Proof: By Euclid
(related to Proposition: 5.13: Relative Sizes of Proportional Magnitudes)
 For since there are some equal multiples of $C$ and $E$, and other random equal multiples of $D$ and $F$, (for which) the multiple of $C$ exceeds the (multiple) of $D$, and the multiple of $E$ does not exceed the multiple of $F$ [Def. 5.7] , let them have been taken.
 And let $G$ and $H$ be equal multiples of $C$ and $E$ (respectively), and $K$ and $L$ other random equal multiples of $D$ and $F$ (respectively), such that $G$ exceeds $K$, but $H$ does not exceed $L$.
 And as many times as $G$ is (divisible) by $C$, so many times let $M$ be (divisible) by $A$.
 And as many times as $K$ (is divisible) by $D$, so many times let $N$ be (divisible) by $B$.
 And since as $A$ is to $B$, so $C$ (is) to $D$, and the equal multiples $M$ and $G$ have been taken of $A$ and $C$ (respectively), and the other random equal multiples $N$ and $K$ of $B$ and $D$ (respectively), thus if $M$ exceeds $N$ then $G$ exceeds $K$, and if ($M$ is) equal (to $N$ then $G$ is also) equal (to $K$), and if ($M$ is) less (than $N$ then $G$ is also) less (than $K$) [Def. 5.5] .
 And $G$ exceeds $K$.
 Thus, $M$ also exceeds $N$.
 And $H$ does not exceeds $L$.
 And $M$ and $H$ are equal multiples of $A$ and $E$ (respectively), and $N$ and $L$ other random equal multiples of $B$ and $F$ (respectively).
 Thus, $A$ has a greater ratio to $B$ than $E$ (has) to $F$ [Def. 5.7] .
 Thus, if a first (magnitude) has the same ratio to a second that a third (has) to a fourth, and a third (magnitude) has a greater ratio to a fourth than a fifth (has) to a sixth, then the first (magnitude) will also have a greater ratio to the second than the fifth (has) to the sixth.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"