Equal Sized Triangles on Same Base have 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

 * Euclid-I-39.png

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 $BC$.