We have shown that in an isosceles triangle, the angles opposite the congruent sides are congruent. This was the Isosceles Triangle Theorem which we proved two weeks ago.

In this post, we are gong to learn a slightly related theorem: a theorem that states that if in a triangle, two sides are not congruent, then the angles opposite these sides are not congruent and the angle opposite to the larger side is the larger angle.

In the triangle above, $latex \overline{AB}$ is greater than $latex \overline{BC}$. We will show that they are not equal and $latex \angle C > \angle B$.

Theorem

If in a triangle, two sides are not congruent, then the angles opposite these sides are not congruent and the angle opposite to the larger side is the larger angle.

Given

Triangle $latex ABC$ with $latex \overline{AB} > \overline{AC}$.

Proof

Locate point $latex D$ on $latex \overline{AB}$ such that $latex \overline{AC}$ is congruent to $latex \overline{AD}$. Draw $latex \overline{CD}$. The resulting figure is shown below.

By the Isosceles Triangle Theorem, $latex \angle 1\cong \angle 2$.

Now, by the Exterior Angle Theorem,

$latex m \angle 2 = m \angle B + m \angle 3$.

Since both the measures of $latex \angle B$ and $latex \angle 1$ are positive, $latex m \angle 2 > m \angle B$.

Also, by the Angle Addition Postulate,

$latex m \angle BCA = m \angle 1 + m \angle 3$.

Again, since both angles 1 and 3 are positive, $latex m \angle BCA > m \angle 1$ (Why?).

This means that $latex m \angle BCA > m \angle B$ which is what we want to show.