202 GEOMETR®Y 
Principles two sides DE, DF of another triangle DEF, each ‘to. for thé dase, and the opposite angle is the vertex ; but Princ 
—\~" each, and if the angle BAC contained by the former 
Fig. 32. 
Fig. 33. 
Fig. 34. 
Fig. 36. 
Fig. 35. 
be greater than the angle EDF contained by the latter; 
the base BC of the cease which has the greater angle 
shall be greater than 
angle. 
sae 
Make the angle CAG = D, take AG “DE ot AB; 
and join CG; and because the triangles CAG, DEF) 
have an angle‘of the’ one equal to an angle of the other,: 
and the sides which contain these’ angles ‘equal; CG 
shall be equal to EF, (6.).. Now «there may be three) 
cases, according as the ‘point Gifalls: without the tri-) 
angle ABC, or on the side BC; or within the triangle. 
Cass Ist, (Fig. $2.): Because GC is less than GI -+ 
IC, and AB less than AI +. IB, therefore GC 4 AB is) 
less than GI + AI + IC + IB,. that is, GO. 4 AB 
AG + BC; from these unequal sums: take away AB; 
or its equal AG, and there remains GC — BC; but 
GC=EF, therefore EF — BC. 
Casz 2d, (Fig. 33.) If the point G fall on BC, it is 
evident that GC, or its equal EF, is less than BC. 
Case 3d, (Fig. 34,) Lastly, if the point G fall with- 
in the triangle, by Prop. 9. we have AG + GC = 
AB+BC; and taking away AG from one of these un- 
equals, and AB=AG from the other, we have GC = 
BC. ; 
Prov. XI. Turon. 
Two triangles are equal when the three sides of the 
one are equal to. the three. sides of the other, each to 
each, 
Let, the side AB= DE, AC = DF, and BC = EF, 
then shall the angle A= D,, B=E, C=F. _ For if the 
angle A were greater than D; then, as the sides AB, 
AC are equal .to the, sides DE, DF; each to each, it 
would follow (10.) that the, side BC, would be greg: 
than EF: And ifthe angle A were less than D, then 
BC would be less than EF: but BC is equal to EF, 
therefore the angle A can neither be greater nor less 
than D ; therefore A=D. In like manner it may be 
proved that B=E and that C—F. 
Scuoxium. It appears that in two equal triangles, 
the equal angles are eppamtc to the equal sides, for, the 
ne A and_D are. opposite to the equal sides 
Prop. XII. Tueror. 
In an isosceles triangle, the angles opposite to the 
equal sides are equal. 
Let the side AB=AC, then shall the angle C = B. 
For suppose AD to be drawn from the vertex A to D, 
the middle of the base BC: Then the two triangles 
ADB, ADC will have the sides of the one equal to the 
sides of the other, each to each, viz. AB=AC, BD=CD, 
and AD common to both; therefore (1 1.) the angle 
ABD shall be equal to the angle ACD. 
Cor. An equilateral triatigle is also equiangular. 
Scuotium. — It — by ‘the demonstration, that 
the angles BAD, CAD are equal ; also that the angles 
BDA, CDA are equal, and consequently right angles 
(Def. 9.): Therefore a straight linedrawn from the ver- 
tex of an isosceles triangle to the middle es the base bi- 
sects the vertical angle, and is perpendicular to the base. 
Note. In a triangle not isosceles, any side is taken 
‘base EF of the. other» trie. 
in the isosceles triangle, the base is the side which; is : 
not equal to the ; tf} it) 2Gha ee 
INT AER ta =H 
~ Prov. XIII. Tankon, A th a 
jet ons Tie! rently auabioghytaets 
Conversely, if two angles of a triangle are equal, the 
opposite sides are equal, and the triangle is isosceles, 
IST'S FFs Ue e arty ths: ane 
Let the angle ABC = ACB; the side AC shall be Fig. 36. 
equal to the side AB. it Gales 
For if the sides are not let AB be the greater — 
of the two. Take BD= Rae oin DC: The a ’ 
DBC is by hypothesis equal to ACB, and the two si 
DB, BC are equal to the two sides AC, CB ; therefore 
(6.) the triangle DBC is equal to ACB: now this is im- 
possible, (Ax. 9) for the triangle DBC is only a part 
of, the triangle ACB, therefore AB is not unequal to AC, 
that is AB=AC. (pe 
>) ‘Prop. XIV. Treor. - a ” 
Of two sides of a triangle, that is the greater whic 
is opposite to the angle; and conversely, ees 
angles of a triangle, that is the greater which is oppo- 
site to the greater side. . 
1. Let the angle ACB be greater than B; the side fig, 37. 
AB opposite to the angle C is greater than the side AC 
opposite to B. For make BCD=B; then in the tri- 
angle BCD, because the angles DCB and B are Steel 
we have DC=DB (12.); but AC is less than AD+ 
(8.), and AD4+DC= AD + DB=AB; therefore AB 
is greater than AC. ' 
2 De let the sitciee be ig than rota te 
angle ACB opposite to AB shall , = spo than B 
which, is buipsore to AC. For if ACB could be less 
than B; then AB would be less than AC, which is eon- 
trary to the hypothesis of the proposition: And if 
ACB could be equal to B, then AB would be equal to 
AC, which is also contrary to the hypothesis ; therefore 
ACB must be greater than ABC. alin 
Prop. XV. Teor. 
From a point A without a straight line DE, only ne Fig. 3% 
perpendicular ‘ean be drawn to that line 
For suppose it possible to draw two, AB, AC; pro- 
duce one of them, so that BF may beseqtial to AB, and 
joii FC; and becausé AB=BF; and BO is:common to 
the'triangles ABC, FBC, sarap ebincaeaae 
equal; the angle ACB is equal’ to: FCB (6:): 
AC and ‘CF must form one’ contitived line (3.); and so, 
through two points: A, F, two. straight lines AF.and 
ACF may be drawn, that do not'coincide; which is im- 
possible: THerefore it is’ equally impossible’ that two 
perpendiculars ean: be:drawn from: sanie point to 
thie samerstraight litie)! esfgin ott, wlsverly sige 
Scrorium.. Through the same point C, in’a straight Fig. 2 
line AB'(Fig: 24.),”it is impossible yto-draw twovper= 
pendiculars to that line : Ror/if @Bsand):CD could be = 
both perpendicular to AB; then ECB and DCB would 4 
be both right angles, and equal to one another (1.) ? 
which is absurd (Ax) 9.) © 0 5 "S 
Phor. XVI, Turor. te aaa 
DALI 
If from a point A without a straight line ‘DE,,aper- Fig. S¢ 
