[asy][asy] import olympiad; size(7cm); defaultpen(fontsize(10)+linewidth(0.4)); dotfactor *= 1.5; pair B = dir(135), A = dir(210), C = dir(330), I = incenter(A,B,C), A0 = circumcenter(B,I,C), C0 = circumcenter(A,I,B), P = extension(A,B,I,I+dir(0)), Q = extension(B,C,I,I+dir(0)), S = intersectionpoint(B--B+dir(A--B)*100,circumcircle(P,Q,A0)), R = intersectionpoint(B--B+dir(C--B)*100,circumcircle(P,Q,A0)), C1 = P+dir(C0--P)*abs(A-P)*abs(B-P)/abs(C0-P), A1 = Q+dir(A0--Q)*abs(B-Q)*abs(C-Q)/abs(A0-Q), M = dir(270); draw(A--B--C--A); draw(unitcircle); draw(circumcircle(P,Q,A0)); draw(R--B--S--R); draw(C0--C1^^A0--A1); draw(P--Q); dot("$A$", A, dir(210)); dot("$C$", C, dir(330)); dot("$B$", B, dir(15)); dot("$P$", P, dir(220)); dot("$Q$", Q, dir(270)); dot("$S$", S, dir(105)); dot("$R$", R, dir(135)); dot("$A_0$", A0, dir(30)); dot("$C_0$", C0, dir(165)); dot("$M$", M, dir(270)); [/asy][/asy] If $BA = BC$, the problem is trivial by symmetry, so we assume otherwise. Claim: Lines $\overline{A_0Q}$ and $\overline{C_0P}$ meet at $M$, the midpoint of minor arc $AC$. Proof. Let $C_1 = \overline{C_0P} \cap (ABC)$ and $A_1 = \overline{A_0Q} \cap (ABC)$. By angle bisector theorem $$\frac{BC_1}{AC_1} = \frac{BP}{AP} = \frac{BQ}{CQ} = \frac{BA_1}{CA_1}.$$But Reim's forces $AC_1A_1C$ to be an isosceles trapezoid, so $AC_1 = CA_1$ and the above equality can only occur when $BA_1 = BC_1 \implies A_1 = C_1$. This is none other than the midpoint of arc $AC$. $\square$ To finish, note that $$\tfrac 12 \widehat{C_0S} = \angle C_0PB = \tfrac 12 (\widehat{BC_0} + \widehat{AM}) = \tfrac 12 \widehat{MC_0} = \angle MA_0C_0 = \tfrac 12 \widehat{C_0Q},$$so $C_0S = C_0Q$. Similarly $A_0P = A_0R$ and the conclusion follows. $\blacksquare$

i have another way to show $\overline{A_0Q}$ and $\overline{C_0P}$ meet at $M$, the midpoint of minor arc $AC$ when $BA$ not equal to$BC$ As $PQ$ is parallel to $AC$, we have $(ABC)$ is tangent to $(BPQ)$ By Radical Axis on $(ABC)$ ,$(BPQ)$ , $(C_0PQA_0)$ we get $C_0A_0$, tangent to $(ABC)$ at $B$ and $PQ$ meet at a point say D. If $BA$ is not equal to $BC$ , $D$ is a unique point( $C_0A_0$, tangent to $(ABC)$ at $B$ are fixed lines) Hence we have a unique pair PQ satisfying the condition. (Namely, the parallel to AC passing through D intersection the sides) easy to see by angle chasing that $P_0$=$C_0M \cap BA$ and $Q_0$=$A_0M \cap BC$ is one such pair.Hence it is the unique pair.. Rest is same

another way to show that $A_0Q \cap C_0P =L$ where $L$ is the midpoint of arc $BC$ let $A_0L\cap BC=Q'$ and $C_0L \cap BA=P'$ by Pascal on $C_0LA_0ABC$ we have $P'-I-Q'$ are collinear let $PL' \cap AC = P_2$ note that $P'P_2 $ bisects $AI $ and $AI$ bisects $P'P_2$ so $IP' \parallel AC$ $\angle C_0P'Q'=180 - \angle C_0A_0L$ so $$P'=P$$and $$Q'=Q$$

WLOG assume $BA<BC$. Let $B_0$ be the midpoint of minor arc $AC$ Suppose the lines $C_0P$ and $A_0Q$ meet $(ABC)$ at $D$ and $G$ respectively. By Reim's theorem we have $GD\|AC$. Meanwhile let $H$ be the reflection of $C_0$ about the perpendicular bisector of $AC$, then $C_0HDG$ is an isosceles trapezoid. Notice that $\angle CGA_0=\angle CAA_0=\angle BAA_0=\angle QCA_0$, hence $\triangle A_0GC\sim\triangle A_0CQ$, similarly $\angle C_0AP\sim\triangle C_0DA$. Therefore, $$\frac{DA}{AP}=\frac{C_0D}{C_0A}\text{ and }\frac{GC}{CQ}=\frac{A_0G}{A_0C}$$dividing them yields $$\frac{CQ}{AP}=\frac{C_0D}{A_0G}\cdot\frac{A_0C}{C_0A}=\frac{C_0D}{A_0G}\cdot\frac{A_0B}{C_0B}$$Since $\frac{BC}{AB}=\frac{CQ}{AP}$ we have $$\frac{HG}{A_0G}=\frac{C_0D}{A_0G}=\frac{BC\cdot C_0B}{AB\cdot A_0B}=\frac{\sin A\sin\frac{C}{2}}{\sin C\sin\frac{A}{2}}=\frac{\cos\frac{A}{2}}{\cos\frac{C}{2}}=\frac{\sin\angle B_0A_0H}{\sin\angle A_0HB_0}=\frac{HB_0}{A_0B_0}$$Therefore, if $G$ does not lie inside minor arc $A_0H$ then $D=G=B_0$, otherwise $G$ lies in minor arc $A_0H$, but this is impossible since this will imply $B$ lies between $A$ and $P$ while $C$ lies between $B$ and $Q$, which clearly implies that $P$ and $Q$ are not parallel. Therefore $D=G=B_0$, hence $$\angle A_0C_0P=180^{\circ}-\angle A_0CB_0=180^{\circ}-\angle A_0QC=\angle B_0QC=\angle A_0QR=\angle A_0C_0R$$Hence $A_0P=A_0R$ and similarly $C_0S=C_0Q$, so the desired equality easily follows. [asy][asy] size(6cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.780174070005856, xmax = 31.761192814362794, ymin = -24.37603248401924, ymax = 15.978836692314914; /* image dimensions */ /* draw figures */draw(circle((-0.2353458098947867,-5.960744898653019), 9.65817590908592), linewidth(0.8)); draw((-9.481966295969796,-3.1714131727111976)--(6.635657151479008,0.826720032795258), linewidth(0.8)); draw(circle((-1.7930462417413757,0.3187930041496454), 8.443993768194856), linewidth(0.8)); draw((-2.834296714945611,8.698341214486854)--(-8.673670455820556,-10.659153039801398), linewidth(0.8)); draw((-8.673670455820556,-10.659153039801398)--(8.290799010489291,-10.497838284586946), linewidth(0.8)); draw((8.290799010489291,-10.497838284586946)--(-8.023918379906798,6.017675627537095), linewidth(0.8)); draw((-7.294465645281537,-6.087096106946906)--(3.8291942616484356,-5.981321477298818), linewidth(0.8)); draw((-9.481966295969796,-3.1714131727111976)--(-0.14351058613534587,-15.618484188079124), linewidth(0.8)); draw((-0.14351058613534587,-15.618484188079124)--(6.635657151479008,0.826720032795258), linewidth(0.8)); /* dots and labels */dot((-8.673670455820556,-10.659153039801398),dotstyle); label("$A$", (-10.058842773975188,-10.017401646381675), NE * labelscalefactor); dot((-4.667808313199872,2.6202595907783923),dotstyle); label("$B$", (-6.081962512821927,3.4574438043938223), NE * labelscalefactor); dot((8.290799010489291,-10.497838284586946),dotstyle); label("$C$", (8.192944529843992,-11.73373944330045), NE * labelscalefactor); dot((-9.481966295969796,-3.1714131727111976),linewidth(4pt) + dotstyle); label("$C_0$", (-12.22863313374837,-2.972950611504269), NE * labelscalefactor); dot((6.635657151479008,0.826720032795258),linewidth(4pt) + dotstyle); label("$A_0$", (7.523154170070811,0.5736584175317505), NE * labelscalefactor); dot((-0.14351058613534587,-15.618484188079124),linewidth(4pt) + dotstyle); label("$B_0$", (-1.477153789381308,-16.840890936570958), NE * labelscalefactor); dot((-7.294465645281537,-6.087096106946906),linewidth(4pt) + dotstyle); label("$P$", (-8.91691928459894,-7.2428641550582975), NE * labelscalefactor); dot((3.8291942616484356,-5.981321477298818),linewidth(4pt) + dotstyle); label("$Q$", (4.239307191318315,-7.0870688223740075), NE * labelscalefactor); dot((-2.834296714945611,8.698341214486854),linewidth(4pt) + dotstyle); label("$S$", (-3.1934915863000843,9.448380684526398), NE * labelscalefactor); dot((-8.023918379906798,6.017675627537095),linewidth(4pt) + dotstyle); label("$R$", (-9.30532861923036,6.308738373089613), NE * labelscalefactor); dot((8.956560068472669,-2.996081643900217),linewidth(4pt) + dotstyle); label("$H$", (9.699972839333649,-2.7334314838483307), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

This might sound stupid but isn't it possible for R and S to lie on the extension of AB past A and BC past B