Question

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

Answer 1

We have to draw $AD$ perpendicular to $BC$.

Since $△ABC$ is isosceles, $AB=AC.......\left(1\right)$

Now, $AD$ perpendicular to $BC$

$\therefore \angle ADB=\angle ADC={90}^o.......\left(2\right)$

In $△ABD$ and $△ACD$,

$\angle ADB=\angle ADC={90}^o$ [From $\left(2\right)$]

$AB=AC$ [From $\left(1\right)$]

$AD=AD$ [common]

$△ABD\cong △ACD$ [$RHS$ Congruence Rule]

$\therefore \angle B=\angle C$ [by $CPCT$]