Question

O is the circumcentre of the triangle ABC and OD is perpendicular on BC. Prove that ∠BOD = ∠A.

Answer 1

Given:

O is the circumcenter of triangle ABC.

D is the foot of the perpendicular from O to BC.

So OD ⟂ BC.

We have to prove that   ∠BOD = ∠A

OB = OC = OA

Since O is the circumcenter, all these are radii of the circumcircle of triangle ABC.

OD ⟂ BC ⇒ D is midpoint of BC

In a circle, the perpendicular drawn from the center to any chord bisects that chord.

BC is a chord and OD ⟂ BC; hence,

BD = DC

D is the midpoint of BC.

OD bisects ∠BOC

OB = OC (radii)

BD = DC (from Step 2)

O and D lie on perpendicular bisector of BC

Therefore, OD is the perpendicular bisector of chord BC; hence, it bisects the angle at the centre subtended by BC.

∠BOD = ∠COD

Using the Central Angle Theorem

Chord BC subtends:

At center: ∠BOC

At circumference (on ΔABC): ∠A

The angle made at the center is twice the angle made at the circumference by the same chord.

∠BOC = 2∠A     ...[OD bisects ∠BOC]

Therefore,

` angleBOD = 1/2 angleBOC`    ...[Substitute ∠BOC = 2∠A]

`⇒ angleBOD = 1/2 xx (2 angleA)`

⇒ angle BOD = ∠A

`angleBOD = angleA`   ...[Hence proved]