This was a really nice problem . Here is a solution that I and amar_04 found. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.52, xmax = 29.28, ymin = -15.3, ymax = 5.04; /* image dimensions */ pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw((8.44,2.98)--(5.58,-6.24), linewidth(0.8) + dtsfsf); draw((5.58,-6.24)--(19.12,-6.08), linewidth(0.8) + dtsfsf); draw((19.12,-6.08)--(8.44,2.98), linewidth(0.8) + dtsfsf); draw(circle((12.359919441201725,-6.99943271169598), 6.822319669495047), linewidth(1.4)); draw((6.340428295488688,-3.788549341116885)--(19.12,-6.08), linewidth(0.8)); draw((5.58,-6.24)--(12.164915848398643,-0.17990052307974658), linewidth(0.8)); draw(circle((12.315918707971942,-3.275870662125787), 7.35925699864144), linewidth(1.4)); draw((8.44,2.98)--(7.585828269060439,-8.913681231111852), linewidth(0.8)); draw((6.340428295488688,-3.788549341116885)--(12.164915848398643,-0.17990052307974658), linewidth(0.8)); draw((5.5260993251428205,-0.4373822755701926)--(19.12,-6.08), linewidth(0.8)); draw((5.5260993251428205,-0.4373822755701926)--(7.585828269060439,-8.913681231111852), linewidth(0.8)); draw((2.3215520114998403,-6.278504555255542)--(6.340428295488688,-3.788549341116885), linewidth(0.8)); draw((2.3215520114998403,-6.278504555255542)--(5.58,-6.24), linewidth(0.8) + dtsfsf); draw((8.44,2.98)--(9.252672071943666,-1.9842249320983159), linewidth(0.8)); draw((9.252672071943666,-1.9842249320983159)--(10.637108098235617,-10.441082348369939), linewidth(0.8)); draw((2.3215520114998403,-6.278504555255542)--(10.637108098235617,-10.441082348369939), linewidth(0.8)); draw(circle((11.30461833939784,-5.296139196043708), 5.1880641825949905), linewidth(0.8) + linetype("4 4")); draw(circle((13.524465434065673,-5.626656640637814), 5.613869189787343), linewidth(0.8) + linetype("4 4")); /* dots and labels */ dot((8.44,2.98),dotstyle); label("B", (8.52,3.18), NE * labelscalefactor); dot((5.58,-6.24),dotstyle); label("A", (5.22,-6.58), NE * labelscalefactor); dot((19.12,-6.08),dotstyle); label("$C$", (19.26,-6.2), NE * labelscalefactor); dot((12.164915848398643,-0.17990052307974658),dotstyle); label("$A_1$", (12.16,0.08), NE * labelscalefactor); dot((6.340428295488688,-3.788549341116885),linewidth(4pt) + dotstyle); label("$C_1$", (5.82,-3.66), NE * labelscalefactor); dot((7.933388312510872,-4.074176227915555),linewidth(4pt) + dotstyle); label("$P$", (7.44,-4.3), NE * labelscalefactor); dot((7.585828269060439,-8.913681231111852),linewidth(4pt) + dotstyle); label("$Q$", (7.66,-8.76), NE * labelscalefactor); dot((5.5260993251428205,-0.4373822755701926),linewidth(4pt) + dotstyle); label("$T$", (5.7,-0.46), NE * labelscalefactor); dot((9.252672071943666,-1.9842249320983159),linewidth(4pt) + dotstyle); label("$M$", (9.36,-2.38), NE * labelscalefactor); dot((2.3215520114998403,-6.278504555255542),linewidth(4pt) + dotstyle); label("$D$", (2.28,-6.12), NE * labelscalefactor); dot((10.637108098235617,-10.441082348369939),linewidth(4pt) + dotstyle); label("$E$", (10.72,-10.28), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Solution(with amar_04): Firstly we'll give a defination of a few points. Let $A_1C_1 \cap AC=D$. And let $BM \cap \odot(ABC)=E$ and $QC_1\cap \odot(ABC)=T$. We'll continue with a phantom points based approach i.e. assume $TC \cap A_1C_1=M$ . Then we'll prove that $M$ is infact the midpoint of $A_1C_1$. Firstly note that to prove $BM$ is the median in $\Delta BC_1A_1$ we need to prove that its the $B$-symmedian of $\Delta BAC$ since they are oppositely similiar so the median and the symmedian get swapped. So we can restate the problem as follows. Restated problem wrote: In a triangle $\Delta ABC$ Let $C_1$ be any random point on $AB$. Let $\odot(AC_1C) \cap BC=A_1$. Now let $A_1A \cap CC_1=P$ and let $BP \cap \odot(ABC)=Q$.Let $QC_1 \cap \odot(ABC)=T$ and finally let $TC \cap A_1C_1=M$. Then prove that $BM$ is the $B$-symmedian in $\Delta ABC$. By well known property of symmedians we need to prove $(B,E;A,C)=-1$. Taking perpsectivity at $Q$ we have $(B,E;A,C) \overset{Q}{=} (BQ \cap AC,QE \cap AC;A,C)$. Now note that we already have $(D,QB \cap AC;A,C)=-1$ . So we just need to prove $D-Q-E$ collinear. Note that by radical axis theorem we are left to prove $A_1C_1QE$ cyclic. Now we have that since $\angle CEM\equiv \angle CEB =\angle CAB \equiv \angle CAC_1=180^\circ - \angle CA_1C\equiv 180^\circ-\angle CA_1M \implies CA_1ME$ cyclic. Now note that $\angle C_1QE \equiv \angle TQE=180^\circ -\angle TCE=180^\circ-\angle MCE=180^\circ-\angle MA_1E \equiv 180^\circ-\angle C_1A_1E \implies C_1A_1QE$ cyclic as desired.$\blacksquare$

Let $QC_1$ meet circumscribed circle at $T$ and Let $CT$ meet $A_1C_1$ at $M$. Let $BM$ meet circumscribed circle at $S$. Note that $BA_1C_1$ and $BCA$ are similar so we need to prove $BM$ is symmedian in $BAC$. Note that $AC_1A_1C$ is cyclic so we need to prove $C_1QSA_1$ is cyclic. Note that $\angle CA_1M = \angle CA_1C_1 = \angle 180 - \angle CAB = \angle 180 - \angle CSB = \angle 180 - \angle CSM \implies CSMA_1$ is cyclic so $\angle SA_1C_1 = \angle SA_1M = \angle SCM = \angle SCT = \angle 180 - \angle SQT = \angle 180 - \angle SQC_1 \implies SQC_1A_1$ is cyclic. we're Done.