I will show that you can solve this problem with almost only bash, but I will not write full computations. If $AB=AC$ then $A-I-L$ are collinear, which means we don't need deal with this case. Assume $AB<AC$. Let $O'$ be center of $(AEF)$, $N'=O'N\cap (AEF)$, $H=BE\cap (AEF), J=CF\cap (AEF)$. And let $G$ be the intersection point of line $BC$ with line that touchs to $(AEF)$ at point $A$. I will show that $G$ is center of $(AIL)$. Claim 1: It's enough to prove that $GA=GI$. Proof: Assume we get that $GA=GI$. Let $\angle NLA=\alpha$. Since $O'A\perp GA$ we get $\angle GAI=90-\angle O'AN'=90-\angle O'N'A=90-\angle NLA=90-\alpha \implies \angle GAI=\angle GIA=90-\alpha \implies \angle AGI=2\alpha$. Since $GA=GI$ and $\angle AGI=2\angle ALI$ we get $L\in (G,GI) \implies GA=GI=GL$ as desired. Claim 2: $G-H-J$ are collinear. Proof: Just apply Pascal to $AAFJHE$. Now I will show that we can compute $GA$ and $GI$ only using values of $AB,AC,BC$. Obviously if we can find the value of $GB$ only using values of $AB,AC,BC$, then you can compute $AG$ and $GI$ with cosine law in $\triangle AGC$ and $\triangle IGC$ only using values of $AB,AC,BC$. From Menelaus' Theorem we get $\frac{GB}{GC}\cdot\frac{CJ}{JI}\cdot \frac{IH}{HB}=1 \implies 1+\frac{BC}{GB}=\frac{GC}{GB}=\frac{CJ}{JI}\cdot \frac{IH}{HB}=\frac{CE\cdot CA\cdot FI}{CF\cdot AI\cdot IN'}\cdot \frac{AI\cdot IN'\cdot BE}{IE\cdot BF\cdot BA}=\frac{CE\cdot CA\cdot FI\cdot BE}{CF\cdot IE\cdot BF\cdot BA}$. Now obviously we can calculate all of $CE, FI, BE, CF, IE, BE$ as a value of $AB,BC,CA$. So $BG$ can be expressed as a values of $AB,BC,CA$ and now finally using $\mathbb{BORING}$ calculations we are done!

$\textbf{Lemma:}$ Given $\triangle ABC$, $I$ is an arbitrary point lying on the $\angle A$ interior angle bisector. $AD,BE,CF$ cut $BC,CA,AB$ at $D,E,F$ respectively. $K$ is the intersection of $BC$ and the tangent of $(AEF)$ at $A$. Then, $K$ lies on the perpendicular bisector of $AI$ and $KI \parallel EF$. [asy][asy] /* 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 = -629.8980273228005, xmax = 274.8320470180955, ymin = -140.13279929450852, ymax = 291.7002962181313; /* image dimensions */ /* draw figures */ draw((-107.32699609661678,64.52429658494809)--(-96.76449206870984,-43.18897941331275), linewidth(0.8)); draw((-96.76449206870984,-43.18897941331275)--(79.44744091958583,-40.12237380001673), linewidth(0.8)); draw((79.44744091958583,-40.12237380001673)--(-107.32699609661678,64.52429658494809), linewidth(0.8)); draw((-199.48911099925027,-44.97668974424143)--(-62.2065897932942,-4.506567410255455), linewidth(0.8)); draw((-101.52720396504891,5.379740620797153)--(-36.15887563651245,24.649959589570397), linewidth(0.8)); draw((-107.32699609661678,64.52429658494809)--(-199.48911099925027,-44.97668974424143), linewidth(0.8)); draw((-199.48911099925027,-44.97668974424143)--(-96.76449206870984,-43.18897941331275), linewidth(0.8)); draw((-107.32699609661678,64.52429658494809)--(-62.2065897932942,-4.506567410255455), linewidth(0.8)); draw((-36.15887563651245,24.649959589570397)--(-96.76449206870984,-43.18897941331275), linewidth(0.8)); draw((-101.52720396504891,5.379740620797153)--(79.44744091958583,-40.12237380001673), linewidth(0.8)); draw((-276.9168868359384,-46.32416064464626)--(-107.32699609661678,64.52429658494809), linewidth(0.8)); draw((-276.9168868359384,-46.32416064464626)--(-62.2065897932942,-4.506567410255455), linewidth(0.8)); draw((-276.9168868359384,-46.32416064464626)--(-101.52720396504891,5.379740620797153), linewidth(0.8)); draw((-276.9168868359384,-46.32416064464626)--(-199.48911099925027,-44.97668974424143), linewidth(0.8)); draw(circle((-75.5550374158921,37.783250638459556), 41.5275919928171), linewidth(0.8) + linetype("2 2")); draw((-62.2065897932942,-4.506567410255455)--(-37.59578328152455,-42.15926995254478), linewidth(0.8)); /* dots and labels */ dot((-107.32699609661678,64.52429658494809),linewidth(3pt) + dotstyle); label("$A$", (-117.92412727484104,74.45910706453336), NE * labelscalefactor); dot((-96.76449206870984,-43.18897941331275),linewidth(3pt) + dotstyle); label("$B$", (-97.39218561703154,-57.34271196463124), NE * labelscalefactor); dot((79.44744091958583,-40.12237380001673),linewidth(3pt) + dotstyle); label("$C$", (82.09672371414182,-36.14844960818266), NE * labelscalefactor); dot((-62.2065897932942,-4.506567410255455),linewidth(2pt) + dotstyle); label("$I$", (-62.95150928780271,-18.928111443568188), NE * labelscalefactor); dot((-36.15887563651245,24.649959589570397),linewidth(3pt) + dotstyle); label("$E$", (-33.809398547685994,28.758978858441115), NE * labelscalefactor); dot((-101.52720396504891,5.379740620797153),linewidth(3pt) + dotstyle); label("$F$", (-109.31395819253383,-8.3309802653439), NE * labelscalefactor); dot((-199.48911099925027,-44.97668974424143),linewidth(3pt) + dotstyle); label("$K$", (-194.7533283169669,-54.03110847143615), NE * labelscalefactor); dot((-276.9168868359384,-46.32416064464626),linewidth(3pt) + dotstyle); label("$X$", (-287.47822612642915,-54.03110847143615), NE * labelscalefactor); dot((-175.44697162415653,-16.411395015296577),linewidth(3pt) + dotstyle); label("$L$", (-180.18227294690857,-8.3309802653439), NE * labelscalefactor); dot((-73.977056412375,13.501370614053018),linewidth(3pt) + dotstyle); label("$G$", (-71.56167837010992,17.499526981577805), NE * labelscalefactor); dot((-37.59578328152455,-42.15926995254478),linewidth(3pt) + dotstyle); label("$D$", (-35.13403994496403,-38.13541170409972), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Proof Let $G=AD \cap EF$, $X=EF \cap BC$, $L$ is the midpoint of $XG$. Note that $\angle XAG=90^0$ and $(XG,FE)=-1 \Rightarrow $ $LA^2=LG^2=\overline{LF} \cdot \overline{LE}$. $\Rightarrow$ $LA$ is tangent to $(AEF)$ $\Rightarrow $ $\overline{A,L,K}$. We have $-1=K(DG,AI)=K(XG,LI) \Rightarrow KI \parallel EF.$ $\Rightarrow \angle KIG =\angle LGA= \angle LAG$ $\Rightarrow KA=KI$ $\blacksquare$. Back to the main problem [asy][asy] /* 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 = -431.7422061583515, xmax = 749.5390132082467, ymin = -455.2172484353356, ymax = 108.61536871475244; /* image dimensions */ /* draw figures */ draw((-62.13744620510897,-62.155189705137104)--(-23.622005760752018,-277.9676023525891), linewidth(0.8)); draw((-23.622005760752018,-277.9676023525891)--(233.73480619052387,-300.8970851452036), linewidth(0.8)); draw((233.73480619052387,-300.8970851452036)--(-62.13744620510897,-62.155189705137104), linewidth(0.8)); draw(circle((6.3512380236876185,-116.22313080683959), 87.25847879930626), linewidth(0.8)); draw((-62.13744620510897,-62.155189705137104)--(30.549081517303698,-218.82174473579252), linewidth(0.8)); draw((-8.062008112908734,-30.16326525863605)--(30.549081517303698,-218.82174473579252), linewidth(0.8)); draw((-218.7822810663895,-260.5795870889024)--(-23.622005760752018,-277.9676023525891), linewidth(0.8)); draw((-218.7822810663895,-260.5795870889024)--(26.906954732918543,-201.0258703435953), linewidth(0.8)); draw((-62.13744620510897,-62.155189705137104)--(-218.7822810663895,-260.5795870889024), linewidth(0.8)); draw((73.67075932601836,-171.7400126896869)--(-23.622005760752018,-277.9676023525891), linewidth(0.8)); draw((-39.20630173318949,-190.64458398849644)--(233.73480619052387,-300.8970851452036), linewidth(0.8)); draw((-62.13744620510897,-62.155189705137104)--(-8.062008112908734,-30.16326525863605), linewidth(0.8)); draw((-218.7822810663895,-260.5795870889024)--(30.549081517303698,-218.82174473579252), linewidth(0.8)); draw((-8.062008112908734,-30.16326525863605)--(20.764484160284,-202.28299635504317), linewidth(0.8) + linetype("2 2")); draw((-39.20630173318949,-190.64458398849644)--(73.67075932601836,-171.7400126896869), linewidth(0.8)); /* dots and labels */ dot((-62.13744620510897,-62.155189705137104),linewidth(3pt) + dotstyle); label("$A$", (-59.02463255146732,-56.55645624792673), NE * labelscalefactor); dot((-23.622005760752018,-277.9676023525891),linewidth(3pt) + dotstyle); label("$B$", (-30.48709211288918,-295.23406718876157), NE * labelscalefactor); dot((233.73480619052387,-300.8970851452036),linewidth(3pt) + dotstyle); label("$C$", (237.5928332192085,-296.09884114144575), NE * labelscalefactor); dot((30.549081517303698,-218.82174473579252),linewidth(3pt) + dotstyle); label("$I$", (34.37095433842478,-235.56466445355284), NE * labelscalefactor); dot((73.67075932601836,-171.7400126896869),linewidth(3pt) + dotstyle); label("$E$", (85.39261754679175,-173.30093986029158), NE * labelscalefactor); dot((-39.20630173318949,-190.64458398849644),linewidth(3pt) + dotstyle); label("$F$", (-54.70076278804639,-203.56802820423803), NE * labelscalefactor); dot((-8.062008112908734,-30.16326525863605),linewidth(3pt) + dotstyle); label("$N$", (-4.5438735323635955,-24.559819998611918), NE * labelscalefactor); dot((26.906954732918543,-201.0258703435953),linewidth(3pt) + dotstyle); label("$L$", (30.911858527688036,-198.37938448813293), NE * labelscalefactor); dot((-218.7822810663895,-260.5795870889024),linewidth(3pt) + dotstyle); label("$J$", (-233.70897099367292,-258.91356117602584), NE * labelscalefactor); dot((6.3512380236876185,-116.22313080683959),linewidth(3pt) + dotstyle); label("$K$", (15.345927379372684,-113.63153712508289), NE * labelscalefactor); dot((20.764484160284,-202.28299635504317),linewidth(3pt) + dotstyle); label("$Y$", (4.968639947162451,-203.56802820423803), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $K$ be the circumcenter of $(AEF)$, and $Y$ be the midpoint of the minor arc $EF$. Using inversion with center $I$, power $k=P_{I / (AEF)}$: $(AEF) \leftrightarrow (AEF)$ , $A \leftrightarrow Y$, $L\leftrightarrow N$, $(ALI) \leftrightarrow NY$. Note that $M$ is the midpoint of the major arc of $(AEF)$ so $MN$ passes through $K$. $\Rightarrow (ALI)$ and $(AEF)$ are two orthogonal circles. Let $J$ be the circumcenter of $(AIL)$ then $L$ is the intersection of the tangent line wrt $A$ of $(AEF)$ and the perpendicular bisector of $AI$. Applying the above lemma, we get that $J$ lies on $BC$. $\blacksquare$

Phorphyrion wrote: In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$. Projective geo is love, projective geo is life...(but...(i dont wanna go in details abt my life rn lol)) Re-define the point $T$ as $M$, now let $BI \cap (AEF)=L \ne E$ and let $CI \cap (AEF)=A_1$ and let $BI \cap (ABC)=P$ and let $CI \cap (ABC)=Q$ and FINALLY let $PQ \cap BC=G$, now we need these 2 claims: Claim 1: $G,L,A_1$ are colinear. Proof: Instahead of considering $\triangle ABC$ we will concider a fixed circle $\omega$ with a fixed line $\ell$ intersecting it at $B,C$, and a moving point $A$ on $\omega$ (the rest of the labelling will be the same), now consider the maps $f: A \to G$ and $g: A \to G_1$ where $G_1$ is the intersection of $LA_1$ with $BC$, both maps are projective becuase by $\sqrt{bc}$ inversion there is a projective map sending $A \to \infty_{BC}$ and the rest its just maps on $\ell$ (i.e. $A \to \infty_{BC} \to B\infty_{BC} \to BG \to G$ and $A \to \infty_{BC} \to C\infty_{BC} \to CG \to G$) so now we need to show that $G \equiv G_1$ for 3 cases and then we get it for every $A$ on $\omega$. The cases $A \equiv B$ and $A \equiv C$ are basicaly the same thing $I$ goes with $A$ to either $B$ or $C$ and either $\angle ABC$ or $\angle ACB$ go to $0$ so either $F$ or $E$ also go with $A, I$ to either $B$ or $C$ and also becuase $I$ is going into either $B$ or $C$ we have that either $E$ or $F$ go to either $B$ or $C$ means that one of the $P,Q$ go with $A$ to $B$ or $C$ while the other is being midpoint of any of the arcs $BC$ in $\omega$ meaning that $G$ also goes to $A$ and $G_1$ also goes to $A$ becuase $L,A_1$ also go to $A$ since $E,F$ also go with $A$ to $B$ or $C$. Now the last case i'm taking is $A$ being the midpoint of any of the arcs $BC$ (i.e. $\triangle ABC$ isosceles) but in this case its just an easy symetry argument to see that both $G,G_1$ go to $\infty_{BC}$. Since it holds for 3 cases we have it for every $A$ in $\omega$. Claim 2: $AG$ is tangent to $(AEF)$ Proof: Let the tangent from $A$ to $(AEF)$ hit $BC$ at $G'$ then by pascal on the circle formed by the points $A,LM,A_1,E,F$ gives that $G',B,C$ are colinear hence $G=G'$ Finishing: Now noticed that $P,Q$ by I-E lemma is the perpendicular bisector of $AI$ so $GA=GI$ but also projecting we get $$-1=(N, AI \cap (AEF); E,F) \overset{I}{=} (T, A; L, A_1) \implies GT \; \text{tangent to} (EALMA_1F)$$Hence $GI=GA=GT$ which means that $G$ is the center of $(AIT)$ and since it lies on $BC$ we are done

chystudent1-_-'s solution is much better than mine, but take my solution as an excursion to this incredible configuration. Lemma. Let $I$ be the incenter of $\triangle ABC$. Let $E=BI\cap AC$ and $F=CI\cap AB$. Let $T=BC\cap EF$, let $S=(ABC)\cap (AEF)$. Let $M$ be the midpoint of $BC$, $W$ be the midpoint of $EF$, $P$ midpoint of arc $BC$ and $Q$ midpoint of arc $EF$. Then, tangent from $A$ to $(AEF)$, $BC$ and perpendicular from $D$ to $AI$ concur at a point. Proof. Claim. $SDMP$ is cyclic. Proof. By Newton-Gauss, $M,W,D$ are collinear. Since $S$ is the center of similarity of $\triangle AEF$ and $\triangle ABC$, we get that if $Z=WQ\cap MP$, $SWMZ$ and $SQPZ$ are cyclic. Hence, $S$ is the Miquel point of complete quad $PQMWDZ$, therefore $S$ lies on $(PMD)$, as desired. Let $R=(PMDS)\cap BC$. Since $\angle RMP=90^\circ$, we get that $RD\perp AI$. Also, $$\angle TAS=\angle P'AS=\angle P'PS=\angle MPS=\angle MRS=\angle TRS,$$we get that $ASTR$ is cyclic. Hence $\angle RAS=\angle RTS=\angle BTS=\angle BFS=\angle AFS$, which means that $AR$ is tangent to $(AEF)$, as desired. Lemma proven. [asy][asy]import olympiad; import geometry; size(12cm);defaultpen(fontsize(10pt)); pair O,A,B,C,I,F,E,D,M,P,S,T,Q,W,R; O=(0,0);A=dir(162);B=dir(238);C=dir(302);M=midpoint(B--C);I=incenter(A,B,C);E=extension(B,I,A,C);F=extension(C,I,A,B);D=midpoint(A--I);P=2*foot(O,A,I)-A;S=intersectionpoints(circumcircle(A,B,C),circumcircle(A,E,F))[1];R=extension(S,-P,B,C);T=extension(E,F,B,C);W=midpoint(E--F); Q=2*foot(circumcenter(A,E,F),A,I)-A; draw(A--B--C--cycle);draw(circumcircle(A,B,C),royalblue); draw(circumcircle(A,E,F),royalblue);draw(B--T--A);draw(circumcircle(S,P,M),heavygreen+dashed);draw(E--T^^A--P^^A--R^^B--E^^C--F);draw(M--D,dotted); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(M)); dot("$W$",W,dir(W)); dot("$I$",I,dir(I)); dot("$R$",R,dir(R)); dot("$Q$",Q,dir(Q)); dot("$P$",P,dir(P)); dot("$T$",T,dir(T)); dot("$S$",S,dir(S)); [/asy][/asy] Let us return to the original problem. Let $\omega$ be the circle centered at $R$ and with radius $RA=RI$. Let $L'=(AEF)\cap \omega$ and let $N=(AEF)\cap L'I$, I contend that $N\equiv N'$ and thus $L'\equiv L$. Indeed, $$\measuredangle QAN' =\measuredangle QN' A+\measuredangle AQN' =\measuredangle QAR+\measuredangle AL' N' =\measuredangle IAR+\measuredangle IL' A=90^\circ,$$thus $N\equiv N'$, as desired. $\blacksquare$

Pretty easy for its position. Reflect $I$ across $BC$ and call it $I'$. We want to prove that $AILI'$ is cyclic. Let $X,Y$ be the midpoints of the arcs $AC, AB$. $XY$ passes through the midpoint $M$ of $AI$. If $T=XY \cap BC$ then it is "well known" that $XY \perp AI, TA$ is tangent to $(AEF)$.We are done because $\angle ALI=\angle OAI=\angle ATM=\angle IDM =\angle II'A$ where $O$ is the circumcircle of $AEF$.

*I just realised that post 3 used the same lemma.

Solved with Om245. What an insane problem! Looks like we overcomplicated a bit but nevermind. First of all, define the following additional points, $M$ is the midpoint of arc $BAC$, $O$ is the intersection of lines $MX$ and $BC$, $D$ is the intersection of lines $LI$ and $BC$ and $Z$ is the intersection of lines $EF$ and $BC$. Notice that by the nature of the definition, we have that $A-N-M$. We first prove the following initial claims. Claim : Points $X$ , $O$ , $L$ and $D$ are concyclic. Proof : Simply note that, \begin{align*} \measuredangle XOD &= \measuredangle MOC \\ &= \measuredangle OMC + \measuredangle MCO\\ &= \measuredangle XMC + \measuredangle MCB\\ &= \measuredangle XAC + \measuredangle MAB\\ &= \measuredangle XAB + \measuredangle MAC\\ &= \measuredangle XAB + \measuredangle MBC\\ &= \measuredangle XAB + \measuredangle BCM\\ &= \measuredangle XAB + \measuredangle BAM\\ &= \measuredangle XAM\\ &= \measuredangle XAN \\ &= \measuredangle XLN \end{align*}which proves the claim. Claim : Points $A$ , $X$ , $O$ and $Z$ are concyclic. Proof : Since it is well known that $Z$ also lies on $AM$, \[\measuredangle AZO = \measuredangle AZC = \measuredangle MAC + \measuredangle ACZ = \measuredangle BCM + \measuredangle ACB = \measuredangle ACM = \measuredangle AXM\]from which the claim follows. Claim : Line $OA$ is tangent to $(AEF)$ at $A$. Proof : Note that, \[\measuredangle XAO = \measuredangle ZXO = \measuredangle XZC = \measuredangle XEA\]where the final equality is due to the fact that $X$ is clearly the Miquel Point of complete quadrilateral $BCEFAZ$. Now we can get into the real nitty gritty of the problem. Let $Q$ be the intersection of lines $NX$ and $EF$. Then, we can show the following properties of $Q$. Claim : Point $Q$ lies on $(AOX)$. Proof : To see why, simply note that \[\measuredangle XQF = \measuredangle FXN + \measuredangle EFX = \measuredangle FEN + \measuredangle EAX \]\[ =\measuredangle NFE + \measuredangle EAX = \measuredangle NAE + \measuredangle EAX = \measuredangle NAX = \measuredangle XAZ \]from which the claim is clear. Claim : Lines $QI$ and $BC$ are parallel. Proof : First note that, \[\measuredangle OAZ = \measuredangle AXN = \measuredangle AOQ \]from which it is clear that $OQ \parallel AN$. Now, using this we can observe that, \[\measuredangle QAI = \measuredangle QAX + \measuredangle XAI = \measuredangle QZX + \measuredangle XAI = \measuredangle EZX + \measuredangle XAI\]\[= \measuredangle ECX + \measuredangle XAC + \measuredangle CAI = \measuredangle AXC + \measuredangle CAI = \measuredangle ABC + \measuredangle CAI = \measuredangle BCA + \measuredangle IAB \]Let $G$ be the intersection of the line through $A$ parallel to $BC$ and $H$ the intersection of lines $AI$ and $BC$. Now since clearly, \[\measuredangle IAQ = \measuredangle BCA + \measuredangle IAB = \measuredangle GAI\]if $Y = AI \cap (AEF)$ , $R = AQ \cap (AEF)$ and $S = AG \cap (AEF)$, we get \[-1=(N,Y;R,S)\overset{A}{=}(Z,H;Q,G)\]Now since lines $AG$ and $BC$ parallel and it is well known that $(A,I;H,AI \cap BC)=-1$, it follows that $QI \parallel BC$ which proves the claim. As a result, it follows that $\measuredangle QIA =\measuredangle BCA + \measuredangle IAB = \measuredangle IAQ$ so $Q$ lies on the perpendicular bisector of $AI$. Then, \[\measuredangle IQO = \measuredangle ZOQ = \measuredangle OQA\]since $ZOQA$ is an isoecles trapezoid. Thus, $\triangle AOQ \cong \triangle OQI$ which implies that $OA=OI$. Now, let $P$ be the intersection of line $AW$ and $(AOX)$, we can then prove the following final properties. Claim : Point $P$ also lies on line $QI$. Proof : For this, simply note that, \[\measuredangle PQA = \measuredangle PAQ + \measuredangle QPA = \measuredangle NXA + \measuredangle WAQ \]Then, \begin{align*} \measuredangle WAQ &= \measuredangle WAX + \measuredangle XAQ\\ &= \measuredangle WAX + \measuredangle XCA\\ &= \measuredangle WAE + \measuredangle CAX + \measuredangle XCA\\ &= \measuredangle WAE + \measuredangle CBA \\ &= \measuredangle AWE + \measuredangle WEA + \measuredangle CBA \\ &= \measuredangle AFE + \measuredangle MXA + \measuredangle CBA \end{align*}and \begin{align*} \measuredangle NXA &= \measuredangle FXA + \measuredangle NXF\\ &= \measuredangle FEA + \measuredangle NEF \end{align*}Now, \[\measuredangle NXA + \measuredangle WAQ = \measuredangle AFE + \measuredangle MXA + \measuredangle CBA + \measuredangle FEA + \measuredangle NEF\]which are all angles known in terms of $\angle A$ , $\angle B$ and $\angle C$ so you can compute and check that this angle is $2\measuredangle BCA + \measuredangle CAB = 2\measuredangle IAQ $ which implies that $P$ indeed lies on line $IQ$ as desired. Now we are almost there. Claim : Points $P$ , $A$ ,$ L$ and $I$ are concyclic. Proof : This is quite easy to see since, \[\measuredangle PIL = \measuredangle ODL = \measuredangle WXL = \measuredangle WAL\]from which the claim is clear. We finally note that, \[\measuredangle APO = \measuredangle AZO = \measuredangle AXW = \measuredangle OAP\]which implies that $OP=OI$. But, this means that $O$ must be the circumcenter of $\triangle AIP$. Combining this with the previous observation that $AIPL$ is cyclic, this implies that $O$ is the center of $(AIL)$, which finishes the problem.

Wait, this has a pretty quick solution Phorphyrion wrote: In triangle $ABC$, the angle bisectors are $BE$ and $CF$ (where $E, F$ are on the sides of the triangle), and their intersection point is $I$. Point $N$ lies on the circumcircle of $AEF$, and the angle $\angle IAN$ is right. The circumcircle of $AEF$ meets the line $NI$ a second time at the point $L$. Show that the circumcenter of $AIL$ lies on line $BC$. Let $A'$ be the reflection of $A$ in $BC$. It suffices to show $A, A', I, L$ are concyclic or $\angle (AL, LI) = \angle AA'I$ or $\angle ALN = \angle AI_aO$ where $I_a$ is the $A$-excentre and $O$ the circumcentre in $\triangle ABC$, since $\triangle AA'I \sim \triangle AI_aO$. It is well-known that $OI_a \perp EF$ and since $AN \perp AI_a$, we see that $\angle AI_aO = \angle (AN, EF) = \tfrac{1}{2}(\angle AFE-\angle AEF) = \angle ALN$ proving the claim.