Question

$ABC$ is an isosceles triangle with $AB=AC$. Draw $AP\perp BC$. Show that $\angle B=\angle C$

Answer 1

In $\Delta ABP$ and $\Delta ACP,$

$\angle APB=\angle APC$ (Both equal to ${90}^0$)

$AB=AC$ [$\because \Delta ABC$ is an isosceles triangle.]

$AP=AP$ (Common Side)

By R.H.S. criterion of congruence,

$△ABP\cong △ACP$

$\Rightarrow \angle B=\angle C$ (CPCT)