Question

$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using RHS congruence rule, prove that the triangle $ABC$ is isosceles.

Answer 1

In $\Delta BEC$ and $\Delta CFB$
$\angle BEC=\angle CFB$ (each ${90}^{\circ }$)
$BC=CB$ (common)
$BE=CF$ (given)
$\therefore \Delta BEC\cong \Delta CFB$ (by RHS congruency)
$\Rightarrow \angle BCE=\angle CBF$ (by CPCT)
$\therefore AB=AC$ (Sides opposite to equal angles of a triangle are equal)
Hence, $\Delta ABC$ is isosceles triangle.