If someone needs diagram.

[asy][asy] size(10cm); pair A = dir(110); pair X = dir(200); pair Y = dir(340); pair C = 0.7*X+0.3*Y; pair D = 0.35*X+0.65*Y; pair B = intersectionpoints(A--X,circumcircle(A,C,D))[1]; pair E = intersectionpoints(A--Y,circumcircle(A,C,D))[1]; draw(A--X--Y--A--cycle); draw(circumcircle(A,C,D)); pair AA = 2*foot(A,X,Y)-A; pair P = extension(E,X,B,Y); pair Q = intersectionpoints(E--X,circumcircle(A,C,D))[1]; pair R = intersectionpoints(B--Y,circumcircle(A,C,D))[1]; pair M = intersectionpoints(circumcircle(AA,X,Y),circumcircle(P,Q,R))[0]; pair N = intersectionpoints(circumcircle(AA,X,Y),circumcircle(P,Q,R))[1]; pair K = extension(C,M,D,N); draw(B--Y); draw(E--X); draw(circumcircle(C,Q,X),dashed); draw(circumcircle(D,R,Y),dashed); draw(M--K); draw(N--K); draw(B--C); draw(D--E); draw(circumcircle(P,Q,R)); draw(circumcircle(AA,X,Y)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$R$",R,dir(R)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$A'$",AA,dir(AA)); dot("$K$",K,dir(K)); [/asy][/asy] Claim: $DRYN$ is cyclic. Proof: We use directed angles modulo $180^{\circ}$. Let $N'$ be the second point of intersection of $(DRY)$ and $(PQR)$. Then we have $$\measuredangle N'DY=\measuredangle N'RY=\measuredangle N'RP=\measuredangle N'QP$$which implies that $\measuredangle N'DX=\measuredangle N'QX$, and $XQDN'$ is cyclic. Now we have $$\measuredangle XN'Y=\measuredangle XN'D+\measuredangle DN'Y=\measuredangle XQD+\measuredangle DRY=\measuredangle EQD+\measuredangle DRB=\measuredangle EAD+\measuredangle DAB=\measuredangle EAB=\measuredangle XA'Y$$which implies that $N'$ lies on $(A'XY)$, and that in fact $N'\equiv N$. $\square$ Now let $K'$ be the point of intersection of $DN$ and $(PQR)$. We have $$\measuredangle QMK'=\measuredangle QNK'=\measuredangle QND=\measuredangle QXD=\measuredangle QXC=\measuredangle QMC$$which means that $M, C, K'$ are collinear and we are done.

Let $Z,N',M'\in (PQR)$ be such points that $ZP\parallel XY,N'\in (DRY),M'\in (CQX);$ by Reim's $Z=CM'\cap DN',$ so it's suffice to prove $N'=N$ (and analogously $M'=M$). Indeed: $$\measuredangle QN'D=\measuredangle QM'C=\measuredangle QXD\implies N'\in (QXD),$$$$\measuredangle YN'X=\measuredangle YN'D+\measuredangle DN'X=\measuredangle BRD+\measuredangle DQE=\measuredangle XAY\implies N'=N.$$