### Prove that the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the opposite sides are equal.

Step by Step Explanation:
1. Let $\triangle ABC$ be an isosceles triangle with $\angle B$ = $\angle C$. Now, let us draw the perpendiculars from $\angle B$ and $\angle C$ to the opposite sides.

Thus, $BD \perp AC$ and $CE \perp AB.$
2. We need to prove that $BD = CE.$
3. In $\triangle BCD$ and $\triangle BCE,$ we have \begin{aligned} & BC = BC && [\text{Common}] \\ & \angle BDC = \angle CEB && [\text{Each 90} ^ \circ] \\ & \angle BCD = \angle CBE && [ \text{As} \space ABC \space \text{is an isosceles triangle.]} \\ & \therefore \space \triangle BCD \cong \triangle BCE && [\text{By AAS criterion}] \end{aligned}
4. As the corresponding parts of congruent triangles are equal, we have $$BD = CE$$
5. Thus, the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the opposite sides are equal.