The Angle-Side Inequality

A bidirectional proof: the longer side of a triangle is opposite the larger angle, and the larger angle is opposite the longer side.

Keywords: angle-side inequality, unequal sides, unequal angles, isosceles triangle, exterior angle

Prerequisites: isosceles-triangle-proof, exterior-angle-proof · Difficulty: beginner

The angle-side inequality is bidirectional:

The longer side is opposite the larger angle, and vice versa.

Strategy: for the forward direction, place $D$ on $AB$ with $AD=AC$; the isosceles triangle $ACD$ and the exterior-angle theorem give $\angle C>\angle B$. The converse follows by contradiction.

Step 1: Part 1: Longer Side → Larger Angle

Given: $AB>AC$ in triangle $ABC$. We want to prove $\angle C>\angle B$.

Place point $D$ on $AB$ such that $AD=AC$. Since $AB>AC$, point $D$ lies between $A$ and $B$.

Step 2: Isosceles Base Angles

Since $AD=AC$, triangle $ACD$ is isosceles. By the isosceles triangle theorem, its base angles are equal: $\angle ACD=\angle ADC$.

Step 3: Conclude $\angle C>\angle B$

The angle $\angle ADC$ is an exterior angle of triangle $BDC$, so by the exterior angle theorem, $\angle ADC>\angle B$.

Also $\angle ACB>\angle ACD$, since $\angle ACB$ contains $\angle ACD$.

Chaining: $\angle C>\angle ACD=\angle ADC>\angle B$.

Step 4: Part 2: Larger Angle → Longer Side

Now for the converse. Given: $\angle C>\angle B$. We want to prove $AB>AC$.

Step 5: Proof by Contradiction

We argue by contradiction.

If $AB=AC$, then $\angle B=\angle C$ by the isosceles triangle theorem — contradicting $\angle C>\angle B$.

If $AB<AC$, then $\angle C<\angle B$ by the forward direction — again a contradiction.

Therefore $AB>AC$.

Step 6: The Angle-Side Inequality

We have proven both directions:

The longer side is always opposite the larger angle.

In any triangle:

The longer side is always opposite the larger angle. This is the key ingredient in the triangle inequality proof. ∎