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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.
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 ]