Greater Angle of Triangle Subtended by Greater Side

Theorem
In any triangle, the greater angle is subtended by the greater side.

Proof


Let $$\triangle ABC$$ be a triangle such that $$\angle ABC$$ is greater than $$\angle BCA$$.

Suppose $$AC$$ is not greater than $$AB$$.

If $$AC$$ were equal to $$AB$$, then by Isosceles Triangles have Two Equal Angles, $$\angle ABC = \angle BCA$$, but they're not so it isn't.

If $$AC$$ were less than $$AB$$, then by Greater Side of Triangle Subtends Greater Angle it would follow that $$\angle ABC$$ is less than $$\angle BCA$$, but it's not so it isn't.

So $$AC$$ must be greater than $$AB$$

Hence the result.

Note
This is Proposition 19 of Book I of Euclid's "The Elements".

This theorem is the Converse of Proposition 18: Greater Side of Triangle Subtends Greater Angle.