a square in a given circle.
Let be the centre of the given circle.
It is required to inscribe a square in the circle.
Draw the two diameters A C and B D JL to each other.
Join AB, BC, CD, and DA.
Then A B C D is the square required.
For, the A ABC, BCD, etc., are rt. A, § 204
(being inscribed in a semicircle) ,
and the sides AB, BC, etc., ar^ equal, § 181
(in the same O equal arcs are subtended by equal chords) ;
.*. the figure A B CD is a square, § 127
(having Us sides equal and its A rt. A ).
Q. E. F.
390. Corollary. By bisecting the arcs AB, BC, etc., a
regular polygon of 8 sides may be inscribed ; and, by continuing
the process, regular polygons of 1G, 32, G4, etc., sides may be
inscribed.
>TRUCTIONS. 227
Proposition XVI. Problem.
891. To inscribe in a given circle a regular hexagon
v
Let be the centre of the given circle.
It is required to inscribe in the given O a regular hexagon.
From draw any radios, as OC.
From C as a centre, with a radius equal to C,
describe an arc intersecting the circumference at F.
Draw OF*xi& C F.
Then C F \& a side of the regular hexagon required.
For the A F C is equilateral, Cons.
and equiangular, § 112
.*. the Z FO C is J of 2 rt. A, or, £ of 4 rt. A . § 98
.*. the arc FC is \ of the circumference A B C F,
.'. the chord FC, which subtends the arc FC, is a side
of a regular hexagon ;
and the figure CFD, etc., formed by applying the radius
six times as a chord, is the hexagon required.
Q. E. F.
392. Corollary 1. By joining the alternate vertices A, C,
/), an equilateral A is inscribed in a circle.
393. Cor. 2. By bisecting the arcs A B, B C, etc., a regu-
lar polygon of 12 sides may be inscribed in a circle ; and, by
continuing the process, regular polygons of 24, 48, etc., sides
may be inscribed.
228
GEOMETRY. — BOOK V.
Proposition XVII. Problem.
394. To inscribe in a given circle a regular decagon.
Let be the centre of the given circle.
If is required tot tki given O a regular decagon.
Draw the radius 0,
and divide it in extreme and mean ratio, so that shall
be to OS as S is to SO. § 311
From as a centre, -with a radius equal to OS,
describe an arc intersecting the circumference at B.
Draw BO, BS, and B 0.
Then B is a side of the regular decagon required.
For
and
00
( Jons.
Cons.
then
OS : : OS : SO,
BO=0 8.
Substitute for S its equal B 0,
: BO :: BO : SO.
Moreover the Z B = Z S B,
.'.the A 00 B and B OS are similar,
ng an A oftheone equal to an Z. of the other, and the including sides
p r op o rtional).
But the A 00 B is i § 160
(ifs sides C and OB being radii of the same circle).
.'. the ABO S, which is similar to the A OB, is isoscel
Id en.
§284
CONSTRUCTIONS. 229
and BS = BG. §114
Bat OS = BC, Cons.
.'.OS = BS, Ax. 1
■\ the A S B is isosceles,
and theZ = Z S B 0, §112
(tein# opposite equal sides).
But the ZCSB = ZO+ZSBO, § 105
estcrior Z of a A is t nf the two opposite interior A ).
.\theZ CSB = 2Z 0.
Z SO B(= Z C8B) °~2Z 0, § 112
and Z OBC (= Z SCJi) = 2 Z 0. §112
.-. the sum of the a give* regular polygon.
J9' CD
Let A BCD, etc., be the given regular polygon, and
C D' E' the given circle.
ft m required tc C D' E' a regular polygov
si in i In r t<> A B C D.
mi (a the centre <-i* the polygon A BC D, etc.
draw OB and OC.
•in 0' the centra of the O C D' & t
draw VCsad o 1 D\
making the /.<)' = £ 0.
Draw 6" 7/.
Then CD' will be a side of the regular polygon required.
For each polygon will have as many sides as the A
(=Z 0') is contained times in 4 rt. A.
.'. the polygon < u D' E', etc. is similar to the polygon
r/y /;. etc., § 372
{two regular polygons o/tkt same number of sides are similar).
Q. E. F.
'Sd'2 GEOMETRY. BOOK V.
Proposition XX. Problem.
400. To circumscribe about a circle a regular polygon
similar to a given inscribed regular polygon.
BMC
Ha
/?T%#
y
"""-. /V
pV
.....*:4'.'. aL/ the one equal respectively to a side and
cent A of the otJier).
(being homologous A of equal & ).
In like maimer ire may prove Z C = Z. D, etc
.-. the polygon .1 BCD, etc., is equiangular.
Since the ... B II M, CM R, etc, are isosceles, § 241
/// to a Q.arc equal),
the sidee A'//, A'JA Clf, CR t etc, are equal,
{being homologous sides of equal isosceles &).
. . the sides AB>BC,CD, etc. are equal, Ax. 6
and the polygon A BC D t etc La equilateral
Therefore the circumscribed polygon is regular and similar
to the given inscribed polygon. § 372
Q. E F.
A* denote the radius of a regular inscribed polygon,
/• the apothem, a one side, A one angle, and C the angle at the
(■••litre ; Bh0W that
1. In a tegular inscribed triangle a = R V^, r = \ R,
A = 60°, C = 120°.
2. In an inscribed equate a = R v 7 ^, r =* \R ^2, A = 90°,
C = 90°.
3. In a regular inscribed hexagon a = R, r = ^ R V^3,
A = 120°, (7 = 60°.
R 0/5 - 1)
I. In a regular inscribed decagon a = g >
r = J /* ^10 + 2 V6, ^ = U4°, (7 = 36°.
234 GEOMETRY. BOOK V.
Proposition XXI. Problem.
401. To find the value of the chord of one-half an arc,
in terms of the chord of the whole arc and the radius oj the
circle,
D
Let A B be the chord of arc A B and A D the chord
of one- half the arc A />.
It is required to find the value of A D in terms of A B and
11 (radius).
From D draw J) II through the centre 0,
and draw A.
II I) is J_ to the choid A B at its middle point C, § 00
{two points, and D, equally distant from the extremities, A and B, de-
termine the position of a ± to the middle point of A B).
The Z HA D is a rt. Z, § 204
/ inscribed in a semicircle),
.\A~Jf = DHX DC, §289
{tlie square on one side of a rt. A is equal to /In' product of the hypotenuse by
the adjacent segment made by the J_ let fall from the vertex of tJie rt. Z ).
Now D II = 2 //,
and DC = DO-CO = B-CO;
.\AD l - -111(11- CO).
■ 0N8TBUCTION8. 235
Since A C is a rt. A,
i !'■
4 A- - AB\
= V4 IP - AR* .
2
h. the equation AH? = 2R (R— CO),
substitute
for C its value V4 *» - ^
2
then
V 2
2 8» — W^4JP-
-J-5'V
. J - \ 2 A'- - R NilP- A tf\ .
Q. E. F.
( loROLLART. It* \\v take the radius equal to unity,
equation A D - t/-J A'- - £ ^4 # 2 - i ~i* 2 ) becomes
AD = ^2- V^-Jtf 2 .
236
GEOMETRY. BOOK V.
Proposition XXII. Problem.
403. To compute the ratio of the circumference of a
circle to its diameter, approximately.
Since
§370
Let C be the circumference and R the radius of a
circle.
, = G
2 It'
C
Wll.'ll R = 1, 7T = - '
// i$ required to find the numerical ''due of tt.
We make the following -computations by the use <>1 tqe
l
formula obtained in the last proposition,
Nh.
A 1) = 1/2 - V^4 - ii 2? 2 ,
when .4 /? is a side of a regular hexagon :
In a polygon of
Form of Computation. Length of Bide. Perimeter.
12 .1 1) = y/ 2 - y/J^P .517G3809 G.211G5708
2 I J /> = ^ 2-^4 — (.51763809) 2 .26105238 6.26525722
H .-1 /) = ^ 2^4 -(.26 1052 38^ .13080626 6.27870041
96 A I) — ^ 2 — ^4 — (.13080626)2 .06543817 6.28206396
192 AD = ^ 2 — ^i — (.06543* I 7 g .08272346 6.28290510
AD = > ]'2 — )Ji — (.0327234^ .01636228 6.28311544
768 A D = ^2 - V4 — (.01636228)2 .00818121 6.28816941
Eenoe we may consider 6.28317 as approximately the cir
cumference of a O whose radius is unity.
7r, which equals — ,
6. 28317
2
.-. 7t = 3.1 U59 nearly.
ISOPERIMETRICAL POLYGONS. 237
On Isoperimetrical Polygons. — Supplementary.
H' L I)i;f. Isoperimetrical figures are figures which have
aqua] perimet
>. Def. Among magnitudes of the same kind, that
whirl i is greatest is a Maximum, and that which is smallest
Minimum,
I'h us the diameter of a circle is the maximum among all
inscribed straight lines; and a perpendicular is the minimum
among all straight lines drawn from a point to a given straight
line.
Proposition XXIII. Theori
406. Of all triangle* having two ride* respectively equal,
f/taf in wkiek these B .il'f angle is the maxi-
A
E hv E
Let the triangles A BC and EBC have the sides AB
and BC equal respectively to KB and BC] and
let the angle ABC be a right angle.
m AABOAEBC.
From A'. Lei fall the _L E D.
The A ABC and EBC, having the same base B C, are to
each other as their altitudes A B and ED, § 326
(& } t n /se are to each otJier as their altitudes).
Nnw ED is
A
A A'
Let A BCDB be a polygon inscribed in a circle, and
A' I> ' <" J)' /;•' be a polygon, equilateral with re-
spect to ABODE, but which cannot be inscribed
in a circle.
We are to prove
the polygon A It C D E > the polygon A' B' C l D' E'.
Draw the diameter A II.
Join C// and J) II.
:i C D 1 (=CD) construct the A C II' D' = A C H D y
and draw A 1 II'.
Xuw the polygon A BOB > the polygon A'B'C'IP, § 407
fans fonncd of sides all given but one, the polygon inscribed in a
tied side, for its diameter, is the maximum).
And the polygon A E D II > the polygon A' E' D' H'. § 407
Add these two inequalities, then
the polygon ABC II BE > the polygon A' B f C H l D 1 E' .
Take away from the two figures the equal A CHD and
t ■ II //.
Then the polygon ABCDE>the polygon A* B' C D' E'.
Q. E. D
240
GEOMETRY. BOOK V.
Proposition XXVI. Theorem.
409. Of all triangles hiving the same base and equal
perimeters, the isosceles triangle is the maximum.
,\II
Let the AACB and ADB have equal perimeters,
and let the AACB be isosceles.
FPi are to prove AACB>AADB.
Draw the J§ OJBmd 1> I'.
AACB CE
AABD 1)F y
(& having the same bast err fa ttuk other as their altitudes).
Produce A C to //, making Off** AC.
D»W 1/ II.
The Z A B JI is a rt. Z, for it will be inscribed in the
semicircle drawn from C as a centre, with fch< radios CB.
ISOPERIMETRICAL POLYGONS. 241
From (7 let fall the LCK;
and from D as a centre, with a radius equal to D B,
describe an arc cutting // B produced, at ^.
Draw D P and A P,
and let fall the ± DM.
Since .\!l = AC+CB = AD + DB,
and AP A P.
.-. BH> BP. §56
BK = \BH, §113
(a _L tfnriOTi /rowi /Ac rcrtec o/a» isosceles A fcisccte /Ac tec)>
and BM=hBP. §113
But '•/-' »*; § 185
(II* compi-chcndcd between \\s are equal);
and DF=BM, §136
.-. CE> DF.
.:AACB> AADB.
Q. E. D.
242
GEOMETRY. BOOK V.
Proposition XXVII. Theorem.
410. The maximum of isoperimetrical polygons of the
same number of sides is equilateral.
Let ABCD, etc., be the maximum of isoperimetrical
polygons of any given number of sides.
If, an to prove A B, BC, C D, etc., equal.
Draw A C.
The A ABC must be the maximum of all the A which
are formed upon A C with a perimeter equal to that of A ABC.
Otherwise, a greater A A KC could be substituted for A ABC,
without changing the perimeter of the polygon.
But this is inconsistent with the hypothesis that the poly-
gon ABCD, etc., is the maximum polygon.
.*. the A A B C, is isosceles, § 409
{<>/ all A having the same base and equal perimeters, tlie isosceles A is the
maximum ».
In like manner it may be proved that B C = CD, etc.
Q. E. D.
111. Corollary. The maximum of isoperimetrical poly-
gons of the same number of sides is a regular polygon.
For, it is equilateral, § 410
(the maximum of isoperimetrical polygons of the same number of tides is
equilateral).
Also it can be inscribed in a O, § 408
(the maximum of all polygons formed of given sides can be inscribed in a O).
Hence it is regular, § 364
(tin equilateral polygon inscribed fa a is regular).
ISOPERIMETRICAL POLYGONS.
243
Proposition XXYIII. Theorem.
412. Ofiiqperimeirieal regular polygons, that is greatest
7/ ias the greatest number of ride*.
Let Q be a regular polygon of three sides, and Q' be
a regular polygon of four sides, each having the
same perimeter.
We are to prove Q' > Q.
1 n any side A B of Q, take any point D.
The polj may be considered an irregular polygon
of fi in which the sides A D and D B make with each
r an A equal to two rt. A .
Then the irregular polygon Q, of four sides is less than the
r isoperimetrical polygon Q' of four sides, § 411
naximwm of isoperimetrical polygons of the same number of sides is a
regular polygon).
In like manner it may be shown that Q f is less than a
regular isoperimetrical polygon of five sides, and so on.
Q. E. D.
413. Corollary. Of all isoperimetrical plane figures the
circle ifl the maximum.
244
GEOMETRY.
BOOK Y.
Proposition XXIX. Theorem.
414. If a regular polygon be constructed with a given
area, its perimeter will be the less the greater the number
of its sides.
Let Q and Q' be regular polygons having the same
area, and let Q' have the greater number of sides.
We are to prove the perimeter of Q> the perimeter of Q'.
Let Q" be a regular polygon having the same perimeter as
Q' y and the same number of sides as Q.
Th.-n Cis> us, thai is t In greatest which has the greatest
>/>'S).
But Q = Q',
.'. £is><2".
.*. the perimeter of Q is > the perimeter of Q".
But the perimeter of Q' = the perimeter of Q", Cons.
.-. the perimeter of Q is > that of Q'.
Q. E. D.
415. Corolla by. The circumference of a circle is less than
the perimeter of any other plane figure of equal area.
SYMMETRY.
245
On Symmetry. — Supplementary.
416. Two points are Symmetrical when they are situated
nil opposite sides of, and at equal distance* ./font, a fixed point,
line, or plane, taken as an object of reference.
417. When a point is taken as an object of reference, it is
called the Centre of Symmetry . when a line is taken, it is called
the Axis of Symmetry ; when a plane is taken, it is called the
Pin ry.
418. Two points are symmetrical with re-
spect to b autre bisect the sti.
tine terminated by these points. Thus, J\ p
are symmetrica] with respect to C, if C b
the straight line PP.
419. The distance of either of the two symmetrical points
from the called the Rodin s of Symmetry,
Thus either OP 01 CP ifl the radius of symmetry.
420. Txoo points are symmetrieal with
set to an axis, if the axis bisect at right
- the straight line terminated by these
points. Thus, 1\ P are symmetrical with re-
in the axis XX', if XX' bisect PP at
right angles.
1 2 1 . Two points are symmetrical with
ct to a plane, if the plane bisect at
right angles the straight line terminated by
these points. Tims P, P' are symmetrical
with respect to MN, if M N bisect P P' at
right angles.
M L
Pt
rN
246
GEOMETRY.
-BOOK V.
422. Two plane figures are symmetrical with respect to a
centre, an axis, or a plane, if every point of either figure have
its corresponding symmetrical point in the other.
B>
Fig. 1.
Fig. 2.
Fig. 3.
Thus, the lines A B and A' B' are symmetrical with respect
to the centre C (Fig. 1), to the axis X X' (Fig. 2), to the plane
MX (Fig. 3), if every point of either have its corresponding
symmetrical point in the other.
V I
c<
.V
W D'
ML
Fig. 6.
Also, the triangles A B D and A' B 1 D' are symmetrical with
■t to the centre C (Fig. 4), to the axis XX' (Fig. 5), to the
plane MX (Fig. G), if every point in the perimeter of either
have it.s corresponding symmetrica] point in the perimeter of the
other.
423. Def. In two symmetrical figures the corresponding
Symmetrical points and lines arc called homologous.
SYMMETRY.
247
Two symmetrical figures with respect to a centre can be
brought into coincidence by revolving one of them in its own
plane about the centre, every radius of symmetry revolving
through two right angles at the same time.
Two symmetrica] figures with respect to an axis can be
brought into coincidence by the revolution of either about the
axis until it conns into the plane of the other.
424. Dip. A single figure is a lymmetrical figure, either
when it can be divided by an axis, or plane, into two figures
symmetrica] with respect to that axis or plane; or, when it has
atre such that every straight line drawn through it cuts the
perimeter of the figure in two points which are symmetrical
with respect to that centre.
Fig. 1.
Thus, Fig. 1 is a symmetrica] figure with respect to the
divided by X A" into figures ABC D and AB'C'D
which are symmetrica] with reaped to XX'.
And, Fi- 2 is a symmetrica] figure with respect to the
e 0, if the centre bisect every straight line drawn
through it and terminated by the perimeter.
v such straight line is called a diameter.
The circle is an illustration of a single figure symmetrical
with respect to its centre as the centre of symmetry, or to any
diai;. of symmetry.
248 GEOMETRY. BOOK V.
Proposition XXX. Theorem.
1:25. Two equal and parallel lines are symmetrical with
respect to a centre.
A B>
B A'
Let AB and A 1 B' be equal and parallel lines.
We are to prove A B and A' B' symmetrical.
Draw A A' and B />', and through the point of their inter-
section C, draw any other line 11(11', terminated in AB and
A'B'.
In the A CAB md C A' h>
A 11 = A'B', Hyp.
also, A A and B = A A' and B' respectively, §68
(firing off. -inf. A),
.-.A CAB = A CA'B'; § 107
,\ CA and CB = CA' and C B' respectively,
(//> ///'/ homologoxu sides of equal &).
Now in the A A C II and A' C II'
AC = A'C,
A A and A C H '■■ A A' and A'C H* respectively,
.'.A A GII = A A'C IV, § 107
[hewing a ride and two adjj, A of the one equal respectively to a side and two
adj. A of the other),
.'. CII=G //',
(being homologous sides of equal A ).
.*. //' is the symmetrical point of //.
r»»it // is any point in A B ;
.'. every point in A B has its symmetrica] point in A' B'.
.*. A B and A' B' are symmetrical with respect to G as a
centre of symmetry.
Q. E. D.
426. Corollary. If the extremities of one line be re-
spectively the symmetricals of another line with respect to the
same centre, the two lines are symmetrical with respect to that
centre.
SYMMETRY.
249
Proposition XXXI. Theorem.
I*7« Tf a figure he q // with respect to two axes
vendicular to each other, it is symmetrical with respect
to tin w as a centre.
Y
JD
V
Let the figure ABODE FO // be symmetrical to the
two axes X X' % ) V which intersect at 0.
Wi u try of the figure,
I be any pouri in the perimeter of the figure.
Draw IKL ± I tid IMN± to )' V.
J«n £0, o.\\ and AM/.
Now a/- /■.
7 /////i respect to X X ! ).
Bui A'/- 02f,
(lis comprehended between \\s ar
.\ A' A - 0JT.
.-. KLOM k a O,
(having two sides equal and parulhl).
.\ LO is equal and parallel to A r J/,
'<7 opposite sides of a CJ).
In like manner we may prove OiV equal and parallel to AT if.
Hence the points Z, 0, and ^V are in the same straight line
drawn through the point IJ to KM.
Also L0 = 0N,
{since each is equal to KM).
lit line L X, drawn through 0, is bisected at 0.
.'. is the centre of symmetry of the figure. § 424
O. E. D.
§420
§ 135
Ax. 1
§ 136
§ 134
250
GEOMETRY.
BOOK V.
Exercises.
1. The area of any triangle may be found as follows : From
half the sum of the three sides subtract each side severally, mul-
tiply together the half sum and the three remainders, and extract
the square root of the product.
Denote the sides of the tri-
angle A B C by a, b, c, the alti-
ii f'l + b + c .
tude by p, and by s.
Show that
a 2 = b 2 +-c*-'2cXAD,
AD
and show that
i *
P
2c
p =
= V (b+c + a)(b + c— a) (a + b~c) (a— b + c)
2e
Hence, show that area of A A B 0, which La equal to
t X p
= \