Proof: (Euclid)
(related to Proposition: 1.18: Angles and Sides in a Triangle I)
 For since $AC$ is greater than $AB$, let $AD$ be made equal to $AB$ [Prop. 1.3], and let $BD$ have been joined.
 And since angle $ADB$ is external to triangle $BCD$, it is greater than the internal and opposite (angle) $DCB$ [Prop. 1.16].
 But $ADB$ (is) equal to $ABD$, since side $AB$ is also equal to side $AD$ [Prop. 1.5].
 Thus, $ABD$ is also greater than $ACB$.
 Thus, $ABC$ is much greater than $ACB$.
 Thus, in any triangle, the greater side subtends the greater angle.
 (Which is) the very thing it was required to show.
∎
Thank you to the contributors under CC BYSA 4.0!
 Github:

 nonGithub:
 @Fitzpatrick
References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"