Merely angle chasing though still a nice problem. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(14cm); 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 = -14.714143005565347, xmax = 17.51683230788338, ymin = -5.707943268010838, ymax = 15.577057961714761; /* image dimensions */ pen rvwvcq = rgb(0.08235294117647059,0.396078431372549,0.7529411764705882); pen wvvxds = rgb(0.396078431372549,0.3411764705882353,0.8235294117647058); pen sexdts = rgb(0.1803921568627451,0.49019607843137253,0.19607843137254902); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); /* draw figures */ draw((0.10373395672146964,8.58670357555119)--(-7.42,-0.05), linewidth(0.4) + rvwvcq); draw((-7.42,-0.05)--(6.1,-0.01), linewidth(0.4) + rvwvcq); draw((6.1,-0.01)--(0.10373395672146964,8.58670357555119), linewidth(0.4) + rvwvcq); draw(circle((-0.6650027560220105,1.6609315354395524), 6.9683049936919215), linewidth(0.4)); draw((-6.079726060992161,9.17824096744086)--(-0.6650027560220108,1.6609315354395526), linewidth(0.4) + wvvxds); draw((-0.6650027560220108,1.6609315354395526)--(4.805869458563446,8.136876300351975), linewidth(0.4) + wvvxds); draw((4.805869458563446,8.136876300351975)--(-6.079726060992161,9.17824096744086), linewidth(0.4) + rvwvcq); draw((-0.7671706351990542,7.296109170430735)--(-0.6650027560220108,1.6609315354395526), linewidth(0.4) + wvvxds); draw(circle((-6.07972606099216,9.17824096744086), 6.211690130444907), linewidth(0.4) + sexdts); draw(circle((4.805869458563446,8.136876300351975), 4.723602741043731), linewidth(0.4) + sexdts); draw(circle((-0.7671706351990537,7.296109170430733), 5.636103729833257), linewidth(0.4) + linetype("4 4") + green); draw(circle((-4.24274754823292,1.596065683582963), 3.578332764979994), linewidth(0.4) + wrwrwr); draw((-0.6650027560220108,1.6609315354395526)--(0.10373395672146964,8.58670357555119), linewidth(0.4) + wrwrwr); draw((-0.7671706351990542,7.296109170430735)--(-6.079726060992161,9.17824096744086), linewidth(0.4) + wvvxds); draw((-0.6650027560220108,1.6609315354395526)--(-7.42,-0.05), linewidth(0.4) + wvvxds); draw((-6.079726060992161,9.17824096744086)--(-7.4775798542010055,3.1258778711388624), linewidth(0.4) + rvwvcq); draw((4.805869458563446,8.136876300351975)--(6.03498468878035,3.575988381750521), linewidth(0.4) + rvwvcq); draw((0.10373395672146964,8.58670357555119)--(-7.4775798542010055,3.1258778711388624), linewidth(0.4) + dbwrru); draw((-7.696052277076719,15.175955852283623)--(6.03498468878035,3.575988381750521), linewidth(0.4) + dbwrru); draw((-6.2195074810318935,5.868633015458621)--(4.7333272528854256,6.067212007679287), linewidth(0.4) + linetype("4 4") + wvvxds); draw((-7.696052277076719,15.175955852283623)--(-7.42,-0.05), linewidth(0.4) + dbwrru); draw((5.868850043640822,12.739320773775246)--(6.1,-0.01), linewidth(0.4) + dbwrru); draw((0.10373395672146964,8.58670357555119)--(5.868850043640822,12.739320773775246), linewidth(0.4) + dbwrru); /* dots and labels */ dot((0.10373395672146964,8.58670357555119),dotstyle); label("$A$", (0.18745077571744034,8.795995623041119), NE * labelscalefactor); dot((-7.42,-0.05),dotstyle); label("$B$", (-7.326133729170932,0.1522340617071246), NE * labelscalefactor); dot((6.1,-0.01),dotstyle); label("$C$", (6.19413253867834,0.19409247120511006), NE * labelscalefactor); dot((-7.4775798542010055,3.1258778711388624),dotstyle); label("$D$", (-7.38892134341791,3.3334731835540183), NE * labelscalefactor); dot((6.03498468878035,3.575988381750521),linewidth(4pt) + dotstyle); label("$E$", (6.1104157196823685,3.7520572785338726), NE * labelscalefactor); dot((5.868850043640822,12.739320773775246),linewidth(4pt) + dotstyle); label("$F$", (5.9429820816904275,12.898119753843693), NE * labelscalefactor); dot((-7.696052277076719,15.175955852283623),linewidth(4pt) + dotstyle); label("$G$", (-7.61914259565683,15.242190685730877), NE * labelscalefactor); dot((-6.079726060992161,9.17824096744086),linewidth(4pt) + dotstyle); label("$O_1$", (-5.9866646252354005,9.340154946514929), NE * labelscalefactor); dot((4.805869458563446,8.136876300351975),linewidth(4pt) + dotstyle); label("$O_2$", (4.896521844240794,8.314623913814286), NE * labelscalefactor); dot((-0.6650027560220108,1.6609315354395526),linewidth(4pt) + dotstyle); label("$O$", (-0.5869297999952888,1.8265704416265425), NE * labelscalefactor); dot((-0.7671706351990542,7.296109170430735),linewidth(4pt) + dotstyle); label("$P$", (-0.6915758237402522,7.4565265191055845), NE * labelscalefactor); dot((4.7333272528854256,6.067212007679287),linewidth(4pt) + dotstyle); label("$H$", (4.812805025244823,6.242632643664007), NE * labelscalefactor); dot((-6.2195074810318935,5.868633015458621),linewidth(4pt) + dotstyle); label("$I$", (-6.133169058478349,6.03334059617408), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof: Let $\odot(O_1O_2O) \cap \odot(ABC)=\{I,H\}$. Notice that since $IH$ is the radical axis of $\odot(O_1O_2O)$ and $\odot(ABC) \implies IH \perp OP$ . So we just need to show $IH \perp GD$. Similiarly $IH \perp EF$. Invert about the $\odot(ABC)$. Notice that under this inversion $\odot(O_1O_2O) \mapsto IH$ and $\ell_b \mapsto \odot(ODB)$. So we just need to show that $\odot(O_1O_2O)$ and $\odot(ODB)$ are orthagonal $\iff PO$ tangent to $\odot(ODB)$. Now a quick angle chase follows. Its quite easily observable that $AB \cap OO_1 \in \odot(ODB)$ since $OO_1$ is the perpendicular bisector of $AD$. Let this point be $J$. Now note that $$\angle POJ\equiv \angle POO_1=90^\circ-\angle O_1O_2O=90^\circ-\frac{\angle AO_2E}{2}=90^\circ-\angle EFA=90^\circ-\angle ADG=90^\circ-\angle ACB=\angle OBA\equiv \angle OBJ$$. Done $\blacksquare$.

\[ \angle FAO_2 = 90^\circ- \angle AEF = 90^\circ-\angle AGD = \angle DAO_1 \implies O_1, \ A, \ O_2 \text{ collinear}. \] Let $OO_1$ meet $\ell_B$ at $X$. Note that \[ \angle DXO = 90^\circ - \angle XDA = 90^\circ - \angle AFE = 90^\circ - \angle AO_2O = \angle O_1OP. \] Done!

EMC 2019 Junior P3 wrote: Let $ABC$ be a triangle with circumcircle ω. Let $l_B$ and $l_C$ be two lines through the points $B$ and $C$, respectively, such that $l_B || l_C$. The second intersections of $l_B$ and $l_C$ with ω are $D$ and $E$, respectively. Assume that $D$ and $E$ are on the same side of $BC$ as $A$. Let $DA$ intersect $l_C$ at $F$ and let $EA$ intersect $l_B$ at $G$. If $O$, $O_1$ and $O_2$ are circumcenters of the triangles $ABC$, $ADG$ and $AEF$, respectively, and $P$ is the circumcenter of the triangle $OO_1O_2$, prove that $l_B || OP || l_C$. Proposed by Stefan Lozanovski, Macedonia Proof. By angle chasing we get that: $$\triangle ABC\cup\{O\}\sim\triangle AGD\cup\{O_1\}\sim\triangle AEF\cup\{O_2\}\sim\triangle OO_1O_2\cup\{P\}$$Let $X=l_B\cap OO_1$, $$\angle DXO=90^{\circ}-\angle GDA=90^{\circ}-\angle O_1O_2O=\angle O_1OP.\blacksquare$$[asy][asy] 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 = -14.63, xmax = 16.09, ymin = -9.19, ymax = 5.77; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0); pen qqwuqq = rgb(0,0.39215686274509803,0); pen qqqqcc = rgb(0,0,0.8); draw((-3.21,0.27)--(-6.81,-6.21)--(2.63,-5.33)--cycle, linewidth(1.2) + zzttqq); /* draw figures */ draw((-3.21,0.27)--(-6.81,-6.21), linewidth(1.2) + zzttqq); draw((-6.81,-6.21)--(2.63,-5.33), linewidth(1.2) + zzttqq); draw((2.63,-5.33)--(-3.21,0.27), linewidth(1.2) + zzttqq); draw(circle((-2.20578947368421,-4.527894736842105), 4.901860125195394), linewidth(1.2) + qqwuqq); draw(circle((-6.188226063482703,-0.14064761694228195), 3.006403490968577), linewidth(1.2) + qqwuqq); draw(circle((1.3189612112660292,0.8944680856951436), 4.571810368246262), linewidth(1.2) + qqwuqq); draw((-6.188226063482703,-0.14064761694228195)--(1.3189612112660292,0.8944680856951436), linewidth(1.2) + qqqqcc); draw((1.3189612112660292,0.8944680856951436)--(-2.20578947368421,-4.527894736842105), linewidth(1.2) + qqqqcc); draw((-2.20578947368421,-4.527894736842105)--(-6.188226063482703,-0.14064761694228195), linewidth(1.2) + qqqqcc); draw(circle((-2.2984329915507815,-0.6108775649326091), 3.918112600020137), linewidth(1.2) + qqwuqq); draw((-7.021867365798136,2.7478648254027176)--(2.5866719363144957,-3.4980664401567716), linewidth(1.2)); draw((2.377590113586171,5.342024143843986)--(-6.884376028424571,-3.065341990175062), linewidth(1.2)); draw((-7.021867365798136,2.7478648254027176)--(-6.81,-6.21), linewidth(1.2) + red); draw((2.63,-5.33)--(2.377590113586171,5.342024143843986), linewidth(1.2) + red); draw((-6.884376028424571,-3.065341990175062)--(-2.20578947368421,-4.527894736842105), linewidth(1.2)); draw((-6.974024096374227,0.7250259672793623)--(-6.188226063482703,-0.14064761694228195), linewidth(1.2)); draw((-2.2984329915507815,-0.6108775649326091)--(-2.20578947368421,-4.527894736842105), linewidth(1.2) + red); /* dots and labels */ dot((-3.21,0.27),dotstyle); label("$A$", (-3.21,0.63), NE * labelscalefactor); dot((-6.81,-6.21),dotstyle); label("$B$", (-7.23,-6.51), NE * labelscalefactor); dot((2.63,-5.33),dotstyle); label("$C$", (2.73,-5.85), NE * labelscalefactor); dot((-6.884376028424571,-3.065341990175062),dotstyle); label("$D$", (-6.81,-2.87), NE * labelscalefactor); dot((2.5866719363144957,-3.4980664401567716),linewidth(4pt) + dotstyle); label("$E$", (2.67,-3.33), NE * labelscalefactor); dot((2.377590113586171,5.342024143843986),linewidth(4pt) + dotstyle); label("$F$", (2.45,5.45), NE * labelscalefactor); dot((-7.021867365798136,2.7478648254027176),linewidth(4pt) + dotstyle); label("$G$", (-7.15,2.95), NE * labelscalefactor); dot((-6.188226063482703,-0.14064761694228195),linewidth(4pt) + dotstyle); label("$O_1$", (-6.71,-0.33), NE * labelscalefactor); dot((1.3189612112660292,0.8944680856951436),linewidth(4pt) + dotstyle); label("$O_2$", (1.39,1.05), NE * labelscalefactor); dot((-2.20578947368421,-4.527894736842105),linewidth(4pt) + dotstyle); label("$O$", (-2.37,-5.05), NE * labelscalefactor); dot((-2.2984329915507815,-0.6108775649326091),linewidth(4pt) + dotstyle); label("$P$", (-2.69,-0.81), NE * labelscalefactor); dot((-6.974024096374227,0.7250259672793623),linewidth(4pt) + dotstyle); label("$X$", (-6.89,0.89), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

This problem is also EMC 2019 Junior division P3.

∠O1AD = 90 - ∠DGA = 90 - ∠AEF = ∠O2AF so O1,A,O2 are collinear. Let OO1 meet BG at S. ∠POO1 = 90 - ∠O1O2O = ∠AFE = ∠ADG = ∠DSO1 so AP || BG.

Since $DG||EF \implies \triangle ADG \sim \triangle AEF$ , Hence homotety centered at $A$ sends $ADG$ to $AEF\implies O_1,A,O_2$ are collinear We have $ADBC , ABCE$ are cyclic quadrilaterals also $O_1AOD$ and $O_2AOE$ are kites. In order to these facts angle chasing gives us $\triangle ADG \sim \triangle AEF \sim \triangle ABC \sim \triangle O_1O_2O $ Let $X=DG \cap OO_1$ More angle chasing gives us $\angle DXO = \angle XOP \implies l_B || OP$ so we are done

Ibrahim_K wrote: Since $DG||EF \implies \triangle ADG \sim \triangle AEF$ , Hence homotety centered at $A$ sends $ADG$ to $AEF\implies O_1,A,O_2$ are collinear We have $ADBC , ABCE$ are cyclic quadrilaterals also $O_1AOD$ and $O_2AOE$ are kites. In order to these facts angle chasing gives us $\triangle ADG \sim \triangle AEF \sim \triangle ABC \sim \triangle O_1O_2O $ Let $X=DG \cap OO_1$ More angle chasing gives us $\angle DXO = \angle XOP \implies l_B || OP$ so we are done nice solution!

Here is a complex bash. Engoy Big thanks to hukilau17 for fixing a mistake [asy][asy] import graph; size(9.6cm); 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 = -11., xmax = 13., ymin = 2., ymax = 17.; /* image dimensions */ pen xdxdff = rgb(0.49019607843137253,0.49019607843137253,1.); pen qqffff = rgb(0.,1.,1.); draw((5.180588010544329,16.270199824261166)--(0.78,4.2)--(-10.18365810131105,4.2)--cycle, linewidth(0.4) + qqffff); draw((5.180588010544329,16.270199824261166)--(12.29222650614523,11.342257110655025)--(10.42,4.2)--cycle, linewidth(0.4) + qqffff); draw((1.1458590202529157,14.949708805663773)--(-4.701829050655524,13.0358703308184)--(5.6,9.28)--cycle, linewidth(0.4) + blue); /* draw figures */ draw(circle((5.6,9.28), 7.002770880158796), linewidth(0.4) + linetype("4 4") + xdxdff); draw((xmin, 0.*xmin + 11.342257110655025)--(xmax, 0.*xmax + 11.342257110655025), linewidth(0.4) + linetype("4 4")); /* line */ draw((xmin, 0.*xmin + 4.2)--(xmax, 0.*xmax + 4.2), linewidth(0.4) + linetype("4 4")); /* line */ draw((5.180588010544329,16.270199824261166)--(0.78,4.2), linewidth(0.4) + qqffff); draw((0.78,4.2)--(-10.18365810131105,4.2), linewidth(0.4) + qqffff); draw((-10.18365810131105,4.2)--(5.180588010544329,16.270199824261166), linewidth(0.4) + qqffff); draw((5.180588010544329,16.270199824261166)--(12.29222650614523,11.342257110655025), linewidth(0.4) + qqffff); draw((12.29222650614523,11.342257110655025)--(10.42,4.2), linewidth(0.4) + qqffff); draw((10.42,4.2)--(5.180588010544329,16.270199824261166), linewidth(0.4) + qqffff); draw((1.1458590202529157,14.949708805663773)--(-4.701829050655524,13.0358703308184), linewidth(0.4) + blue); draw((-4.701829050655524,13.0358703308184)--(5.6,9.28), linewidth(0.4) + blue); draw((5.6,9.28)--(1.1458590202529157,14.949708805663773), linewidth(0.4) + blue); draw((-0.23557750471511035,9.28)--(5.6,9.28), linewidth(0.4)); /* dots and labels */ dot((5.6,9.28),linewidth(1.pt) + dotstyle); label("$O$", (5.708413069207604,9.3229312870982), NE * labelscalefactor); dot((10.42,4.2),linewidth(1.pt) + dotstyle); label("$B$", (10.525148420448259,4.247231669661814), NE * labelscalefactor); dot((5.180588010544329,16.270199824261166),linewidth(1.pt) + dotstyle); label("$A$", (5.2940702432944295,16.314966474383017), NE * labelscalefactor); dot((12.29222650614523,11.342257110655025),linewidth(1.pt) + dotstyle); label("$C$", (12.389691137057545,11.394645416664071), NE * labelscalefactor); dot((-1.092226506145229,11.342257110655025),linewidth(1.pt) + dotstyle); label("$E$", (-0.9987614252619101,11.394645416664071), NE * labelscalefactor); dot((0.78,4.2),linewidth(1.pt) + dotstyle); label("$D$", (0.8916777179669491,4.247231669661814), NE * labelscalefactor); dot((-10.18365810131105,4.2),linewidth(1.pt) + dotstyle); label("$G$", (-10.088407168732179,4.247231669661814), NE * labelscalefactor); dot((3.383944546651063,11.342257110655027),linewidth(1.pt) + dotstyle); label("$F$", (3.4813203799242904,11.394645416664071), NE * labelscalefactor); dot((1.1458590202529157,14.949708805663773),linewidth(1.pt) + dotstyle); label("$O1$", (1.2542276906409768,14.994248716784773), NE * labelscalefactor); dot((-4.701829050655524,13.0358703308184),linewidth(1.pt) + dotstyle); label("$O2$", (-4.5983647253826145,13.077913146936343), NE * labelscalefactor); dot((-0.23557750471511035,9.28),linewidth(1.pt) + dotstyle); label("$P$", (-0.1441793468159875,9.3229312870982), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Angle chasing gives us $\triangle ABC \sim \triangle AGD \sim \triangle AEF$ Let $(ABC)$ be the unit circle And $l_b\parallel l_c \parallel \mathbb{R}-axis$ From the similarities we obtain: $\frac{o_1-a}{o-a} = \frac{-\frac{1}{c}-a}{b-a} \Longleftrightarrow o_1 =\frac{a(bc+1)}{c(b-a)} \Longrightarrow \bar{o_1} = \frac{bc+1}{(a-b)}$ $\frac{o_2-a}{o-a} = \frac{-\frac{1}{b}-a}{c-a} \Longleftrightarrow o_2 =\frac{a(bc+1)}{b(c-a)} \Longrightarrow \bar{o_2} = \frac{bc+1}{(a-c)}$ We can now compute $p$: $p = \frac{o_1o_2(\bar{o_1}-\bar{o_2})}{\bar{o_1}o_2-o_1\bar{o_2}}=\frac{\frac{a(bc+1)}{c(b-a)}\frac{a(bc+1)}{b(c-a)}( \frac{bc+1}{(a-b)}-\frac{bc+1}{(a-c)})}{\frac{bc+1}{(a-b)}\frac{a(bc+1)}{b(c-a)}-\frac{a(bc+1)}{c(b-a)}\frac{bc+1}{(a-c)}}=\frac{a(bc+1)}{(a-b)(a-c)}$ But notice that: $\overline{\frac{a(bc+1)}{(a-b)(a-c)}} = \frac{\frac{1}{a}(\frac{1}{bc}+1)}{(\frac{1}{a}-\frac{1}{b})(\frac{1}{a}-\frac{1}{c})}\overset{\times a^2bc}{=}\frac{a(1+bc)}{(b-a)(c-a)}$ From which we conclude that $p \in \mathbb{R}$ which is equivalent to $l_b \parallel l_c \parallel \overline{OP}$