Question

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

Answer 1

Data: In ∆ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect each other. O is the centre of the circle. 

To Prove: Angle bisector of ∠A and perpendicular bisector of BC intersect at D. 

Construction: Join OB, OC. 

Proof: Angle subtenced at the Centre 

= 2 × angle subtended in the circumference. 

∠BOC = 2 × ∠BAC 

In ∆BOE and ∆COE, 

∠OEB = ∠OEC = 90° (∵ OE⊥BC) 

∴ BO = OC (radii) OE is common. 

∴ ∆BOE ≅ ∆COE (RHS postulate) 

But, ∠BOE + ∠COE = ∠BOC 

∠BOE + ∠BOE = ∠BOC 

2∠BOE = ∠BOC 

2∠BOE = 2∠BAC 

∴ ∠BOE = ∠BAC 

But, ∠BOE = ∠COE = ∠BAC 

∠BAD = ∠BAC 

∠BAD = ∠BOE 

∠BAD = ∠BOD 

∴ ∠BOD = 2∠BAD 

∴ The angle subtended by an arc at the centre is double the angle subtended by it at any point on the circumference. 

∴ Angle bisector of ∠A and perpendicular bisector of BC intersect at D.