Equal Sized Triangles on Same Base have Same Height

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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:

Equal triangles which are on the same base, and on the same side are also in the same parallels.

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


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 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 theorem 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