# In the figure, O

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)

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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