Solution: We begin by constructing a diagram:
$[asy] import olympiad; unitsize(13); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); 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 = -10.252427827191417, xmax = 15.150472936867494, ymin = -3.4167712881064825, ymax = 11.241299387760378; /* image dimensions */ pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); /* draw figures */ draw(circle((0.2801492142784242,1.5), 4.481884633503346)); draw(circle((3.2738086301390403,7.2491118286162095), 3.3297516312587083)); draw((-3.454001382502029,3.9785901614824524)--(4,4)); draw((4,4)--(4,-1)); draw((4,-1)--(-3.420838326674272,-1.0456127964221371)); draw((-3.420838326674272,-1.0456127964221371)--(-3.454001382502029,3.9785901614824524)); draw((-3.454001382502029,3.9785901614824524)--(0.06810107679554245,8.14949230836156)); draw((-3.454001382502029,3.9785901614824524)--(-5.709089178789237,5.8523028157505035)); draw((-5.709089178789237,5.8523028157505035)--(-2,10)); draw((0.06810107679554245,8.14949230836156)--(-2,10)); draw((0.06810107679554245,8.14949230836156)--(4,4)); draw((0.06810107679554245,8.14949230836156)--(2.3992779710964074,10.461967575808181)); draw((-3.454001382502029,3.9785901614824524)--(4,-1)); draw((4,4)--(2.3992779710964074,10.461967575808181)); draw((0.06810107679554245,8.14949230836156)--(-5.709089178789237,5.8523028157505035)); draw((4,-1)--(-2,10)); draw((2.3992779710964074,10.461967575808181)--(-3.420838326674272,-1.0456127964221371)); draw((2.3992779710964074,10.461967575808181)--(6.428589254168415,6.183933374028034)); draw((6.428589254168415,6.183933374028034)--(4,4)); draw(circle((-2.820834296659645,7.001753244155373), 3.108577255193373)); draw((-5.709089178789237,5.8523028157505035)--(6.428589254168415,6.183933374028034)); draw((-3.454001382502029,3.9785901614824524)--(0.1949568576384766,5.981074885605413)); draw((4.000284475054789,3.9995766660700363)--(0.1949568576384766,5.981074885605413)); draw((0.06810107679554245,8.14949230836156)--(0.1949568576384766,5.981074885605413)); /* dots and labels */ dot((6.428589254168415,6.183933374028034),dotstyle); label("$C_{2}$", (6.494915365823923,6.349748653166799), NE * labelscalefactor); dot((4,-1),dotstyle); label("$C_{1}$", (4.074012290397943,-0.8300529335417423), NE * labelscalefactor); dot((-3.454001382502029,3.9785901614824524),dotstyle); label("$B$", (-3.387675270846515,4.144405440621219), NE * labelscalefactor); dot((4,4),dotstyle); label("$C$", (4.04084923457019,4.277057663932232), NE * labelscalefactor); dot((0.06810107679554245,8.14949230836156),dotstyle); label("$A$", (0.12760864689531848,8.322950474918107), NE * labelscalefactor); dot((2.3992779710964074,10.461967575808181),dotstyle); label("$A_{1}$", (2.4987671385796686,10.71069049451633), NE * labelscalefactor); dot((-3.420838326674272,-1.0456127964221371),dotstyle); label("$B_{2}$", (-3.354512215018762,-0.8797975172833719), NE * labelscalefactor); dot((-5.709089178789237,5.8523028157505035),dotstyle); label("$B_{1}$", (-5.908067513755754,6.3994932369084285), NE * labelscalefactor); dot((-2,10),dotstyle); label("$A_{2}$", (-1.7792670632004874,10.196663129186156), NE * labelscalefactor); dot((4.000284475054789,3.9995766660700363),linewidth(4pt) + dotstyle); dot((0.1949568576384766,5.981074885605413),linewidth(4pt) + dotstyle); label("$X$", (0.24367934229245447,6.117607262372528), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Claim 1: Circumcircles of $B_1A_2AB, AA_1C_2C,$ and $BCC_1B_2$ mutually intersect at some point $X.$

Proof: Start by letting $X$ be the intersection of two circumcircles such that $X \neq A.$ By some simple cyclic quad angle properties, we see $\angle AXC = 180^{\circ} - \angle AA_1C, \angle BXA = 180^{\circ} - \angle AB_1B.$ It follows that $\angle CXB = 180^{\circ} - BC_1C,$ and so our claim is proven.

Claim 2: Lines $B_1C_2, C_1A_2,$ and $A_1B_2$ concur at $X.$

Proof: We proceed with angle chasing. Note that $\angle AA_1C_2= \angle AXC_2 = 90^{\circ}, \angle B_1A_2A = \angle AXB_1 = 90^{\circ},$ meaning $B_1, X, C_2,$ are collinear. Then, we may similarly obtain $\angle BB_2C_1 = \angle BXC_1 = 90^{\circ}, \angle A_2B_1B = \angle A_2XB = 90^{\circ},$ meaning $A_2, X, C_1$ are collinear. Finally, we have $\angle A_1C_2C = A_1XC=90^{\circ}, \angle CC_1B_2 = \angle B_2XC = 90^{\circ},$ meaning $A_1, X , B_2$ are collinear. Putting these three results together, it is clear that $B_1C_2, C_1A_2,$ and $A_1B_2$ are concurrent at $X. \qquad \square$