# 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 = 2BAC

BAC = 1/2(BOC)

BAC = 1/2(180° - 2θ)

BAC = (90° - θ)

BAC + θ = 90°

BAC + OBC = 90°

Hence, proved.

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