| Hence | JA | : AC :: OD : DF [iv.]; | |||||||||
| but | AC | : AB :: DF : DE [iv.]. | |||||||||
| Hence | JA | : AB :: OD : DE (ex æquali); | |||||||||
| but | AB | = AG, and DE = DL; | |||||||||
| therefore | JA | : AG :: OD : DL. | |||||||||
| Again, | JA | : AG :: AI : square AH [i.], | |||||||||
| and | OD | : DL :: DN : square DM [i.]. | |||||||||
| Hence | AI | : AH :: DN : DM; | |||||||||
| therefore | AI | : DN :: AH : DM [V. xvi.]; | |||||||||
| hence | △ABC : △DEF :: AB2 : DE2. |
Exercises.
1. If one of two similar triangles has its sides 50 per cent. longer than the homologous sides of the other; what is the ratio of their areas?
2. When the inscribed and circumscribed regular polygons of any common number of sides to a circle have more than four sides, the difference of their areas is less than the square of the side of the inscribed polygon.
PROP. XX.—Theorem.
Similar polygons may be divided (1) into the same number of similar triangles; (2) the corresponding triangles have the same ratio to one another which the polygons have; (3) the polygons are to each other in the duplicate ratio of their homologous sides.
Dem.—Let ABHIJ, CDEFG be the polygons, and let the sides AB, CD be homologous. Join AH, AI, CE, CF.
1. The triangles into which the polygons are divided are similar. For, since the polygons are similar, they are equiangular, and have the sides about their equal angles proportional [Def. i.]; hence the angle B is equal to D, and AB : BH :: CD : DE; therefore [vi.] the triangle ABH is equiangular to CDE; hence the angle BHA is equal to DEC; but BHI is equal to DEF (hyp.); therefore the angle AHI is equal to CEF. Again, because the polygons are similar, IH : HB :: FE : ED; and since the triangles ABH, CDE are similar, HB : HA :: ED : EC; hence (ex aequali) IH : HA :: FE : EC, and the angle IHA has been proved to be equal to the angle FEC; therefore the triangles IHA, FEC are equiangular. In the same manner it can be proved that the remaining triangles are equiangular.