Let there is a circle having centre at O.

Let PA and PB are the tangents drawn from an external point P to the circle.

Again, let AOBP is a quadrilateral.

Since, radius is perpendicular to the tangent at the point of contact,

So, ∠OAP = 90 ......1

and ∠OBP = 90 ......2

Add equation 1 and 2, we get

∠OAP + ∠OBP = 90 + 90

=> ∠OAP + ∠OBP = 180

Since the sum of any pair of opposite angles of a quadrilateral is 180

So, quadrilateral AOBP is cyclic.