Let $E$ be a point such that $A$ & $E$ lie on opposite sides of $BC$, and $\triangle EB_2C_2 \cong \triangle A_1B_1C_1$. Now, $d(E,B_2C_2) = d(A_1,B_1C_1) = \frac{1}{2}d(A,BC) = d(B_3,BC) = d(C_3,BC) \Rightarrow \overline {B_3EC_3}$ is parallel to $BC$. And, $B_2E = A_1B_1 = \frac{1}{2}AB = BB_3 \Rightarrow BB_2EB_3$ is an isosceles trapezium $\Rightarrow E$ lies on $\odot (BB_2B_3)$. Similarly, $E$ lies on $\odot (CC_2C_3) \Rightarrow \odot (BB_2B_3)$ and $\odot (CC_2C_3)$ meet once at $E$. Now, Let $D$ be the second intersection point of $\odot (BB_2B_3)$ and $\odot (CC_2C_3)$. Note that $A_1B \cdot A_1B_2 = A_1C \cdot A_1C_2 \Rightarrow A_1$ lies on $DE$. Thus, $\measuredangle BDC = \measuredangle BDA_1+\measuredangle A_1DC = \measuredangle BB_3E + \measuredangle EC_3C = \measuredangle BAC \Rightarrow D$ lies on $\odot (ABC)$. $\blacksquare$

Let $D$ be the point outside $\triangle ABC$ such that $\triangle DB_2C_2\cong \triangle A_1B_1C_1.$ Notice that since $\measuredangle BB_2D=\measuredangle A_1B_1C_1=\measuredangle B_3BB_2$ and $BB_3=A_1B_1=DB_2,$ it follows that $BB_2DB_3$ must be an isosceles trapezoid, and similarly $CC_2DC_3$ must also be an isosceles trapezoid. This implies that $B_3,D,C_3$ are collinear since $B_3D\parallel BC\parallel DC_3,$ but Miquel's Pivot Theorem applied to $\triangle AB_3C_3$ then implies that $(ABC),(BB_2DB_3),(CC_2DC_3)$ concur, so we're done. $\blacksquare$ [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 = -9, xmax = 9, ymin = -6.36, ymax = 6.8; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); draw((-0.92,6.32)--(-3.28,-0.78)--(2.56,-0.78)--cycle, linewidth(1) + rvwvcq); /* draw figures */ draw((-0.92,6.32)--(-3.28,-0.78), linewidth(1) + rvwvcq); draw((-3.28,-0.78)--(2.56,-0.78), linewidth(1) + rvwvcq); draw((2.56,-0.78)--(-0.92,6.32), linewidth(1) + rvwvcq); draw((-4.46,-4.33)--(4.3,-4.33), linewidth(1) + wrwrwr); draw(circle((-2.55,-2.993760563380281), 2.3310160514200198), linewidth(1) + wrwrwr); draw(circle((1.83,-3.3392253521126753), 2.6613031399854195), linewidth(1) + wrwrwr); draw(circle((-0.36,2.191633802816902), 4.166174199195713), linewidth(1) + wrwrwr); draw((-2.1,2.77)--(0.82,2.77), linewidth(1) + wrwrwr); draw((-3.28,-0.78)--(-4.46,-4.33), linewidth(1) + wrwrwr); draw((4.3,-4.33)--(2.56,-0.78), linewidth(1) + wrwrwr); draw((-1.82,-0.78)--(-0.4541333817000375,-1.9734768036968984), linewidth(1) + wrwrwr); draw((-0.4541333817000375,-1.9734768036968984)--(1.1,-0.78), linewidth(1) + wrwrwr); draw((-2.1,2.77)--(-0.36,-0.78), linewidth(1) + wrwrwr); draw((0.82,2.77)--(-0.36,-0.78), linewidth(1) + wrwrwr); draw((-1.82,-0.78)--(-0.64,-4.33), linewidth(1) + wrwrwr); draw((1.1,-0.78)--(-0.64,-4.33), linewidth(1) + wrwrwr); draw((-0.64,-4.33)--(-0.36,-0.78), linewidth(1) + wrwrwr); draw((-0.36,-0.78)--(-0.4541333817000375,-1.9734768036968984), linewidth(1) + wrwrwr); /* dots and labels */ dot((2.56,-0.78),dotstyle); label("$C$", (2.64,-0.58), NE * labelscalefactor); dot((-3.28,-0.78),dotstyle); label("$B$", (-3.2,-0.58), NE * labelscalefactor); dot((-0.92,6.32),dotstyle); label("$A$", (-0.84,6.52), NE * labelscalefactor); dot((-2.1,2.77),linewidth(4pt) + dotstyle); label("$C_{1}$", (-2.02,2.92), NE * labelscalefactor); dot((-4.46,-4.33),linewidth(4pt) + dotstyle); label("$B_{3}$", (-4.38,-4.18), NE * labelscalefactor); dot((0.82,2.77),linewidth(4pt) + dotstyle); label("$B_{1}$", (0.9,2.92), NE * labelscalefactor); dot((4.3,-4.33),linewidth(4pt) + dotstyle); label("$C_{3}$", (4.38,-4.18), NE * labelscalefactor); dot((-0.36,-0.78),linewidth(4pt) + dotstyle); label("$A_{1}$", (-0.28,-0.62), NE * labelscalefactor); dot((-1.82,-0.78),linewidth(4pt) + dotstyle); label("$B_{2}$", (-1.74,-0.62), NE * labelscalefactor); dot((1.1,-0.78),linewidth(4pt) + dotstyle); label("$C_{2}$", (1.18,-0.62), NE * labelscalefactor); dot((-0.4541333817000375,-1.9734768036968984),linewidth(4pt) + dotstyle); dot((-0.64,-4.33),linewidth(4pt) + dotstyle); label("$D$", (-0.56,-4.18), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

Claim 1: $B_1-B_2-B_3$ and $C_1-C_2-C_3$ and these two lines concur at point $L$ on $\overline{AA_1}$ with $AL:LA_1=3$. proof: The collinearity thing is a just a direct application of Menelaus. For the second part, Suppose the two lines cut $\overline{AA_1}$ at points $L_1,L_2$ and again use Menelaus to show that $AL_1:L_1A_1 = AL_2:L_2A_1 = 3$. $\square$ [asy][asy] size(250); pair A=dir(115),B=dir(-150),C=dir(-30),A1=1/2*(B+C),B1=1/2*(A+C),C1=1/2*(A+B),B2=1/2*(B+A1),C2=1/2*(C+A1),B3=2*B-C1,C3=2*C-B1; draw(unitcircle^^circumcircle(A,C1,C3)^^circumcircle(A,B1,B3),red); pair T=2*foot(A,(0,0),circumcenter(A,C1,C3)) - A; dot("$T$",T,dir(-70)); draw(B3--A--C3^^B--C,magenta); draw(C1--C3^^B1--B3^^A--A1,royalblue); pair Bp=1/2*(B1+C),Cp=1/2*(C1+B),L=extension(A,A1,B3,B2); dot("$L$",L); dot("$B'$",Bp,dir(Bp)); dot("$C'$",Cp,dir(Cp)); draw(Bp--Cp,dotted); draw(circumcircle(B,B2,B3)^^circumcircle(C,C2,C3),dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$A_1$",A1,dir(A1)); dot("$B_1$",B1,dir(B1)); dot("$C_1$",C1,dir(C1)); dot("$B_2$",B2,dir(B2)); dot("$C_2$",C2,dir(C2)); dot("$C_3$",C3,dir(C3)); dot("$B_3$",B3,dir(B3)); [/asy][/asy] Claim 2: $\odot(ABC),\odot(AB_1B_3),\odot(AC_1C_3)$ concur at some point $T$ (say). proof: Let points $B' \in \overline{AC}$ and $C' \in \overline{AB}$ be such that $AB' : B'C = AC' : C'B = 3$. Let $\Phi$ be the $\sqrt{ AB \cdot AC \cdot 3/4}$ inversion at $A$ followed by reflection in angle bisector of $\angle BAC$. Note that $\Phi$ swaps $$B \longleftrightarrow B' ~,~ C \longleftrightarrow C' ~,~ B_1 \longleftrightarrow B_3 ~,~ C_1 \longleftrightarrow C_3.$$In particular, $\Phi$ also swaps $$ \odot(ABC) \longleftrightarrow \overline{B'C'} ~,~ \odot(AB_1B_3) \longleftrightarrow \overline{B_3B_1} ~,~ \odot(AC_1C_3) \longleftrightarrow \overline{C_3C_1}.$$So it suffices to show lines $B'C',B_3B_1,C_3C_1$ concur, or equivalently, $L \in \overline{B'C'}$. But this is true since $AB':AC = AC':AB = AX:AA_1 = 3/4$, so all the three points $B',C',L$ lie on the image of line $BC$ under homothety at $A$ with ratio $3/4$. $\square$ Finally, we obtain that $T$ is the Miquel point of both $\{\overline{BC},\overline{CA},\overline{AB},\overline{B_1B_3} \} ~,~ \{\overline{BC},\overline{CA},\overline{AB},\overline{C_1C_3} \}$ which respectively imply $T$ lies on $\odot(BB_2B_3),\odot(CC_2C_3)$. This completes the proof of the problem. $\blacksquare$

Another super quick solution Let $X$ be the point on $(ABC)$ such that $ABCX$ is harmonic We claim that $X$ is the point of intersection of the two circles To show that Take a homothety centered at $B$ with ratio 2 Let $A'$ denote the reflection of $A$ in $B$ Let $B'$ denote the reflection of $B$ in $X$

It suffices to prove $A_1BA'B'$ cyclic, but that follows immediately from the fact Click to reveal hidden text

see https://www.geogebra.org/calculator/ps2n2hc2 for a diagram [asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(400); 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 = -10.262987331478948, xmax = 15.405031660239903, ymin = -9.343720889579343, ymax = 2.979952440759948; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen ffvvqq = rgb(1,0.3333333333333333,0); draw((-1.452665173787087,0.6825754779367356)--(-1.88,-2.11)--(6.18,-4.33)--cycle, linewidth(0) + zzttqq); draw((2.15,-3.22)--(-1.88,-2.11)--(-0.1290178746053927,-6.197566960219872)--cycle, linewidth(0.4) + ffvvqq); draw((2.15,-3.22)--(-1.452665173787087,0.6825754779367356)--(-2.3073348262129127,-4.902575477936735)--cycle, linewidth(0.4) + ffvvqq); /* draw figures */ draw(circle((2.658051508963212,-1.3754526296200495), 4.597115545353998), linewidth(0.4)); draw((xmin, 1.9974055104769148*xmin + 3.5841369009369695)--(xmax, 1.9974055104769148*xmax + 3.5841369009369695), linewidth(0.4)); /* line */ draw((xmin, -0.2754342431761787*xmin-2.627816377171216)--(xmax, -0.2754342431761787*xmax-2.627816377171216), linewidth(0.4)); /* line */ draw((xmin, -1.3182592870896241*xmin-5.477986715673744)--(xmax, -1.3182592870896241*xmax-5.477986715673744), linewidth(0.4)); /* line */ draw((xmin, 1.9974055104769144*xmin + 3.584136900936968)--(xmax, 1.9974055104769144*xmax + 3.584136900936968), linewidth(0.4)); /* line */ draw(circle((-0.177483573174941,-3.7995124323378477), 2.3985442340372454), linewidth(0.4)); draw((xmin, 6.534865184485054*xmin + 10.175546546831901)--(xmax, 6.534865184485054*xmax + 10.175546546831901), linewidth(0.4)); /* line */ draw((xmin, -0.5944917617830955*xmin-6.274267023795542)--(xmax, -0.5944917617830955*xmax-6.274267023795542), linewidth(0.4)); /* line */ draw((xmin, -2.3344424257321785*xmin-6.4987517603764955)--(xmax, -2.3344424257321785*xmax-6.4987517603764955), linewidth(0.4)); /* line */ draw((xmin, 1.3065132105361095*xmin-6.029003402652635)--(xmax, 1.3065132105361095*xmax-6.029003402652635), linewidth(0.4)); /* line */ draw((-2.3073348262129127,-4.902575477936735)--(2.15,-3.22), linewidth(0.4)); draw((-0.1290178746053927,-6.197566960219872)--(1.2267071859579342,-5.744059360184837), linewidth(0.4)); draw((1.2267071859579342,-5.744059360184837)--(2.15,-3.22), linewidth(0.4)); draw((-1.452665173787087,0.6825754779367356)--(2.15,-3.22), linewidth(0.4)); draw((2.15,-3.22)--(-1.88,-2.11), linewidth(0.4) + ffvvqq); draw((-1.88,-2.11)--(-0.1290178746053927,-6.197566960219872), linewidth(0.4) + ffvvqq); draw((-0.1290178746053927,-6.197566960219872)--(2.15,-3.22), linewidth(0.4) + ffvvqq); draw((2.15,-3.22)--(-1.452665173787087,0.6825754779367356), linewidth(0.4) + ffvvqq); draw((-1.452665173787087,0.6825754779367356)--(-2.3073348262129127,-4.902575477936735), linewidth(0.4) + ffvvqq); draw((-2.3073348262129127,-4.902575477936735)--(2.15,-3.22), linewidth(0.4) + ffvvqq); /* dots and labels */ dot((-1.452665173787087,0.6825754779367356),dotstyle); label("$A$", (-1.3793577835350157,0.8630024208243642), NE * labelscalefactor); dot((-1.88,-2.11),dotstyle); label("$B$", (-1.7951872517366465,-1.9154944803410896), NE * labelscalefactor); dot((6.18,-4.33),dotstyle); label("$C$", (6.2567833598040234,-4.1458525370589365), NE * labelscalefactor); label("$d$", (0.34066410766263916,2.1671948438203934), NE * labelscalefactor); label("$f$", (-0.2641787551760966,2.5263202936308944), NE * labelscalefactor); label("$g$", (-10.111776615769264,-0.138768570752296), NE * labelscalefactor); dot((-2.7331241756590305,-1.8750203883420535),linewidth(4pt) + dotstyle); label("$D$", (-2.664648867067329,-1.7264810857039838), NE * labelscalefactor); label("$h$", (-6.123593988926349,2.5263202936308944), NE * labelscalefactor); label("$i$", (-0.2641787551760966,2.5263202936308944), NE * labelscalefactor); dot((-1.0045089373026963,-4.153783480109936),linewidth(4pt) + dotstyle); label("$X$", (-0.925725636405964,-3.9946418213492523), NE * labelscalefactor); dot((-2.3073348262129127,-4.902575477936735),dotstyle); label("$A'$", (-2.229918059401988,-4.712892720970254), NE * labelscalefactor); dot((-0.1290178746053927,-6.197566960219872),dotstyle); label("$B'$", (-0.05626402107528121,-6.017085143966283), NE * labelscalefactor); dot((2.15,-3.22),linewidth(4pt) + dotstyle); label("$A_1$", (2.2307980540336887,-3.0684761876274345), NE * labelscalefactor); label("$e$", (-1.3793577835350157,-2.1234092144419057), NE * labelscalefactor); label("$j$", (-0.9824296547970954,2.5263202936308944), NE * labelscalefactor); label("$k$", (-10.111776615769264,-0.5734993784176391), NE * labelscalefactor); label("$l$", (-3.83653191381738,2.5263202936308944), NE * labelscalefactor); label("$m$", (6.426895414977418,2.5263202936308944), NE * labelscalefactor); label("$n$", (0.019341336779560767,-4.334865931696043), NE * labelscalefactor); dot((1.2267071859579342,-5.744059360184837),linewidth(4pt) + dotstyle); label("$G$", (1.3046324203118742,-5.6012556757646506), NE * labelscalefactor); 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Redefine $B_2$ and $C_2$ as the midpoints of $B_1B_3$ and $C_1C_3$ respectively. We show that it indeed is the same point as in the problem. Firstly, $BB_2$ and $CC_2 \parallel B_1$. So, $B,B_2,C_2,C$ are collinear. Secondly, $BB_2 = \frac12 B_1C_1 = CC_2$. Which is enough to show that $B_2$ and $C_2$ are the same points as defined in the problem. This means, $B_2C \parallel B_3C_3$ and hence, $BC \parallel B_3C_3.$ Now let $B_3C_3 \cap (BB_2B_3) = X.$ We prove that $X \in (CC_2C_3)$ as well. Firstly, as $A_1B_2 = A_1C_2$, $XB_2 = B_3B = BC_1 = A_1B_1$, and \[\measuredangle A_1B_2X = \measuredangle BB_3X = \measuredangle ABC = \measuredangle B_1A_1C_2\] we see that $\triangle A_1B_2X \cong \triangle A_1B_1C_2.$ Thus $XA_1 = C_2B_1$ and $\measuredangle B_2A_1X = \measuredangle B_1C_2A_1 \implies \measuredangle C_2A_1X = \measuredangle CC_2B_1.$ Combining with $C_2A_1 = C_2C$ we again see that $\triangle A_1C_2X \cong \triangle B_1C_2C.$ Thus, $C_2X = CC_3$ which implies that $CC_2XC_3$ is an isosceles trapezoid as well. This gives $B_3,X,C_3$ collinear as claimed. Now Miquel Theorem on $\triangle AB_3C_3$ finishes.