Question

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

Answer 1

In $△ABD$ and $△ACD$, we have

$DB=DC$ $\mid$ Given

$\angle ADB=\angle ADC$ $\mid$ since $AD\perp BC$

$AD=AD$ $\mid$ Common

$\therefore$ by SAS criterion of congruence, we have.

$△ABD\cong △ACD$

$\Rightarrow AB=AC$ $\mid$ Since corresponding parts of congruent triangles are equal

Hence, $△$ $ABC$ is isosceles.