Given that,
O is the circumcentre of triangle ABC and OD perpendicular BC
To prove :
∠BOD = ∠A
Proof :
In triangle OBD and triangle OCD, we have
∠ODB = ∠ODC (Each 90°)
OB = OC (Radii)
OD = OD (Common)
By R.H.S rule,
ΔODB ≅ ΔODC
∠BOD = ∠COD (By c.p.c.t) ...(i)
By degree measure theorem,
∠BOC = 2∠BAC
2∠BOD = 2∠BAC [From (i)]
∠BOD = ∠BAC
Hence, proved