Proof: By Euclid

For let the equal multiples $G$ and $H$ have been taken of $A$ and $D$ (respectively), and the other random equal multiples $K$ and $L$ of $B$ and $E$ (respectively), and the yet other random equal multiples $M$ and $N$ of $C$ and $F$ (respectively).
And since as $A$ is to $B$, so $D$ (is) to $E$, and the equal multiples $G$ and $H$ have been taken of $A$ and $D$ (respectively), and the other random equal multiples $K$ and $L$ of $B$ and $E$ (respectively), thus as $G$ is to $K$, so $H$ (is) to $L$ [Prop. 5.4].
And, so, for the same (reasons), as $K$ (is) to $M$, so $L$ (is) to $N$.
Therefore, since $G$, $K$, and $M$ are three magnitudes, and $H$, $L$, and $N$ other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, thus, via equality, if $G$ exceeds $M$ then $H$ also exceeds $N$, and if ($G$ is) equal (to $M$ then $H$ is also) equal (to $N$), and if ($G$ is) less (than $M$ then $H$ is also) less (than $N$) [Prop. 5.20].
And $G$ and $H$ are equal multiples of $A$ and $D$ (respectively), and $M$ and $N$ other random equal multiples of $C$ and $F$ (respectively).
Thus, as $A$ is to $C$, so $D$ (is) to $F$ [Def. 5.5] .
Thus, if there are any number of magnitudes whatsoever, and (some) other (magnitudes) of equal number to them, (which are) also in the same ratio taken two by two, then they will also be in the same ratio via equality.
(Which is) the very thing it was required to show.
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