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

Given:-

$ABC$ is an isosceles $△$.

$AB=AC$

$AD$ is altitude.

$\angle ADB=\angle ADC=90°$

To prove:-

(i) $AD$ bisects $BC$, i.e., $BD=CD$

(ii) $AD$ bisects $\angle A$, i.e., $\angle BAD=\angle CAD$

Proof:-

In $△ADB$ and $△ADC$,

$\angle ADB=\angle ADC\left[\text{Each}90°\right]$

$AB=AC\left[\text{Given}\right]$

$AD=AD\left[\text{Common}\right]$

By R.H.S congruency,

$△ADB\cong △ADC$

By C.P.C.T.

$BD=CD$

$\angle BAC=\angle CAD$

Hence proved.