Question

A circle intersects the side `AD` of a parallelogram `ABCD` at P and BC produced at Q. Prove that square `PDQC` is cyclic .

Answer 1

`because AB| |DC`
`therefore angle1 +angle3 =180^(@)` (sum of co-interior angles is `180^(@)`) ....(1)
and `angle3+angle2=180^(@)`
(sum of opposite angles of a cyclic quadrilateral )....(2)
`therefore` From eqs. (1) and (2) , we get
`angle1 + angle3 =angle 3 +angle 2`
`rArrangle1 =angle2`
But these are the angles subtended by line segment joining two points P and C lying on the same side of PC.
So, square ` PDQC` is cyclic.