[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(9cm); 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 = -13.05963259905671, xmax = 43.67228357297989, ymin = -23.753350291558267, ymax = 13.374293689755277; /* image dimensions */ pen wwffqq = rgb(0.4,1,0); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen ffqqff = rgb(1,0,1); pen dtsfsf = rgb(0.8274509803921568,0.1843137254901961,0.1843137254901961); /* draw figures */ draw(circle((4.92,-2.52), 6), linewidth(1.2) + wwffqq); draw(circle((14.08,-2.26), 5), linewidth(1.2) + wwffqq); draw((2.997985934927978,3.1638245868134702)--(19.06883677826532,-2.5939275367907593), linewidth(0.8) + wvvxds); draw((0.17759004911692422,-6.19553368339418)--(12.763637257037924,2.5636074808115747), linewidth(0.8) + wvvxds); draw(circle((12.226622916033177,3.615401900667733), 9.23967870650329), linewidth(1.2) + wwffqq); draw((12.48242249820904,-0.2341930599037824)--(4.087512019909805,-19.10222025481696), linewidth(0.8) + rvwvcq); draw((2.997985934927978,3.1638245868134702)--(4.087512019909805,-19.10222025481696), linewidth(0.8) + rvwvcq); draw((4.087512019909805,-19.10222025481696)--(19.06883677826532,-2.5939275367907593), linewidth(0.8) + wvvxds); draw(circle((4.655407255945439,0.7923146942287302), 8.29945009542299), linewidth(1.2) + wwffqq); draw((0.17759004911692422,-6.19553368339418)--(17.037256936136718,-16.999217794493273), linewidth(0.8) + wvvxds); draw((17.037256936136718,-16.999217794493273)--(12.763637257037924,2.5636074808115747), linewidth(0.8) + rvwvcq); draw((2.997985934927978,3.1638245868134702)--(10.185761689759875,-5.3960656853864135), linewidth(0.8) + ffqqff); draw((10.185761689759875,-5.3960656853864135)--(19.06883677826532,-2.5939275367907593), linewidth(0.8) + blue); draw((0.17759004911692422,-6.19553368339418)--(10.185761689759875,-5.3960656853864135), linewidth(0.8) + blue); draw((10.185761689759875,-5.3960656853864135)--(12.763637257037924,2.5636074808115747), linewidth(0.8) + ffqqff); draw((7.60542118821784,-1.0262044717317693)--(17.037256936136718,-16.999217794493273), linewidth(0.8) + rvwvcq); draw(circle((10.325901639344261,-2.3665573770491783), 3.032747895185789), linewidth(0.8) + wwffqq); draw(circle((9.813652583294603,-13.44022178068754), 8.052758066195993), linewidth(0.8) + wwffqq); draw((7.60542118821784,-1.0262044717317693)--(12.48242249820904,-0.2341930599037824), linewidth(0.8) + ffqqff); draw((4.087512019909805,-19.10222025481696)--(17.037256936136718,-16.999217794493273), linewidth(0.8) + dtsfsf); draw((2.997985934927978,3.1638245868134702)--(0.17759004911692422,-6.19553368339418), linewidth(0.8) + dtsfsf); draw((12.763637257037924,2.5636074808115747)--(19.06883677826532,-2.5939275367907593), linewidth(0.8) + dtsfsf); draw((10.014144947119508,0.6501241707126855)--(10.185761689759875,-5.3960656853864135), linewidth(0.8) + ffqqff); /* dots and labels */ dot((10.014144947119508,0.6501241707126855),linewidth(4pt) + dotstyle); label("$P$", (9.866778826965804,0.9984123626507618), NE * labelscalefactor); dot((10.185761689759875,-5.3960656853864135),linewidth(4pt) + dotstyle); label("$R$", (10.012806925220724,-6.302992550095264), NE * labelscalefactor); dot((2.997985934927978,3.1638245868134702),dotstyle); label("$A_1$", (2.236810693146209,3.480890032984411), NE * labelscalefactor); dot((19.06883677826532,-2.5939275367907593),linewidth(4pt) + dotstyle); label("$B_1$", (19.468126287226823,-2.907839265668362), NE * labelscalefactor); dot((0.17759004911692422,-6.19553368339418),dotstyle); label("$A_2$", (-2.12820210000608955,-6.850597918551216), NE * labelscalefactor); dot((12.763637257037924,2.5636074808115747),linewidth(4pt) + dotstyle); label("$B_2$", (12.787340792064214,3.18883383647457), NE * labelscalefactor); dot((4.087512019909805,-19.10222025481696),linewidth(4pt) + dotstyle); label("$C_1$", (1.368528454621843,-19.59154949129303), NE * labelscalefactor); dot((12.48242249820904,-0.2341930599037824),linewidth(4pt) + dotstyle); label("$D_1$", (12.166721374480801,0.15875079768496883), NE * labelscalefactor); dot((17.037256936136718,-16.999217794493273),linewidth(4pt) + dotstyle); label("$C_2$", (17.16818373971183,-17.401128017469222), NE * labelscalefactor); dot((7.60542118821784,-1.0262044717317693),linewidth(4pt) + dotstyle); label("$D_2$", (6.946216861867394,-0.49837564446217353), NE * labelscalefactor); dot((3.562083104913849,-8.364318754657345),linewidth(4pt) + dotstyle); label("$D$", (1.7114020124747005,-9.858484269556374), NE * labelscalefactor); dot((14.895054755681423,-7.1931212984520325),linewidth(4pt) + dotstyle); label("$E$", (15.561874658907701,-7.580738409825819), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] We break the problem into some claims. Let $D= A_1C_1 \cap A_2C_2,E=C_2B_2 \cap C_1B_1$. Claim 1: $A_1,A_2,D,R,P$ are concyclic. Similiarly $B_2,B_1,E,R,P$ are concyclic. Proof: For the first part we'll prove that $D \in \odot(A_1A_2PR)$. Observe that $\angle DA_2R=\angle A_2B_2R$ (alternate segment theorem since $C_2A_2$ is tangent to $\odot (A_2RB_2)$ ) $\equiv PB_2R =\angle PB_1R \equiv A_1B_1R =\angle C_1A_1R$(again by alternate segment theorem since $C_1A_1$ is tangent to $\odot(RA_1B_1)$ ) $\equiv \angle DA_1R \implies D \in \odot(A_1A_2PR)$ . Done with the first part. For the second part observe that $\angle EB_1R=\angle B_1A_1R$ (by alternate segment theorem since $C_1B_1$ is tangent to $\odot (A_1RB_1)$ ) $\equiv PA_1R =\angle PA_2R \equiv B_2A_2R =\angle C_2B_2R$(again by alternate segment theorem since $C_2B_2$ is tangent to $\odot(RA_2B_2)$ ) $\equiv \angle EB_2R \implies E \in \odot(B_1B_2PR)$ . Done with the second part too. $\square$ Claim 2: $C_1,C_2,E,R,D$ are concyclic. Proof: By claim 1 since we have that $A_1,A_2,D,R,P$ and $B_2,B_1,E,R,P$ as concyclic $\implies \angle C_1DC_2=\angle A_2DA_1=\angle A_2PA_1=\angle B_2PB_1=\angle B_2EB_1=\angle C_1EC_2 \implies C_1DEC_2$ as cyclic. Now by miquels theorem in triangle $A_1B_1C_1$ for the points $D,E,P$ we know that $\odot(C_1DE),\odot(A_1DP),\odot(B_1PE)$ are concurrent at a point. Now we know that $\odot(A_1DP) \cap \odot(B_1PE)=R$ and so we have $R \in \odot(C_1DE)\implies C_1,C_2,E,R,D$ are concyclic. We are done with our claim. $\square$. Now back to the main problem. By claim 1 we have that $\angle D_2RD_1=\angle C_1RC_2=\angle C_1DC_2=\angle A_1DA_2=\angle A_1PA_2=180-\angle D_2PD_1 \implies PD_2D_1R$ is cyclic. Now by reims theorem we just need to prove that $D_2D_1 \parallel C_1C_2$. Now observe that $\angle D_1D_2R=\angle RPD_1=\angle RPB_1=\angle RB_2B_1=180-\angle REB_1=\angle REC_1=\angle RC_2C_1\implies D_1D_2 \parallel C_1C_2$. We are done$\blacksquare$.

I had a inversion solution for this as well, but submitted this as this was finer

Attachments:
Sharygin_P8-1.pdf (159kb)

This is from one of the problems which I liked the most this year. Sharygin 2020 CR P8 wrote: Two circles meeting at points $P$ and $R$ are given. Let $\ell_1$, $\ell_2$ be two lines passing through $P$. The line $\ell_1$ meets the circles for the second time at points $A_1$ and $B_1$. The tangents at these points to the circumcircle of triangle $A_1RB_1$ meet at point $C_1$. The line $C_1R$ meets $A_1B_1$ at point $D_1$. Points $A_2$, $B_2, C_2, D_2$ are defined similarly. Prove that the circles $D_1D_2P$ and $C_1C_2R$ touch. Let the two circles be $\omega_1$ and $\omega_2$ where $A_1,A_2\in\omega_1$ and $B_1,B_2\in\omega_2$. Let $A_2C_2\cap A_1C_1=X$ and $B_2C_2\cap B_1C_1=Y$. Claim 1:- $X\in\omega_1$ and $Y\in\omega_2$.

Proof

Claim 2:- $(C_2,X,Y,C_1)=\omega_3$ are concyclic.

Proof

Claim 3:- $R\in\omega_3$.

Proof

Now notice that triangles $A_1RA_2, C_1RC_2,B_1RB_2$ are all spirally similar. Claim 4:- $R,P,D_2,D_1$ are concyclic.

Proof

Claim 5:- $D_1D_2\| C_1C_2$.

Proof

To finish it off notice that the Homothety($\mathcal H$) mapping $D_1$ to $C_1$ maps $D_2$ to $C_2$, so $\mathcal H$ maps $\odot(D_1RD_2)$ to $\odot(C_1RC_2)$, this forces that $\odot(D_1D_2P)$ and $\odot(C_1C_2R)$ are internally tangent at their Insimillicenter $R$. $\blacksquare$

Solution: Claim 1: $P,D_2,D_1,R $ are concyclic . Proof: As $A_1,B_1,C_1,D_1$ and $A_2,B_2,C_2,D_2$ are defined similarly , by rotating $l_1$ to $l_2$ , we get that $\angle A_2D_2R=\angle A_1D_1R=\angle PD_1R$(can also be proved by simple angle chasing). $\implies P,D_2,D_1,R $ are concyclic (1) Let $A_1C_1 \cap A_2C_2=E$ and $B_1C_1 \cap B_2C_2=F$. Claim 2: $E$ lies on $\odot (A_1A_2RP)$ and $F$ lies on $\odot (B_1B_2RP)$. Now ,$\angle EA_1R=\angle C_1A_R=\angle A_1B_1R=\angle PB_R=\angle PB_2R=\angle C_2A_2R=\angle EA_2R$. $\implies E$ lies on $\odot (A_1A_2RP)$ . Similarly ,$F$ lies on $\odot (B_1B_2RP)$. Claim 3: $C_1,C_2,E,F,R$ are concyclic. Proof: $\angle C_1EC_2=\angle A_2EA_1=\angle A_2RA_1=\angle A_2PA_1=\angle B_2PB_1=\angle B_2RB_1=\angle B_2FB_1=\angle C_1FC_2$ --- (Alternate segment theorem) $\implies C_1,C_2,E,F$ are concyclic. From (1) , $\angle A_1PA_2=\angle D_1RD_2=\angle C_1RC_2=\angle C_1EC_2=\angle C_1FC_2$. $\implies C_1,C_2,E,F,R$ are concyclic. So , now we have to prove that $\odot (C_1ER)$ is tangent to $\odot (D_1RP)$ at $R$. Let $X$ be a point on the tangent $m$ of $\odot (D_1RP)$ as shown in the figure . $\angle XRC_1=\angle C_1ER=180^\circ-\angle A_1ER=\angle A_1PR=180^\circ-\angle RPD_1=180^\circ - \angle XRD_1$. $\implies \angle XRD_1=\angle RPD_1 \implies m$ is also tangent to $\odot (D_1RP)$ --(alternate segment theorem) . $\implies \odot (D_1D_2P)$ touch $\odot (C_1C_2R)$. $\blacksquare$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(24.583cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ real xmin = -4.94, xmax = 32.88, ymin = -12.44, ymax = 5.08; /* image dimensions */ pen uuuuuu = rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666); pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); draw((9.879351295774278,3.7535404154921497)--(11.010550906263507,-2.58835323884082)--(17.90726308856711,-3.596849678504043)--cycle, linewidth(2.) + red); draw((5.042302367942616,-1.9235362342299978)--(11.010550906263507,-2.58835323884082)--(13.734763158575062,3.3103894741991278)--cycle, linewidth(2.) + green); draw(shift((11.294473690624159,-0.24226072969218018)) * scale(2.363210149132414, 2.363210149132414)*unitcircle, linewidth(2.) + yellow); draw(shift((10.109817480268866,-10.031235810679732)) * scale(7.497187584880577, 7.497187584880577)*unitcircle, linewidth(2.) + yellow); /* draw figures */ draw(shift((8.3,0.2)) * scale(3.888701582790843, 3.888701582790843)*unitcircle, linewidth(2.)); draw(shift((14.8,-0.76)) * scale(4.207469548315236, 4.207469548315236)*unitcircle, linewidth(2.)); draw((xmin, -0.9156042422632377*xmin + 12.79911637271189)--(xmax, -0.9156042422632377*xmax + 12.79911637271189), linewidth(1.6) + dotted); /* line */ draw((xmin, 0.6021224408707762*xmin-4.959619643624101)--(xmax, 0.6021224408707762*xmax-4.959619643624101), linewidth(1.2) + dotted); /* line */ draw((9.879351295774278,3.7535404154921497)--(11.010550906263507,-2.58835323884082), linewidth(2.) + red); draw((11.010550906263507,-2.58835323884082)--(17.90726308856711,-3.596849678504043), linewidth(2.) + red); draw((17.90726308856711,-3.596849678504043)--(9.879351295774278,3.7535404154921497), linewidth(2.) + red); draw(shift((15.118216950753512,1.4161610993987332)) * scale(5.736641475482178, 5.736641475482178)*unitcircle, linewidth(2.) + red); draw((xmin, 2.24134166795624*xmin-18.389461296104205)--(xmax, 2.24134166795624*xmax-18.389461296104205), linewidth(2.) + dotted + red); /* line */ draw((xmin, 0.5563614884108452*xmin-13.559761223823832)--(xmax, 0.5563614884108452*xmax-13.559761223823832), linewidth(2.) + dotted + red); /* line */ draw((xmin, 1.1513304011993875*xmin-15.265135231175464)--(xmax, 1.1513304011993875*xmax-15.265135231175464), linewidth(2.) + linetype("2 2") + red); /* line */ draw((5.042302367942616,-1.9235362342299978)--(11.010550906263507,-2.58835323884082), linewidth(2.) + green); draw((11.010550906263507,-2.58835323884082)--(13.734763158575062,3.3103894741991278), linewidth(2.) + green); draw((13.734763158575062,3.3103894741991278)--(5.042302367942616,-1.9235362342299978), linewidth(2.) + green); draw(shift((8.516322364359212,2.1419864803980992)) * scale(5.347643394866506, 5.347643394866506)*unitcircle, linewidth(2.) + green); draw((xmin, -0.8545075849451722*xmin + 2.3851493847639738)--(xmax, -0.8545075849451722*xmax + 2.3851493847639738), linewidth(2.) + dotted + green); /* line */ draw((xmin, -4.4663021422422995*xmin + 64.65399159253356)--(xmax, -4.4663021422422995*xmax + 64.65399159253356), linewidth(2.) + dotted + green); /* line */ draw((xmin, -1.5664164740207618*xmin + 14.658755088774566)--(xmax, -1.5664164740207618*xmax + 14.658755088774566), linewidth(2.)); /* line */ draw((11.010550906263507,-2.58835323884082)--(11.700878829951963,2.085742077799767), linewidth(2.)); draw((xmin, -0.1210194326325502*xmin-1.2558626151929975)--(xmax, -0.1210194326325502*xmax-1.2558626151929975), linewidth(2.)); /* line */ /* dots and labels */ dot((11.700878829951963,2.085742077799767),linewidth(4.pt) + uuuuuu); label("$P$", (11.64,2.48), NE * labelscalefactor,uuuuuu); dot((11.010550906263507,-2.58835323884082),linewidth(4.pt) + uuuuuu); label("$R$", (10.94,-3.18), NE * labelscalefactor,uuuuuu); dot((9.879351295774278,3.7535404154921497),linewidth(4.pt) + xdxdff); label("$A_1$", (9.16,3.4), NE * labelscalefactor,xdxdff); dot((5.042302367942616,-1.9235362342299978),linewidth(4.pt) + xdxdff); label("$A_2$", (4.8,-2.46), NE * labelscalefactor,xdxdff); dot((13.734763158575062,3.3103894741991278),linewidth(4.pt) + uuuuuu); label("$B_2$", (14.,2.94), NE * labelscalefactor,uuuuuu); dot((17.90726308856711,-3.596849678504043),linewidth(4.pt) + uuuuuu); label("$B_1$", (17.8,-4.18), NE * labelscalefactor,uuuuuu); dot((2.866324560318224,-11.965048625176625),linewidth(4.pt) + uuuuuu); label("$C_1$", (2.22,-11.84), NE * labelscalefactor,uuuuuu); dot((13.577716011800359,0.36730199206199127),linewidth(3.pt) + uuuuuu); label("$D_1$", (13.62,-0.26), NE * labelscalefactor,uuuuuu); dot((17.240416424562156,-12.346917217637715),linewidth(4.pt) + uuuuuu); label("$C_2$", (17.44,-12.38), NE * labelscalefactor,uuuuuu); dot((9.046817005531995,0.48767189385807097),linewidth(4.pt) + uuuuuu); label("$D_2$", (8.84,-0.06), NE * labelscalefactor,uuuuuu); dot((6.710472307848439,-3.349000100857051),linewidth(4.pt) + uuuuuu); label("$E$", (6.8,-3.18), NE * labelscalefactor,uuuuuu); dot((15.57216619863247,-4.89600765977163),linewidth(4.pt) + uuuuuu); label("$F$", (15.66,-4.74), NE * labelscalefactor,uuuuuu); dot((14.217875777977447,-2.976501875083908),xdxdff); label("$X$", (14.3,-2.78), NE * labelscalefactor,xdxdff); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* re-scale y/x */ currentpicture = yscale(1.5384615384615383) * currentpicture; /* end of picture */ [/asy][/asy]

basically trivial by spiral similarity??? [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(13cm); 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, xmax = 11, ymin = -9.64, ymax = 9.64; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); pen zzttff = rgb(0.6,0.2,1); pen fuqqzz = rgb(0.9568627450980393,0,0.6); draw((-3.07,-2.92)--(-5.99,0)--(-0.11,5.88)--cycle, linewidth(1) + ccqqqq); draw((-3.07,-2.92)--(-0.2528896419858224,-0.7683028249129576)--(4.079980011956577,-6.441114093790823)--cycle, linewidth(1) + zzttff); /* draw figures */ draw(circle((-3.07,0), 2.92), linewidth(1) + blue); draw(circle((2.81,0), 6.565119953207253), linewidth(1) + blue); draw(circle((-1.59,1.48), 4.642240838215959), linewidth(1) + ccqqqq); draw(circle((0.5049900059782886,-4.680557046895411), 3.984986155335767), linewidth(1) + zzttff); draw((-8.970273972602746,8.860273972602748)--(-3.07,-2.92), linewidth(1) + ccqqqq); draw((7.625211325567886,0.7578383840010742)--(-3.07,-2.92), linewidth(1) + zzttff); draw(circle((-2.098898875011621,0), 3.0772450982906276), linewidth(1) + fuqqzz); draw(circle((-0.13,5.920273972602744), 9.316332106074661), linewidth(1) + fuqqzz); draw((-3.07,-2.92)--(-5.99,0), linewidth(1) + ccqqqq); draw((-5.99,0)--(-0.11,5.88), linewidth(1) + ccqqqq); draw((-0.11,5.88)--(-3.07,-2.92), linewidth(1) + ccqqqq); draw((-3.07,-2.92)--(-0.2528896419858224,-0.7683028249129576), linewidth(1) + zzttff); draw((-0.2528896419858224,-0.7683028249129576)--(4.079980011956577,-6.441114093790823), linewidth(1) + zzttff); draw((4.079980011956577,-6.441114093790823)--(-3.07,-2.92), linewidth(1) + zzttff); draw((-8.970273972602746,8.860273972602748)--(-5.99,0), linewidth(1) + ccqqqq); draw((-8.970273972602746,8.860273972602748)--(-0.11,5.88), linewidth(1) + ccqqqq); draw((-8.970273972602746,8.860273972602748)--(7.625211325567886,0.7578383840010742), linewidth(1) + fuqqzz); draw((7.625211325567886,0.7578383840010742)--(4.079980011956577,-6.441114093790823), linewidth(1) + zzttff); draw((7.625211325567886,0.7578383840010742)--(-0.2528896419858224,-0.7683028249129576), linewidth(1) + zzttff); draw((-5.0188988750116215,0.9711011249883785)--(0.46269787423381503,-1.7051860570650057), linewidth(1) + fuqqzz); /* dots and labels */ dot((-3.07,2.92),dotstyle); label("$P$", (-3.39,3.12), NE * labelscalefactor); dot((-3.07,-2.92),linewidth(4pt) + dotstyle); label("$R$", (-3.77,-3.3), NE * labelscalefactor*1/9); dot((-5.99,0),dotstyle); label("$A_{1}$", (-6.63,0.02), NE * labelscalefactor*1/12); dot((-0.11,5.88),linewidth(4pt) + dotstyle); label("$B_{1}$", (-0.37,6.3), NE * labelscalefactor); dot((-0.2528896419858224,-0.7683028249129576),dotstyle); label("$A_{2}$", (-0.17,-0.56), NE * labelscalefactor); dot((4.079980011956577,-6.441114093790823),linewidth(4pt) + dotstyle); label("$B_{2}$", (4.15,-6.28), NE * labelscalefactor); dot((-8.970273972602746,8.860273972602748),linewidth(4pt) + dotstyle); label("$C_{1}$", (-8.89,9.02), NE * labelscalefactor*3); dot((7.625211325567886,0.7578383840010742),linewidth(4pt) + dotstyle); label("$C_{2}$", (7.23,1.2), NE * labelscalefactor); dot((-5.0188988750116215,0.9711011249883785),linewidth(4pt) + dotstyle); label("$D_{1}$", (-5.59,1), NE * labelscalefactor*1/1.9); dot((0.46269787423381503,-1.7051860570650057),linewidth(4pt) + dotstyle); label("$D_{2}$", (0.79,-1.9), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] First, it's well-known that there is a spiral similarity sending $A_1A_2$ to $B_1B_2$ centered at $R.$ Since spiral similarities come in pairs, this means that $A_1B_1$ is also sent to $A_2B_2$ by another spiral similarity $\phi$ also centered at $R.$ As a result, we observe that we must also have $\phi(C_1)=C_2$ (simply observe that $\angle B_1A_1C_1=\angle C_1B_1A_1=\angle B_1RA_1=\angle B_2RA_2=\angle C_2B_2A_2=\angle B_2A_2C_2\implies \triangle C_1A_1B_1\stackrel{+}{\sim}\triangle C_2A_2B_2$) and $\phi(D_1)=D_2,$ which means that $\frac{RD_1}{RC_1}=\frac{RD_2}{RC_2}\implies \triangle RD_1D_2\stackrel{+}{\sim}\triangle RC_1C_2.$ However, since $R,D_1,C_1$ and $R,D_2,C_2$ are given to be collinear, we must have $D_1D_2||C_1C_2$ and we're done by homothety.

Here's my solution

Attachments:
IFS 2020 P8 SOL.pdf (29kb)

My solution is attached.

Attachments:
P8.pdf (77kb)
P8_diagram.pdf (3kb)

Let $X=\overline{B_1C_1}\cap\overline{B_2C_2}$ and $Y=\overline{A_1C_1}\cap\overline{A_2C_2}.$ Claim: $X$ lies on $(PRA_1).$ Proof. Since $R$ is the center of the spiral similarity $\overline{A_2B_2}\mapsto\overline{A_1B_1},$ $$\measuredangle XB_2P=\measuredangle C_2B_2A_2=\measuredangle B_2RA_2=\measuredangle B_1RA_1=\measuredangle XB_1A_1.$$$\blacksquare$ Similarly, $Y$ lies on $(PRB_1).$ Claim: $X$ and $Y$ lie on $(C_1C_2R).$ Proof. First, $$\measuredangle C_2XC_1=\measuredangle B_2XB_1=\measuredangle B_2PB_1=\measuredangle A_2PA_1=\measuredangle A_2YA_1=\measuredangle C_2YC_1$$so $XYC_2C_1$ is cyclic. Then, $$\measuredangle C_2YR=\measuredangle A_2YR=\measuredangle A_2PR=\measuredangle B_2XR$$so $XRYC_2$ is cyclic. $\blacksquare$ Claim: $R$ lies on $(PD_1D_2).$ Proof. Notice $$\measuredangle D_2RD=\measuredangle C_2RC_1=\measuredangle C_2XC_1=\measuredangle B_2XB_1=\measuredangle B_2PB_1=\measuredangle D_2PD_1.$$$\blacksquare$ Claim: $\measuredangle D_1RC_2=\measuredangle A_1PR+\measuredangle RXC_2.$ $(*)$ Proof. Note $$\measuredangle D_1RC_2=\measuredangle D_1RD_2=\measuredangle D_1PD_2=\measuredangle A_1PR+\measuredangle RPB_2=\measuredangle A_1PR+\measuredangle RXC_2.$$$\blacksquare$ Consider the line $\ell$ tangent to $(PD_1RD_2)$ at $R.$ By $(*),$ $$\measuredangle (\ell,\overline{RC_2})=\measuredangle D_1RC_2-\measuredangle (\overline{D_1R},\ell)=\measuredangle D_1RC_2-\measuredangle A_1PR=\measuredangle RXC_2$$and $\ell$ is tangent to $(C_2YRX)$ at $R.$ $\square$