Proof: By Euclid

For since $ABCD$ is a parallelogram, $AD$ is equal to $BC$ [Prop. 1.34].
So, for the same (reasons), $EF$ is also equal to $BC$.
So $AD$ is also equal to $EF$.
And $DE$ is common.
Thus, the whole (straight line) $AE$ is equal to the whole (straight line) $DF$.
And $AB$ is also equal to $DC$.
So the two (straight lines) $EA$, $AB$ are equal to the two (straight lines) $FD$, $DC$, respectively.
And angle $FDC$ is equal to angle $EAB$, the external to the internal [Prop. 1.29].
Thus, the base $EB$ is equal to the base $FC$, and triangle $EAB$ will be equal to triangle $DFC$ [Prop. 1.4].
Let $DGE$ have been taken away from both.
Thus, the remaining trapezium $ABGD$ is equal to the remaining trapezium $EGCF$.
Let triangle $GBC$ have been added to both.
Thus, the whole parallelogram $ABCD$ is equal to the whole parallelogram $EBCF$.
Thus, parallelograms 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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