`PA` and `PB` are tangents to the circle with centre `O` from an external point `P`, touching the circle at `A` and `B` respectively. Show that the quadrilateral `AOBP` is cyclic.

Explanation

GIVEN `PA` and `PB` are tangents to the circle with centre `O` from an external point `P`.
TO PROVE Quadrilateral `AOBP` is cyclic.
PROOF We know that the tangent at any point of a circle is perpendicular to radius through the point of contact
`:. PAbotOA`, i.e., `/_OAP=90^(@)`.........`(i)`
and `PBbotOB`, i.e., `/_OBP=90^(@)`..........`(ii)`
Now, the sum of all the angles of a quadrilateral is `360^(@)`.
`:./_AOB+/_OAP+/_APB+/_OBP=360^(@)`
`implies/_AOB+/_APB=180^(@)` [using `(i)` and `(ii)`]
`:.` quadrilateral `OAPB` is cyclic [since both pairs of opposite angles have the sum `180^(@)`]