Proof: By Euclid

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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