Proof: (Euclid)
(related to Proposition: 1.19: Angles and Sides in a Triangle II)
 For if not, $AC$ is certainly either equal to, or less than, $AB$.
 In fact, $AC$ is not equal to $AB$.
 For then angle $ABC$ would also have been equal to $ACB$ [Prop. 1.5].
 But it is not.
 Thus, $AC$ is not equal to $AB$.
 Neither, indeed, is $AC$ less than $AB$.
 For then angle $ABC$ would also have been less than $ACB$ [Prop. 1.18].
 But it is not.
 Thus, $AC$ is not less than $AB$.
 But it was shown that ($AC$) is not equal (to $AB$) either.
 Thus, $AC$ is greater than $AB$.
 Thus, in any triangle, the greater angle is subtended by the greater side.
 (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"