[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -0.5000000000000006, xmax = 15.00000000000000, ymin = -6.000000000000000, ymax = 6.000000000000000; /* image dimensions */ /* draw figures */ draw((4.300000000000004,4.780000000000006)--(2.160000000000002,-5.180000000000005)); draw((2.160000000000002,-5.180000000000005)--(14.40000000000002,-5.180000000000005)); draw((14.40000000000002,-5.180000000000005)--(4.300000000000004,4.780000000000006)); draw(circle((6.281199176994883,-1.850225888688387), 3.329774111311619)); draw(circle((5.636478819992440,0.3073772698420307), 2.246201506037988)); draw(circle((4.307171280146841,-3.445166269785066), 1.734833730214942)); draw((0.2601472526100818,-0.1710740086230200)--(7.213659851571188,1.906727512708028)); draw((0.2601472526100818,-0.1710740086230200)--(4.307171280146841,-5.180000000000005)); draw((0.2601472526100818,-0.1710740086230200)--(14.40000000000002,-5.180000000000005)); /* dots and labels */ dot((4.300000000000004,4.780000000000006),dotstyle); label("$A$", (4.380000000000004,4.900000000000006), NE * labelscalefactor); dot((2.160000000000002,-5.180000000000005),dotstyle); label("$B$", (1.600000000000002,-5.880000000000006), NE * labelscalefactor); dot((14.40000000000002,-5.180000000000005),dotstyle); label("$C$", (14.48000000000002,-5.060000000000006), NE * labelscalefactor); dot((3.025721162109383,-1.150755712799334),dotstyle); label("$C'$", (3.200000000000004,-1.060000000000001), NE * labelscalefactor); dot((4.886449638939712,-1.809903399624514),dotstyle); label("$S$", (4.960000000000005,-1.680000000000002), NE * labelscalefactor); dot((3.440396030151498,0.7792263833219115),dotstyle); label("$C_1$", (2.820000000000003,0.8800000000000010), NE * labelscalefactor); dot((7.213659851571188,1.906727512708028),dotstyle); label("$B_1$", (7.300000000000008,2.020000000000002), NE * labelscalefactor); dot((2.611046294067268,-3.080737808920580),dotstyle); label("$C_2$", (1.760000000000002,-3.440000000000004), NE * labelscalefactor); dot((4.307171280146841,-5.180000000000005),dotstyle); label("$A_2$", (3.880000000000004,-5.900000000000007), NE * labelscalefactor); dot((0.2601472526100818,-0.1710740086230200),dotstyle); label("$X$", (-0.2400000000000003,-0.1000000000000001), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $X$ be the intersection of lines $B_1C_1$ and $CC'$, and let $Y$ be the intersection of lines $A_2C_2$ and $CC'$; let $S$ be the tangency point of the incircle of $\triangle ACC'$ with side $CC'$, and let $T$ be the tangency point of the incircle of $\triangle BCC'$ with side $CC'$. We claim that $S$ and $T$ are coincident. Let $BC = a, CA = b, AB = c, CC' = d$, and let $s, s_2, s_3$ be the semiperimeters of $\triangle ABC, \triangle ACC', \triangle BCC'$, respectively. Note that $C'S = s_2 - b$ and $C'T = s_3 - a$. Since $C'A = s-a$ and $C'B = s-b$, we have \[s_2 - b = \frac{(s-a)+b+d}{2}-b = \frac{s-a-b+d}{2}\] and \[s_3 - a = \frac{(s-b)+a+d}{2}-a = \frac{s-a-b+d}{2}\] so $C'S = C'T$ and it follows that $S$ and $T$ are coincident. Segments $AS$, $B_1C'$, and $CC_1$ concur at the Gergonne point of $\triangle ACC'$. Therefore, $(XC'SC)$ is harmonic. Similarly, $(YC'SC)$ is also harmonic. Thus $X$ and $Y$ are coincident and so we are done.

Since $AC-AC'=BC-BC'= \frac{BC+AC-AB}{2}$ so $\frac{AC'+CC'-AC}{BC'+CC'-BC}= \frac{AC+CC'-AC'}{BC+CC'-BC'}$ or $\frac{C_1C'}{B_1C}= \frac{C_2C'}{A_2C} \qquad (1)$. $A_2C_2$ intersects $CC'$ at $T$. Applying Menelaus theorem to $T,C_2,A_2$ and $\triangle BCC'$, we have $\frac{C_2C'}{C_2B} \cdot \frac{A_2B}{A_2C} \cdot \frac{TC}{TC'}=1$. Since $A_2B=C_2B$ so $\frac{C_2C'}{A_2C}= \frac{TC'}{TC} \qquad (2)$. From $(1)$ and $(2)$ we obtain $\frac{C_1C'}{C_1A} \cdot \frac{B_1A}{B_1C} \cdot \frac{TC}{TC'}=1$. Therefore $T,C_1,B_1$ are collinear. Thus, $A_2C_2,CC'$ and $C_1,B_1$ are concurrent. $\blacksquare$

[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(250); 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 = -5.957928530913223, xmax = 10.21467595818816, ymin = -3.090201226264757, ymax = 7.210659535964139; /* image dimensions */ /* draw figures */ draw((-4.085044755962515,-1.875357696567000)--(8.797358506874113,-1.445100613132377), linewidth(1.200000000000002)); draw((-0.5417511276773911,6.552619290711188)--(8.797358506874113,-1.445100613132377), linewidth(1.200000000000002)); draw((-0.5417511276773911,6.552619290711188)--(-4.085044755962515,-1.875357696567000), linewidth(1.200000000000002)); draw(circle((0.6763891401299050,1.403820534491368), 3.118413170655879), linewidth(1.200000000000002)); draw(circle((-1.132561062239664,0.1580046197630245), 1.933674652967049), linewidth(1.200000000000002)); draw(circle((0.1492868679241661,3.631760612704826), 1.769042567475847), linewidth(1.200000000000002)); draw((0.6763891401299053,1.403820534491368)--(-0.5417511276773911,6.552619290711188), linewidth(1.200000000000002)); draw((0.6763891401299053,1.403820534491368)--(-4.085044755962515,-1.875357696567000), linewidth(1.200000000000002)); draw((-2.198300329264843,2.612398689792463)--(8.797358506874113,-1.445100613132377), linewidth(1.200000000000002)); draw((-4.733632345997233,3.547959536475384)--(1.299964637752037,4.975431805953194), linewidth(1.200000000000002)); draw((-4.733632345997233,3.547959536475384)--(-1.068014598362894,-1.774592445723790), linewidth(1.200000000000002)); draw((-3.465966337631038,3.080179113133924)--(1.778552969195272,4.320965645180931), linewidth(1.200000000000002)); draw((-3.465966337631038,3.080179113133924)--(-0.2637533890869617,-1.569499617742118), linewidth(1.200000000000002)); draw((-2.198300329264843,2.612398689792463)--(-3.465966337631038,3.080179113133924), linewidth(1.200000000000002)); /* dots and labels */ dot((-4.085044755962515,-1.875357696567000),dotstyle); label("$B$", (-4.565920319801211,-2.406851740809768), NE * labelscalefactor); dot((8.797358506874113,-1.445100613132377),dotstyle); label("$C$", (8.898595467682261,-1.293245171920158), NE * labelscalefactor); dot((-0.5417511276773911,6.552619290711188),dotstyle); label("$A$", (-1.098554412122196,6.426073089701005), NE * labelscalefactor); dot((-2.198300329264843,2.612398689792463),dotstyle); label("$C'$", (-2.794273505658648,2.250048456364966), NE * labelscalefactor); dot((-2.915106818019288,0.9074232558265302),dotstyle); label("$C_2$", (-3.831952353942149,0.7821125246468433), NE * labelscalefactor); dot((0.6763891401299053,1.403820534491368),dotstyle); label("$I$", (0.7743293628285122,1.566698970909978), NE * labelscalefactor); dot((1.299964637752037,4.975431805953194),dotstyle); label("$B_1$", (1.407060367879427,5.135301839397138), NE * labelscalefactor); dot((-1.068014598362894,-1.774592445723790),dotstyle); label("$A_2$", (-1.326337573940526,-2.457470221213841), NE * labelscalefactor); dot((-4.733632345997233,3.547959536475384),dotstyle); label("$X$", (-5.350506766064346,3.515510466466796), NE * labelscalefactor); dot((1.778552969195272,4.320965645180931),dotstyle); label("$H$", (1.432369608081464,3.667365907679016), NE * labelscalefactor); dot((-1.616083688137953,3.517837808444829),dotstyle); label("$K$", (-1.478193015152745,3.034634902628101), NE * labelscalefactor); dot((-0.2637533890869617,-1.569499617742118),dotstyle); label("$M$", (-0.2621125246468421,-1.419791372930341), NE * labelscalefactor); dot((-2.436716272122118,1.585686955600970),dotstyle); label("$N$", (-2.212160981011807,1.187060367879429), NE * labelscalefactor); dot((-3.465966337631038,3.080179113133924),dotstyle); label("$Y$", (-3.376386030305490,3.237108824244393), NE * labelscalefactor); dot((-1.481493840510397,4.317374123758396),dotstyle); label("$C_1$", (-2.237470221213843,4.502570834346224), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Denote the point contact of incircle $ ACC' $ and $ CC' $ is $ A' $, the point contact of incircle $ BCC' $ and $ CC' $ is $ B' $. Let s be the semiperimeter of $ ABC $. We have $AC' = s - a $ implying $ AC' + a = s = BC' + b.$ $C'A' = \frac{1}{2} (AC' + CC' - b) = \frac{1}{2} (AC' + a + CC' - a - b) = \frac{1}{2}(BC' + b + CC' - a - b) = C'B'$ So that $A' \equiv B'$ We have three radical axes of incircles $ ABC, ACC', BCC' $ are $HK, MN, CC'$ concur at $Y.$ Consider the homothety with center $C'$ and ratio 2, it send $HK $ to $B_1C_1$, $MN$ to $A_2C_2$ and $CC'$ to $CC'$ so that $B_1C_1, A_2C_2, CC'$ concur at point $X.$

My solution: Let $\Gamma$ is the incircle of $\triangle ACC'$ and $\Gamma \cap CC'=X$ $\Longrightarrow $ $CX=CB_1$ $...(1)$ Let $w$ is the incircle of $\triangle BCC'$ and $w \cap CC'=X'$ $\Longrightarrow $ $CX'=CA_2$ $...(2)$ By $...(1)$ and $...(2)$ we get $X=X'$ $\Longrightarrow $ $\Gamma$ and $w$ are tangent in $X$ By easy angle chasing we can find that: $C_2,A_2,B_1,C_1$ are cyclic on the circunference $\Omega$ By radical axis in: $\Gamma,w,\Omega$ we get: $B_1C_1,CC',A_2C_2$ are concurrent...

Let the incircle of triangle $ABC$ touches sides $BC,CA$ at points $A',B'$so we khow that if the lines$B1C1,CC',A2C2$ are concurent at $X$ we know that $C2A2//C'A'$ and $C1B1//C'B'$ so by tales teorm we must prove that $(CB'/CB1)=(CA'/CA2)$ so we must prove that $A'A2=B'B1$ we khow that $A'A2=C'C2$ and we know that$ B'B1=C1C'$ so we are done