See pic for reference.
Suppose angle bisector of angle A meets circumcircle with centre O at P, and OP is joined to meet BC at D. ...(i)
Angle BAP = Angle CAP (Because AP is angle bisector)
Angle BOP = 2 Angle BAP
Angle COP = 2 Angle CAP
So, Angle BOP = Angle COP
Therefore in triangles BDO and CDO,
BO = CO (Radii of the same circle)
Angle BOP = Angle COP (Proven above)
DO = DO (Common side
So triangles BDO and CDO are congruent. (SAS criterion)
So, BD = CD and angle BDO = angle CDO (CPCTE)
But angles BDO and CDO form linear pair.
So, angle BDO + angle CDO = 180 degree
So, angle BDO = half of 180 = 90 degree
With BD = CD and angle BDO = 90 degree,
OD is perpendicular bisector of BC
OP is perpendicular bisector of CD ...(ii)
By (i) and (ii),
P lies on angle bisector of angle A and perpendicular bisector of BC and is a point on the circle.