Question -
Answer -
Letperpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of sideBC intersect it at E.
Perpendicular bisector of side BC will pass throughcircumcentre O of the circle. ∠BOC and ∠BAC are theangles subtended by arc BC at the centre and a point A on the remaining part ofthe circle respectively. We also know that the angle subtended by an arc at thecentre is double the angle subtended by it at any point on the remaining partof the circle.
∠BOC = 2 ∠BAC = 2 ∠A … (1)
In ΔBOE and ΔCOE,
OE = OE (Common)
OB = OC (Radii of same circle)
∠OEB = ∠OEC (Each 90° as OD ⊥ BC)
∴ ΔBOE ≅ ∠COE (RHScongruence rule)
∠BOE = ∠COE (By CPCT) … (2)
However, ∠BOE + ∠COE = ∠BOC
⇒ ∠BOE +∠BOE = 2 ∠A [Usingequations (1) and (2)]
⇒ 2 ∠BOE = 2 ∠A
⇒ ∠BOE = ∠A
∴ ∠BOE = ∠COE = ∠A
The perpendicular bisector of side BC and angle bisectorof ∠A meet at point D.
∴ ∠BOD = ∠BOE = ∠A … (3)
Since AD is the bisector of angle ∠A,
∠BAD = 
⇒ 2 ∠BAD = ∠A … (4)From equations (3) and (4), we obtain
∠BOD = 2 ∠BAD
This can be possible only when point BD will be a chord ofthe circle. For this, the point D lies on the circum circle.
Therefore,the perpendicular bisector of side BC and the angle bisector of ∠A meet on the circum circle of triangle ABC.