Proof: By Euclid

For (if) not, then, if possible, let $AHC$ be [$ABCD$'s] diagonal.
And producing $GF$, let it have been drawn through to (point) $H$.
And let $HK$ have been drawn through (point) $H$, parallel to either of $AD$ or $BC$ [Prop. 1.31].
Therefore, since $ABCD$ is about the same diagonal as $KG$, thus as $DA$ is to $AB$, so $GA$ (is) to $AK$ [Prop. 6.24].
And, on account of the similarity of $ABCD$ and $EG$, also, as $DA$ (is) to $AB$, so $GA$ (is) to $AE$.
Thus, also, as $GA$ (is) to $AK$, so $GA$ (is) to $AE$.
Thus, $GA$ has the same ratio to each of $AK$ and $AE$.
Thus, $AE$ is equal to $AK$ [Prop. 5.9], the lesser to the greater.
The very thing is impossible.
Thus, $ABCD$ is not not about the same diagonal as $AF$.
Thus, parallelogram $ABCD$ is about the same diagonal as parallelogram $AF$.
Thus, if from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole.
(Which is) the very thing it was required to show.
∎

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