The circumcenter of the triangle ABC is O. Prove that ∠OBC + ∠BAC = 90°.

Let ABC be the triangle whose circumcenter is O.

∠OBC = ∠OCB = θ (opposite angles of equal sides

In ΔBOC, using the angle sum property of tringle, sum of all angles is 180°, we have:
∠BOC + ∠OBC + ∠OCB = 180°
⇒ ∠BOC + θ + θ = 180°

⇒ ∠BOC = 180° -2θ

Also, in a circle, angle subtended by an arc at the center is twice the angle subtended by it at any other point in the remaining part of the circle.
∠BOC = 2∠BAC

⇒ ∠BAC = 1/2(∠BOC)

⇒ ∠BAC = 1/2(180° - 2θ)

⇒ ∠BAC = (90° - θ)

⇒ ∠BAC + θ = 90°

⇒ ∠BAC + ∠OBC = 90°

Hence, proved.