In quadrilateral $ABCD$ sides $BC$ and $AD$ are parallel. In each of the four vertices we draw an angular bisector. The angular bisectors of angles $A$ and $B$ intersect in point $P$, those of angles $B$ and $C$ intersect in point $Q$, those of angles $C$ and $D$ intersect in point $R$, and those of angles $D$ and $A$ intersect in point S. Suppose that $PS$ is parallel to $QR$. Prove that $|AB| =|CD|$. [asy][asy] unitsize(1.2 cm); pair A, B, C, D, P, Q, R, S; A = (0,0); D = (3,0); B = (0.8,1.5); C = (3.2,1.5); S = extension(A, incenter(A,B,D), D, incenter(A,C,D)); Q = extension(B, incenter(A,B,C), C, C + incenter(A,B,D) - A); P = extension(A, S, B, Q); R = extension(D, S, C, Q); draw(A--D--C--B--cycle); draw(B--Q--C); draw(A--S--D); dot("$A$", A, SW); dot("$B$", B, NW); dot("$C$", C, NE); dot("$D$", D, SE); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(270)); dot("$R$", R, dir(90)); dot("$S$", S, dir(90)); [/asy][/asy] Attention: the figure is not drawn to scale.