o is circumcentre of a ∆ABC
prove OBA + ACB
90°
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Answer:
O is the centre of circumscribed circle.
OB = OC = radii
⇒ ∠OBC = ∠OCB = x
∴ x + x + ∠BOC = 180° (Angle sum property of △OBC)
2x + ∠BOC = 180°
∠BOC = 180° - 2x
Also, ∠BOC = 2∠BAC
180° - 2x = 2∠BAC
⇒ 90 - x = ∠BAC
∴ ∠BAC + ∠OBC = (90° - x) + x
∠BAC + ∠OBC = 90°
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