In any triangle ABC,if the angle bisector of

                                       

 

Let perpendicular bisector of side BC and angle bisector of A meet at point D.
 Let perpendicular bisector of side BC intersects it at E.

Perpendicular bisector of side BC will pass through circum centre O of circle. Now, BOC and BAC are the angles subtended by arc BC at the centre and a point A on the remaining part of the circle respectively.
We also know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
BOC = 2 BAC = 2A                 ... (1)
In BOE and COE

OE = OE                                   (common)
OB = OC                                  (radii of same circle)

OEB = OEC                       (Each 90^o as OD  BC)

BOE  COE                (RHS congruence rule)

BOE = COE            (by CPCT)    ... (2)
But BOE + COE = BOC

BOE +BOE = 2 A        [Using equations (1) and (2)]
BOE = 2A

BOE = A

 BOE = COE = A

The perpendicular bisector of side BC and angle bisector of A meet at point D.

BOD = BOE = A                ... (3)

Since AD is the bisector of angle A

BAD =  
 2BAD = A                    ... (4)
From equations (3) and (4), we have
BOD = 2 BAD
It is possible only if BD will be a chord of the circle. For this the point D lies on circum circle.  
Therefore, the perpendicular bisector of side BC and angle bisector of A meet on the circum circle of triangle ABC.