Proof: By Euclid
(related to Proposition: 5.20: Relative Sizes of Successive Ratios)
 And if ($A$ is) equal (to $C$ then $D$ will also be) equal (to $F$).
 And if ($A$ is) less (than $C$ then $D$ will also be) less (than $F$).
 For since $A$ is greater than $C$, and $B$ some other (magnitude), and the greater (magnitude) has a greater ratio than the lesser to the same (magnitude) [Prop. 5.8], $A$ thus has a greater ratio to $B$ than $C$ (has) to $B$.
 But as $A$ (is) to $B$, [so] $D$ (is) to $E$.
 And, inversely, as $C$ (is) to $B$, so $F$ (is) to $E$ [Prop. 5.7 corr.] .
 Thus, $D$ also has a greater ratio to $E$ than $F$ (has) to $E$ [Prop. 5.13].
 And for (magnitudes) having a ratio to the same (magnitude), that having the greater ratio is greater [Prop. 5.10].
 Thus, $D$ (is) greater than $F$.
 Similarly, we can show that even if $A$ is equal to $C$ then $D$ will also be equal to $F$, and even if ($A$ is) less (than $C$ then $D$ will also be) less (than $F$).
 Thus, if there are three magnitudes, and others of equal number to them, (being) also in the same ratio taken two by two, and (if), via equality, the first is greater than the third, then the fourth will also be greater than the sixth.
 And if (the first is) equal (to the third then the fourth will also be) equal (to the sixth).
 And (if the first is) less (than the third then the fourth will also be) less (than 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"