Proof: By Euclid
(related to Proposition: 6.23: Ratio of Areas of Equiangular Parallelograms)
 For let $BC$ be laid down so as to be straighton to $CG$.
 Thus, $DC$ is also straighton to $CE$ [Prop. 1.14].
 And let the parallelogram $DG$ have been completed.
 And let some straight line $K$ have been laid down.
 And let it be contrived that as $BC$ (is) to $CG$, so $K$ (is) to $L$, and as $DC$ (is) to $CE$, so $L$ (is) to $M$ [Prop. 6.12].
 Thus, the ratios of $K$ to $L$ and of $L$ to $M$ are the same as the ratios of the sides, (namely), $BC$ to $CG$ and $DC$ to $CE$ (respectively).
 But, the ratio of $K$ to $M$ is compounded out of the ratio of $K$ to $L$ and (the ratio) of $L$ to $M$.
 Hence, $K$ also has to $M$ the ratio compounded out of (the ratios of) the sides (of the parallelograms).
 And since as $BC$ is to $CG$, so parallelogram $AC$ (is) to $CH$ [Prop. 6.1], but as $BC$ (is) to $CG$, so $K$ (is) to $L$, thus, also, as $K$ (is) to $L$, so (parallelogram) $AC$ (is) to $CH$.
 Again, since as $DC$ (is) to $CE$, so parallelogram $CH$ (is) to $CF$ [Prop. 6.1], but as $DC$ (is) to $CE$, so $L$ (is) to $M$, thus, also, as $L$ (is) to $M$, so parallelogram $CH$ (is) to parallelogram $CF$.
 Therefore, since it was shown that as $K$ (is) to $L$, so parallelogram $AC$ (is) to parallelogram $CH$, and as $L$ (is) to $M$, so parallelogram $CH$ (is) to parallelogram $CF$, thus, via equality, as $K$ is to $M$, so (parallelogram) $AC$ (is) to parallelogram $CF$ [Prop. 5.22].
 And $K$ has to $M$ the ratio compounded out of (the ratios of) the sides (of the parallelograms).
 Thus, (parallelogram) $AC$ also has to (parallelogram) $CF$ the ratio compounded out of (the ratio of) their sides.
 Thus, equiangular parallelograms have to one another the ratio compounded out of (the ratio of) their sides.
 (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"