Greater Side of Triangle Subtends Greater Angle

Proof

 * Euclid-I-18.png

Let $\triangle ABC$ be a triangle such that $AC$ is greater than $AB$.

Let $AD$ be made equal to $AB$.

Let $BD$ be joined.

Then $\angle ADB$ is an exterior angle of the triangle $\triangle BCD$.

Therefore $\angle ADB$ is greater than $\angle ACB$.

As $AD = AB$, the triangle $\triangle ABD$ is isosceles.

From Isosceles Triangle has Two Equal Angles, $\angle ADB = \angle ABD$.

Therefore $\angle ABD$ is greater than $\angle ACB$.

Therefore, as $\angle ABC = \angle ABD + \angle DBC$, it follows that $\angle ABC$ is greater than $\angle ACB$.

Hence the result.