Given:
O is the circumcentre of ∆ABC
D is the mid-point of the base BC
In ∆BOD and ∆COD,
OB = OC (radius)
BD = CD (D is the mid-point of the base BC)
OD = OD (common)
By SSS property,
∆BOD ≅ ∆COD
Therefore, ∠BOD = ∠COD (by C.P.C.T.)
⇒ ∠BOC = ∠BOD + ∠COD
⇒ ∠BOC = 2∠BOD ..(1)
We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠BOC = 2∠BAC
⇒ 2∠BOD = 2∠BAC
⇒ ∠BOD = ∠BAC
Hence, ∠BOD = ∠BAC.