218 
Proportion of the mean ; and ifthe rectangle contained by the ex- 
of Figures. tremes be equal to the square of the mean, the three 
al: straight lines are proportionals, 
Prov. XXIII. Tueor. 
If four-straight lines be proportionals, and also other 
four, the rectangles contained by the corresponding 
terms shall be proportionals ; that is, if A: B::C:D, 
and E: F:: G: H; then supposing rectangles construct- 
éd, having these lines for their sides, as in the figure, 
AXxE:BxXF::CxG:DxH. 
Fig, LLL. 
On the line B construct another rectangle which shall 
have E for its altitude ; and on the line D construct a 
second rectangle that shall have G for its altitude. 
Then, by Prop. 3, 
A:Br:AxE:BxXE,andC:D::CxG:DxG; 
but A:B::C:D, by hypothesis, 
therefore AXE: BkKE::CxG:DxG; 
Now, BXE:BxF::E:F,andDxG:DxH:: 
G:H, and by hypothesis E: F:: G: H; 
therefore Bk E: Bx F::DxG:DxH; 
but it was shewn that . 
AxXE:BxE::CxG:DxG 
therefore, ex aque, AKE: BX F::CxG:DxH. 
Cor. Hence the squares of four proportional straight 
lines are also proportionals. 
| Prop. XXIV. Teor. 
Two triangles which have an angle of the one equal 
to an angle of the other, are to each other as the rectangle 
of the sides about the equal angles: That is, the triangle 
ABC is to the triangle ADE as ABX AC to ADx 
AE: 
Fig. 112. 
Draw BE; the triangles ABE, ADE have a com- 
mon vertex E, therefore ABE: ADE:: AB: AD 
(Cor. 6.) ; but AB: AD:: ABx AE: AD x AE 
(3.) therefore, 
trian. ABE ; trian. ADE :: ABK AE: ADx AE. 
In like mammer it may be demonstfated, that 
trian. ABC : trian. ABE:: ABX AC: ABKX AE; 
‘therefore (8. 3.) 
‘trian. ABC : trian. ADE:: ABX AC: ADXAE. 
Cor. 1. Therefore the two triangles are equivalent, 
if AB x AC=AD~x AE; orif AB: AD:: AE: 
AC (22.) 
Cor. 2. Two'parallelograms which have an angle of 
the one equal to an angle of the other, will be to each 
other as the rectangles contained by the sides about 
these angles: For the parallelograms are the doubles of 
‘triangles which have an angle and two sides common 
with those of the parallelogram, 
‘Prop. XXV. Tueor. 
‘Similar triangles are to each other as the squares ef 
their homologous sides. 
Let the angle A=D, and the angle B=E, then, 
AB: DE:: AC: DF (18.) 
and AB: DE:: AB: DE, 
& the terms of the two last ratios are identical, there- 
ore, 
AB: DE?:: ACXAB: DFx DE (23.) © 
(ay BAC : trian. EDF: ; AC x AB: DF x DE 
24. 
Fig. 108. 
GEOMETRY. en's 
“12. 3, the triangle ABC is to the triang 
Therefore trian. ABC’: trian. EDF :: AB*: DE*. . Propo 
Prop, XXVI. Turor,. 
Similar polygons are composed of the same number 
of triangles, which are similar, each to each, and simi- 
larly situated. 
In the poly. ABCDE, draw from an angle A the Fig. 1 
diagonals AC AD); and in the other poly, on F 
draw in like manner from the angle “md ich is homo. 
logous to A, the diagonals FH, FI. And since the 
polygons are similar, the angle B is equal to its homo- 
gone angle G, (Def. 2.) and besides, AB: BC:: FG: 
GH ; therefore the triangles ABC, FGH are similar, 
(20.), and the-angle BCA= GHF;; these equal angles 
being taken from the equal angle BCD, GHiJ, the re- 
mainders ACD, FHI are equal; but since the triangles 
ABC, FGH are similar, we have AC: FH: : BC: 
GH, and because of the similitude of the poiygons we 
have BC:GH::CD: HI; therefore AC: FH:: 
CD : HI. Now it has been shewn that the angle ACD 
=FH1I; therefore the triangles ACD, FHI are simi- 
lar (20.) .In like manner, it may be demonstrated, 
that the remaining triangles of the two polygons are 
similar ; therefore the polygons are composed of the 
same number of similar triangles similarly situated. - 
Prop. XXVII. Tueor. 
The circumferences or medenaioes of similar poly- 
ons are to one another as their homologous sides; and 
eir areas are as the squares of their homologous sides. 
1. For by the nature of similar figures, AB : FG : : wig. 
BC:GH::CD:HI,&c. Therefore, AB is to FG as 
AB 4+ BC + CD, &c. the perimeter of the first figure 
toFG + ch + HI, &e. the perimeter of the second 
figure (12. 3.) 
2. And because the triangles ABC, FGH are si- 
milar, ABC: FGH:: AC?:-FH? (25.), and in like 
manner, because the triangles ACD, FHI are similar, 
ACD: FHI:: AC’: FH?; therefore, ABC: FGH:: 
ACD: FHI. In the same way it may be shewn 
that ACD: FHI: : ADE: FIK, and so on, if the 
polygons consist of more triangles. Hence, PL ENe- 
the sum of the triangles ABC, ACD, ADE, or the 
polygon ABCDE, to the sum of the triangles FGH, 
FHI, FIK, or the polygon FGHIK: But the le 
ABC is to the triangle FGH as AB? to FG* (25.) ; 
therefore the similar polygons are as the squares of their 
homologous sides. , : 
Cor. 1. If three similar figures have their homolo- 
gous sides equal to the three sides of a right angled 
triangle ; the figure made on the side opposite to the 
right angle shall be equal to the other two. For the 
figures are proportional to the squares on their homo-_ 
logous sides ; anid since the square on the side opposite 
to the right angie is equal to the squares on the other 
two sides, the figure on the former shall be equal to 
those on the latter. } , 
Cor. 2. Similar polygons have to each other thedu» 
plicate ratio of their homologous sides.» ForletL bea 
third proportional to the hanelngies sides AB, FG, 
then (Det. 11. 3.) AB has to L the duplicate. ratio. of 
ABto FG; but AB: L:: AB*: ABXL(3.); or, 
since AB x L= FG? (Cor. 22.), AB: L:: AB*: 
FG? ;: ABCDE : FGHIK; therefore. the figure 
