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