Equal Sized Triangles on Equal Base have Same Height

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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:

Equal triangles which are on equal bases and on the same side are also in the same parallels.

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


Proof

Euclid-I-40.png

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 theorem 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.


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