Proof: By Euclid
(related to Proposition: 1.41: Parallelograms and Triagles)
 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.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"