Equal Sized Triangles on Equal Base are 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

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 $BD$.

$\blacksquare$

Historical Note

This is Proposition 40 of Book I of Euclid's The Elements.

This is the (partial) converse of Proposition 38 of Book I.

It is also appparent from the original manuscript that this proposition was a later addition by an editor who believed that there should be a proposition related to Proposition 39 in the same way that Proposition 38 is related to Proposition 37, and so on.