In any triangle ABC, if the angle bisector of $\angle A$and perpendicular bisector of BCintersect, prove that they intersect on the circumcircle of the triangle ABC

Solution: Angle bisector$\angle A$meets the circumcircle of triangleABC at point P
from diagram
$\angle BAD=\angle CAD$
so, ABD is a Cyclic Quadrilateral
$\angle DBC=\angle DAC=\frac{1}{2}\angle A$
$\angle DCB=\angle BAD=\frac{1}{2}\angle A$
$\angle DBC=\angle DCB$
$In△DCB$
BP and CD are equal
D lies on perpendicular bisector of BC