Proof: By Euclid

For let $A$ make $G$ (by) multiplying $C$.
Therefore, since $A$ has made $G$ (by) multiplying $C$, and has made $E$ (by) multiplying $D$, the number $A$ has made $G$ and $E$ by multiplying the two numbers $C$ and $D$ (respectively).
Thus, as $C$ is to $D$, so $G$ (is) to $E$ [Prop. 7.17].
But, as $C$ (is) to $D$, so $A$ (is) to $B$.
Thus, also, as $A$ (is) to $B$, so $G$ (is) to $E$.
Again, since $A$ has made $G$ (by) multiplying $C$, but, in fact, $B$ has also made $F$ (by) multiplying $C$, the two numbers $A$ and $B$ have made $G$ and $F$ (respectively, by) multiplying some number $C$.
Thus, as $A$ is to $B$, so $G$ (is) to $F$ [Prop. 7.18].
But, also, as $A$ (is) to $B$, so $G$ (is) to $E$.
And thus, as $G$ (is) to $E$, so $G$ (is) to $F$.
Thus, $G$ has the same ratio to each of $E$ and $F$.
Thus, $E$ is equal to $F$ [Prop. 5.9].
So, again, let $E$ be equal to $F$.
I say that as $A$ is to $B$, so $C$ (is) to $D$.
For, with the same construction, since $E$ is equal to $F$, thus as $G$ is to $E$, so $G$ (is) to $F$ [Prop. 5.7].
But, as $G$ (is) to $E$, so $C$ (is) to $D$ [Prop. 7.17].
And as $G$ (is) to $F$, so $A$ (is) to $B$ [Prop. 7.18].
And, thus, as $A$ (is) to $B$, so $C$ (is) to $D$.
(Which is) the very thing it was required to show.
∎

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