Proof: By Euclid
(related to Proposition: 1.39: Triangles of Equal Area III)
 For let $AD$ have been joined.
 I say that $AD$ and $BC$ are parallel.
 For, if not, let $AE$ have been drawn through point. A parallel to the straight line $BC$ [Prop. 1.31], and let $EC$ have been joined.
 Thus, triangle $ABC$ is equal to triangle $EBC$.
 For it is on the same base as it, $BC$, and between the same parallels [Prop. 1.37].
 But $ABC$ is equal to $DBC$.
 Thus, $DBC$ is also equal to $EBC$, the greater to the lesser.
 The very thing is impossible.
 Thus, $AE$ is not parallel to $BC$.
 Similarly, we can show that neither (is) any other (straight line) than $AD$.
 Thus, $AD$ is parallel to $BC$.
 Thus, equal triangles which are on the same base, 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"