If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

Explanation

The bisectors of opposite angles `angleA` and `angleC` of a cyclic qudarilateral ABCD intersect the circle at the point P and Q , respectively. We have to prove that PQ is a diameter of the circle.Join AQ and DQ. ltbr. Since, opposite angles of a cyclic quadrilateral are supplementary, so in cyclic qudarilateral ABCD, we have `angle DAB+DCB=180^@`
`therefore (1)/(2)angleDAB+(1)/(2)angleDCB=(1)/(2)(180^@)`
`rArr angle1 +angle2=90^@`
`therefore` (`because AP` and `CQ` are the bisectors of `angleA` and `anlgeC` respectively).
`angle1 to angle3 =90^@ (because angle2 =angle3) `
(`because angle2` and `angle3` are angles in the same segment of a circle with chord QD)
`rArrangle PAQ=90^@`
`therefore` So, ` anlgePAQ` is in a semi-circle.
Hence, PQ is a diameter of circle.