Question

In given figure $AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$.
Show that, $AD$ bisects $\angle A$

Answer 1

As $△ABC$ is an isosceles triangle, $AB=AC$

$AD$ is the altitude, $\angle ADC=\angle ADB={90}^o$

In $△ADB$ and $△ADC$,

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

$AB=AC$ and $AD=AD$ [common]

$\therefore △ADB\cong △ADC$ [RHS Congruency]

Hence by CPCT,

$\Rightarrow BD=DC$

$\therefore \angle BAD=\angle DAC$

i.e, $AD$ bisects $\angle A$ $\left[henceproved\right]$