# Proof: By Euclid

• For let $BC$ be laid down so as to be straight-on to $CG$.
• Thus, $DC$ is also straight-on 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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