Question

$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$.
Show that, $AD$ bisects $BC$

Answer 1

Given,

$△ABC$ is an isosceles triangle,

$\Rightarrow AB=AC$

$\Rightarrow AD$ is the altitude

$\therefore \angle ADC=\angle ADB={90}^{\circ }$

To prove,

$AD$ bisects $BC$

Proof,

In $△ADB$ and $△ADC$

$\angle ADC=\angle ADB={90}^{\circ }$

$AB=AC$ [GIVEN]

$AD=AD$ [COMMON SIDE]

$\therefore △ADB\cong △ADC$ [by $SAS$ congruence]

by CPCT,$BD=DC$

Hence $AD$ bisects $BC$