Question

In the given figure, O is the centre of the circle. PA and PB. Show that AOBP is a cyclic quadrilateral [CBSE 2014]

Answer 1

We know that the radius and tangent are perperpendular 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^∘
Also, ∠OBP + ∠OAP = 180^∘
Since, the sum of the opposite angles of the quadrilateral is 180^∘
Hence, AOBP is a cyclic quadrilateral.