Let $AQ\cap BS=X$,$BR\cap AS=Y$,$QA_1\cap AS=Z$,$RB_1\cap BS=T$. By easy angle we get that $X,Y$ lie on $(ABC)$ and $Z,T$ lie on $(A_1B_1C)$. Again some easy angle chasing yields that $(QXZS)$,$(RYTS)$ are cyclic so $\triangle QXZ\sim \triangle RYT$ $\implies$ $\frac{QX}{QZ}=\frac{RY}{RT}$ we also have $\frac{QA}{QA_1}=\frac{SA}{SA_1}=\frac{SB}{SB_1}=\frac{RB}{RB_1}$ So multiplying above equalities and using Coaxality Lemma we are done.

Indeed a Beautiful Problem All-Russian MO 2019 Grade 10 P4 wrote: Let $ABC$ be an acute-angled triangle with $AC<BC.$ A circle passes through $A$ and $B$ and crosses the segments $AC$ and $BC$ again at $A_1$ and $B_1$ respectively. The circumcircles of $A_1B_1C$ and $ABC$ meet each other at points $P$ and $C.$ The segments $AB_1$ and $A_1B$ intersect at $S.$ Let $Q$ and $R$ be the reflections of $S$ in the lines $CA$ and $CB$ respectively. Prove that the points $P,$ $Q,$ $R,$ and $C$ are concyclic. Solution: Let $D$ be the radical center WRT $\odot (ABC), $ $\odot (A_1B_1C),$ and $\odot (AA_1B_1B)$ $\implies$ $\angle CAB$ $=$ $\angle A_1B_1C$ $=$ $\angle A_1PD$ $=$ $AA_1PD$ is cyclic, By Power of Point $\implies$ $BB_1PD$ is also cyclic. Let $BR $ $\cap$ $AB_1$ $=$ $E$ and $AQ$ $\cap $ $A_1B$ $=$ $F$, then simple angle chasing shows $E,F$ $\in $ $\odot (ABC)$. Now, Let $AB_1$ $\cap$ $\odot (A_1B_1C)$ $=$ $F'$ $$\implies AA_1Q=\angle AA_1S=\angle AB_1B=\angle F'A_1C \text{ and } \angle AF'A_1=\angle ACB=\angle AFA_1 $$Hence, $F'$ is the reflection of $F$ over $AC$ $\implies$ $F'SQF$ is a cyclic quadrilateral. Now since, $P$ is the miquel point of $CB_1BADA_1$ $\implies$ $P$ is the image of $S$ WRT $\odot (AA_1B_1B)$, suppose, say $O_1$ is the center of $\odot (AA_1B_1B)$ $\implies$ $S$ is the orthocenter WRT $\Delta O'CD$ $\implies$ $SP$ $\perp$ $CD$ $\implies$ if $X,Y$ are the foot from $S$ to $BC,CA$, then, $P$ $\in$ $\odot (XSYC)$. Let $PS$ $\cap $ $\odot (ABC)$ $=$ $G$ $\implies$ $YS$ $||$ $AG$ $\implies$ $P$ $\in$ $\odot (QSF'F)$ $\implies$ $QFPS$ is cyclic. Similarly, it can be shown that $REPS$ is also cyclic $\implies$ $\angle RPQ$ $=$ $\angle AFB$ $+$ $\angle AEB$ $=$ $2 \angle ACB$ $=$ $\angle RCQ$ $\implies$ $PQRC$ is cyclic!

Nice and easy . Reformulate the problem as follows. Quote: Let $ABCD$ be a cyclic quadrilateral with Miquel point $M$. $E=AB\cap CD$ and $P=AC\cap BD$. Let $X,Y$ be reflections of $P$ across $AB, CD$. Prove that $E,P,X,Y$ are concyclic. Let $X_1, Y_1$ be midpoints of $PX, PY$ (a.k.a. feet from $P$ from $AB, CD$). Obviously $E,M,P,X_1,Y_1$ are concyclic. Claim : $MP$ is $P$-symmedian of $\triangle PXY$. Proof : It suffices to show that $(X_1, Y_1; P, M) = -1$. Indeed, this follows from projecting at $P$ onto $BC$ (which gives $E(B, C; P, F)=-1$ where $F=AD\cap BC$). Let tangents from $X,Y$ to $\odot(PXY)$ meet at $T$. By the Claim, $T, P, M$ are colinear. However, as $E$ is circumcenter of $\odot(PXY)$, we get $\angle EXT = \angle EYT =90^{\circ}$ but $\angle EMT=90^{\circ}$ so $E, M, X, Y$ lies on the circle with diameter $ET$.

Quote: Let $ABCD$ be a cyclic quadrilateral with Miquel point $M$. $E=AB\cap CD$ and $P=AC\cap BD$. Let $X,Y$ be reflections of $P$ across $AB, CD$. Prove that $E,P,X,Y$ are concyclic. Because $\measuredangle{MDY}=\measuredangle{MAP}$ and $$\frac{MA}{MD}=\frac{AB}{CD}=\frac{AP}{DP}=\frac{AP}{DY},$$we have that $\triangle{MAP} \sim \triangle{MDY} \implies \measuredangle{DAM}=\measuredangle{YMD}. $ Analogously, $\measuredangle{XMB}=\measuredangle{PMC}.$ Using that $MP$ bisects angles $\angle{BMD}$ and $\angle{AMC}$ we have $$\measuredangle{XMY}=\measuredangle{XMB}+\measuredangle{DMY}+2\measuredangle{PMD}=2\measuredangle{AMP}+2\measuredangle{PMD}=2\measuredangle{AMD}=2\measuredangle{AED}=\measuredangle{XEY}$$

Cute problem. Quote: Let $ABCD$ be a cyclic quadrilateral with Miquel point $M$. $E=AB\cap CD$ and $P=AC\cap BD$. Let $X,Y$ be reflections of $P$ across $AB, CD$. Prove that $E,M,X,Y$ are concyclic. Solution. Let $X'$ and $Y'$ be the feet from $P$ on $AB$ and $CD$ respectively. As $\angle PME = 90^{\circ}$, $E,M,X_1,P,Y_1$ are concyclic. Note that the reflection of $P$ in $M$ lies on $\odot(PXY)$. Also, $-1=(EM\cap BC,EP\cap BC;B,C)=(M,P;X_1,Y_1)$ which gives $MP$ is the $P-$ symmedian of $\triangle PX'Y'$, so $M$ is the $P-$ Dumpty point of $\triangle PXY$. This gives $E,M,X,Y$ are concyclic and we are done. $~\square$

Suppose line $RS$ and line $AC$ intersect at point $M$ and line $QS$ and line $BC$ intersect at point $N$. Then $\angle CQR=\angle CRQ=90^{\circ} -\frac{\angle QCR}{2}=90^{\circ} -\angle ACB=\angle QNC=\angle CMR$, thus $M,Q,C,R,N$ are concyclic. Also $\angle A_1 MS=\angle B_1 NS$ and $\angle A_1 SM=\angle B_1 SN$,so $\triangle A_1 MS\sim \triangle B_1 NS$,so $\frac{MA_1}{NB_1}=\frac{SA_1}{SB_1}=\frac{AA_1}{BB_1}$,so for similar triangle $PAA_1$ and $PBB_1$,$M$ and $N$ are in the similar position. Thus $\triangle PMA_1\sim \triangle PNB_1$,so $P,M,N,C$ are concyclic. So $P,Q,M,N,R,C$ are concyclic.Done.

Let $Q_1$ and $R_1$ be the projections of $S$ to $AC$ and $BC$. Since $\dfrac{A_1Q_1}{Q_1A}=\dfrac{B_1R_1}{R_1B}$, there is a rotational similarity with center $P$ taking $Q_1R_1$ to $AB$, so $C,S,P,Q_1,R_1$ are concyclic. Also there is a rotational similarity with center $P$ taking $QS$ to $SR$, so $PS$ is the bisector of $\angle QPR$. Since $\angle CPS=\angle CQ_1S=90^\circ$, $CP$ is the external bisector of $\angle QPR$, which, combined with $CQ=CS=CP$, gives the desired result.

Let $M$ be the midpoint of $SR$, $N$ the midpoint of $SQ$. We know $\angle SMC=90,\angle SNC=90$. Note that $P=(A_1B_1C)\cap (ABC)$ is the Miquel Point of cyclic quad $ABB_1A$, so $\angle SPC=90$. Therefore, $M,P,N \in (CS)$. Let $D=CS\cap AB, T=A_1B_1\cap AB$. We know $(AB;DT)=-1$ by Ceva-Menelaus. Projecting through $C$ onto $CS$ gives $(NM;SP)=-1$, i.e. $SP$ is the $S-$symmedian of $\triangle SMN$. Also, since $\angle SMC=\angle SNC=90$, $C$ is the $S-$antipode in $(CS)$. We want to show $P,Q,R,C$ are cyclic. Perform an inversion at $S$ with radius $\sqrt{SM\cdot SN}$ and reflect over the angle bisector of $\angle MSN$. Let $X_1,X_2,X_3$ are the midpoints of $SM,SN,MN$, and $X_4$ the foot from $S$ to $MN$. Then $\{R,X_1\},\{Q,X_2\},\{P,X_3\},\{C,X_4\}$ swap. We know $X_1X_2X_3X_4$ cyclic since they lie on the nine-point circle of $\triangle SMN$. Inverting back finishes.

Well-known that $\angle CPS = 90^{\circ}$. Let $Q',R'$ be the midpoints of $SQ,SR$ respectively. $T=CS \cap \odot(ABC)$ So, $S,P,Q',R',C$ are concyclic. By angle chasing, $\triangle SQ'P \overset{+}{\sim} \triangle TAP$. Assume that $B$ be the midpoint of $TB'$. Then, $\angle PBB' = \angle PAT$ and $PB/BB' = PB / BT = PA / AT$ (since $(P,T;A,B) = C(A_1B_1 \cap AB,S;A,B)=-1)$. Hence, $\triangle PAT \overset{+}{\sim} \triangle PBB'$. Similarly, $\triangle PBT \overset{+}{\sim} \triangle PAA'$. Hence, $\angle A'PB' = \angle A'PA + \angle APT + \angle TPB + \angle BPB' = 2 \angle APB$. But $\triangle PA'B' \overset{+}{\sim} \triangle PQR$. Therefore, $\angle QPR = 2 \angle APB = \angle QCR$.

Becoming good at Angel Chasing !

Let $\Phi$ be the inversion at $C$ fixing $\odot(ABB_1A_1)$. Then $T := \Phi(P) = \overline{AB} \cap \overline{A_1B_1}$ ; $M := \Phi(S)$ is Miquel point of lines $\{\overline{AA_1},\overline{BB_1},\overline{AB_1},\overline{BA_1}\}$ ; $U,V := \{\Phi(P),\Phi(R)\}$ are reflections of $M$ in $\overline{AA_1},\overline{BB_1}$, respectively. Let $W,X$ be reflection of $M$ in $\overline{AB_1},\overline{BA_1}$, respectively ; $H_A,H_B$ be orthocenters of $\triangle ASA_1,\triangle BSB_1$, respectively. [asy][asy] size(300); pair A= dir(-170),B=dir(-10),B1=dir(110),A1=dir(150),C=extension(A,A1,B,B1),S=extension(A,B1,B,A1),T=extension(A,B,A1,B1),Q=2*foot(S,A,A1)-S,R=2*foot(S,B,B1)-S,M=2*foot(S,circumcenter(S,A,A1),circumcenter(S,B,B1))-S,P=2*foot(C,circumcenter(C,A,B),circumcenter(C,A1,B1))-C,U=2*foot(M,A,A1)-M,V=2*foot(M,B,B1)-M,HA=orthocenter(A,S,A1),HB=orthocenter(B,S,B1),X=2*foot(M,B,A1)-M,W=2*foot(M,A,B1)-M; draw(unitcircle^^circumcircle(A,B,C)^^circumcircle(C,A1,B1),green); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$A_1$",A1,dir(190)); dot("$B_1$",B1,dir(40)); dot("$C$",C,dir(C)); dot("$S$",S,dir(-60)); dot("$Q$",Q,dir(190)); dot("$R$",R,dir(R)); dot("$M$",M,dir(-50)); dot("$T$",T,dir(T)); dot("$U$",U,dir(-90)); dot("$P$",P,dir(P)); dot("$V$",V,dir(V)); dot("$H_A$",HA,dir(-90)); dot("$H_B$",HB,dir(110)); dot("$X$",X,dir(90)); dot("$W$",W,dir(-90)); draw(circumcircle(C,P,R),brown+dotted); draw(T--V,brown+linewidth(0.9)); draw(T--C--B^^U--C--V^^A--C--M,fuchsia); draw(A1--B--T--B1--A,purple); draw(U--M--V^^W--M--X,orange); draw(circle(1/2*(A1 + B1),1/2*abs(A1-B1))^^circle(1/2*(A+B),1/2*abs(A-B)),cyan); clip((-2,-0.5)--(2,-0.5)--(2,2)--(-2,2)--(-2,-0.5)--cycle); [/asy][/asy] By Simson Line of $M$, we obtain points $U,V,W,X,H_A,H_B$ are collinear. Now observe $H_A,H_B,T$ lie on radical axes of $\odot(AB),\odot(A_1B_1)$, implying points $T,U,V$ are collinear. $\blacksquare$

Since $P$ is a center of spiral similarity $AA_1\mapsto BB_1$ and $QAA_1\stackrel{-}{\sim}SAA_1\stackrel{-}{\sim}SBB_1\stackrel{-}{\sim}RBB_1$ it follows $$\measuredangle QPR=\measuredangle QPS+\measuredangle SPR=2\measuredangle A_1PB_1=2\measuredangle A_1CB_1=\measuredangle QCR\implies C\in \odot (PQR)\quad \blacksquare$$

Nice problem but you guys are overcomplicating this. We claim that $\triangle PQA\stackrel{+}{\sim}\triangle PSB$. First, we claim that $\angle QAP=\angle SBP$. Note that $$\angle QAP=\angle QAC-\angle PAC=\angle SAA_1-\angle PBC=\angle SBB_1-\angle PBC=\angle SBP.$$Now, we bash sides using the fact that $P$ is the center of spiral mapping $AA_1\to BB_1$. $$\frac{PA}{PB}=\frac{AA_1}{BB_1}=\frac{SA}{SB}=\frac{QA}{SB}$$So we get the desired similarity. Thus, $\angle QPA=\angle SPB$ so $\angle QPS=\angle APB=\angle ACB$. Similarly, we have that $\angle RPS=\angle ACB$. Now, we finish the problem by the following angle chasing: $$\angle QPR=\angle QPS+\angle RPS=2\angle ACB=2(\angle ACS+\angle BCS)=\angle QCS+\angle RCS=\angle QCR.$$Hence, $P,Q,R,C$ are concyclic, as desired.

lol one-liner? Notice $A_1SN \sim BRM$ so $P : B_1R \rightarrow A_1S$ and hence $P : A_1B_1 \rightarrow RS$ but also $P : A_1B_1 \rightarrow QS$ and hence $P$ is dumpty point in $QSR$ (since both spirals hv same ratio) and therefore $\measuredangle QPR = 2(180 - \measuredangle QSR) = \measuredangle QCR$

Let $X = AQ \cap BA_1$, $Y = BR \cap AB_1$. Then both $X$ and $Y$ lie on $(ABC)$. Also since $P$ is the Miquel point of $ABB_1A_1$, we have $SP \perp PC$. Therefore, \[ \measuredangle SPX = 90^\circ - \measuredangle CPX = 90^\circ - \measuredangle CAX = \measuredangle SQX, \]and so $SPXQ$ is cyclic. Similarly, $SPYR$ is also cyclic, so this gives \[ \measuredangle RPQ = \measuredangle RPS + \measuredangle SPQ = \measuredangle RYS + \measuredangle SXQ = 2 \measuredangle BCA = \measuredangle RCQ \]and we are done.

Let $U, V$ be second intersections of $AS, BS$ with $(CA_1B_1)$. We have $$\angle{SB_1R} = 2\angle{AB_1B} = 2\angle{AA_1B} = 360^{\circ} - 2\angle{BA_1C} = \angle{CB_1V} + 180^{\circ} - \angle{AB_1C} = 180^{\circ} - \angle{SB_1V}$$Then $V, B_1, R$ are collinear. We also have the familiar result that $SP \perp CP$. Hence $\angle{RSP} = \angle{BCP} = \angle{RVP}$ or $S, R, P, V$ lie on a circle. Similarly, we have $S, Q, U, P$ lie on a circle. From this, we have $$\angle{QPR} = \angle{QPS} + \angle{SPR} = \angle{QUS} + \angle{SVR} = 2\angle{ACB} = \angle{QCR}$$or $Q, C, P, R$ lie on a circle