Proof: By Euclid
(related to Proposition: 5.10: Relative Sizes of Magnitudes on Unequal Ratios)
 For if not, $A$ is surely either equal to or less than $B$.
 In fact, $A$ is not equal to $B$.
 For (then) $A$ and $B$ would each have the same ratio to $C$ [Prop. 5.7].
 But they do not.
 Thus, $A$ is not equal to $B$.
 Neither, indeed, is $A$ less than $B$.
 For (then) $A$ would have a lesser ratio to $C$ than $B$ (has) to $C$ [Prop. 5.8].
 But it does not.
 Thus, $A$ is not less than $B$.
 And it was shown not (to be) equal either.
 Thus, $A$ is greater than $B$.
 So, again, let $C$ have a greater ratio to $B$ than $C$ (has) to $A$.
 I say that $B$ is less than $A$.
 For if not, (it is) surely either equal or greater.
 In fact, $B$ is not equal to $A$.
 For (then) $C$ would have the same ratio to each of $A$ and $B$ [Prop. 5.7].
 But it does not.
 Thus, $A$ is not equal to $B$.
 Neither, indeed, is $B$ greater than $A$.
 For (then) $C$ would have a lesser ratio to $B$ than (it has) to $A$ [Prop. 5.8].
 But it does not.
 Thus, $B$ is not greater than $A$.
 And it was shown that (it is) not equal (to $A$) either.
 Thus, $B$ is less than $A$.
 Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater.
 And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser.
 (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"