Question

In the given figure, $AB=AC$. Then, can we say that AD bisects angle A

Answer 1

Given, $AB=AC$
$\angle ABC=\angle ACB$ (Isosceles triangle property)
Now, in $△DPB$ and $△DQC$,
$BD=DC$ (Given)
$\angle DBP=\angle DCQ$ (Proved above)
$\angle DPB=\angle DQC$ (Each ${90}^{\circ }$)
thus, $△DPB\cong △DQC$ (ASA rule)
Hence, $PB=QC$ (By cpct)
We know, $AB=AC$
Hence, $AB-PB=AC-QC$
$AP=AQ$ (I)
Now, In $△APD$ and $△AQD$,
$\angle APD=\angle AQD$ (each ${90}^{\circ }$)
$AP=AQ$ (From I)
$AD=AD$ (Common)
Thus, $△APD\cong △AQD$ (SAS postulate)
Hence, $\angle DAP=\angle DAQ$ (By cpct)

$AD$ bisects $\angle A$