# Proof: By Euclid

• 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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