### Prove that the sides opposite to equal angles of a triangle are equal.

1. Let $ABC$ be an isosceles triangle with $\angle B = \angle C.$
2. We need to prove that $AB = AC$
3. Let us construct an angle bisector $AD$ of angle $A.$
In $\triangle ABD$ and $\triangle ACD,$ we have \begin{aligned} & \angle B = \angle C && \text{[Given]} \\ & \angle BAD = \angle CAD && \text{[By construction]} \\ & \ AD = AD && \text{[Common]} \\ & \therefore \space \triangle ABD \cong \triangle ACD && [\text{By AAS citerion}] \end{aligned}
4. As the corresponding parts of congruent triangles are equal, we have $$AB = AC.$$