Geometry problem - Circle

Let PAQ produce and BSR produce to meet at point X.
Angle XQR = Angle PSR ( ext. angle of cyclic quad.)
Join PR.
Angle RQS = Angle RPS ( angles in the same segment of circle PQRS)
So angle XQS = angle XQR + angle RQS = angle PSR + angle RPS.... (1)
Angle QAB + Angle BAP = 180 (adj. angles on a st. line)
Angle PSR + angle RPS + angle SRP = 180 ( angle sum of triangle)
So angle QAB + angle BAP = angle PSR + angle RPS + angle SRP
But angle BAP = angle SRP (angle in the same segment of circle PARB)
Therefore, angle QAB = angle PSR + angle RPS .............. (2)
From (1) and (2)
angle XQS = angle QAB
so QS//AB ( corresponding angle equal)