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.