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 congruentÂ 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.

Sometimes we forget the complexity behind some of the things that seem simple or which we do automatically! I did not know this theorem, but it seems so obvious that angle C is greater than angle B.

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