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