Proof: By Euclid
(related to Proposition: 1.17: The Sum of Two Angles of a Triangle)
- For let $BC$ have been produced to $D$.
- And since the angle $ACD$ is external to triangle $ABC$, it is greater than the internal and opposite angle $ABC$ [Prop. 1.16].
- Let $ACB$ have been added to both.
- Thus, the (sum of the angles) $ACD$ and $ACB$ is greater than the (sum of the angles) $ABC$ and $BCA$.
- But, (the sum of) $ACD$ and $ACB$ is equal to two right angles [Prop. 1.13].
- Thus, (the sum of) $ABC$ and $BCA$ is less than two right angles.
- Similarly, we can show that (the sum of) $BAC$ and $ACB$ is also less than two right angles, and further (that the sum of) $CAB$ and $ABC$ (is less than two right angles).
- Thus, for any triangle, (the sum of) two angles taken together in any (possible way) is less than two right angles.
- (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"