Proof: By Euclid
(related to Proposition: 1.36: Parallelograms on Equal Bases and on the Same Parallels)
 For let $BE$ and $CH$ have been joined.
 And since $BC$ is equal to $FG$, but $FG$ is equal to $EH$ [Prop. 1.34], $BC$ is thus equal to $EH$.
 And they are also parallel, and $EB$ and $HC$ join them.
 But (straight lines) joining equal and parallel (straight lines) on the same sides are (themselves) equal and parallel [Prop. 1.33] [thus, $EB$ and $HC$ are also equal and parallel].
 Thus, $EBCH$ is a parallelogram [Prop. 1.34], and is equal to $ABCD$.
 For it has the same base, $BC$, as ($ABCD$), and is between the same parallels, $BC$ and $AH$, as ($ABCD$) [Prop. 1.35].
 So, for the same (reasons), $EFGH$ is also equal to the same (parallelogram) $EBCH$ [Prop. 1.34].
 So that the parallelogram $ABCD$ is also equal to $EFGH$.
 Thus, parallelograms which are on equal bases and between the same parallels are equal to one another.
 (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"