Let $AQ,BR$ cut $BC,AD$ at $E,F,$ respectively. Since $\angle BAE=\angle CBP$ and $AB=BC,$ then $\triangle ABE \cong \triangle BCP$ by ASA. Thus, $BE=CP$ $\Longrightarrow$ $CE=DP.$ Likewise, $\triangle BAF \cong \triangle ADP$ gives $DF=CP$ $\Longrightarrow$ $\triangle PCE \cong \triangle FDP$ by SAS $\Longrightarrow$ $\angle EPC=\angle PFD.$ Since $EQPC$ and $FRPD$ are both cyclic, due to their right angles at $C,Q$ and $D,R,$ then $\angle PRD=\angle PFD=\angle EPC=\angle EQC$ $\Longrightarrow$ $ARQS$ is cyclic $\Longrightarrow$ $S$ is on the circle with diameter $AB.$

Dear MLs M.T.

Attachments:
505386.doc (28kb)

If you set $ABCD$ as only a rhombus, the statement remains true!

Dear Mathlinkers and M.T. you can see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=362839 or http://perso.orange.fr/jl.ayme vol. 7 Miniatures géométriques p. 10-12 for an original synthetic proof. M.T. : your interesting figure lead for a circularly thema which can offer nice results... perhaps a new thema... Sincerely Jean-Louis

More general Let $ABCD$ be a quadrilateral. $AC$ cuts $BD$ at $E$. $(O)$ is circumcircle of triangle $ECD$. $P$ is a point on $AB$. $PC,PD$ cut $(O)$ again at $M,N$. Prove that $AN$ intersects $BM$ on $(O)$.

You can prove your generalization by using the converse of Pascal's theorem to hexagon $NDEGMC$ ($G$ is where $\overline{AM}$ cuts $\overline{BN}$)

Here the video solution at this problem, by my partner Cristian.

There is another approach.Let O be the intersection of the diagonals AC and BD. Then we have cyclic quadritedrals DPOR and CQOP ,and then easy by angle chasing get <RSQ=<QAR=<QBR=90-<APB,so we are done.

This is Serbia TST 2004, you can see at Imomath