Greater Side of Triangle Subtends Greater Angle

Theorem
In any triangle, the greater side subtends the greater angle.

Proof


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

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

From Isosceles Triangles have 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.

Note
This is Proposition 18 of Book I of Euclid's "The Elements".

This theorem is the converse of Proposition 19: Greater Angle of Triangle Subtended by Greater Side.