Question

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

Answer 1

Given,

$ABC$ is an isosceles triangle with $AB=AC$.
To Prove : $\angle B=\angle C$
Construction: Draw $AP\perp BC$.

Proof: In $\Delta ABC,AP\perp BC$ and $AB=BC$.
$\therefore$ In $\Delta ABP$ and $\Delta ACP$
$\angle APB=\angle APC={90}^0(\because AP\perp BC)$
Hypotenuse $AB$ = Hypotenuse $AC$
$AP$ is common.
As per $RHS$ postulate,
$\Delta ABP\cong \Delta ACP$
$\therefore \angle ABP=\angle ACP$ [by CPCT]
$\therefore \angle ABC=\angle ACB$
$\therefore \angle B=\angle C$. $\left(Henceproved\right)$