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Question
In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is cyclic quadrilateral.
Solution
We know that the radius and tangent are perpendicular at their point of contact
∵ ∠OBP = ∠OAP = 90°
Now, In quadrilateral AOBP
∠APB +∠AOB + ∠OBP + ∠OAP = 360° [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90° + 90° = 360°
⇒∠APB + ∠AOB =180°
Since, the sum of the opposite angles of the quadrilateral is 180°
Hence, AOBP is a cyclic quadrilateral
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