In any triangle ABC, if the angle bisector of ∠A and perpendicular b

1.	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» 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.

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»

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.