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