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