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Question

O is the centre of circle. PA and PB are tangent segment. Show that the quadrilateral AOBP is cyclic


Solution


In quadrilaterial PAOB,
Angle A = Angle B = 90 degrees
(As PA and PB are tangents to the circle)
Sum of the angles of quadrilateral PAOB = 360
Angle P + Angle A + Angle O + Angle B = 360
Angle P + Angle O + 90 + 90 = 360
Angle P + Angle O = 180
Since, the opposite angles of quadrilateral PAOB is 180, it is a cycline quadrilateral.

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