# 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.

In the words of Euclid:

(*The Elements*: Book $\text{I}$: Proposition $40$)

## 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 Parallel Line 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 bigger 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 proof is Proposition $40$ of Book $\text{I}$ of Euclid's *The Elements*.

It is the (partial) converse of Proposition $38$: Triangles with Same Base and Same Height have Equal Area.

It is also apparent 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 $39$: Equal Sized Triangles on Same Base have Same Height in the same way that Proposition $38$: Triangles with Equal Base and Same Height have Equal Area is related to Proposition $37$: Triangles with Same Base and Same Height have Equal Area, and so on.

## Sources

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