Next, let the angle FBD be equal to the angle EBG, and let the sides about the equal angles be reciprocally proportional, namely, DB to BE as GB is to BF: the parallelogram AB shall be equal to the parallelogram BC.
For, let the same construction be made.
Then, because DB is to BE as GB is to BF, [Hypothesis
and that DB is to BE as the parallelogram AB is to the parallelogram FE, [VI. 1.
and that GB is to BF as the parallelogram BG is to the parallelogram 'FE'; [VI. 1.
therefore the parallelogram AB is to the parallelogram FE as the parallelogram BC is to the parallelogram FE; [V. 11.
therefore the parallelogram AB is equal to the parallelogram BC. [V. 9.
Wherefore, equal parallelograms &c. q.e.d.
PROPOSITION 15. THEOREM.
Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional; and triangles which have one angle of the one equal to one angle of the other, and their sides about the equM angles reciprocally proportional, are equal to one another.
Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE: the sides of the triangles about the equal angles shall be reciprocally proportional; that is, CA shall be to AD as EA is to AB.
Let the triangles be placed so that the sides CA, AD may be in the same straight line,
