Given: O is the circumcenter of the triangle ABC and D is the midpoint of BC.

To prove: ∠BOD = ∠A

Join OB and OC.

In ΔOBD and ΔCD:

OD = OD (common side)

DB = Dc (D is the midpoint of BC)

OB = OC (Both are radius of the circle)

By SSS congruence rule, ΔOBD ≅ ΔOCD.

∴ ∠BOD = ∠COD = x (say) (By CPCT)

Since, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle, we have:

2∠BAC = ∠BOC

⇒ 2∠BAC = ∠BOD + ∠DOC

⇒ 2∠BAC = x + x

⇒ 2∠BAC = 2x

⇒ ∠BAC = x

⇒ ∠BAC = ∠BOD

Hence, proved.