Question

DP = DQ
Answer: DP = DQ
If true then enter 1 else if False enter 0

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, $DP=DQ$ (By cpct)