# In any triangle A

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

Perpendicular bisector of side BC will pass through circum centre O of the circle

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

BOC = 2 BAC = 2 A (i)

In ΔBOE and ΔCOE,

OE = OE (Common)

OB = OC (Radii of same circle)

OEB = OEC (Each 90° as OD perpendicular to BC)

ΔBOE COE (RHS congruence rule)

BOE = COE (By CPCT) (ii)

However,

BOE + COE = BOC

BOE + BOE = 2 A [From (i) and (ii)]

2 BOE = 2 A

BOE = A

BOE = COE = A

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

BOD = BOE = A (iii)

Since AD is the bisector of angle A

From (iii) and (iv), we get

Hence, proved

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