Proof: By Euclid

For let $AD$ have been joined.
I say that $AD$ is parallel to $BE$.
For if not, let $AF$ have been drawn through $A$ parallel to $BE$ [Prop. 1.31], and let $FE$ have been joined.
Thus, triangle $ABC$ is equal to triangle $FCE$.
For they are on equal bases, $BC$ and $CE$, and between the same parallels, $BE$ and $AF$ [Prop. 1.38].
But, triangle $ABC$ is equal to [triangle] $DCE$.
Thus, [triangle] $DCE$ is also equal to triangle $FCE$, the greater to the lesser.
The very thing is impossible.
Thus, $AF$ is not parallel to $BE$.
Similarly, we can show that neither (is) any other (straight line) than $AD$.
Thus, $AD$ is parallel to $BE$.
Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels.
(Which is) the very thing it was required to show.
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