» BD meets the base between its extremities; and in the 
other, it meets the base produced. 
: Prop, XVIII. Turon. 
hPa “Two equian have their homologous sides 
proportionals, and the triangles are similar, 
Let ABO) DCE be two triangles which have their 
angles equal, eaeh to each, viz) BAC=CDE, ABC= 
DCE, and ACB=DEC; the homologous ‘sides, or the 
“sides adjacent to the equal angles, shall be ae og 
als: that is, BC: CE:: AB: DC:: AC: DE. 
Place the homologous sides BC, CE in the same di- 
rection, and because the angles B and E are together 
less than two right angles, the lines BA, ED shall meet 
if produced, (Schol. 21. 1.) ; letthem meet in F, Then 
‘since BCE is a straight line, and the angle BCA=E ; 
AC is parallel to FE (2 Cor. 20.1.) In like manner, 
because the angle DCE = B; CD is parallel to FB ; 
therefore ACDF is a parallelogram. 
In the triangle BFE, the line AC is parallel to FE, 
~wherefore BC: CE:: BA: AF (15.); or-since AF = 
‘CD, BC:CE::BA:CD. Again, in the same tri- 
“angle BFE, CD is parallel to BF, therefore BC : CE:: 
FD: DE; that is, because FD = AC, BC:CE:: 
.AC:DE. Since then it appears that the ratio of BC 
to CE is equal to the ratio of BA to CD, and also to 
the ratio of AC to DE, it follows that BA:CD:: 
AC: DE;; therefore the homologous sides are propor- 
tionals ; and because’the triangles are equiangular, they 
are similar, (Def. 2.) 
Scuotrum. It may be remarked that the homolo- 
gous sides are opposite to the equal angles. 
Prop. XIX. Tueor. 
7 , Two triangles which have their homologous sides 
__. _ proportionals, are equiangular and similar. 
ig. 107. Sup that BC: EF:: AB: DE:: AC: DF; the 
. trian Re ABC, DEF have their angles equal, viz. 
- A=D, B=E, C=F. 
"At the point E, make theangle FEG = B, and at F, 
“make the angle EFG = C, then G shall be equal to 
A (1 Cor. 24. 1.), ere GEF, ABC shall be 
£ equi ; therefore, by the preceding theorem, 
i - BC: E :: AB: EG; but by thesis, BC: EF:: 
| -AB: DE; thereforeEG=DE. In like manner we 
have BC; EF::AC:FG; but by hypothesis, BC: 
EF:: AC: DF;; therefore FG= DF. Thus it appears 
| ‘that the triangles DEF,'GEF have their three sides 
wea al, each to each ; therefore they are equal (11.1.) 
: But, by construction, the triangle GEF is equiangular 
to the triangle ABC ; therefore also the triangles 
* DEF, ABC, are equiangular and similar. 
~~ « 
Prop, XX. Tueor. 
Two triangles which have-an angle of the one equal 
~ to an angle of the other, and the sides about them pro- 
portionals, are similar. 
Let the angle A = D, and suppose that AB: DE: : 
‘AC: DF; the triangle ABC is similar to DEF. 
"Take AG= DE, and draw GH parallel to BC; 
‘the angle AGH shall be’ equal to ABC (3 Cor. 21.1.) 
and the triangle AGH equiangular to. the triangle 
) )  'yorex. Farr. vA 
-GEOMETRY. 
217 
ABC; therefore AB: AG:: AC: AH; but by hy- Proportion 
hesis, AB: DE:: AC: DF, and by construction, bet a 
AG= DE, therefore AH = DF. The two triangles ~ “~~ 
AGH, DEF have therefore an angle of the one equal 
to an angle of the other, and the sides containing these 
angles equal ; therefore they are equal (6. 1.) ; but the 
triangle AGH is similar'to ABC; therefore DEF is 
“also similar to ABC, ' 
, Prop. XXI. 
In a right angled os if from the right angle A Fig. 109. 
a perpendicular AD be drawn to the hypothenuse. 
“1. The two'triangles ABD, ADC are similar to'the 
whole triangle ABC, and to each other. 
2. Each side AB or AC is a mean Ay. stern: be~« 
tween the hypothenuse BC, and the adjacent segment 
is “The perpen dicular AD 
$. The icular is a mean proportional be- 
tween the two segments BD, DC. © i Ben 
THeEor. 
First, The triangles BAD and BAC hive the com- 
mon angle B; and, besides, the right angle BDA is 
equal to the right angle BAC; therefore the third 
angle BAD of the one, is equal to the third angle € of 
the other (1 Cor: 24, 1.) ; therefore the two triangles 
are equiangular and similar. In like manner it may be 
demonstrated, that the triangle DAC is’ equiangular, 
and similar to the triangle BAC; therefore the three 
triangles are equiangular, and similar’to one another. 
Secondly, Since the triangle BA D is similar to BCA, 
their homologous ‘sides ' are rtionals, that is, 
CB: BA:: BA: BD (18.) In like manner, because 
of the similar triangles CAD, CBA, we have BC: 
CA:: CA: CD ; therefore each side is a mean propor- 
tional between the hypothenuse and its segment adja- 
cent to that side. 
Thirdly, Because of the similar triangles BDA, 
ADC, we have BD: DA::DA: DC: so that the 
perpendicular isa mean proportional between the seg- 
ments of the hypothenuse. 
Prop. XXII.. Tueor. 
If four sermight lines be proportionals, the rectangle 
contained by the extremes is equal to the rectangle con- 
tained by the means ; and conversely, if the rectangle 
contained by the extremes be equal to the rectangle con- 
men by the means, the four straight lines are propor- 
tionals, 
Let HA, HB, HC, HD be four straight lines, pro- 
ee ; the rectangle HA x HD= rectangle HB 
x HC. 
Let HA, HB be placed in a straight line, and HC, 
HD also in a sttaight line perpendicular to AB, and 
construct the rectangles P=HA x HD, Q=HC x HB, 
and R>HD x HB; then HA:HB::P:R, and-HC: 
HD::Q:R, (3.) but by hypothesis, HA: HB:: HC: 
HD; therefore P: R::Q:R, and hence P=Q, that is 
HA x HD=HB-» HC. 
Again, if P=Q, that is, if HAxHD=HB x HC, 
then HA: HB:: HC: HD; for the same construction 
being made, we have P: R::Q:R; but P:R::HA: 
ig we R:: HC: AD; therefore HA :HB:: HC: 
Fig. lo. 
Cor, If three straight lines be proportionals, the cee- 
tangle contained by the extremes is equal to the square 
2B 
