In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Given: Angle bisector of angle A and perpendicular bisector of BC intersect.Let the bisector of A meet the circumcircle at E.In△ABEand△AEC,∠BAE=∠CAEAE Angle bisector of A∴BE=EC (1) converse of the angle subtended by the chord of equal length at the same point are equalLet D be the midpoint of BC and now connect ED.In△BDEand△CDE,BE=CEfrom eq. (1)BD=CD...Dis mid point ofBCDE=DE...common side∴△BDE≅△CDE...SSS test of congruence⇒∠BDE=∠CDE.c.p.c.t.Also,∠BDE+∠CDE=180∘...angles in linear pair∴∠BDE=∠CDE=90∘Therefore,DEis the right angled bisector ofBC.Hence, they intersect on the circumcircle of triangle ABC