Question

AP bisects $\angle BAC$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△$s, BDP and CDP,
$DP=DP$ (Common)
$DB=DC$ (Given, ABC is an isosceles triangle)
$\angle DBC=\angle DCB$ (Given, Isosceles triangle preperty)
Thus, $△DBP\cong △DCP$ (SAS postulate)
Hence, $\angle BDP=\angle CDP$ (By cpct)
Thus, PD bisects $\angle BDC$