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

Question

In any triangle ABC, if the angle bisector of &#x2220; A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.

Philoid · Accepted Answer

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

∠BAD =

2 ∠BAD = A (IV)

From (iii) and (iv), we get

∠BOD = 2 ∠BAD

Hence, proved