Let ABC, DCE, be two triangles, which have their angles equal, each to each; viz. BAC equal to CDE, ABC equal to DCE, and ACB equal to DEC; the homologous sides, or the sides adjacent to the equal angles, shall be proportionals; that

is, BC CE: BA: CD, and BA: CD :: AC: DE.

:

B

Place the homologous sides BC, CE, in a straight line; and, because the angles B and E are together less than two right angles, the lines BA and ED shall meet, if produced (76): let them meet in F. Then, since BCE is a straight line, and the angle BCA equal to E, AC is parallel to EF. In like manner, because the angle DCE is equal to B, the straight line CD is parallel to BF; therefore ACDF is a parallelogram.

156. In the triangle BFE, the straight line AC is parallel to FE; wherefore BC: CE :: BA: AF (152). Again, in the same triangle, BFE, CD is parallel to BF; therefore, BC: CE :: FD: DE; but, by substituting CD for its equal AF, in the first set of proportionals, and AC for its equal FD in the second set, we have.... CE: BA CD by the first, CE: AC

and ...

BC
BC

DE by the second,

therefore, by equality, BA: CD :: AC: DE;

therefore the homologous sides are proportionals; and, because the triangles are equiangular, they are similar.

SCHOLIUM. It may be remarked that the homologous sides are opposite to the equal angles.

THEOREM 59.

157. Two triangles, which have their homologous sides proportionals, are equiangular and similar.

Suppose that BC: EF:: AB: DE

and that.... AB : DE :: AC: DF

the triangles ABC, DEF, have their angles equal : viz. A equal to D, B equal to E, and C equal to F. At the point E make the angle FEG equal to B, and at the

E

B

point F make the angle EFG equal to C, then G shall be equal to A (80), and the triangles GEF, ABC, shall be equiangular; therefore,

by the preceding prop. BC : EF :: AB : EG

and, by hypothesis, .. BC: EF:: AB: DE; therefore EG is equal to DE. In like manner. BC EF:: AC: FG

and, by hypothesis, .. BC: EF :: AC: DF; therefore FG is equal to DF. Thus, it appears that, the triangles DEF, GEF, have their three sides equal, each to each, therefore they are equal (58). But, by construction, the triangle GEF is equiangular to the triangle ABC; therefore, also, the triangles DEF, ABC, are equiangular and similar.

THEOREM 60.

158. Two triangles which have an angle of the one equal to an angle of the other, and the sides about them proportionals, are similar.

Let the angle A equal D, and suppose that AB: DE: AC: DF, the triangle ABC is similar to DEF.

B

H

C

E

Take AG equal to DE, and draw GH parallel to BC, the angle AGH shall be equal to ABC (75), and the triangle AGH equiangular to the triangle ABC; therefore AB: AG :: AC: AH; but AG is equal to DE;

therefore........ AB : DE :: AC: AH,

but, by hypothesis, AB : DE :: AC: DF; therefore AH is equal to 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 (53); but the triangle AGH is similar to ABC.

THEOREM 61.

159. A perpendicular, AD, drawn from the right angle, A, of a right-angled triangle, upon the hypothenuse, or longest side, BC, will divide that triangle into two others, which will be similar to each other, and to the whole.

The triangles BAD and BAC have the common 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 C of the other (80); therefore the B

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.

THEOREM 62.

160. The square described upon the hypothenuse, or longest side, is equal to the squares described upon the other two sides.

From the right angle C draw CD, perpendicular to the hypothenuse AB; then the triangle ABC is divided into two triangles, ADC, CDB, which are similar to one another, and to the whole triangle ABC (159);

A

C

therefore, by the similar triangles, ABC, CBD,..AB: BC :: BC : BD ; again, by the similar triangles, BAC, CAD .... AB : AC :: AC : AD; therefore, reducing the first to an equation, AB × BD = BC2

...

and, reducing the second analogy to an equation, AB × AD = AC2

B

then, adding these two equations ...... AB × (AD + BD) = AC2 + BC2 but, since AB is equal to the sum of the two lines AD, DB, therefore AB2 = AC2 + BC2,

THEOREM 63.

161. 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.

Suppose the two triangles joined, so as to have a common angle, and let the two triangles be ABC, ADE. Draw the straight line BE.

B

D

Now the triangle ABE tria. ADE :: AB: AD;

Therefore the triangle ABE : tria. ADE:: AB × AE: AD × AE.

Or, alternately, the triangle ABE: AB × AE :: the triangle ADE: AD × AE. In like manner, the triangle ABE: ABAE :: the triangle ABC: AB × AC. Therefore, by equality, the tria. ABC : tria. ADE :: AB × AC: AD × AE.

THEOREM 64.

162. Similar triangles are to one another as the squares of their homologous sides.

Let the angle A be equal to the angle D, and

the angle B equal to E.

Then AB DE: AC: DF (155)

and AB DE :: AB : DE.

therefore, by multiplying the corresponding terms, we have AB2: DE2 :: AC x AB: DF x DE.

D

But the triangle BAC : triangle EDF :: AC × AB : DF × DE (162). Therefore the triangle ABC : triangle DEF :: AB2 : DE3.

Or thus, let a signify a triangle; then (162)—

F

163. Similar polygons are composed of the same number of triangles, which are similar, and similarly situated.

In the polygon ABCDE, draw from

any angle, A, the diagonals AC, AD; B

and, in the other polygon, FGHIK, draw,

in like manner, from the angle F, which

is homologous to A, the diagonals FH, FI.

E

Since the polygons are similar, the angle B is equal to its homologous angle

G; and, besides, AB : BC :: FG: GH; therefore, the triangles ABC and FGH

are similar (158), and the angle BCA is equal to GHF; these equal angles being taken from the equal angles BCD, GHI, the remainders ACD, FHI, are equal: but, since the triangles ABC and FGH are similar, we have AC: FH :: BC: GH; and, because of the similitude of the polygons, we have BC: GH:: CD: HI; therefore, AC: FH:: CD: HI. Now it has been shown that the angle ACD is equal to FHI; therefore the triangles ACD, FHI, are similar (158).

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.

THEOREM 66.

164. The perimeters of similar polygons are to one another as their homologous sides.

AB: FG:: BC: GH.

BC: GH:: CD: HI.

B

H

Therefore, AB : FG :: AB + BC + CD :: FG+

K

GH+HI (131); wherefore AB is to FG as the perimeter of the polygon ABCD is to the perimeter of the polygon FGHIK.

THEOREM 67.

I

165. The areas of similar polygons are as the squares of their homologous sides.

Let the polygons be ABCDE and FGHIK; from any angle, A, draw the diagonals AC, AD; and, from the homologous angle F, draw the diagonals FH, FI; then the triangles ABC, ACD, ADE, are respectively equal, and similar to the triangles FGH, FHI, FIK. Therefore the triangle ABC : triangle FGH :: AC2 : FH3 And .... the triangle ACD: triangle FHI :: AC2 : FH3. Therefore the triangle ABC : triangle FGH :: ACD: FHI.

K

In the same manner it may be demonstrated that the triangle ACD : triangle FHI :: ADE: FIK, and so on, if the polygons consist of more triangles.