Dem.—1. Let the equal angles be so placed as to be vertically opposite, and that AC, CD may form one right line; then it may be demonstrated, as in the last Proposition, that BC, CE form one right line. Join BD.
Now since the triangles ACB, DCE are equal,
| ACB | : BCD :: DCE : BCD [V. vii.]; | ||||||||||
| but | ACB | : BCD :: AC : CD [i.], | |||||||||
| and | DCE | : BCD :: EC : CB [i.]. | |||||||||
| Therefore | AC | : CD :: EC : CB; |
that is, the sides about the equal angles are reciprocally proportional.
2. If AC : CD :: EC : CB, to prove the triangle ACB equal to DCE.
Dem.—Let the same construction be made, then we have
| AC : CD | :: triangle ACB : BCD [i.], | ||||||||||
| and | EC : CB | :: triangle DCE : BCD [i.]; | |||||||||
| but | AC : CD | :: EC : CB (hyp.). |
Therefore the triangle
Hence the triangle ACB = DCE [V. ix.]—that is, the triangles are equal.