Question

The given figure shows two isosceles triangles ABC and DBC with common
base BC. AD is extended to intersect BC at point P. Then, $\Delta ABP\equiv \Delta ACP$
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△$s, ABP and ACP,
$AP=AP$ (Common)
$AB=AC$ (Given, ABC is an isosceles triangle)
$\angle ABC=\angle ACB$ (Isosceles triangle porperty)
Thus, $△ABP\cong △ACP$ (SAS postulate)