Answer :

Given : PA and PB are tangents to a circle with center O

To show : AOBP is a cyclic quadrilateral.

Proof :

OB ⏊ PB and OA ⏊ AP

[Tangent at any point on the circle is perpendicular to the radius through point of contact]

∠OBP = ∠OAP = 90°

∠OBP + ∠OAP = 90 + 90 = 180°

AOBP is a cyclic quadrilateral

[ As we know if the sum of opposite angles in a quadrilateral is 180° then quadrilateral is cyclic ]

Hence Proved.

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