Given:
O is the circumcentre of ∆ABC
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 ...(1)
In ∆OBC ,
OB = OC (radius)
Thus, ∠OBC = ∠OCB ...(2)
Also, ∠OBC + ∠OCB + ∠BOC = 180° (angle sum property)
⇒ 2∠OBC + 2∠BAC = 180° (from (1) and (2))
⇒ ∠OBC + ∠BAC = 90°
Hence, ∠OBC + ∠BAC = 90°.