Question

AP is perpendicular bisector of BC.
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△$s, ABD and ACD,
$AD=AD$ (Common)
$AB=AC$ (Given, ABC is an isosceles triangle)
$DB=DC$ (Given, DBC is an isosceles triangle)
Thus, $△ABD\cong △ACD$ (SSS postulate)
Hence, $PB=PC$ (By cpct)
and, $\angle APB=\angle APC=x$ (By cpct)
Now, $\angle APB+\angle APC=180$
$x+x=180$
$x={90}^{\circ }$
Thus, AP is perpendicular bisector of BC