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 has two parts:

Proof Strategy

Forward: Given $AB > AC$, place point $D$ on $AB$ with $AD = AC$. Triangle $ACD$ is isosceles, so $\angle ACD = \angle ADC$. Since $\angle ADC$ is an exterior angle of triangle $BDC$, $\angle ADC > \angle B$. And $\angle ACB > \angle ACD$ since $\angle ACB$ contains $\angle ACD$. Therefore $\angle C > \angle B$.

Converse: Given $\angle C > \angle B$, we argue by contradiction. If $AB = AC$, then $\angle B = \angle C$ by the isosceles triangle theorem — contradiction. If $AB < AC$, then $\angle C < \angle B$ by the forward direction — contradiction. So $AB > AC$.

The Proof

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:

AB > AC \iff \angle C > \angle B

The longer side is always opposite the larger angle.

Conclusion

In any triangle:

AB > AC \iff \angle C > \angle B

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