Question

$\huge \mathbb \color{cyan}{Question - }$

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.


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Answer 1

$\LARGE{\underline{\frak{\red{Answєr :}}}}$

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.

Hopє this hєlps you !

Answer 2

Answer:

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