If O is the circumcentre of Triangle ABC and OD prependicular to BC. Prove that

**Angle BOD = Angle A**

O is the circumcentre of ∆ABC and OD⊥BC.

∴ O is the point of intersection of perpendicular bisectors of the sides of ∆ABC.

⇒ D is the mid point of BC.

⇒ BD = DC

In ∆OBD and ∆OCD,

OB = OC (Radius of the circle)

OD = OD (Common)

BD = DC (Proved)

∴ ∆OBD ≅ ∆OCD (SSS Congruence criterion)

⇒∠BOD = ∠COD (CPCT)

We know that,** the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.**

∴ ∠BOC = 2 ∠BAC

⇒ 2∠BOD = 2 ∠A [ Using (1) ]

⇒ ∠BOD = ∠A

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