### If the altitudes from the two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.

1. Let $BD$ and $CE$ be the altitudes from the vertices $B$ and $D$ of $\triangle ABC.$
As the altitude of a triangle is perpendicular to the opposite side, we have $$BD \perp AC \text { and } CE \perp AB$$ Also, we are told that the altitudes are of equal length. $$\implies BD = CE$$
2. We need to prove that $AB = AC.$
3. In $\triangle ADB$ and $\triangle AEC$, we have \begin{aligned} &BD = CE &&[\text{Given}] \\ &\angle BAD = \angle CAE &&[\text{Common}]\\ &\angle ADB = \angle AEC = 90^ \circ &&[ BD \perp AC \text { and } CE \perp AB] \\ \therefore \space &\triangle ADB \cong \triangle AEC &&[\text{By AAS criterion}] \end{aligned}
4. As the corresponding parts of congruent triangles are equal, we have $AB = AC .$
Hence, $\triangle ABC$ is isosceles.