Let $O_1$ be the circumcenter of triangle $AOI$ and $O_2$ that of $COI$. Then $O_1O_2$ is perpendicular to, and bisects $OI$. Also, $OO_1I$ and $OO_2I$ are isosceles triangles and have the same base. However, $\angle OO_1I=2\angle OAI=2\angle OCI=\angle OO_2I$ and thus the radii of the circles with centers $(O_1)$ and $(O_2)$ are equal. But $QP$ is a chord of circle $(O_1)$ and corresponds to $\angle QOP=180-\angle BAD=\angle BCD =\angle RCS$, while $RS$ is a chord of circle $(O_2)$ and corresponds to $\angle RCS$. Therefore, $QP=RS.$ Let $I_1$ and $I_2$ be the points of intersection of $QP$ and $RS$ with $AC$, respectively. Then $\angle PI_1C=\angle AIQ+\angle CAD$ and $\angle RI_2A=\angle CIS +\angle BCA.$ Therefore $\angle PI_1C-\angle RI_2I=(180+\angle DBC-\angle BCA)-(\angle QIO+\angle OIS)=0,$ after chasing around some angles. Therefore, $PQ$ and $RS$ are equal and parallel and thus $PQRS$ is a parallelogram.

It seems $I$ may be any point on $\overline{AC}$, as long as the 4 points belong to the sides only, and not to their extensions! Key point is to see that the 2nd common point of the circles $(BROQ), (POSD)$ lies on $\overline{BD}$. Best regards sunken rock

Claim. Radii of circumcircles $AOI$ and $COI$ are equal. Proof. Note that $\measuredangle AIO=\measuredangle CIO\implies \sin{\angle AIO}=\sin{\angle OIC}$, and as $OC=OA$, by the Law of Sines, the claim follows. Thus by the claim we have, \begin{align*}RS&=2r\cdot \sin{\angle RCS}= 2r\cdot \sin{\angle BCD}\\&= 2r\cdot \sin{\angle DAB}=2r\cdot \sin{\angle QAP}=QP. \end{align*}Now, we'll show that $QP\parallel RS$, which is equivalent proving that $\measuredangle ROQ=\measuredangle PQO+\measuredangle ORS$. Indeed,\begin{align*} \measuredangle PQO+\measuredangle ORS&=\measuredangle DAO+\measuredangle OCD\\&=\frac{180^\circ-\measuredangle AOD}{2}+\frac{180^\circ-\measuredangle AOD}{2}\\&=\frac{\measuredangle COB+\measuredangle BOA}{2}\\&=\measuredangle CAB+\measuredangle BCA\\&=\measuredangle IOQ+\measuredangle ROI\\&=\measuredangle ROQ, \end{align*}we are done. [asy][asy]import olympiad; size(12cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,D,I,O,P,Q,R,S; O=(0,0);A=dir(120);B=dir(60);C=dir(330);D=dir(220);I=extension(A,C,B,D);path w=circumcircle(A,I,O);path s=circumcircle(C,I,O);P=intersectionpoints(A--D,w)[1];Q=intersectionpoints(B--2A-B,w)[1];R=intersectionpoints(B--C,s)[0];S=intersectionpoints(D--2C-D,s)[0]; draw(A--B--C--D--cycle,deep);draw(circumcircle(A,B,C));draw(A--C,deep);draw(B--D,deep); draw(w,org);draw(s,org);draw(Q--A,deep);draw(Q--P,light);draw(R--S,light); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$I$",I,dir(I)); dot("$O$",O,S); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); [/asy][/asy]