Answer :

Since tangents to a circle is perpendicular to the radius.

OA AP and OB BP.


OAP = 90° and OBP = 90°


OAP + OBP = 90° + 90° = 180°........... (1)


In quadrilateral OAPB,


OAP + APB + AOB + OBP = 360°


(APB + AOB) + (OAP + OBP) = 360°


APB + AOB + 180° = 360° [From (1)]


APB + AOB = 180°................ (2)


Thus from equations (1) and (2) it can be concluded that the quadrilateral PAOB is a cyclic quadrilateral.


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