# In the given figure, 0 is the center of the circle. PA and PB are tangents. Show that AOBP is a cyclic quadrilateral.

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°

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

Hence Proved.

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