If O is the circumcentre of ΔABC and ∠OBC = 450, then the ∠BAC is equal to.

In any triangle, the circumcenter is equidistant from all vertices. In triangle ABC, OB = OC = OA (all are radii of circumcircle). Given ∠OBC = 45°, and OB = OC, triangle OBC is isosceles, so ∠OCB = 45°. By central angle theorem, ∠BOC = 2 × ∠BAC (central angle is twice inscribed angle subtending same arc). Using triangle angle sum: ∠BOC = 180° - 45° - 45° = 90°. Thus ∠BAC = 90°/2 = 45°. Wait, that contradicts the answer. Let me recalculate: Actually, angle at center is twice angle at circumference, so ∠BOC = 2∠BAC, and ∠BAC = ∠BOC/2. But more directly, if O is circumcenter, then ∠BOC = 2∠A, ∠AOC = 2∠B, ∠AOB = 2∠C. With OB = OC, ∠OBC = ∠OCB = 45°. Then ∠BOC = 180° - 2×45° = 90°. Since ∠BOC = 2∠BAC, we get ∠BAC = 90°/2 = 45°. This matches option B.