Let $O$ be the intersection point of $AC$ and $BD$. As $\angle SOD=90^\circ=\angle BPD$, $\angle OSD=90^\circ-\angle SDO=\angle PBD$. Furthermore, $\angle SAD=45^\circ=\angle BAC=\angle BPC$. By the law of sines:\[\frac {DA}{DS}=\frac {\sin \angle ASD}{\sin \angle SAD}=\frac {\sin \angle OSD}{\sin \angle SAD}=\frac {\sin \angle PBR}{\sin \angle BPR}=\frac {RP}{RB}\,.\] Calculating the power of $R$ WRT $\omega$, we get: $DR \cdot RB=CR \cdot RP$. Thus, $\frac {DA}{DS}=\frac {RP}{RB}=\frac {DR}{CR}$ and $CB \cdot CR=DA \cdot CR=DS \cdot DR$. Noting that $\angle BCR=\angle BCP=\angle BDP=\angle RDS$, we obtain that $\triangle CBR$ and $\triangle DSR$ have equal areas. But $\triangle CRB \cong \triangle ARB$.

I met this problem few years ago and I have used a small trick to avoid trigo; it took me some time to recall it, but here it is: Clearly $\triangle ARB\cong\triangle CBR\ (\ 1\ )$. Adding to its area the area of $\triangle CDR$ we get: $[CRB]+[CRD]=[BCD]=R^2\ (\ 2\ )$. Adding to $[RDS]$ the same $[CDR]$ we get $[RDS]+[CDR]=[CDSR]$. $CDSR$ having perpendicular diagonals, its area is $[CDSR]=\frac{CS\cdot DR}{2}\ (\ 3\ )$, so we need $CS\cdot DR=2R^2$. The last relation we get from $\triangle CDR\sim\triangle SCD\implies \frac{DR}{CD}=\frac{CD}{CS}$, done. Here $R$ is the circumradius the square. Best regards, sunken rock

Let $O$ be the center of $ABCD$ and let $D_{\infty}$ be the point at infinity of $AC$ (also point at infinity of the tangent of $\omega$ at D). $C(O,B,R,D) \equiv C(A,B,P,D)=D(A,B,P,D_{\infty}) \equiv D(A,O,S,D_{\infty}) \Longrightarrow$ $\frac{OB}{OD} \cdot \frac{RD}{RB}=\frac{OA}{OS} \Longrightarrow OS \cdot RD=OA \cdot RB \Longrightarrow [DSR]=[ARB].$

Dear Mathlinkers, there is a typo in this problem: P is the midpoint of the shorter arc AB... Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, there is a typo in this problem: P is the midpoint of the shorter arc AB... Sincerely Jean-Louis Special Case

Dear Mathlinkers, my figure after toying desn't give the result... Make a figure and verify again... Perhaps I am wrong but make a tentative... Sincerely Jean-Louis

Trigonometry handles this problem quite easily. Let $\phi=\angle ACP $, then $\angle ADP = \phi$ too. Also let $r$ be the length of the radius of the circle. In the right-angled triangle $OCR$ we have $tan\phi = \frac{OR}{OC}=\frac{OR}{r}$ Thus $OR = r tan\phi $. Consequently $BR = OB - OR = r - r tan\phi = r(1-tan\phi)$ So the area of triangle $ARB$ is: $$[ARB] = \frac{1}{2} \cdot AB \cdot BR \cdot \sin(\angle ABR)=\frac{1}{2} r\sqrt{2}\cdot r(1-tan\phi) \frac{\sqrt{2}}{2}=\frac{1}{2}r^2(1-tan\phi)$$On the other hand, in the right-angled triangle $ODS$ we have: $tan(45^\circ-\phi) = \frac{OS}{OD}=\frac{OS}{r}$, so $OS=rtan(45^\circ-\phi)$ Also $DR=OD+OR=r+r tan\phi = r(1+tan\phi)$ So the area of the triangle $DSR$ is: $$[DSR]=\frac{1}{2}\cdot OS \cdot DR=\frac{1}{2}rtan(45^\circ-\phi)r(1+tan\phi) = \frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)$$Finally, using the trigonometrical formula: $$ tan(45^\circ-\phi) = \frac{tan 45^\circ - tan\phi}{1+tan 45^\circ \cdot tan\phi} = \frac{1-tan\phi}{1+tan\phi} $$So the area of the triangle $DSR$ is: $$[DSR]=\frac{1}{2}r^2tan(45^\circ-\phi)(1+tan\phi)=\frac{1}{2}r^2\frac{1-tan\phi}{1+tan\phi}(1+tan\phi)=\frac{1}{2}r^2(1-tan\phi)$$Thus the areas of $ARB$ and $DSR$ are equal.

Let $M$ be the intersection point of lines $PD$ and $AB$ and let $N$ be the projection of $R$ onto line $AB$. Let $O$ be the circumcenter of circle $\omega$. Now $W.L.O.G$, assume that side length of square $ABCD$ = 1. Since $\angle DPC$ = $\angle DAC$ = $45^\circ$ = $\angle ABC$, $PBRM$ is cyclic. So, $\angle DRM$ = $\angle DPB$ = $90^\circ$, implying that $MR \parallel AC$. So $\triangle DSO \sim \triangle DMR \implies DO / DR = SO / MR \implies SO.DR = DO.MR \implies$ area of $\triangle DSR = 1/2.(1/\sqrt2).MR ---> (1)$ In $\triangle MRB$, $\angle BMR = \angle BAC = 45^\circ = \angle MBR \implies \triangle MRB$ is an isosceles right triangle. So $RN$ = $MR / \sqrt2$ Hence area of $\triangle ARB = 1/2.(1).(MR / \sqrt2) ---> (2)$ From $(1)$ and $(2)$ above, $[DSR]=[ARB].$

Attachments:

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20III.pdf p. 18... Sincerely Jean-Louis

Claim. $DC$ is tangent to $(PSC)$. Proof. We have $$\measuredangle SPC=\measuredangle DPC=\measuredangle DBC=45^\circ=\measuredangle ACD=\measuredangle SCD,$$the claim follows. Now, \begin{align*} S_{DSR}&=S_{ARB}\Longleftrightarrow \\ DS\cdot PR\cdot \sin{\angle DPR}\cdot \frac{1}{2}&=AB\cdot BR\cdot \sin{\angle ABR}\cdot \frac{1}{2}\Longleftrightarrow \\ DS\cdot PR&=AB\cdot BR\Longleftrightarrow \\ \frac{RB}{RP}&=\frac{DS}{DC}\Longleftrightarrow \\ \frac{\sin{\angle RPB}}{\sin{\angle PBR} }\cdot &=\frac{\sin{\angle DCS}}{\sin{\angle DSC}}\Longleftrightarrow \\ \sin{\angle PCD}&=\sin{\angle DSC}, \end{align*}which is true by the claim, we are done. [asy][asy] size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair O,A,B,C,D,P,R,S; O=(0,0);B=dir(45);A=dir(135);D=dir(225);C=dir(315);path w=circumcircle(A,B,C);P=dir(100);R=extension(P,C,D,B);S=extension(P,D,A,C); draw(A--B--C--D--cycle,deep);draw(w,deep);draw(A--C,deep);draw(B--D,deep);draw(P--D,deep);draw(P--C,deep);draw(D--S--R--cycle,org);draw(A--R--B--cycle,light); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$P$",P,dir(P)); dot("$R$",R,dir(R)); dot("$S$",S,dir(S)); [/asy][/asy]