Proof: By Euclid
(related to Proposition: 6.26: Parallelogram Similar and in Same Angle has Same Diameter)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"