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 perperpendular at their point of contact

<OBP = <OAP = 90°

Now, In quadrilateral AOBP

<APB +<AOB+ <OBP + <OAP = 360°

=> <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.