Hmm nice problem, i still haven't solved this one.

Let $TI\cap(ABC)$ be $V$. $V$ is the $A-$ mixtilinear touchpoint.(well known) By $Reim's$ it suffices to show that $AV//IJ$. We have $2$ main claims: 1.Let $R$ be the intersection of $EF$ and the $A-$ symmedian of $ABC$. Then,$ASRK$ is cyclic. Proof:

2.$R,I,M$ are collinear. Proof:

*Actually the same collinearity holds for any pair of isogonal lines through $A$. Back to the main proof: $MI$ is known to be parallel to the $A-$ Nagel line and so is $IR$ due to the Lemma. Now reflecting $(ASRK)$ over $AI$ we deduce that $IJ$ is antiparallel to the $A-$ Nagel line meaning that it is parallel to $AV$ meaning that there is nothing left to prove.. Edit:adding a figure for further use.Hope you enjoy the colours

@above: Wait. Reim's on which circles? I tried to identify the circles you are referring to but was unable to. Could you please clarify.

Plops wrote: @above: Wait. Reim's on which circles? I tried to identify the circles you are referring to but was unable to. Could you please clarify. Yes of course.To the hypothetical circle $TJIX$ and $(ABC)$. (I forgot to add $X$ in the diagram though)

Sorry but I have one more question for @above: can you clarify how you got $A(B.C,M,D')=A(C,B,R',D')$. Thanks.

Let line $AI$ meet $BC$ at $D$, and $\Gamma=\odot(ABC)$ at $N$. It is well-known that $AT, EF, BC$ concur at some point $Y$. Now we have the complete quadrilateral $AFCYBE$, and $S$ is its Miquel point. Thus $\measuredangle SYK=\measuredangle SBE=\measuredangle SNA,$ so $(SYNK)$ are cyclic. Hence if we let $Z=YN \cap \odot(ABC)$ and $U=AZ \cap EF$, we have $$\measuredangle AJK=\measuredangle AS'K=-\measuredangle ASK=-\measuredangle ASN-\measuredangle NSK=-\measuredangle AZN-\measuredangle NYK=-\measuredangle AUK.$$Furthermore, it turns out that $-1=A(YD,BC)=N(YD,BC)=(ZA,BC),$ so $AZ$ and $AJ$ are isogonal WRT $\angle BAC$. Hence $J$ and $U$ are symmetrical WRT $AI$. Now, by Reim's, it suffices to prove that $IJ$ is parallel to $AV$, where $V$ is the $A$-mixtillinear incircle touchpoint. If $P$ is the touchpoint of the $A$-excircle with line $BC$, then this is equivalent with showing $AP \parallel UI$. It is well-known that $AP \parallel MI$, so it remains to prove that $M, I, U$ are collinear. But we have $$(FE,KU)=(AC,AB;AD,AZ)=(AB,AC;AD,AM)=A(BC, DM)=I(BC,DM),$$so $U, I, M$ are collinear. We're done.

Let $R,P$ be the intersections of $A$-symmedian and $EF$ , $(AEF)$ $SR$ bisects $FSE$ and $SRKA$ is cyclic which then gives that $R$ ,$J$ are symmetrical wrt $AI$. Since $AP$ is symmedian we have that $\frac{FP}{EP}=\frac{sin(FAP)}{sin(EAP)}=\frac{AB}{AC}$ then get that: $\frac{RF}{RE}=\frac{AF\cdot FP}{AE\cdot EP}=\frac{AF}{AE}\cdot \frac{AB}{AC} $ ,$BE , CF$ are bisectors so $\frac{AF}{AC}=\frac{FB}{BC}$ and $\frac{AB}{AE}=\frac{BC}{CE}$ , and the above finally gives that $\frac{FR}{RE}=\frac{FB}{CE}$ However $S$ is spiral similarity center of $BF$ , $CE$ and we also have that $\frac{SF}{SE}=\frac{BF}{CE}$ thus $SR$ is bisector. $M,I,R$ are collinear Let $D,D'$ are the intersecrions of $AT$ with $BC$ , $MI$ and $R'$ that of $MI,EF$ (we wish to show that $R'\equiv R$) Then $(I,D', R',M)=D(I,D', R',M)=-1$ and since $AI,AD'$ are orthogonal $AI$ bisects $R'AM$ implying $R'\equiv R$. Let $N$ be the midpoint of minor arc $BC$ then : $NI^2=NB^2=NM\cdot NT$ which gives that $\angle JIA= \angle NIM = \angle NTI $ the rest is really simple . $\angle ITX=\angle NTI + \angle NTX = \angle JIA +\angle IAJ=\angle IJX$ , hence $TIJX$ is cyclic.

So this is sort've an outside question but I've been searching for a long time for a different way to characterize the center of $(TJIX)$ rather than the intersection of the perpendicular bisectors of $TI$ and $IX$. If anyone knows another way, that would be awesome.

Interesting fact: if $AI\cap BC=D$ then $TJIXD$ is cyclic.

Also I don’t understand why the cross ratio condition Iminsl used to show AZ and AJ are isogonal. Can someone please explain.

Plops wrote: Also I don’t understand why the cross ratio condition Iminsl used to show AZ and AJ are isogonal. Can someone please explain. From $(ZA,BC)=-1$, it follows that $AZ$ is the $A$-symmedian of triangle $ABC$. Note that $AJ$ is the $A$-median, so the result follows.

Groupsolved with Arindam Bhattacharyya, Aatman Supkar and Kazi Aryan Amin among others. We begin by showing a nice lemma. Lemma 1. Let $Y$ be a point on $BC$, let $E,F$ be the feet of the angle bisectors from $B,C$ and let $I$ be the incenter. Now, if $Z$ is a point on $EF$ such that $AY,AZ$ are isogonal then $Y,I,Z$ are collinear. Proof. Animate $Y$ on $BC$, then $Y\mapsto YI\mapsto YI\cap EF$ is projective and so is $Y\mapsto AY\mapsto AZ\mapsto Z$. So, we need to show that these projective maps are equivalent. Thus, we need 3 cases for $Y$. Let $Y=B,C,AI\cap BC$ and we are done. Now, let $Z$ be a point on the $A$ symmedian and line $EF$. Now, we show that $ASZK$ is cyclic. Claim 1. $SZ$ is angle bisector of $\angle ESF$. Proof. Let $U=AZ\cap (AEF)$ and $V=AZ\cap (ABC)$. Now, $\frac{EZ}{ZF}=\frac{AE\cdot EU}{AF\cdot FU}=\frac{AE}{AF}\cdot \frac{CV}{BV}=\frac{AE}{AF}\cdot\frac{AC}{AB}=\frac{AE\cdot FB}{AF\cdot EC}\cdot\frac{AC}{AB}\cdot \frac{EC}{FB}=\frac{EC}{FB}=\frac{SE}{SF}$. Thus, the claim follows. Last result is by spiral similarity. Now, $\measuredangle SZK=\measuredangle SZE= \measuredangle SEZ+\measuredangle ZSE=\measuredangle SEF+\measuredangle ZSE=\measuredangle SAF+\measuredangle FAK=\measuredangle SAK$. Thus, we get that $S,A,K,Z$ cyclic. Let $M$ be the midpoint of $BC$ and $A'$ be the ex-touch point on $BC$. Now, by Lemma 1, we have $MIZ$ collinear but note that $IM\parallel AA'$. Now, reflect $Z,S,A',I,K$ in $AI$, thus $Z'$ lies on $AX$ and $(AKS')$, thus $Z'\equiv J$. Now, also notice that $IZ'\parallel AA"$ but $AA"$ is the line joining $A$ and the mixtilinear touch-point $T_A$. Thus, $IZ'\parallel AT_A$. Now, we see that $AT_AXT$ is cyclic, $A,J,X$ collinear, $TIT_A$ collinear and $AJ\parallel AT_A$, thus by Reim's theorem we are done. We are lazy so you may have the .ggb at https://www.geogebra.org/geometry/rfe9gvwv

Let $MI$ meet $EF$ at $Q$. CLAIM 1. $AQ$ lies on the $A$-symmedian Proof. We will use barycentric coordinates. We have $I=(a:b:c)$, $M=(0:\frac{1}{2}:\frac{1}{2})$, $E=(a:0:c),F=(a:b:0)$. Hence the equation of line IM is $$(b-c)x-ay+az=0$$while the equation of line FE is $$-bcx+axy+abz=0$$Solving for $Q$, we have $b^2z=c^2y$ and hence $Q$ lies on the $A$-symmedian. CLAIM 2. $SQ$ bisects $\angle FSE$. Proof. Notice that $$\frac{\sin\angle CAM}{\sin\angle BAM}=\frac{\sin\angle CAM}{\sin\angle AMC}\cdot\frac{\sin\angle BAM}{\sin\angle BMA}=\frac{CM}{AC}\cdot\frac{AB}{BM}=\frac{AB}{AC}$$hence $$\frac{FQ}{QE}=\frac{FQ}{QA}\cdot\frac{QA}{QE}=\frac{\sin\angle FAQ}{\sin\angle EAQ}\cdot\frac{\sin\angle AEF}{\sin\angle AFE}=\frac{\sin\angle CAM}{\sin\angle BAM}\cdot\frac{AF}{AE}=\frac{AB}{AC}\cdot\frac{AF}{AE}$$Meanwhile by spiral sim. lemma, $S$ is the center of spiral similarity sending $FE$ to $BC$. Therefore $$\frac{SF}{SE}=\frac{FB}{EC}=\frac{FB}{AF}\cdot\frac{AE}{EC}\cdot\frac{AF}{AE}=\frac{BC}{AC}\cdot\frac{AB}{BC}\cdot\frac{AF}{AE}=\frac{AB}{AC}\cdot\frac{AF}{AE}$$Therefore we prove CLAIM 2 by angle bisector theorem CLAIM 3. $A,S,Q,K$ are concyclic. Proof. Let $AK$ meet $(AEF)$ at $G$ again. Then by shooting lemma $$\frac{GQ}{GS}=GF^2=\frac{GK}{GA}$$Hence $A,S,Q,K$ are concyclic as desired. Now let $AI$,$TI$, $JI$ meet $BC$ again at $D$, $Z$ and $L$ respectively. Let $Y$ be the projection of $I$ on $BC$. Let $AI$ meet $(ABC)$ again at $N$. CLAIM 4. $\angle JIT=\angle YID$ Proof. It suffices to show that $\angle ZIY=\angle DIL$ Since $N$ is the circumcenter of $\triangle IBC$, $IY$ and $IN$ are isogonal w.r.t. $\angle IBC$. Meanwhile, by CLAIM 3, $Q$ is the reflection of $J$ in $AI$. Now $$\angle DIL=\angle MIN$$Since $\angle TBN=\angle TCN=90^{\circ}$, $TI$ is the $I$-symmedian. Hence $IZ$ and $IM$ are also isogonal lines. Therefore $$\angle MIN=\angle ZIY$$as deisred. Now we finish by noticing that $$\angle JXT=\angle TXA=\frac{B-C}{2}=90^{\circ}-\angle ADB=\angle YID=\angle JIT$$This completes the proof.

Let $M$ and $N$ be the midpoints of $BC$ and arc $BC$. Let $\overline{AX}$ cut $(AFE)$ at $L$. We first claim that $L,K,S$ are collinear. Since $\triangle FLE\sim\triangle BXC$, by the ratio lemma, we just need to check that \[\frac{FS}{ES}\cdot\frac{BX}{XC}=\frac{FK}{EK}\Longleftrightarrow \frac{BF}{CE}\cdot\frac{CA}{AB}=\frac{FA}{AE}\Longleftrightarrow\frac{BF}{FA}\cdot\frac{AE}{CE}=\frac{AB}{AC}\]which is obvious from the angle bisector theorem. Now by spiral similarity, it is easy to see that $\triangle SBF\sim\triangle SXL\sim\triangle SCE$. Let the $A$-symmedian meet segment $EF$ at $J'$. Claim 1: $J'$ is the reflection of $J$ over $\overline{AI}$. Proof. Since $\overline{AJ'}$ and $\overline{AX}$ are isogonal, $\measuredangle ASX=\measuredangle ACX=\measuredangle (AJ',BC)$. Meanwhile, due to the similarities, $\measuredangle LSX=\measuredangle FSB=\measuredangle (EF,BC)$. Subtracting the two equations show that $\measuredangle ASL=\measuredangle AJ'K$ and since $\measuredangle AJK=-\measuredangle ASK$, we see that $J$ is the reflection of $J'$ as desired. $\blacksquare$ Claim 2: $J',I,M$ are collinear. Proof. Let $\overline{J'I}$ meet $\overline{BC}$ at $M'$. By the ratio lemma, \[\frac{FJ'}{J'E}=\frac{FA}{AE}\div\frac{FL}{LE}=\frac{FA}{AE}\div\frac{BX}{XC}=\frac{FA}{AE}\cdot\frac{AB}{AC}.\]By the ratio lemma again on $\triangle FIE$ and $\triangle BIC$, \[\frac{FJ'}{J'E}\div\frac{FI}{IE}=\frac{CM'}{M'B}\div\frac{CI}{IB}.\]Therefore, \begin{align*} \frac{CM'}{M'B}&=\frac{FJ'}{J'E}\cdot\frac{IE}{IF}\cdot\frac{CI}{IB}\\ &=\frac{FA}{AE}\cdot\frac{AB}{AC}\cdot\frac{IE}{IF}\cdot\frac{CI}{IB}\\ &=\frac{FA}{AE}\cdot\frac{AB}{AC}\cdot\frac{EA}{AB}\cdot\frac{CA}{AF}\\ &=1 \end{align*}and so $M'=M$. $\blacksquare$ Let $\overline{TI}$ meet $\Gamma$ again at $P$. Since $NI^2=NM\cdot NT$, $\triangle NIM\sim\triangle NTI$, \[\measuredangle JIA=\measuredangle AIJ'=\measuredangle NIM=\measuredangle ITN=\measuredangle PAI\]and thus $IJ\parallel AP$. Therefore, $\measuredangle IJX=\measuredangle PAX=\measuredangle PTX=\measuredangle ITX$ and we're finally done. [asy][asy] defaultpen(fontsize(10pt)); size(12cm); pen mydash = linetype(new real[] {5,5}); pair A = dir(150); pair B = dir(220); pair C = dir(320); pair I = incenter(A,B,C); pair E = extension(B,I,A,C); pair F = extension(C,I,A,B); pair M = midpoint(B--C); pair X = 2*foot(circumcenter(A,B,C),A,M)-A; pair S[] = intersectionpoints(circumcircle(A,E,F),circumcircle(A,B,C)); pair K = extension(A,I,E,F); pair N = 2*foot(circumcenter(A,B,C),A,I)-A; pair X1 = 2*foot(X,M,N)-X; pair T = 2*foot(circumcenter(A,B,C),N,M)-N; pair J1 = extension(A,X1,E,F); pair J = 2*foot(J1,A,I)-J1; pair L = extension(A,X,S[1],K); pair P = 2*foot(circumcenter(A,B,C),I,T)-T; draw(A--B--C--cycle, black+1); draw(A--X); draw(A--N); draw(A--X1); draw(A--S[1]); draw(A--P); draw(B--E, dotted); draw(C--F, dotted); draw(S[1]--L, mydash); draw(J1--M, mydash); draw(E--F); draw(N--T, dotted); draw(T--P, dotted); draw(I--J); draw(circumcircle(A,B,C)); draw(circumcircle(A,E,F)); draw(anglemark(B,A,X1,5)); draw(anglemark(X,A,C,5)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(10)); dot("$F$", F, dir(180)); dot("$I$", I, dir(270)); dot("$J$", J, dir(60)); dot("$J'$", J1, dir(240)); dot("$K$", K, dir(60)); dot("$L$", L, dir(60)); dot("$M$", M, dir(45)); dot("$N$", N, dir(270)); dot("$P$", P, dir(P)); dot("$S$", S[1], dir(180)); dot("$T$", T, dir(90)); dot("$X$", X, dir(270)); dot("$X'$", X1, dir(270)); [/asy][/asy]

geogebra solution link

Nice! Claim 1: If $AP,AQ$ are isogonal and $P \in EF$ and $Q \in BC$ then $P-I-Q$

Claim 2: $ASZK$ is cyclic.

Next, notice $Z,J$ are simply the reflections of each other in $AI$, so $IJ \parallel AT_A$ and we finish by reims.

We will prove that $\measuredangle IJA = \measuredangle ITX$. Let $(ASK)$ intersect $EF$ at $J' \neq K$. Claim 1. $J$ and $J'$ are reflections across $AI$. Proof. By Shooting Lemma, $SJ'$ passes through the midpoint $W$ of arc $EF$ not containing $A$. Also let $AX$ intersect $(AEF)$ at $Y \neq A$. Since $\measuredangle J'AK = \measuredangle J'SK = \measuredangle WSK$ and $\measuredangle KAJ = \measuredangle WAY = \measuredangle WSY$, it suffices to check that $S,K,Y$ are collinear. Indeed, \[ \frac{SF}{SE} \cdot \frac{YF}{YE} = \frac{BF}{CE} \cdot \frac{XB}{XC} = \frac{AF \cdot \frac{BC}{AC}}{AE \cdot \frac{BC}{AB}} \cdot \frac{AC}{AB} = \frac{AF}{AE} = \frac{KF}{KE} \]which proves the claim. Claim 2. $J', I, M$ are collinear. Proof. Let the line through $I$ parallel to $BC$ meet $EF$ at $G$. By perspectivity at $I$, it suffices to check that $(G,J';F,E)=-1$. This is true because \[ \frac{GF}{GE} = \frac{IF \sin \angle FIG}{IE \sin \angle GIE} = \frac{IF}{IE} \cdot \frac{\sin \angle ICB}{\sin \angle IBC} = \frac{IF}{IC} \cdot \frac{IB}{IE} = \frac{BF}{BC} \cdot \frac{CB}{CE} = \frac{BF}{CE} = \frac{SF}{SE} = \frac{J'F}{J'E}. \] Now we can finish. Let $TI$ intersect $(ABC)$ at the $A$-mixtilinear touchpoint $T_a$, and let $N$ be the $A$-extouch point. Then \[ \measuredangle ITX = \measuredangle T_aAX = \measuredangle J'AN = \measuredangle AJ'I = \measuredangle IJA \]because it is well-known that $IM \parallel AN$.

Solved with stillwater_25 who honestly carried this time, coming up with both key ideas to the solution, with me merely filling in the gaps. This problem is a beautiful configuration I had explored before but for the first time we realized that it also has connections with the symmedian and median. Denote by $P$ the intersection of the $A-$symmedian and line $\overline{EF}$. Let $Z= \overline{EF} \cap \overline{BC}$. Let $L = \overline{AT} \cap (AEF)$ and let $N' = \overline{AI} \cap (AEF)$. [asy][asy] /* Converted from GeoGebra by User:Azjps using Evan's magic cleaner https://github.com/vEnhance/dotfiles/blob/main/py-scripts/export-ggb-clean-asy.py */ import geometry; pair foot(pair P, pair A, pair B) { return foot(triangle(A,B,P).VC); } pen pri; pri=RGB(24, 105, 174); pen sec; sec=RGB(217, 165, 179); pen tri; tri=RGB(126, 123, 235); pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair A = (1.86750,0.08192); pair B = (4.,-7.); pair C = (15.,-7.); pair E = (7.14735,-2.76532); pair F = (3.09501,-3.99459); pair N = (9.5,-9.27286); pair M = (9.5,-7.); pair T = (9.5,6.30922); pair X = (11.62974,-8.97611); pair S = (1.79431,-2.63189); pair Sp = (4.54078,-0.39106); pair J = (5.08961,-2.90775); pair P = (4.14963,-3.67467); pair Np = (5.37518,-4.21726); pair L = (3.63386,1.52308); pair Z = (-6.81246,-7.); pair Tp = (-5.76499,-6.14537); pair Xp = (8.78183,-11.29970); pair Q = intersectionpoint(line(L,S),line(E,F)); pair K = intersectionpoint(line(A,N),line(E,F)); pair I = intersectionpoint(line(B,E),line(C,F)); size(15.cm); filldraw((path)(A--B--C--cycle), white+0.1*pri, pri); filldraw(circumcircle(A,B,C), tfil, tri); draw(T--Z,pri); draw(Z--B,pri); filldraw(circumcircle(A,E,F), tfil, sec); draw(A--X,pri); draw(A--Xp,pri); draw(Z--E,pri); draw(L--Q,pri); draw(A--N,tri); filldraw(circumcircle(A,S,P), tfil, pri+dotted); filldraw(circumcircle(T,X,J), tfil, sec+dotted); dot("$A$", A, dir((8.021, 21.540))); dot("$B$", B, dir((7.805, 21.071))); dot("$C$", C, dir((7.753, 21.071))); dot("$E$", E, dir((8.272, 17.148))); dot("$F$", F, dir((9.177, 18.342))); dot("$I$", I, dir(10)); dot("$N$", N, dir((7.779, 17.933))); dot("$M$", M, dir((7.779, 16.723))); dot("$T$", T, dir((7.779, 18.346))); dot("$X$", X, dir((7.838, 16.518))); dot("$S$", S, dir(S)); dot("$K$", K, dir(330)); dot("$S'$", Sp, dir((8.072, 16.668))); dot("$J$", J, dir((9.708, 18.349))); dot("$P$", P, dir((8.058, 16.783))); dot("$N'$", Np, dir(270)); dot("$L$", L, dir((9.638, 16.548))); dot("$Z$", Z, dir((8.668, 16.723))); dot("$T'$", Tp, dir((8.264, 18.213))); dot("$X'$", Xp, dir((7.860, 18.453))); dot("Q",Q,dir(Q)); [/asy][/asy] First note that from the Ceva/Menalaus picture we have that $(B,C;Z,AI \cap BC)=-1$. But also note that, \[-1=(TN;BC) \overset{A}{=}(AT \cap BC , AI \cap BC;B,C)\]which implies that $Z = AT \cap BC$, so lines $\overline{AT}$ , $\overline{EF}$ and $\overline{BC}$ concur at $Z$. We now work towards showing that point $P$ is the reflection of $J$ across the $\angle A-$bisector. We first make the following simple observations. Claim : The tangent to $(ABC)$ at $A$ , line $\overline SL$ and $\overline{EF}$ concur at a point $Q$. Proof : Let the $A-$tangent to $(ABC)$ intersect $\overline{EF}$ at $Q_1$ and let $Q_2 = SL \cap EF$. Note that $S$ is the Miquel point of $BCEF$. Thus, \[\measuredangle SZQ_1 = \measuredangle SZF = \measuredangle SBF = \measuredangle SBA = \measuredangle SAQ_1\]which implies that $SAZQ_1$ is cyclic. Further, \[\measuredangle AZQ_2 = \measuredangle LAE + \measuredangle AEF = \measuredangle FEL + \measuredangle AEF = \measuredangle AEL = \measuredangle ASL\]which implies that $SAZQ_2$ is also cyclic. But this means that $Q_1 \equiv Q_2$ proving the claim. Claim : Points $S$ , $P$ and $N'$ are collinear. Proof : Note that due to the spiral similarity at $S$ , $L$ is the major arc midpoint of $EF$ in $(AEF)$. Let $X' = AP \cap (ABC)$. Then, \[-1=(BC;AX')\overset{A}{=}(FE;QP)\overset{S}{=}(F,E;L,SP \cap (AEF))\]which implies that $N' = SP \cap (AEF)$. Thus, points $S$ , $P$ and $N'$ are indeed collinear, as desired. Claim : Points $A$ , $S$ , $P$ and $K$ are concyclic. Proof : This is a pretty straightforward angle chase. Note that, \[\measuredangle PKA = \measuredangle FEA + \measuredangle EAK = \measuredangle EFA + \measuredangle KAE = \measuredangle ESA + \measuredangle N'SE = \measuredangle N'SA = \measuredangle PSA \]which implies the claim. By definition lines $\overline{AJ}$ and $\overline{AP}$ are reflections of each other across the $\angle A-$bisector. Further note that due to the above claims, both $AS'JK$ and $ASPK$ are cyclic, which must imply that $P$ is the reflection of $J$ across the $\angle A-$bisector. We make a couple of final observations before attacking the final result. Claim : Lines $\overline{BC}$ and $\overline{QI}$ are parallel. Proof : Let $M_b$ and $M_c$ denote the $AB$ and $AC$ minor arc midpoints respectively. Note that by Pascal's Theorem on hexagon $AABM_bM_cC$ it follows that points $E$ , $F$ and $M_bM_c \cap AA$ are collinear. Thus, $\overline{M_bM_c}$ passes through $Q$. Further note that if $T_a$ denotes the $A-$mixtillinear intouch point, since it is well known that points $T$ , $I$ and $T_a$ are collinear, \[-1=(NT;BC)\overset{I}{=}(T_aA;M_bM_c)\]which implies that the tangents to $(ABC)$ at $A$ and $T_a$ intersect on $\overline{M_bM_c}$. Since we showed that the tangent to $(ABC)$ at $A$ intersects $\overline{M_bM_c}$ at $Q$, it follows that $Q$ is in particular to the intersection of the tangents to $(ABC)$ at $A$ and $T_a$. Finally if $U$ and $V$ are the intersections of line $\overline{QI}$ with $(ABC)$, \[-1=(AT_a;UV) \overset{I}{=}(NT;VU)\]which implies that $UV \perp MN$ and indeed $UV \parallel BC$, which implies the claim. Note that it immediately follows that $P$ , $I$ and $M$ are collinear since, \[-1=(EF;PQ)\overset{I}{=}(B,C;PI \cap BC , P_\infty)\] The problem then reduces all the way down to the following result. Claim : Let $M'$ denote the reflection of $M$ across the $\angle A-$bisector. Then circle $(IM'N)$ is tangent to $\overline{TI}$. Proof : Let $X_a$ denote the $A-$extouch point. It is well known that $AT_a$ and $AX_a$ are isogonal and that $AX_a \parallel MI$. Thus, \[\measuredangle (\overline{TI},\overline{IM'}) = \measuredangle AIT + \measuredangle NIM' = \measuredangle AIT + \measuredangle MIN = 90^\circ + \measuredangle ATI + \measuredangle X_aAN = 90^\circ + \measuredangle ATT_a + \measuredangle NAT_a = 90^\circ + \measuredangle ATT_a + \measuredangle NTT_a = \measuredangle ANT\]which implies the tangency. Now, let $T'$ and $X'$ denote respectively the reflections of $T$ and $X$ across the $\angle A-$bisector. Then note that as a result of the above claim (reflected), $(IMN)$ is tangent to $\overline{IT'}$. Thus, using the fact that $P$ , $I$ and $M$ are collinear we have that, \[\measuredangle T'X'P = \measuredangle T'X'A = \measuredangle AXT = \measuredangle INM = \measuredangle T'IP\]which implies that $(T'X'IP)$ is cyclic. Reflecting across the $\angle A-$bisector we have that $TXIJ$ is indeed cyclic, as claimed.