Question

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

Answer 1

We know that the radius and tangent are perpendicular at their point of contact
∵ ∠OBP = ∠OAP = 90°
Now, In quadrilateral AOBP
∠APB +∠AOB + ∠OBP + ∠OAP = 360° [Angle sum property of a quadrilateral]
⇒ ∠APB + ∠AOB + 90°  + 90°  = 360°
⇒∠APB + ∠AOB =180°
Since, the sum of the opposite angles of the quadrilateral is 180°
Hence, AOBP is a cyclic quadrilateral