Proof: By Euclid

For let the equal multiples $D$ and $E$ have been taken of $A$ and $B$ (respectively), and the other random multiple $F$ of $C$.
Therefore, since $D$ and $E$ are equal multiples of $A$ and $B$ (respectively), and $A$ (is) equal to $B$, $D$ (is) thus also equal to $E$.
And $F$ (is) different, at random.
Thus, if $D$ exceeds $F$ then $E$ also exceeds $F$, and if ($D$ is) equal (to $F$ then $E$ is also) equal (to $F$), and if ($D$ is) less (than $F$ then $E$ is also) less (than $F$).
And $D$ and $E$ are equal multiples of $A$ and $B$ (respectively), and $F$ another random multiple of $C$.
Thus, as $A$ (is) to $C$, so $B$ (is) to $C$ [Def. 5.5] .
So I say that $C$¹ also has the same ratio to each of $A$ and $B$.
For, similarly, we can show, by the same construction, that $D$ is equal to $E$.
And $F$ (has) some other (value).
Thus, if $F$ exceeds $D$ then it also exceeds $E$, and if ($F$ is) equal (to $D$ then it is also) equal (to $E$), and if ($F$ is) less (than $D$ then it is also) less (than $E$).
And $F$ is a multiple of $C$, and $D$ and $E$ other random equal multiples of $A$ and $B$.
Thus, as $C$ (is) to $A$, so $C$ (is) to $B$ [Def. 5.5] .
Thus, equal (magnitudes) have the same ratio to the same (magnitude), and the latter (magnitude has the same ratio) to the equal (magnitudes).

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