Let angle bisector of ∠A intersect circumcircle of ∆ABC at D. Join DC and DB.

∠BCD = ∠BAD [Angles in the same segment]

⇒ ∠BCD = ∠BAD 1/ 2 ∠A [AD is bisector of ∠A] ...(i)

Similarly ∠DBC = ∠DAC 1/ 2 ∠A ... (ii)

From (i) and (ii) ∠DBC = ∠BCD

⇒ BD = DC [sides opposite to equal angles are equal]

⇒ D lies on the perpendicular bisector of BC.

Hence, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ∆ABC **Proved.**