Equal Sized Triangles on Same Base are Same Height

Theorem

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

Proof

Let $ABC$ and $DBC$ be equal-area triangles which are on the same base $BC$ and on the same side as it.

Let $AD$ be joined.

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

Then, by Construction of a Parallel we draw $AE$ parallel to $BC$.

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

But $\triangle ABC = \triangle DBC$, which means $\triangle DBC = \triangle EBC$.

But $\triangle DBC$ is bigger than $\triangle EBC$.

From this contradiction we deduce that $AE$ can not be parallel to $BC$.

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

$\blacksquare$

Historical Note

This is Proposition 39 of Book I of Euclid's The Elements.

This is the (partial) converse of Proposition 37 of Book I.