In a trapezoid $ABCD$ with bases $AD$ and $BC$, the bisector of the angle $\angle DAB$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $P$ and $S$, respectively, and the bisector of the angle $\angle BCD$ intersects the bisectors of the angles $\angle ABC$ and $\angle CDA$ at the points $Q$ and $R$, respectively. Prove that if $PS\parallel RQ$, then $AB = CD$.