Let $O$ be the intersection point of $PS$ and $RQ$. Since $M$ is a point on the circumcircle of $ABCD$ we have $\angle QMR = \angle BMA = 45^{\circ} = \angle ACD = \angle QCR$. But $\angle QMR = \angle QCR$ means that $MCQR$ is cyclic. So $\angle SRO = \angle CRQ = \angle CMQ = \angle CMB = 45^{\circ}$ Similarily we get $\angle RSO = 45^{\circ}$. That means that $\angle ROS = 180^{\circ} - \angle RSO - \angle SRO = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$. And that means that $PS$ and $QR$ are perpendicular.

Dear Mathlinkers, after un angle chasing in order to prove that M, C, Q and R are concyclic, we can involve the Reim's theorem to prove QR//BD and PS//AC and we are done... Sincerely Jean-Louis

Dear Mathlinkers, another approach is to think to the Pascal's theorem after considering isoceles trapeze... Sincerely Jean-Louis

sthebedagain wrote: Let $O$ be the intersection point of $PS$ and $RQ$. Similarily we get $\angle RSO = 45^{\circ}$. That means that $\angle ROS = 180^{\circ} - \angle RSO - \angle SRO = 180^{\circ} - 45^{\circ} - 45^{\circ} = 90^{\circ}$. And that means that $PS$ and $QR$ are perpendicular. Hello, could you explain how do you $\angle RSO = 45^{\circ}$ ? Thanks. Here is my solution Consider the cyclic quadrilateral $BCMD$, by intersecting chord, we have $BS \cdot SD = CS \cdot SD$, which equivalent to ${\frac{BS}{SD} = \frac{CS}SM}$. $BCMA$ is also cyclic, so $\angle AMB = \angle ACB = 45^{\circ}$. Now, notice that $\angle RMS \equiv \angle AMB = \angle ACB = \angle ACD \equiv \angle QCS = 45^{\circ}$, hence $CQRM$ is cyclic, so again, by intersecting chord, we get $QS \cdot SM = CS \cdot SR$, in other words$\frac{QS}{SR} = \frac{CS}{SM}$. Combining, we get $\frac{BS}{SD} = \frac{CS}{SM} = \frac{QS}{SR}$, which means that $QR \parallel BD$. Concentrate on $\triangle OAP$, clearly, from cyclic property of $BCMA$, we have $\angle PAO \equiv \angle MAC = \angle MBC \equiv \angle SBC$. But, because $\angle POA = \angle SCB$ then $\triangle PAO \sim \triangle SBC$, so $\frac{OP}{OA} = \frac{OP}{OB} = \frac{CS}{CB}$. It is clear that $\angle QCS = \angle QCB = 45^{\circ}$, so $CQ$ is the angle bisector of $\triangle BCS$, then $\frac{QS}{QB} = \frac{CS}{CB}$. Therefore, $\frac{OP}{OB} = \frac{QS}{QB}$, hence $PS \parallel AC$. Finally, we derive that $\angle (PS, QR) = \angle (AC, BD) = 90^{\circ}$. $\text {Q.E.D}$