Question

$\Delta ABC$ is an isosceles triangle such that $AB=AC$, prove that the altitude $AO$ from $A$ on $BC$ bisects $BC$.

Answer 1

In the right triangles $AOB$ and $AOC$,
$\angle AOB=\angle AOC$ = 90 degree
$AB=AC$ (Since $\Delta ABC$ is an isosceles)
$AO=OA$
Therefore, $\Delta AOB\cong \Delta AOC$ by RHS congruence condition.
So, $BO$ and $CO$ are corresponding parts of congruence triangles.