Hints : Use Droz-Farny Circle theorem and homotety

Erm, I don't know exactly how the Droz-Farn Circle Thm goes (oh well, I found the Wolfram page), but I used good ol' angle chasing and inversion. You need to prove that some sides of the hexagon are parallel to the sides of triangle $XYZ$. Then invert about, say, $X$ and use the fact that $AX$ is a symmedian of $XYZ$ and isogonal conjugates and Ceva's thm.

Actually we don't need Droz-Farny neither Ceva theorem . Let $U$ the second intersection (different of $Z$) of circles $C_1$ and $C_2$ and define $V,W$ as the intersections of lines $IU$ and $AC$, $UJ$ and $BC$, respectively. We see that since $\angle{ZXB}=\angle{ZYX}=\angle{ZMJ}$ since $MZYJ$ is cyclic, so $BC//MJ$, analogously $LI//AC$. Now we see that $\angle{UVY}=\angle{UVC}=\angle{UIL}=\angle{UZL}$ thus $UZYV$ is cyclic, $V\in C_1$ and analogously $W\in C_2$, so by power of point $CZ\cdot CU=CX\cdot CW=CY\cdot CV$ with $CX=CY$, thus $XY//VW$ or $JI//l VW$. If $Q$ is the intersection of lines $LI$ and $MJ$ we have that the triangles $QIJ$ and $CVW$ are homothetic, then the line $QC$ passes throught the intersection $U$ of lines $IV$ and $JW$. But this line is the radical axis of $C_1$ and $C_2$, so $LMJI$ is cyclic. A symple angle chase shows that $LMJI$ is indeed a isosceles trapezoid (just note that $\angle{JIL}=\angle{YZX}=90^\circ-\frac{C}{2}=\angle{MJI}$), then $\angle{YLM}=\angle{KNM}=\angle{ZYX}=90^\circ-\frac{B}{2}$, analogously $MJKN$ is cyclic with $\angle{NMJ}=\angle{KNM}=90^\circ-\frac{B}{2}$, so $MLNK$ is cyclic and finally $MLINKJ$ is cyclic.

We will show $INMJ$ and similar are cyclic. This will end the problem, as if they were different circumferences their radical axis would concur, but thos are $AB, AC, CA$ which don't, contradicition. Now notice how $JMZY$ is cyclic, so $INMJ$ is cyclic iff $IN\parallel ZY.$ This is true because of the following lemma, which solves the problem: $\textbf{Lemma.}$ Let $ABC$ be a triangle and $K$ be a point on the $A-symeddian.$ Then if $E=(ABK)\cap AC$ and $F=(ACK)\cap AB,$ then $EF\parallel BC.$ $\textit{Proof.}$ Denote by $\Psi$ as the inversion around the circumference of center $A$ and radius $\sqrt{AB\cdot AC}$ followed by a reflexion around the bisector of $\angle BAC.$ Then $\Psi(C)=B, \Psi(B)=C,$ and $\Psi(K)=K',$ for some $K'$ in the $A-Median.$ Now $\Psi(E)=E'=CM\cap AB$ and $\Psi(F)=F'=BM\cap AC.$ Clearly by the inversion we have that $E'F'\parallel EF,$ so it suffices to show $E'F'\parallel BC.$ But this is clear applying ceva on $AE'F',$ as $M$ lies on the $A-median,$ like this showing the lemma, and the problem. $\hfill \blacksquare$ $\hfill \square$