Proof: By Euclid
(related to Proposition: 8.05: Ratio of Products of Sides of Plane Numbers)
 For given the ratios which $C$ has to $E$, and $D$ (has) to $F$, let the least numbers, $G$, $H$, $K$, in continued proportion in the ratios $C$ $E$, $D$ $F$ have been taken [Prop. 8.4], so that as $C$ is to $E$, so $G$ (is) to $H$, and as $D$ (is) to $F$, so $H$ (is) to $K$.
 And let $D$ make $L$ (by) multiplying $E$.
 And since $D$ has made $A$ (by) multiplying $C$, and has made $L$ (by) multiplying $E$, thus as $C$ is to $E$, so $A$ (is) to $L$ [Prop. 7.17].
 And as $C$ (is) to $E$, so $G$ (is) to $H$.
 And thus as $G$ (is) to $H$, so $A$ (is) to $L$.
 Again, since $E$ has made $L$ (by) multiplying $D$ [Prop. 7.16], but, in fact, has also made $B$ (by) multiplying $F$, thus as $D$ is to $F$, so $L$ (is) to $B$ [Prop. 7.17].
 But, as $D$ (is) to $F$, so $H$ (is) to $K$.
 And thus as $H$ (is) to $K$, so $L$ (is) to $B$.
 And it was also shown that as $G$ (is) to $H$, so $A$ (is) to $L$.
 Thus, via equality, as $G$ is to $K$, [so] $A$ (is) to $B$ [Prop. 7.14].
 And $G$ has to $K$ the ratio compounded out of (the ratios of) the sides (of $A$ and $B$).
 Thus, $A$ also has to $B$ the ratio compounded out of (the ratios of) the sides (of $A$ and $B$).
 (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"