Question

$BE$ and $CF$ are two equal altitudes of $\Delta ABC$. By using RHS congruency rule, prove that $\Delta ABC$ is an isosceles triangle.

Answer 1

In $\Delta BFC$ and $\Delta CEB$
$\angle BFC=\angle CEB={90}^0$ (given)
hyp. $BC$ = hyp. $BC$ (common)
and altitude $CF$ = altitude $BE$
$\Delta BFC\cong \Delta CEB$
(by RHS congruency rule)
$\angle B=\angle C$
$\Delta ABC$ is an isosceles triangle.
Hence proved.