Q. 133.8( 6 Votes )
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°
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