Question

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

Answer 1

Given that,

$△ABC$ is an isosceles triangle

so,

$AB=AC--\left(i\right)$

Also, AD is the altitude

so, $\angle ADC=\angle ADB=90---\left(ii\right)$

To prove :

$\left(i\right)BD=CD$

$\left(ii\right)\angle BAD=\angle CAD$

Proof :-

In $△ADB$ and $△ADC$

$\angle ADC=\angle ADB=90$

$AB=AC$[from $\left(i\right)$]

$AD=AD$

$△ADB\cong △ADC$

Hence,by CPCT

$BD=DC$

and

$\angle ABC=\angle DAC$

hence,proved