O is the centre of a circle.PA & PB are tangents segments. Show that AOBP is a cyclic quadrilateral.

Given a circle with center O. PA and PB are tangents the circle.

We know that radius is perpendicular to the tangent at the point of contact.

So, ∠PAO = ∠PBO =

 ⇒∠PAO + ∠PBO =

In quadilateral PAOB,

 ∠PAO + ∠PBO + ∠APB + ∠AOB =     (sum of angles of a quadilateral is )

∠APB + ∠AOB = – =

Thus, opposite angles of the quadilateral are supplementary.

We know that a quadilateral is cyclic if its opposite angles are supplementary.

Hence, PAOB is a cyclic quadilateral.