Proof: By Euclid

Let $AD$ have been produced in both directions to $E$ and $F$, and let the (straight line) $BE$ have been drawn through $B$ parallel to $CA$ [Prop. 1.31], and let the (straight line) $CF$ have been drawn through $C$ parallel to $BD$ [Prop. 1.31].
Thus, $EBCA$ and $DBCF$ are both parallelograms, and are equal.
For they are on the same base $BC$, and between the same parallels $BC$ and $EF$ [Prop. 1.35].
And the triangle $ABC$ is half of the parallelogram $EBCA$.
For the diagonal $AB$ cuts the latter in half [Prop. 1.34].
And the triangle $DBC$ (is) half of the parallelogram $DBCF$.
For the diagonal $DC$ cuts the latter in half [Prop. 1.34].
And the halves of equal things are equal to one another.]¹ Thus, triangle $ABC$ is equal to triangle $DBC$.
Thus, triangles which are on the same base and between the same parallels are equal to one another.
(Which is) the very thing it was required to show.

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