# 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.

In the words of Euclid:

(*The Elements*: Book $\text{I}$: Proposition $39$)

## 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 Parallel Line 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 proof is Proposition $39$ of Book $\text{I}$ of Euclid's *The Elements*.

It is the partial converse of Proposition $37$: Triangles with Same Base and Same Height have Equal Area.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 1*(2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions