Proof: By Euclid

For let $AC$ have been joined.
So triangle $ABC$ is equal to triangle $EBC$.
For it is on the same base, $BC$, as ($EBC$), and between the same parallels, $BC$ and $AE$ [Prop. 1.37].
But, parallelogram $ABCD$ is double (the area) of triangle $ABC$.
For the diagonal $AC$ cuts the former in half [Prop. 1.34].
So parallelogram $ABCD$ is also double (the area) of triangle $EBC$.
Thus, if a parallelogram has the same base as a triangle, and is between the same parallels, then the parallelogram is double (the area) of the triangle.
(Which is) the very thing it was required to show.
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