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In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is cyclic quadrilateral.

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In the given figure, O is the centre of the circle. PA and PB are tangents. Show that AOBP is cyclic quadrilateral.

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We know that the radius and tangent are perpendicular at their point of contact

∵ ∠OBP = ∠OAP = 90°

Now, In quadrilateral AOBP

Since, the sum of the opposite angles of the quadrilateral is 180°

Hence, AOBP is a cyclic quadrilateral

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