Equal Sized Triangles on Equal Base have Same Height

Theorem
Triangles of equal area which are on equal bases, and on the same side of it, are also in the same parallels.

Proof

 * Euclid-I-40.png

Let $ABC$ and $CDE$ be equal-area triangles which are on equal bases $BC$ and $CD$, and on the same side.

Let $AE$ be joined.

Suppose $AE$ were not parallel to $BC$.

Then, by Construction of a Parallel we draw $AF$ parallel to $BD$.

So by Triangles with Equal Base and Same Height have Equal Area, $\triangle ABC = \triangle FCD$.

But $\triangle ABC = \triangle DCE$, which means $\triangle FCD = \triangle DCE$.

But $\triangle DCE$ is bugger than $\triangle FCD$.

From this contradiction we deduce that $AF$ can not be parallel to $BD$.

In a similar way, we prove that no other line except $AE$ can be parallel to $BD$.