Proof: By Euclid
(related to Proposition: 1.40: Triangles of Equal Area IV)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"