Question

$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB=AC$. Show that
(i) $AD$ bisects $BC$ (ii) $AD$ bisects $\angle A$

Answer 1

(i) Based on the $△BAD$ and $△CAD$
We know that $AD$ is the altitude so the angle is ${90}^{\circ }$
So we get
$\angle ADB=\angle ADC={90}^{\circ }$
It is given that $AB=AC$ and we know that $AD$ is common
Based on the $RHS$ Congruence Criterion we get
$△BAD\cong △CAD$
So we get $BD=CD(c.p.c.t)$
Therefore, it is proved that $AD$ bisects $BC$.
(ii) We also know that $\angle BAD=\angle CAD(c.p.c.t)$
Therefore, it is proved that $AD$ bisects $\angle A$.