Question

In $△ABC,AD$ is the perpendicular bisector of $BC$ (see given figure). Show that $△ABC$ is an isosceles triangle in which $AB=AC$

Answer 1

As $AD$ is the perpendicular bisector$\angle ADB={90}^0----\left(1\right)$ and $BD=DC----\left(2\right)$

In $\Delta ADB$ and $\Delta ADC$

$\angle ADB=\angle ADC={90}^0$ from $1$

$BD=DC$ From $2$

$AD$ is the common side.

$\Delta ADB\cong \Delta ADC$ [by $SAS$ congruence]

Thus sides $AB=AC$

i.e. $\Delta ABC$ is isosceles triangle.

Hence Proved