Proof: By Euclid

By hypothesis, any multitude of numbers whatsoever, $A$, $B$, $C$, and (some) other (numbers), $D$, $E$, $F$, of equal multitude to them, (which are) in the same ratio taken two by two, (such that) as $A$ (is) to $B$, so $D$ (is) to $E$, and as $B$ (is) to $C$, so $E$ (is) to $F.$
For since as $A$ is to $B$, so $D$ (is) to $E$, thus, alternately, as $A$ is to $D$, so $B$ (is) to $E$ [Prop. 7.13].
Again, since as $B$ is to $C$, so $E$ (is) to $F$, thus, alternately, as $B$ is to $E$, so $C$ (is) to $F$ [Prop. 7.13].
And as $B$ (is) to $E$, so $A$ (is) to $D$.
Thus, also, as $A$ (is) to $D$, so $C$ (is) to $F$.
Thus, alternately, as $A$ is to $C$, so $D$ (is) to $F$ [Prop. 7.13] , via equality.
(Which is) the very thing it was required to show.
∎

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