We let $\ell_a$, $\ell_b$, $\ell_c$ denote the reflections of $\ell = \overline{DEF}$ across the perpendicular bisectors of $\overline{AD}$, $\overline{BE}$, $\overline{CF}$. Claim: The lines $\ell_a$, $\ell_b$, $\ell_c$ intersect at a point $T$ on $\omega$. Proof. This is direct angle chasing: since \begin{align*} \measuredangle(\ell_a, \ell_b) &= \measuredangle(\ell_a, \ell) + \measuredangle(\ell, \ell_b) = 2\measuredangle(\ell_a, \overline{AI}) + 2\measuredangle(\overline{BI}, \ell_b) \\ &= 2\measuredangle(\ell_a, \ell_b) + 2\measuredangle(\overline{BI}, \overline{AI}) \\ \implies \measuredangle(\ell_a, \ell_b) &= 2 \measuredangle AIB = \measuredangle ACB \end{align*}it follows that $\ell_a$ and $\ell_b$ meet on $\omega$, as desired. $\blacksquare$ Let line $TI$ meet $\ell$ at $H$. Then, consider a homothety centered at $T$ which maps $I$ to $H$. It then maps $\triangle ABC$ to a triangle $\triangle PQR$, say, whose circumcircle $\gamma$ is tangent to $\triangle ABC$ at $T$, and whose incenter is $H$. [asy][asy]size(250); defaultpen(fontsize(11pt)); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair T = dir(17); pair I = incenter(A, B, C); pair hom(pair P) { return 1.4*P - 0.4*T; } pair H = hom(I); pair P = hom(A); pair Q = hom(B); pair R = hom(C); pair D = extension(A, I, H, H+dir(I-A)*dir(I-A)/dir(P-A)); pair E = extension(D, H, B, I); pair F = extension(D, E, C, I); filldraw(A--B--C--cycle, invisible, heavycyan); draw(circumcircle(A, B, C), heavycyan); filldraw(P--Q--R--cycle, invisible, heavygreen); draw(circumcircle(P, Q, R), heavygreen); pair U = extension(H, D, P, A); draw(H--T, blue); draw(P--T, pink); draw(Q--T, pink); draw(R--T, pink); draw(E--U, red); draw(A--I, grey); draw(E--I, grey); draw(F--C, grey); draw(P--H, grey); pair Z = hom(dir(40)); pair Y = hom(dir(160)); pair X = hom(dir(270)); draw(X--Y--Z--cycle, red); dot("$A$", A, dir(A)); dot("$B$", B, dir(255)); dot("$C$", C, dir(295)); dot("$T$", T, dir(T)); dot("$I$", I, dir(280)); dot("$H$", H, dir(320)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$R$", R, dir(R)); dot("$D$", D, dir(170)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot(U); dot(Z); dot(Y); dot(X); /* TSQ Source: A = dir 110 B = dir 210 R255 C = dir 330 R295 T = dir 17 I = incenter A B C R280 ! pair hom(pair P) { return 1.4*P - 0.4*T; } H = hom(I) R320 P = hom(A) Q = hom(B) R = hom(C) D = extension A I H H+dir(I-A)*dir(I-A)/dir(P-A) R170 E = extension D H B I F = extension D E C I A--B--C--cycle 0.1 lightcyan / heavycyan circumcircle A B C heavycyan P--Q--R--cycle 0.1 lightgreen / heavygreen circumcircle P Q R heavygreen U .= extension H D P A H--T blue P--T pink Q--T pink R--T pink E--U red A--I grey E--I grey F--C grey P--H grey Z .= hom(dir(40)) Y .= hom(dir(160)) X .= hom(dir(270)) X--Y--Z--cycle red */ [/asy][/asy] Recall that $\ell \equiv \overline{HD}$ is the reflection of line $\overline{APT}$ across $x$. As $\overline{HP} \parallel \overline{AD}$, $x$ is also the perpendicular bisector of $\overline{PH}$. Similarly, $y$ and $z$ are the perpendicular bisectors of $\overline{QH}$ and $\overline{RH}$. Since $H$ is the incenter of $\triangle PQR$, this implies that our perpendicular bisectors determine a triangle with circumcircle $\gamma$, as the meet at the arc midpoints of $\gamma$. This is the desired result. Remark: For me, most of the difficulty in this solution is identifying all the relevant points $T$, $P$, $Q$, $R$, $H$, which can be found from an accurately drawn diagram.

Let $AI, BI, CI$ meet the circumcircle of $ABC$ at $A', B', C'$. To prove this problem it is enough to verify that the vertices $XYZ$ of $\Theta$ obey the property that $A'X$, $B'Y$, $C'Z$ concur on the circumcircle of $ABC$. Let $X'Y'Z'$ be triangle formed with the perpendicular lines from $D, E, F$ to $AI, BI, CI$. One can get that $|IX'| = 2|A'X|$ and $IX'||A'X$. From similar properties and some angle chase we conclude that $A'X, B'Y, C'Z$ meet on the circumcircle of $ABC$.

Let $ T $ be the pole of the Simson line of $ \triangle ABC $ with the direction $ \perp \ell, $ $ T_a, T_b, T_c $ be the intersection of $ AI, BI, CI $ with the parallel $ \tau $ from $ T $ to $ \ell $ and let $ \triangle XYZ $ be the triangle $ \Theta, $ then note that the isogonal conjugate of $ T $ WRT $ \triangle ABC $ is the point at infinity with the direction $ \parallel \ell $ we get $ T $ lies on the perpendicular bisector of $ BT_b, CT_c, $ so consider the homothety $ \mathbf{H}^{B}_{\frac{1}{2}}, \mathbf{H}^{C}_{\frac{1}{2}} $ we conclude that $ TX $ passes through the midpoint $ M_a $ of arc $ BC $ in $ \odot (ABC) $ and $ \frac{TM_a}{XM_a} = \frac{\text{dist} (I, \tau)}{\text{dist} (I, \ell)}. $

Let $A_1=y\cap z,$ $B_1=x\cap z,$ and $C_1=x\cap y.$ Let the reflections $l_a$ and $l_b$ of $l$ in $x$ and $y,$ resspectively meet at $Q.$ $$\measuredangle{BQA}=\measuredangle(l_b,l_a)=\measuredangle{AIB}-\measuredangle(AI,l_a)-\measuredangle(BI,l_b)=\measuredangle{AIB}-\measuredangle(l,AI)-\measuredangle(l,BI)=-2\measuredangle{BIA}=\measuredangle{BCA}$$Hence, $Q$ lies on $\omega,$ which implies that $l_c,$ the reflection of $l$ in $z$ also passes through $Q.$ Let $\delta(X,m)$ denote the distance from point $X$ to line $m.$ $$\delta(C_1, QB)=\delta(C_1,l_b)=\delta(C_1,l)=\delta(C_1,l_a)=\delta(C_1,AQ)$$Thus, $C_1$ lies on the angle bisector of $\angle{AQB}.$ As $A_1B_1C_1$ and the triangle defined by the midpoint of arcs $BC, AC$ and $AB$ are homothetic, this implies that $Q$ is their homothety center, and since $Q$ lies on $\omega \implies$ circle $QA_1B_1C_1$ is tangent to $\omega.$

This was proposed by Mads Christensen, Denmark and lost to G6 by one vote.

wait this would seem really easy for a 3 like this is maximum a 2 - especially considering that the homothety tangency trick has already appeared before in the imo with examples like 2011/6, and once you realise its that kind of problem it's a lot easier

Wonderful problem Let $M, N, K$ be the midpoints of the smaller arc $BC, CA, AB$, respectively. In order to prove the tangency, we first construct the tangency point: Main Lemma: Among all points on $\omega$, there is a unique point $P$ such that $$PM:PN:PK=\frac{1}{ID}:\frac{1}{IE}:\frac{1}{IF}.$$ Proof. Since such $P$ is unique if it exists, so let us only prove its existence. We perform an inversion centered at $I$ with radius $1$. Let $D', E', F'$ denote the image of $D,E,F$ under inversion, respectively. Since $D, E,F$ are collinear, $I, D', E', F'$ are concyclic and moreover, simple angle chasing yields $\triangle D'E'F' \sim \triangle MNK$. Hence, if $P$ is the point such that $\triangle D'E'F' \cap I \sim \triangle MNK \cap P$, then $P$ lies on $\omega$ and $P$ satisfies the desired length property, and we are done. $\blacksquare$ Let $x$ meet $NP, KP$ at $Y,Z$, respectively. A line passing $Y$ and parallel to $MN$ cut $PM$ at $X$. Then, $$\frac{MX}{KZ}=\frac{MX}{NY}\times\frac{NY}{KZ}=\frac{PM}{PN} \times \frac{PN}{PK}=\frac{PM}{PK},$$so $XZ$ is parallel to $MK$. Also, $$\text{dist}(XY, MN)=\text{dist}(YZ, KN) \times \frac{\sin \angle PNM}{\sin \angle PNK}=\frac{ID}{2}\times \frac{PM}{PK}=\frac{IF}{2},$$so $XY$ coincides with $z$. Similarly, $XZ$ coincides with $y$, so $\Theta=\triangle XYZ$. Since $\triangle XYZ$ and $\triangle MNK$ are homothetic with center $P$, $\odot(XYZ)$ touches $\Omega$ at $P$, as desired. Remark: This proof shows that as $l$ traces with its direction fixed, the tangency point of the circumcircle of $\Theta$ and $\Omega$ is fixed too

What a nice problem! I wished that this was IMO 2018 P2/5 as it's very hard to bash but G6 is clearly more elegant. Let $AI, BI, CI$ meet $\Omega$ again at $M_A, M_B, M_C$. Let $X = y\cap z$, $Y=x\cap z$, $Z=x\cap y$. Clearly $\triangle XYZ$ and $\triangle M_AM_BM_C$ are homothetic so lines $XM_A$, $YM_B$, $ZM_C$ are concurrent at $T$. Let lines through $D,E,F$ perpendicular to lines $AI$, $BI$, $CI$ forms a triangle $\triangle A_1B_1C_1$ where $D\in B_1C_1$, $E\in C_1A_1$, $F\in A_1B_1$. The key claim is Claim: Quadrilaterals $IA_1B_1C_1$ and $TM_AM_BM_C$ are homothetic. Proof: Both $B_1C_1$ and $M_BM_C$ are perpendicular to $AI$ by definitions so it suffices to show that $IA_1\parallel M_AX$. Let $I_A$ be the $A$-excenter. Notice that projections from points $I_A, X, A_1$ to each of $BI$ and $CI$ are three equally spaced points. Thus $X$ is the midpoint of $I_AA_1$. Hence $M_AX$ is $I_A$-midline of $\triangle I_AIA_1$ so we are done. To finish, note that $\overline{DEF}$ is $I$-Simson Line of $\triangle A_1B_1C_1$. Thus by the converse, quadrilateral $IA_1B_1C_1$ is cyclic. By the claim, $TM_AM_BM_C$ is cyclic too so by homothety, these circles are tangent at $T$.

jdhua wrote: This was proposed by Mads Christensen, Denmark and lost to G6 by one vote. Actually this problem is well known if I am not mistaken.

psi241 wrote: Let $ABC$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $AI$, $BI$, and $CI$ at points $D$, $E$, and $F$, respectively, distinct from the points $A$, $B$, $C$, and $I$. The perpendicular bisectors $x$, $y$, and $z$ of the segments $AD$, $BE$, and $CF$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$. Solution. Let $M_a$ be midpoint of arc $BC$ and define $M_b,M_c$ similarly. Also, $X=y\cap z$ and define $Y,Z$ similarly. Without loss of generality, let $D$ lie on segment $EF$. Further let $B_1,B_2$ be feet from $M_b$ on $x$ and $z$. We have $\angle B_1M_bM_c = \angle M_cM_bM_a$ as $XYZ$ is similar to $M_aM_bM_c$ and \[M_bB_1:M_bB_2 = ID/2:IF/2 = ID:IF.\]This gives \[\angle YM_bM_c = \angle M_bYB_1 = \angle M_bB_2B_1 = \angle IFD.\]Similarly $\angle M_bM_cZ = \angle DEI$, so $M_bY\cap M_cZ$ lies on $\odot(M_aM_bM_c)$ and we are done. $~\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); 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 = -4.770325951955437, xmax = 14.63584922186477, ymin = -3.2441547295906172, ymax = 7.047480792918943; /* image dimensions */ pair M_a = (0.,6.), M_c = (-2.,0.), M_b = (7.,0.), A = (0.,-2.3333333333333326), B = (-2.2,3.0666666666666664), C = (3.6235294117647054,6.560784313725489), D = (0.,1.209721539810738), F = (-1.4068174324789477,0.6920463287745612), I = (0.,2.333333333333333), X = (0.32322002353142404,4.299389006488415), Z = (-1.2971782775518137,-0.5618058967612973), Y = (5.994614077322754,-0.5618058967612973); draw(M_a--M_c--M_b--cycle, linewidth(2.)); /* draw figures */ draw(circle((2.5,1.8333333333333333), 4.85912657903775), linewidth(1.)); draw(M_a--A, linewidth(1.)); draw(M_c--C, linewidth(1.)); draw(B--M_b, linewidth(1.)); draw(F--(1.6021626384635135,1.7992791205121619), linewidth(1.)); draw(D--F, linewidth(1.)); draw(D--(1.6021626384635135,1.7992791205121619), linewidth(1.)); draw(M_c--(1.7029429765660729,-2.959975671154668), linewidth(1.)); draw((1.7029429765660729,-2.959975671154668)--M_b, linewidth(1.)); draw(Z--X, linewidth(1.)); draw(X--Y, linewidth(1)); draw(Y--Z, linewidth(1.)); /* dots and labels */ dot(M_a,dotstyle); label("$M_a$", (-0.20510277581904354,6.1), NE * labelscalefactor); dot(M_c,dotstyle); label("$M_c$", (-2.4,-0.5), NE * labelscalefactor); dot(M_b,dotstyle); label("$M_b$", (7.127886367731051,-0.06281453716107784), NE * labelscalefactor); dot(A,linewidth(4.pt) + dotstyle); label("$A$", (-0.1,-2.8), NE * labelscalefactor); dot(B,linewidth(4.pt) + dotstyle); label("$B$", (-2.5,3.1821524591170527), NE * labelscalefactor); dot(C,linewidth(4.pt) + dotstyle); label("$C$", (3.692038959907146,6.681626670789546), NE * labelscalefactor); dot((1.6021626384635135,1.7992791205121619),dotstyle); label("$E$", (1.7037013396386818,1.8778029802209413), NE * labelscalefactor); dot(D,dotstyle); label("$D$", (0.06531114053746755,1.368788549432215), NE * labelscalefactor); dot(F,linewidth(4.pt) + dotstyle); label("$F$", (-1.4776388527908604,0.8120540157570455), NE * labelscalefactor); dot(I,linewidth(4.pt) + dotstyle); label("$I$", (0.16075134631035382,2.323190607161077), NE * labelscalefactor); dot(X,linewidth(4.pt) + dotstyle); label("$X$", (0.38344515978042176,4.422875134164573), NE * labelscalefactor); dot(Z,linewidth(4.pt) + dotstyle); label("$Z$", (-1.3503852450936789,-1), NE * labelscalefactor); dot(Y,linewidth(4.pt) + dotstyle); label("$Y$", (6.1,-0.8581495852684627), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Denote the reflections $l_A,l_B,l_C$ of $l$ over $x,y,z$ respectively. Defining $l_B\cap l_C=T$, note $\angle BTC=\angle(l_B,l)+\angle(l,l_C)=2\angle(BI,l)+2\angle(l,CI)=2\angle(BI,CI)$, from which it is clear that $T$ lies on the circumcircle. Similarly, $l_B\cap l_A$ lies on $\omega$, so we must have $l_A\cap l_B\cap l_C=T\in\omega$. Denoting $y\cap z=A'$, and $AI\cap \omega=X$, I claim that $TA'X$ is a line. However, if we look at the triangle formed by $l,l_B,l_C$, we see that $X$ is the midpoint of its arc, so $TI$ bisects the angle formed by $l_B,l_C$. Also, $y,z$ are angle bisectors of the other two vertices. Hence, they all concur at $A'$, the incircle of the triangle in question, and $A'$ must lie on $TX$ as desired. If we define $Y,Z,B',C'$ similar to $X,A'$, note that $XY$ and $A'B'$ are both perpendicular to $CI$, and similar relations give that $XYZ$ and $A'B'C'$ are homothetic. However, we may duplicate the argument above to get that $T$ is their homothetic center. Therefore, $(A'B'C')$ must be tangent to $(XYZT)=\omega$, as desired.

Let $(I) $ cuts $YZ $ , $ZX $, $YZ $ at $G $, $H $ and $J $ respectively. Let $BH $ cuts $CJ $ at $K $, by angle chasing we get $\angle BKC=\angle BAC$. Hence, $K $ lies on $\omega $. $AG ,BH,CJ $ are concurrent at $K $, lies on $\omega $. Let $AG $, $BH $ and $CJ $ be the reflections of $(I) $ wrt $YZ $, $ZX$, $XY $. Applying Collings' theorem note that $(I) $ passes through the orthocenter $\ell $ of $\triangle XYZ $. $(I) $ is the Steiner line of $K $ wrt $\triangle XYZ $. $K $ lies on $\Theta $. Let $K'$ be the reflection of $K $ wrt $YZ $. $BI $ cuts $YZ $ at point $P $, also cuts $\omega $ again $N'$. Note that $Y $, $Z $, $\ell$, $K'$ lies on a circle. And, $\angle KZY=\angle K'ZY=\angle K'\ell Y=\angle IEF=\angle KBM $ Hence, $K $, $Z $, $B $, $P $ lies on a circle. By Reim's theorem $MN'$ and $MN $ are parallel to $ZY $, and $N=N'$. $O $ is the intersection of $AI $ and $\omega$. $XO $, $YM $ and $ZN $ are concurrent at $K $ also note that $YZ||MN $, $ZX||NO $, $XY||OM $. Applying homothety with center $K $, we can prove that $\Theta $ and $\Omega $ are tangent at $K $ $\blacksquare $ Attachments

[asy][asy] unitsize(2inches); pair X=dir(125); pair Y=dir(215); pair Z=dir(-35); pair I=orthocenter(X,Y,Z); pair A=2*foot(I,Y,Z)-I; pair B=2*foot(I,X,Z)-I; pair C=2*foot(I,X,Y)-I; pair L=dir(-80); pair P=(3/5)*X+(2/5)*L; pair E=2*foot(P,B,Y)-B; pair F=2*foot(P,C,Z)-C; pair D=extension(E,F,X,A); draw(X--Y--Z--cycle); draw(circumcircle(X,Y,Z)); draw(X--A); draw(B--Y); draw(C--Z); draw(E--F,dotted); draw(X--L); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(Z)); dot("$I$",I,dir(I)); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$L$",L,dir(L)); dot("$P$",P,dir(P)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); [/asy][/asy] Let $XYZ$ be the arc midpoints of $ABC$. Note that $I$ is then the orthocenter of $XYZ$, and $A,B,C$ are the reflections of $I$ over the sides of $XYZ$. Now, let $PQR$ be the triangle $\Theta$, so $P=y\cap z$, $Q=x\cap z$, $R=x\cap y$. Note that $x\parallel YZ$, $y\parallel XZ$, $z\parallel XY$, so the triangle $PQR$ is homothetic to $XYZ$. Thus, it suffices to show that the center of homothety $XP\cap YQ\cap ZR$ lies on $\omega=(XYZ)$. To do this end, let $L_X=XP\cap YZ$, $L_Y=YQ\cap XZ$, $L_Z=ZR\cap XY$. It suffices to show that $L_X=L_Y=L_Z$. Claim: We have that $L_X$ is the unique point $L$ such that $LY:LZ=IE^{-1}:IF^{-1}$. Proof: Using signed distances, we have \[\mathrm{dist}(P,XZ)=\frac{\mathrm{dist}(B,XZ)+\mathrm{dist}(E,XZ)}{2}=\frac{\mathrm{dist}(E,XZ)-\mathrm{dist}(I,XZ)}{2}=\frac{IE}{2}.\]Thus, \[\frac{LY}{LZ}=\frac{\sin PXY}{\sin PXZ}=\frac{\mathrm{dist}(P,XY)}{\mathrm{dist}(P,XZ)}=\frac{IF}{IE},\]as desired. To finish, note that $LY/LZ=k(LX;YZ)$ for some constant $k$, and the function from the circle to $\mathbb{RP}^1$ defined by $(LX;YZ)$ is bijective, so $LY/LZ$ is bijective. This completes the proof of the claim. $\blacksquare$ The quantity $IE/IF$ is bijective in the variation of the direction of $\ell$ as well, so to solve the problem, it suffices to show that the set of possible values of $ID^{-1}:IE^{-1}:IF^{-1}$ and $LX:LY:LZ$ are identical as $\ell$ and $L$ vary. By the law of sines of triangles $IED$, $IDF$, $IEF$, we see that \[ID^{-1}:IE^{-1}:IF^{-1} = \sin\angle(XA,\ell):\sin\angle(YB,\ell):\sin\angle(ZC,\ell).\]Let $m$ be a line perpendicular to $\ell$. Then, we have that \[\boxed{ID^{-1}:IE^{-1}:IF^{-1} = \sin\angle(YZ,m):\sin\angle(XZ,m):\sin\angle(XY,m)}.\]Let $T$ be an arbitrary fixed point on $\omega$. Then, by the extended law of sines, we have \[LX:LY:LZ = \sin\angle(TX,TL):\sin\angle(TY,TL):\sin\angle(TZ,TL).\]We see that $\angle XTY=\angle XZY$ and symmetric variants, so there is a fixed rotation sending the triple of directions $(YZ,ZX,XY)$ to $(TX,TY,TZ)$. Thus, there is some other fixed point $S$ on the circle such that \[\boxed{LX:LY:LZ = \sin\angle(YZ,SL):\sin\angle(ZX,SL):\sin\angle(XY,SL)}.\]From the two boxed equations, it is clear that both ratios have the same range as $L$ and $\ell$ vary, so we are done.

Reflect $\ell$ with respect to $x,y,z$: let them be $\ell_x, \ell_y, \ell_z$. For obvious reasons, $\ell_x$ passes through $A$. Similarly, $\ell_y$ passes through $B$ and $\ell_z$ passes through $C$. $\textbf{Claim 01.}$ $\ell_x, \ell_y, \ell_z$ concur on $(ABC)$ at point $K$. $\textit{Proof.}$ This is because \[ \measuredangle (\ell_x, \ell_y) = \measuredangle(\ell_x,\ell) - \measuredangle(\ell_y, \ell) = 2 \measuredangle(AI, \ell) - 2 \measuredangle(BI, \ell) = 2 \measuredangle BIA = \measuredangle BCA \]The other is analogously the same. Let $x \cap y = Z$, $y \cap z = X$, $z \cap x = Y$. Denote $d(X,\ell)$ as distance of points $X$ to line $\ell$. Since $\ell, \ell_y, y$ concur, then we have \[ d(X,\ell_y) = d(X,\ell) = d(X,\ell_z) \]This proves that $X$ lies on the angle bisector of $\measuredangle BKC$. Now, consider the midpoint arc of $AB, BC, CA: C_0, A_0, B_0$. Since $A_0$ lies on the angle bisector of $\measuredangle BKC$. Then we have $K,A_0, X$ being collinear. This implies that $A_0B_0C_0$ and $XYZ$ are homothetic. But since $A_0 X \cap B_0 Y \cap C_0 Z = K$. Then $(A_0 B_0 C_0K)$ are cyclic as well, proven.

Let $AI,BI,CI$ meet $\omega$ again at $M,N,P$ respectively. By fact 5 we have that $PM, PN,NM$ are the perpendicular bisector of $BI,AI,CI$ respectively. Without loss of generality assume that $D$ lies between $E$ and $F$. Now let the intersection of $y$ and $z$ be $X$, the intersection of $z$ and $x$ be $Y$ and the intersection of $x$ and $y$ be $Z$, since $ZX\perp BI$ we have $ZX\|BI$, similarly we have $XY\|MN$ and $YZ\|PN$. Hence the two triangles $\triangle ZYX$ and $\triangle PNM$ are homothetic to each other. Let their homothetic center be $Q$ and the ratio of homothety be $1+k$. Now the crucial claim is $\underline{CLAIM.}$$IE=2kd(Q,PM)$ $\underline{Proof.}$Since $ZX$ is the prependicular bisector of $BE$,$PM$ is the perpendicular bisector of $BI$ we have $$IE=BI-2d(B,XZ)$$$$d(B,PM)=\frac{1}{2}BI$$Meanwhile from the homothety centered at $Q$ we have $$d(B,XZ)=d(B,PM)-d(P,ZX)=\frac{1}{2}BI-kd(P,QM)$$Put this into the first relation we have the identity as claimed.(The above proof only deal with the case when $E$ and $F$ are interior points of the segments $BI$ and $CI$, we may use directed length to deal with the general case. Now let the projection of $Q$ on $PM,PN,NM$ be $T,R,S$ since $$\angle TQR=\angle MPN=\angle MBI=\angle MIB=\angle DIE$$and similarly $\angle DIF=\angle RQS$, $\angle EIF=\angle TQS$ From this and the claim above we have that $QTRS$ and $IEDF$ are similar quadrilaterals with similitude $2k$, hence $T,R,S$ are collinear, by the converse of Simpson's theorem we obtain that $Q$ lies on $(PMN)$, therefore a homothety centered at $Q$ sends $(PMN)$ to $(ZYX)$, that is, sending $\Theta$ to $\Omega$, so those two circles are tangent to each other at point $Q$, so we're done.

psi241 wrote: Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. A line $\ell$ intersects the lines $AI$, $BI$, and $CI$ at points $D$, $E$, and $F$, respectively, distinct from the points $A$, $B$, $C$, and $I$. The perpendicular bisectors $x$, $y$, and $z$ of the segments $AD$, $BE$, and $CF$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$. We first present the following lemma: Lemma: Let $TAB, TCD$ be two triangles, and $\ell$ be the line through $A,C.$ Suppose that $\ell \cap TB=M$ lies on the perpendicular bisector $\ell_1$ of $AB,$ and $\ell \cap TD=N$ lies on the perpendicular bisector $\ell_2$ of $CD.$ If $\ell_1 \cap \ell_2=P,$ then $TP$ bisects $\angle DTB.$ Proof: Let $P_1,P_2$ be the projections of $P$ onto $BM,DN$ respectively. We must show $PP_1=PP_2.$ Clearly, $\ell_1$ bisects $\angle BMA$ and $\ell_2$ bisects $\angle CND.$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); 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 = -9.92, xmax = 11.12, ymin = -8.21, ymax = 4.55; /* image dimensions */ /* draw figures */ draw((-2,3)--(-2,-7), linewidth(0.5)); draw((-6,2)--(-2,-6), linewidth(0.5)); draw((-2,2)--(-6,-4), linewidth(0.5)); draw((-3.142857142857143,-1.5714285714285714)--(-0.8571428571428572,-1.8571428571428572), linewidth(0.5)); draw((-6,2)--(-0.8571428571428572,-1.8571428571428572), linewidth(0.5)); draw((-0.8571428571428572,-1.8571428571428572)--(-6,-4), linewidth(0.5)); draw((-8.236923076923077,1.8246153846153845)--(0.9476923076923081,-4.298461538461538), linewidth(0.5)); draw((-7.78,-3.89)--(1.1,0.55), linewidth(0.5)); draw((-3.142857142857143,-1.5714285714285714)--(-2.4571428571428577,-0.6571428571428574), linewidth(0.5)); draw((-3.142857142857143,-1.5714285714285714)--(-2.703296703296704,-2.626373626373627), linewidth(0.5)); /* dots and labels */ dot((-2,3),linewidth(1pt) + dotstyle); label("$\ell$", (-1.92,3.03), NE * labelscalefactor); dot((-6,2),dotstyle); label("$B$", (-6.28,2.29), NE * labelscalefactor); dot((-2,-6),dotstyle); label("$A$", (-1.68,-6.29), NE * labelscalefactor); dot((-2,2),dotstyle); label("$C$", (-1.66,1.87), NE * labelscalefactor); dot((-6,-4),dotstyle); label("$D$", (-6.46,-4.61), NE * labelscalefactor); dot((-2,-2.3333333333333335),linewidth(4pt) + dotstyle); label("$N$", (-1.84,-3.05), NE * labelscalefactor); dot((-2,-1),linewidth(4pt) + dotstyle); label("$M$", (-1.86,-0.55), NE * labelscalefactor); dot((-3.142857142857143,-1.5714285714285714),linewidth(4pt) + dotstyle); label("$P$", (-3.32,-1.27), NE * labelscalefactor); dot((-0.8571428571428572,-1.8571428571428572),linewidth(4pt) + dotstyle); label("$T$", (-0.42,-2.13), NE * labelscalefactor); dot((0.9476923076923081,-4.298461538461538),linewidth(1pt) + dotstyle); label("$\ell_1$", (1.16,-4.55), NE * labelscalefactor); dot((1.1,0.55),linewidth(1pt) + dotstyle); label("$\ell_2$", (1.26,0.09), NE * labelscalefactor); dot((-2.4571428571428577,-0.6571428571428574),linewidth(4pt) + dotstyle); label("$P_1$", (-2.68,-0.23), NE * labelscalefactor); dot((-2.703296703296704,-2.626373626373627),linewidth(4pt) + dotstyle); label("$P_2$", (-2.76,-3.29), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Hence, $PP_1$ reflected over $PM$ coincides with $PP_2$ reflected over $PN.$ So $PP_1=PP_2.$ $\square$ Back to our main problem, let $M,N,P$ be the midpoints of the arcs $BC,CA,AB$ not containing the opposite vertices. Let the perpendicular bisectors of $BE,CF$ meet in $X.$ Define $Y,Z$ similarly. Claim: Let $\ell \cap \{YZ,ZX,XY\}=\{U,V,W\}.$ Then $AU,BV,CW$ concur at a point $T \in \omega.$ Proof: It suffices to show $AU \cap BV=T \in \omega.$ Then $$\angle VUA+\angle BVU=2\angle VUZ+2\angle ZVU=2(\pi-\angle UZV)=2(\pi-(\pi/2-\angle C))=\pi+2\angle C.$$Thus, $\angle ATB=\angle UTV=\angle C$ and so $T \in \omega.$ $\square$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; usepackage("amsmath"); size(12cm); 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 = -9.37120209719099, xmax = 10.106666947660992, ymin = -9.999374873740129, ymax = 7.062441361200569; /* image dimensions */ pen ffwwzz = rgb(1,0.4,0.6); pen qqzzcc = rgb(0,0.6,0.8); pen ttffqq = rgb(0.2,1,0); pen ccqqqq = rgb(0.8,0,0); pen ffxfqq = rgb(1,0.4980392156862745,0); pen wwzzff = rgb(0.4,0.6,1); pen ttffcc = rgb(0.2,1,0.8); pen ccwwff = rgb(0.8,0.4,1); pen ffqqff = rgb(1,0,1); draw((-5.012343934979869,2.9028452864048075)--(-6.06,-4.89)--(4.8,-5.19)--cycle, linewidth(0.3) + ffwwzz); draw((-4.00716555623139,-0.680048935115538)--(4.025909449710277,2.470396792494808)--(0.7487825934487295,-6.27214244011035)--cycle, linewidth(0.5) + wwzzff); draw((-6.875237851081766,-0.8135557545653594)--(3.532483974480068,3.268189123360688)--(-0.7133900752232645,-8.058720723082232)--cycle, linewidth(0.5) + ttffcc); draw((1.4180996959754038,-4.48657499504406)--(1.9128557813097669,-4.6720335227566)--(2.098314309022307,-4.177277437422236)--(1.6035582236879438,-3.9918189097096963)--cycle, linewidth(0.5) + ffqqff); draw((-3.4675937993211168,-1.035992191870934)--(-2.9756969092413126,-0.8430779647996691)--(-3.168611136312578,-0.3511810747198648)--(-3.660508026392382,-0.54409530179113)--cycle, linewidth(0.5) + ffqqff); draw((-2.38862434979392,-1.7675786358396293)--(-2.730936045857668,-1.3650849685420594)--(-3.133429713155238,-1.7073966646058074)--(-2.79111801709149,-2.1098903319033773)--cycle, linewidth(0.5) + ffqqff); /* draw figures */ draw((-5.012343934979869,2.9028452864048075)--(-6.06,-4.89), linewidth(0.3) + ffwwzz); draw((-6.06,-4.89)--(4.8,-5.19), linewidth(0.3) + ffwwzz); draw((4.8,-5.19)--(-5.012343934979869,2.9028452864048075), linewidth(0.3) + ffwwzz); draw(circle((-0.5367868969260705,-1.6656856687237636), 6.39547385960109), linewidth(0.5) + qqzzcc); draw((-5.012343934979869,2.9028452864048075)--(-2.3086721178048952,-3.9910358899870686), linewidth(0.3) + ttffqq); draw((-6.06,-4.89)--(0.4777639658170194,0.6702193361932457), linewidth(0.3) + ttffqq); draw((-2.9825701579728308,-2.272716016839372)--(4.8,-5.19), linewidth(0.3) + ttffqq); draw(circle((0.8850916341135904,-1.3377551680424133), 4.9362696345836925), linewidth(0.3) + linetype("4 4") + blue); draw((5.538185438841105,9.1354829668408)--(-7.53203961595977,-12.728881768506938), linewidth(0.5) + ccqqqq); draw((3.532483974480068,3.268189123360688)--(5.695094028914671,-0.22841533403262995), linewidth(0.5) + linetype("2 2") + ffxfqq); draw((5.695094028914671,-0.22841533403262995)--(-6.875237851081766,-0.8135557545653594), linewidth(0.5) + linetype("2 2") + ffxfqq); draw((-0.7133900752232645,-8.058720723082232)--(5.695094028914671,-0.22841533403262995), linewidth(0.5) + linetype("2 2") + ffxfqq); draw((-4.00716555623139,-0.680048935115538)--(4.025909449710277,2.470396792494808), linewidth(0.5) + wwzzff); draw((4.025909449710277,2.470396792494808)--(0.7487825934487295,-6.27214244011035), linewidth(0.5) + wwzzff); draw((0.7487825934487295,-6.27214244011035)--(-4.00716555623139,-0.680048935115538), linewidth(0.5) + wwzzff); draw((-6.875237851081766,-0.8135557545653594)--(3.532483974480068,3.268189123360688), linewidth(0.5) + ttffcc); draw((3.532483974480068,3.268189123360688)--(-0.7133900752232645,-8.058720723082232), linewidth(0.5) + ttffcc); draw((-0.7133900752232645,-8.058720723082232)--(-6.875237851081766,-0.8135557545653594), linewidth(0.5) + ttffcc); draw((-5.012343934979869,2.9028452864048075)--(5.695094028914671,-0.22841533403262995), linewidth(0.5) + linetype("4 4") + ccwwff); draw((-6.06,-4.89)--(5.695094028914671,-0.22841533403262995), linewidth(0.5) + linetype("4 4") + ccwwff); draw((7.517965639467718,9.875921795680743)--(4.8,-5.19), linewidth(0.5) + linetype("4 4") + ccwwff); label("$\ell$",(2.1610912119732912,5.792145553927612),SE*labelscalefactor); /* dots and labels */ dot((-5.012343934979869,2.9028452864048075),dotstyle); label("$A$", (-5.361052587956759,3.1768306566009357), NE * labelscalefactor); dot((-6.06,-4.89),dotstyle); label("$B$", (-6.730979438937396,-5.291808058552111), NE * labelscalefactor); dot((4.8,-5.19),dotstyle); label("$C$", (5.125114762276854,-5.6903322333828426), NE * labelscalefactor); dot((-2.9825701579728308,-2.272716016839372),linewidth(4pt) + dotstyle); label("$I$", (-3.5676938012184682,-2.2779689863947032), NE * labelscalefactor); dot((-2.3086721178048952,-3.9910358899870686),linewidth(4pt) + dotstyle); label("$D$", (-2.2724902330185923,-4.644206274452172), NE * labelscalefactor); dot((-1.5928835526241125,-2.7936378194193923),linewidth(4pt) + dotstyle); label("$F$", (-1.8739660581878612,-2.452323312883148), NE * labelscalefactor); dot((0.4777639658170194,0.6702193361932457),linewidth(4pt) + dotstyle); label("$E$", (0.7413488391388117,0.2875303890781316), NE * labelscalefactor); dot((0.7487825934487295,-6.27214244011035),linewidth(4pt) + dotstyle); label("$X$", (0.6915333172849703,-6.836089236021196), NE * labelscalefactor); dot((4.025909449710277,2.470396792494808),linewidth(4pt) + dotstyle); label("$Y$", (4.303158651688471,2.62885991620868), NE * labelscalefactor); dot((-4.00716555623139,-0.680048935115538),linewidth(4pt) + dotstyle); label("$Z$", (-4.6885430429299,-1.1571197446832704), NE * labelscalefactor); dot((5.695094028914671,-0.22841533403262995),linewidth(4pt) + dotstyle); label("$T$", (6.0467019165729194,-0.36007139502180724), NE * labelscalefactor); dot((0.7968593461338032,1.204014106471358),linewidth(4pt) + dotstyle); label("$U$", (0.3428246643080805,1.6823650009856923), NE * labelscalefactor); dot((-6.875237851081766,-0.8135557545653594),linewidth(4pt) + dotstyle); label("$P$", (-7.478212266745017,-0.9329498963409839), NE * labelscalefactor); dot((-0.7133900752232645,-8.058720723082232),linewidth(4pt) + dotstyle); label("$M$", (-0.8776556211110335,-8.753986827394092), NE * labelscalefactor); dot((3.532483974480068,3.268189123360688),linewidth(4pt) + dotstyle); label("$N$", (3.630649106661612,3.4757237877239846), NE * labelscalefactor); dot((-1.8474425865336201,-3.2194737587960316),linewidth(4pt) + dotstyle); label("$V$", (-1.4754418833571301,-3.5980803155215013), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: The lines $MX, NY, PZ$ all pass through $T.$ Proof: It suffices to show $PZ$ passes through $T.$ For this, apply the lemma on $\triangle TBE, \triangle TAD,$ with $\ell=UD,$ and the perpendicular bisectors of $EB$ passes through $V \in TB,\ell,$ and the perpendicular bisector of $AD$ passes through $U \in TA, \ell.$ Since these perpendicular bisectors meet at $Z,$ we have, by the lemma, that $TZ$ is the bisector of $\angle ATB,$ and hence passes through $P.$ $\square$ Now, clearly the corresponding sides of $\triangle XYZ$ are parallel to that of $\triangle MNP,$ as their sides are perpendicular to $AI,BI,CI.$ Hence, a homothety at $T$ takes $\triangle XYZ$ to $\triangle MNP$ and thus, $(XYZ)$ is tangent to $\omega$ at $T,$ and we are done. $\blacksquare$

Let the perpendicular bisector of BE and FC intersect at X,BE and AD intersect at Z and FC and AD intersect at Y, Let l intersect XZ,YZ,XY at M,K,N then BM,CN,KA intersect at the point where $(ABC)$ and $\Theta$ are tangent.you can see the solution @belove for more information.

Beautiful !!! This is a hardcore angle chasing solution. Ig my solution is entirely different from the other solutions here. ISL 2018 G5 wrote: Let $ABC$ be a triangle with circumcircle $\Omega$ and incentre $I$. A line $\ell$ intersects the lines $AI$, $BI$, and $CI$ at points $D$, $E$, and $F$, respectively, distinct from the points $A$, $B$, $C$, and $I$. The perpendicular bisectors $x$, $y$, and $z$ of the segments $AD$, $BE$, and $CF$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\Omega$. First we start off with a Lemma which basically kills this problem. Lemma:- $ABC$ be a triangle with Incenter $I$ and a line $(\ell)$ intersects $\{AI,BI,CI\}$ at $\{D,E,F\}$ respectively. Let the Perpendicular bisectors of $AD,BE,CF$ intersect $\ell$ at $\{L,T,K\}$ respectively. Then $\{AL,BT,CK\}$ concurs at a point $X$ which lies on $\odot(ABC)$. Let $AL\cap\odot(ABC)=X$ and $CX\cap\ell=K'$. Let $\angle ADL=\angle LAD=\alpha$. So, $\angle K'FC=\alpha-(A/2+C/2)$ and $\angle XCF=\angle XCB-(C/2)=\angle XAB-(C/2)=\alpha-(A/2+C/2)=\angle K'FC\implies K'\equiv K$. Similarly $L'\equiv L$. Hence, we get that $\{AL,BT,CK\}$ concurs at $X\in\odot(ABC)$. If $\{\ell_a,\ell_b,\ell_c\}$ are the perpendicular bisectors of $\{AD,BE,CF\}$ then define the triangle formed $\{\ell_a,\ell_b,\ell_c\}$ be $\triangle PQR$ where $P,Q,R$ are against $A,B,C$ respectively. Claim:- $X\in\odot(PQR)$ Notice that by Lemma we get that $R$ and $Q$ are the Incenter of $\triangle XLT$ and the $L-\text{Excenter}$ of $\triangle XLK$ respectively. Hence, $\angle LXR=C/2$ and $\angle AXQ=B/2$. So, $\angle RXQ=90^\circ+B/2=\angle BIC=\pi-\angle RPQ\implies X\in\odot(PQR)$. Now we just need to show that $\{\odot(ABC),\odot(PQR)\}$ are tangent to each other at $X$. So for this draw a tangent $(\tau)$ to $\odot(ABC)$ at $X$. We will show that $\tau$ is tangent to $\odot(PQR)$ at $X$ as well. So, $\angle (\tau,XR)=\angle(\tau,XB)-\angle RXB=\angle XCB-(C/2)=\angle XCF+C/2-C/2=\angle XCF=\angle KFC$ and $\angle XQR=\pi-B/2+\angle KFC-C/2-(\pi)/2+(A/2)=\angle KFC=\angle (\tau,XR)=\angle XQR\implies \tau$ is tangent to $\odot(PQR)$ at $X$. Hence, $\{\odot(ABC),\odot(PQR)\}$ are tangent to each other at $X$. $\blacksquare$

Dear amar if you noticed one post befor yours i introduced the lemma,while i leave the rest to the reader,so i dont think is all original. P.s Nice that you post the whole solution,congrats,i really liked it

The pinnacle of all olympiad geometry problems. Suppose that the perpendicular bisectors of $\overline{AD}, \overline{BE}, \overline{CF}$ intersect at points $X,Y,Z$. Let $\ell_A, \ell_B, \ell_C$ be the reflection of $\ell$ across $\overline{YZ}, \overline{XZ}, \overline{XY}$, respectively. By some angle chasing, we find that $\ell_A, \ell_B, \ell_C$ concur on $\Omega$. Moreover, this implies that $\ell$ is the Steiner line of $P$ with respect to $\triangle XYZ$, so $P$ lies on $\Omega$ as well. Let $\overline{PI}$ and $\ell$ meet at a point $H$. Consider the homothety centered at $P$ sending $I$ to $H$. Suppose it sends $\triangle ABC$ to a triangle $\triangle A'B'C'$ with incenter $H$. Note now that $X$ is the circumcenter of $\triangle A'HB'$, so it lies on $(A'B'C')$. Similar results apply to $Y$ and $Z$, so we find that $(XYZ)$ and $(ABC)$ are homothetic at $P$, as desired. Remarks: Moreover, the Simson line of $P$ with respect to $\triangle ABC$ is actually perpendicular to $\ell$. Finally, $H$ is actually the orthocenter of $\triangle XYZ$. Surprisingly, despite the importance of $H$ in the solution, this fact is not actually used.

A moving point solution. Let $X = \ell_B\cap \ell_C$, $Y= \ell_C \cap \ell_A$, $Z = \ell_A \cap \ell_B$. Let $L,M,N$ be the second intersection of $AI,BI,CI$ with $\odot(ABC)$. Then $LM \parallel XY$, $MN \parallel YZ$, $NL \parallel ZX$, thus $LX,MY,NZ$ are concurrent at their homothety center $T$. It suffices to show that $T$ lies on $\odot (ABC)$. We fixed $A,B,C,D$, let $E$ moving on $BI$ has degree 1, then $F = DE \cap CI$ has degree 1. $\ell_A$ is fixe. The midpoint of $BE$,$CF$ has degree 1. As the directions of $BI,CI$ are fixed, their orthogonal directions are fixed as well. Hence $\ell_B, \ell_C$ has degree 1. And $Y=\ell_A \cap \ell_B$, $Z = \ell_C \cap \ell_A$ both have degree 1. Finally the intersection $T = MY \cap NZ$ has degree 2. Finally, the statement $A,N,T,M$ are concyclic has degree $deg(AN) + deg(NT) + deg(TM) +deg(MA) = deg(AN) + deg(NZ) + deg(YM) +deg(MA) =0+1+1+0 =2$. Therefore it suffices to check 3 special cases. Let $E = I$, then $F=I$, $YM$, $ZN$ are the perpendicular bisector of $BI$ and $CI$ intersects at $L \in \odot (ABC)$. Let $E$ such that $DE \parallel CI$, then $F = CI \cap \ell_\infty$. The midpoint of $CF$ is still $F$. Hence $\ell_C =\ell_\infty$ and $Y = \ell_A \cap \ell_\infty$. This implies that $YM$ is parallel to $\ell_A$, which is orthogonal to $AI$, note that $MN \perp AI$ as well, therefore $N$ lies on $MY$. This means $N = MY \cap NZ$ which lies on $\odot (ABC)$. Let $E = BI \cap \ell_\infty$ gives the same reasoning as the second bullet. [EDIT: corrected the typo (see the post below) on the degree of $T$, it does not affect the main proof]

Flash_Sloth wrote: A moving point solution. Let $X = \ell_B\cap \ell_C$, $Y= \ell_C \cap \ell_A$, $Z = \ell_A \cap \ell_B$. Let $L,M,N$ be the second intersection of $AI,BI,CI$ with $\odot(ABC)$. Then $LM \parallel XY$, $MN \parallel YZ$, $NL \parallel ZX$, thus $LX,MY,NZ$ are concurrent at their homothety center $T$. It suffices to show that $T$ lies on $\odot (ABC)$. We fixed $A,B,C,D$, let $E$ moving on $BI$ has degree 1, then $F = DE \cap CI$ has degree 1. $\ell_A$ is fixe. The midpoint of $BE$,$CF$ has degree 1. As the directions of $BI,CI$ are fixed, their orthogonal directions are fixed as well. Hence $\ell_B, \ell_C$ has degree 1. And $Y=\ell_A \cap \ell_B$, $Z = \ell_C \cap \ell_A$ both have degree 1. Finally the intersection $T = MY \cap NZ$ has degree 1. Finally, the statement $A,N,T,M$ are concyclic has degree $deg(AN) + deg(NT) + deg(TM) +deg(MA) = 0+1+1+0 =2$. Therefore it suffices to check 3 special cases. Let $E = I$, then $F=I$, $YM$, $ZN$ are the perpendicular bisector of $BI$ and $CI$ intersects at $L \in \odot (ABC)$. Let $E$ such that $DE \parallel CI$, then $F = CI \cap \ell_\infty$. The midpoint of $CF$ is still $F$. Hence $\ell_C =\ell_\infty$ and $Y = \ell_A \cap \ell_\infty$. This implies that $YM$ is parallel to $\ell_A$, which is orthogonal to $AI$, note that $MN \perp AI$ as well, therefore $N$ lies on $MY$. This means $N = MY \cap NZ$ which lies on $\odot (ABC)$. Let $E = BI \cap \ell_\infty$ gives the same reasoning as the second bullet. Sorry, why $T = MY \cap NZ$ has degree $1$? Isn't it degree $1+1=2$?

Let the perpendicular bisector of $AD$ be $\ell_D$. Define $\ell_E,\ell_F$ likewise. Let the reflection of $\ell$ over line $AD$ be $\ell_A$. Define $\ell_B,\ell_C$ likewise. Let $\ell_E\cap\ell_F=X,\ell_F\cap\ell_D=Y,\ell_D\cap\ell_E=Z$. Claim: Lines $\ell_A,\ell_B,\ell_C$ concur at a point $T$ on $(ABC)$. Solution: By symmetry it suffices to show $\ell_A\cap\ell_C=T$ lies on $(ABC)$, which is true by the following angle chase: \[\measuredangle ATC=\measuredangle ACT+\measuredangle TAC=\measuredangle ACI+\measuredangle ICT+\measuredangle TAI+\measuredangle IAC=\measuredangle AIC+\measuredangle DFI+\measuredangle IDF=\]\[\measuredangle AIC+\measuredangle DIF=2\measuredangle AIC=\measuredangle ACB+\measuredangle BAC=\measuredangle ABC.\] Now, let line $AI$ intersect $(ABC)$ again at $M$, line $BI$ at $N$, and line $CI$ at $P$. Note that by homothety at $I$, $\angle NMP=180^\circ-\angle BIC=\angle ZXY$, so $\triangle XYZ\sim\triangle MNP$. Moreover, the corresponding sides of these triangles are parallel, so $XM,YN,ZP$ concur at a point $T'$. It suffices to show $T'=T$. By symmetry, it would be enough to show $YN,ZP,\ell_A$ concur. The problem is now a sitting duck for the METHOD OF MOVING POINTS. Animate line $\ell_A$ passing through $A$ with degree $1$, then $YN$ has degree $1$ because of the projective map \[\ell_A\mapsto \ell\mapsto \ell\cap CI=F\mapsto \ell_F\mapsto \ell_F\cap\ell_D=Y\mapsto YN.\]Similarly $ZP$ has degree $1$, so it suffices to check $4$ cases for $\ell_A$. $E=I$: Then $XYZ$ and $MNP$ are clearly homothetic with center $M$. $E=B$: By symmetry we can replace this case with the case $\ell$ passes through $A$. Then $\ell_A$ and $\ell$ are isogonal. But similarly to before it suffices to check four values of $E$, so $E=I$ is trivial as before. The case of $\ell$ parallel to one of $AI,BI,CI$ suffices for the last two cases (which will be the next case to check in this approach), so we need to resolve $E=B$. Then $T=C$, so $Z$ is the $C$-excenter. Let $CF$ have midpoint $Q$ so it suffices to show \[\frac{CZ}{CP}=\frac{YZ}{NP}=2\cdot\frac{ZQ}{ZC},\]or equivalently \[CP^2+IZ\cdot CP+PZ^2=(CP+PZ)^2=CZ^2=2ZQ\cdot CP=(IZ+2FP+CF)\cdot CP=\]\[IZ\cdot CP+CP^2+FP\cdot CP.\]But $FP\cdot CP=FP^2+FA\cdot FB=PZ^2$, so we are done. Now $\ell$ is parallel to one of $AI,BI,CI$. WLOG it is parallel to $AI$. Then $\ell_D$ is considered to be the line at infinity, so $Y,Z$ are the points at infinity along lines $\ell_F,\ell_E$ respectively. That is, circle $(XYZ)$ is essentially considered to be the external bisector of $\angle VXU$ where $V$ is the midpoint of $BE$ and $U$ is the midpoint of $CF$. Since the angle between lines $\ell_E,\ell_F$ and line $BC$ is fixed as $E$ varies, it suffices to check $X$ lies on the tangent to $(ABC)$ at $M$. Observe that as $\ell$ moves linearly, so do $E,F,\ell_E,\ell_F,\ell_E\cap MM,\ell_F\cap MM$. Thus it is sufficient to check two values of $E$. At $E=I$ the result is obvious because $X=M$. Compute that the ratio of change in $X^*=\ell_E\cap MM$ on line $MM$ to change in $E$ on line $BI$ is \[\frac 12\frac{1}{\cos\angle IBC}\]and the ratio of change $X^*=\ell_F\cap MM$ on line $MM$ to change in $F$ on line $CI$ is $\frac 12\frac{1}{\cos \angle ICB}$. Since the ratio of change in $F$ on line $CI$ to change in $E$ on line $BI$ is $\frac{\sin\angle IEF}{\sin\angle IFE}$, it suffices to check \[\cos\angle ICB=\cos\angle IBC\cdot \frac{\sin\angle AIB}{\sin\angle AIC}=\cos\angle IBC\cdot \frac{\sin (90^\circ+\angle ICB)}{\sin (90^\circ+\angle IBC)},\]which is trivial.

dame dame

Always use complex numbers... Let $X=y\cap z,Y=z\cap x,Z=x\cap y$ and $M_a,M_b,M_c$ be the arc-midpoints. Since $XY\perp CI\perp M_aM_b\implies XY\parallel M_aM_B$, we know that $\triangle XYZ\sim\triangle M_aM_bM_c$. Now it remains to show that $M_aX,M_bY,M_cZ$ concur on $(ABC)$. It suffices to show that if the $M_aX\cap M_bY=T\in (ABC)$, then $D,E,F$ are collinear. Let $a=p^2,b=q^2,c=r^2$. Then there is a $\lambda\in\mathbb{R}$ such that $x=\lambda t+(1-\lambda)m_a=\lambda t-(1-\lambda)qr$ and similarly for $y,z$. Now $f$ is the reflection of $B$ over $XY$, so $f=\frac{(1-\lambda)(p-q)\bar r-\lambda(1-\lambda)\bar q\bar rt-\lambda(1-\lambda)pr\bar t+(1-\lambda)^2p\bar q+\lambda(1-\lambda)\bar p\bar rt+\lambda(1-\lambda)\bar tqr-(1-\lambda)^2\bar pq}{(1-\lambda)(\bar p\bar r-\bar q\bar r)}\\=\frac{pq(p-q)-\lambda pt-\lambda p^2qr^2\bar t+(1-\lambda)(p^2r-q^2r)+\lambda qt+\lambda \bar tpq^2r^2}{q-p}=\lambda t-pq-(1-\lambda)(p+q)r+\lambda prq^2\bar t$. Analogous formulae hold for $d,e$. Observe that $\frac{d-e}{d-f}=\frac{(p-q)r(t+pq)}{(p-r)q(t+pr)}\in\mathbb{R}$, as desired.

Let $XYZ$ be the vertices of $\Theta$ and $M_AM_BM_C$ be the circumcevian triangle of $I$. Also let $\infty$ be the point at infinity of $\ell$ and $T$ its isogonal conjugate. It suffices to prove that $T-M_A-X$ and its analogues. From here it follows that $T$ is the exsimilicenter of $\Omega$ and $(\Theta)$ and as $T\in\Omega$, the two are tangent to eachother. Now fix $\infty$ and animate $\ell$ linearly on the pencil of lines through $\infty$. It's straightforward to see that $\deg X=1$ and thus $X$ lies on a fixed line. However letting $\ell=I\infty$ we have $X=M_A$ since $M_AM_BM_C$ is the unique triangle with sides bisecting $CI$, $AI$ and $BI$ $\ell=T\infty$ we have $X=T$ since $\angle DAT=\angle \infty AD=\angle TDA\implies T\in x$ and similarly $T\in y$ and $T\in z$ Thus this fixed line is precisely $TM_A$. The analogues follow by symmetry.

somehow, this is new (i think) Solved with rowechen.

It suffices to show $V \in (ABC)$. By definition, $B_1M_B:M_BV$ is symmetric in $A,B,C$, thus so is $$B_1M_B:M_BV=M_{AD}X:XA_2=DI:IA_3,$$and in particular because $D,E,F$ are collinear so are $A_3,B_3,C_3$ by homothety at $I$. Complex bash with $(ABC)$ as the unit circle, $M_A=a$, $M_B=b$, $M_C=c$, $V=v$. We easily get $A_3=a+v+bc\overline{v}$, and because $A_3,B_3,C_3$ are collinear, shifting them by $-(a+b+c+v)$ gives $b+c-bc\overline{v}, \dots$ are collinear. But $b+c-bc\overline{v}$ is the reflection of $V$ over $BC$, and by converse of steiner we're done. Remark. Considering the question with respect to $M_AM_BM_C$ is much more natural - you definitely don't want to deal with $A_1,B_1,C_1$ and look to lengths - orthocenter reflections is now very obvious. btw yall dont mind an extremely clean complex, right?

Let $X = \overline{AI} \cap \Omega \setminus A$ and similar for $Y$ and $Z$. Note that $I$ is the orthocenter of $\triangle XYZ$. Define $\ell'$ to be the line through $I$ parallel to $\ell$, and $T$ to be the Anti-Steiner point of $\ell'$ wrt $\triangle XYZ$. By definition $T \in (ABC)$. I claim that $T$ is the desired tangency point. Let $\triangle X'Y'Z' \equiv \Theta$. Consider the reflection of $\ell'$ over $\overline{YZ}$ (which is $\overline{TA}$ by definition) and the reflection of $\ell$ over $\overline{Y'Z'}$: They are parallel because $\ell \parallel \ell'$ and $\overline{YZ} \parallel \overline{Y'Z'}$. They both pass through $A$. So they must be the same line. It follows that $T$ is the Anti-Steiner point of $\ell$ wrt $\Theta$, which lies on $(X'Y'Z')$. Finally, if $A'$ is the point on $(X'Y'Z')$ with $\overline{X'A'} \perp \overline{Y'Z'}$, then $A'$ lies on the reflection of $\ell$ over $\overline{Y'Z'}$, which is $\overline{TA}$. Since $\triangle XYZ$ and $\triangle X'Y'Z'$ are homothetic, $\triangle ABC$ and $\triangle A'B'C'$ are also homothetic, from where it follows that $T$ is the center of homothety sending $(X'Y'Z')$ to $\Omega$. $\square$

An alternative way to finish after showing $\ell_x,\ell_y,\ell_z$ conur: Let $P,Q,R$ be intersections of $\ell$ with $\overline{YZ},\overline{ZX},\overline{XY}$. As shown previously, lines $AP,BQ,CR$ concur at some point $T \in \Omega$ and it suffices to show $\overline{TX}$ bisects $\angle BTC$. Just note that $X$ is the $T$-excenter of $\triangle AQR$ (a similar thing was in post #7 which considered the distance of $X$ to the sides of $\triangle AQR$, but doing such things in this type of problems seems a bit counter-inuitive to me, so I posted the synthetic way to think about that)

Oh god what have I done Let the vertices of $\Theta$ be $X,Y,Z$, and let the arc midpoints be $M_AM_BM_C$. Since $\triangle XYZ$ and $\triangle M_AM_BM_C$ have parallel sides, it suffices to show $XM_A,YM_B,ZM_C$ concur on $\Omega$. In particular, we show $YM_B$ and $ZM_C$ always meet on $\Omega$. To do this, we sin, fixing $ABCD$ and moving $E$ projectively on $BI$. We see all the other points move projectively, so it suffices to check three cases. When $E=F=I$, we see $YM_B$ and $ZM_C$ are the perpendicular bisecors of $BI$ and $CI$. These lines meet at $M_A$.

. When $E$ is the point such that $CM_BY$ are collinear (clearly such a point always exists since the map $E \mapsto F \mapsto Y$ is bijective), the case is the same as the above. This case is always distinct from the above case, since we've already shown that in the above case we have $BM_BY$ collinear.

Awesome problem! Let $\ell_a,\ell_b,\ell_c$ be our reflections across $AD.BE,CF$ Claim 1:-$\ell_a,\ell_b$ concur at a point $G \in \Omega $ $$\measuredangle AGB=\measuredangle(\ell_a,\ell_b)=\measuredangle(\ell_a,\ell)+\measuredangle(\ell_b,\ell)=2\measuredangle(\ell_a,AI)+2\measuredangle(\ell_b,BI) \cdots \quad\{\star \}$$$$\measuredangle(\ell_a,\ell_b)+\measuredangle(\ell_b,BI)+\measuredangle AGB=90+\frac{\measuredangle C}{2} \cdots \quad\{\clubsuit \}$$From $\clubsuit$ and $\star$ we get that $\measuredangle AGB=\measuredangle ACB \implies G \in \Omega$ Let $A1,B1,CI$ intersect $\Omega$ at $M_a,M_b,M_c$ respectively and also let the perpendicular bisectors intersect at $X,Y,Z$ It's quite trivial that $\triangle M_aM_bM_c$ is homothetic to $\triangle XYZ$ So it suffices to show that $XM_a,YM_c,ZM_b$ concur at $G$ Project $X$ on $\ell_b$ and $\ell_c$ Notice that, $d(\ell_b,X)=d(\ell_c,X)$ which implies $X$ lies on the angle bisector of $\angle BGC$ which implies $X-M_a-G$ are collinear and analogously $XM_a,YM_c,ZM_b$ concur at $G$ which also implies $G \in \odot(XYZ)$ Finally we finish off the problem, using homothety at $G$ sends $\triangle XYZ$ to $\triangle M_aM_bM_c$ and thus it's tangent at G as desired.

Let $\ell_a,\ell_b,$ and $\ell_c$ be the reflections of $\ell$ over $x,y,$ and $z$ and let $x,y,$ and $z$ form triangle $XYZ.$ Also, let $T=\ell_a\cap\ell_c,$ noticing $$\measuredangle CTA=\measuredangle(\ell,\ell_a)+\measuredangle(\ell_c,\ell)=2\measuredangle(\ell,x)+2\measuredangle(z,\ell)=2\measuredangle(z,x)=2\measuredangle AIC=\measuredangle CBA.$$Hence, $T=\ell_a\cap\ell_b\cap\ell_c\cap\Omega.$ Let $M_A,M_B,$ and $M_C$ be the midpoints of the arcs of $(ABC).$ We see $$\delta(Z,\ell_b)=\delta(Z,\ell)=\delta(Z,\ell_a)$$so $Z$ lies on the angle bisector of $\angle ATB$ and therefore on $\overline{TM_C}.$ We know $\overline{M_BM_C}$ and $\overline{YZ}$ are both perpendicular to $\overline{AD}$ so $\triangle M_AM_BM_C$ and $XYZ$ are homothetic at $T.$ $\square$

I hope there's something new. Let $AI,BI,CI$ intersect $\Omega$ again at $M_A,M_B,M_C$ and $y\cap z=X, z\cap x=Y, x\cap y=Z$. Then $M_AX,M_BY,M_CZ$ are concurrent on $T$, which is the homothety center that sends $\triangle XYZ$ to $\triangle M_AM_BM_C$. It suffices to show that $T\in\Omega$. Define $I_A,A'$ such that $II_A,IA'$ are diameter of $(BCI),(EFI)$ respectively. Then $\angle A'EI=\angle A'FI=\angle I_ABI=\angle I_ACI=90^\circ$ so midpoint $X_1$ of a segment $A'I_A$ satisfies $X_1B=X_1C$ and $X_1C=X_1E$, means that $X_1$ coincides with $X$. Note that $M_A$ is the midpoint of $II_A$ we get $M_AX\parallel IA'$. Similarly, if we define $B'$, it holds that $M_BY\parallel IB'$ and we have $T\in\Omega$ by simple angle chasing.

Denote by $\ell_A,\ell_B,\ell_C$ reflections of $\ell$ wrt $x,$ $y,$ $z$ respectively. Claim. $\ell_A,\ell_B,\ell_C$ concur at $X\in \odot (ABC).$ Proof. $\angle (\ell_A,\ell_B)=\angle (\ell_a,\ell)+\angle (\ell,\ell_B)=2\angle (x,y)=2\angle (AI,BI)=\angle (AC,BC),$ so $\ell_A,\ell_B$ concur on $\odot (ABC).$ By similar argument $\ell_C$ passes through the same point. Consider homothety wrt $X,$ which maps $ABCI$ onto $A'B'C'Y,$ where $Y=XI\cap \ell.$ We see that $x,$ $y,$ $z$ are perpendicular bisectors of $A'Y,B'Y,C'Y$ and therefore vertices of $\Theta$ are midpoints of arcs $A'B',B'C',C'A'.$ The conclusion follows.

Prolly one of the best problems solved so far Let the $\triangle PQR \equiv \Theta$ $P\equiv y \cap z$ and so on It is known that the triangle determined by the perpendicular bisectors of $AI, BI, CI$ is the triangle formed by joining midpoints of the minor arcs $BC, CA, AB$. Let this triangle be $\triangle LMN$ Hence clearly $\triangle LMN \sim \triangle PQR$ Note that by symmetry it suffices to show that $MQ \cap NR \in \odot (LMN) \equiv \odot (ABC)$ Draw a circle with diameter $MQ$, and let it intersect $x$ and $z$ at $V$ and $W$. Note that due to the above construction, $MW \perp z$ and $MV \perp x$ But $ID \perp x$ and $IF \perp z \implies \angle VWM=\angle DFI$ $\implies \triangle IDF \sim \triangle MVW$ Due to the cyclic quad $\angle VWM=\angle VQM$ We are done since $x \vert \vert MN \implies \angle QMN=\angle DFI$

NoctNight wrote:

For there to be an anti-Steiner point, the line has to pass through the orthocenter. Your proof is circular reasoning.

Let $A_1$, $B_1$, $C_1$ be $y\cap z$, $z\cap x$, and $x\cap y$, respectively. Let $A_2$, $B_2$, $C_2$ be $AI$, $BI$, $CI$'s intersection of $\omega$, respectively. Since $A_1B_1 \parallel A_2B_2$ due to perpendicularity to $CI$, $\triangle A_1B_1C_1$ is a homothety of $\triangle A_2B_2C_2$. It suffices to show that the center of homothety, which we will denote as $P$, is on $\omega$. $~$ Let the circle with diameter $A_1A_2$ intersect $z$ and $y$ at $G$ and $H$, respectively. Similarly, the circle with diameter $B_1B_2$ intersects $x$ and $z$ at $J$ and $K$, respectively and the circle with diameter $C_1C_2$ intersects $y$ and $x$ at $L$ and $M$. All angles we write are directed. $~$ Note that $\angle A_2HA_1=90^\circ$ therefore $A_2H\parallel IE$. Furthermore, $\angle GA_2H=\angle GA_1H$. Since $IE$ and $IF$ are perpendicular to the two lines making up that angle, $\angle FIE=\angle GA_2H$, thus proving a homothety between $\triangle A_2GH$ and $\triangle IFE$. Analogously, $\triangle B_2JK$ is homothetic to $\triangle IDF$ and $\triangle C_2LM$ to $\triangle IED$. $~$ Therefore, \begin{align*} \angle B_2PA_2 &= \angle A_2A_1B_1 + \angle A_1B_1B_2 \\ &= \angle A_2A_1G + \angle KB_1B_2 \\ &= \angle A_2HG + \angle KJB_2 \\ &= \angle IEF + \angle FDI \\ &= \angle DIE \\ &= \angle B_2C_2A_2 \end{align*}Therefore, $P$ is on $\omega$ as desired.

Let $\ell_a$, $\ell_b$, $\ell_c$ denote the reflections of $\overline{DEF}$ over the three perpendicular bisectors. Claim. $\ell_a$, $\ell_b$, $\ell_c$ concur at some point $K$ on $T$ on $\omega$. Proof. Annoying angle chasing; let $\ell_a$ intersect $\omega$ at $K_A$, and show that $K_A$, $K_B$, $K_C$ are consistent. $\blacksquare$ Let the perpendiculars intersect at $P, Q, R$. Now, let $X = \overline{AK} \cap \overline{DEF}$, which lies on $\overline{QR}$; define $Y$ and $Z$ similarly. Claim. $R$ is the incenter of triangle $KXY$. Proof. Because $\overline{XR}$ and $\overline{YR}$ bisect their respective angles by definition. $\blacksquare$ So $\overline{KR}$ passes through the arc midpoint of $\widehat{AB}$. But letting $MLN$ be the triangle formed by arc midpoints, the triangles $XYZ$ and $MLN$ are homothetic at $K$ as $K, X, M$ are collinear and their sides are parallel (both perpendicular to $\overline{AI}, \overline{BI}, \overline{CI}$). It follows that $K$ takes $(XYZ)$ to $(ABC)$, hence it is the desired tangency point.

Let $l_A, l_B, l_C$ be the reflections of the line $l$ in the lines $x, y$ and $z$, respectively. Then we'll prove the following claim: Claim: $l_A, l_B, l_C$ are concurrent at a point $T$, which lies on $\omega$. Proof. Let $M_A, M_B, M_C$ be the midpoints of arcs $BC, CA$ and $AB$, respectively. Then it's clear that $I$ is the orthocenter of $M_AM_BM_C$. Let $T = l_A \cap \omega$. Then we'll prove that $BT$ is the reflection of $l$ wrt $y$. Note that \begin{align*} \measuredangle(l, M_BB) = \measuredangle(DE, EI) =\\ \measuredangle(DI, EI) + \measuredangle(DE, DI) =\\ \measuredangle(AM_A, BM_B) + \measuredangle(DA, AT) =\\ \measuredangle(M_BM_C, M_AM_C) + \measuredangle(DA, AT) =\\ \measuredangle(M_AM_B, M_BT) + \measuredangle(M_BT, TM_A) =\\ \measuredangle(M_BM_A, M_AT) = \measuredangle(M_BB, BT) \end{align*}, so $BT$ coincides with $l_B$. Thus $l_B$ passes through $T$, completing the claim. $\blacksquare$ Now let $A' = y \cap z$ and define $B', C'$ similarly. Then the reflections of $T$ across the sides of triangle $A'B'C'$ are lies on $l$, so by Simson line, $T$ lies on $(A'B'C')$. Now note that $x \parallel M_BM_C$, $y \parallel M_AM_C$ and $z \parallel M_AM_B$, so triangles $M_AM_BM_C, A'B'C'$ are homothetic, and their circumcircles share common point, namely $T$, hence $\omega, (A'B'C')$ are tangent each other at $T$, as required. $\blacksquare$

very interesting problem i actually tried it a while back and never finished.

as a simpler sketch 1) Guess that the desired concurrency point $P$ is such that its Simson line is parallel to $\ell$. Let $M_A$ be the midpoint of arc $\widehat{BC}$. 2) Vary $\ell$ of a fixed orientation; in this case $D$, $E$, and $F$ move linearly, hence $y\cap z$ moves linearly. 3) Prove that $M_A$ and $P$ both lie on the locus (there exists a line $\ell$ reaching both these points). The first case is trivial ($D=E=F=I$), and the second case requires a bit more work. Done.

TelvCohl wrote: Let $ T $ be the pole of the Simson line of $ \triangle ABC $ with the direction $ \perp \ell, $ $ T_a, T_b, T_c $ be the intersection of $ AI, BI, CI $ with the parallel $ \tau $ from $ T $ to $ \ell $ and let $ \triangle XYZ $ be the triangle $ \Theta, $ then note that the isogonal conjugate of $ T $ WRT $ \triangle ABC $ is the point at infinity with the direction $ \parallel \ell $ we get $ T $ lies on the perpendicular bisector of $ BT_b, CT_c, $ so consider the homothety $ \mathbf{H}^{B}_{\frac{1}{2}}, \mathbf{H}^{C}_{\frac{1}{2}} $ we conclude that $ TX $ passes through the midpoint $ M_a $ of arc $ BC $ in $ \odot (ABC) $ and $ \frac{TM_a}{XM_a} = \frac{\text{dist} (I, \tau)}{\text{dist} (I, \ell)}. $ I cannot understand this solution, lol

Let $M_a$, $M_b$, $M_c$ be the midpoints of arcs $\widehat{BC}, \widehat{CA}, \widehat{AB}$ respectively, and let $X = y \cap z$, $Y = x \cap z$, and $Z = x \cap y$. The sides of $\triangle M_aM_bM_c$ and $\triangle XYZ$ are parallel, so $\triangle M_aM_bM_c \sim \triangle XYZ$, so it suffices to prove that their center of homothety lies on $\Omega$. Let $\ell_a$, $\ell_b$, $\ell_c$ be the reflections of $\ell$ across $x$, $y$, and $z$, respectively. Claim: $\ell_a$, $\ell_b$, $\ell_c$ concur at a point $T$ on $\Omega$. Proof. We have that \begin{align*} \measuredangle(\ell_a, \ell_b) &= \measuredangle(\ell_a, \ell) + \measuredangle(\ell, \ell_b) \\ &= 2\measuredangle(AI, \ell) + 2\measuredangle(\ell, BI) \\ &= \measuredangle ACB,\end{align*}so $\ell_a$ and $\ell_b$ intersect on $\Omega$. Similarly, $\ell_b \cap \ell_c$ and $\ell_c \cap \ell_a$ also lie on $\Omega$. Note that $A \in \ell_a$, $B \in \ell_b$, and $C \in \ell_c$, so either $\ell_a$, $\ell_b$, $\ell_c$ all concur on $\Omega$ or else $\ell_a$, $\ell_b$, $\ell_c$ must correspond to the sides of $\triangle ABC$. To prove the second case is impossible, suppose WLOG that $\ell_b = BC$. Then since $\ell_a$ passes through $A$, we have $\ell_a = AB$. Then $\measuredangle(\ell_a, \ell_b) = \measuredangle(AB, BC) = \measuredangle ACB$, which is impossible because $\measuredangle(AB, BC) = \measuredangle BAC + \measuredangle ACB \neq \measuredangle ACB$ since $0^\circ < \angle BAC < 180^\circ$. Therefore, $\ell_a$, $\ell_b$, $\ell_c$ must concur at a point $T$ on $\Omega$. $\square$ Now note that $Z$ is the $T$-excenter of $\triangle TXY$, so $M_c,$ $Z$, $T$ are collinear. Similarly, $M_B, Y, T$ and $M_A, X, T$ are also collinear. Therefore, $T$ is the center of the homothety sending $\triangle XYZ$ to $\triangle M_aM_bM_c$, so $(XYZ)$ is tangent to $\Omega$ at $T$.

Let $\triangle MNP$ be the circummidarc triangle and let $\Theta=\triangle XYZ$ such that $XYZ,MNP$ are homothetic (by fact $5$) with center $T$. It suffices to show $T\in\Omega$. Fix $D$ and vary $\ell$ with degree $1$. Then $E,F$ have degree $1$, so $y,z$ have degree $1$ and $x$ is fixed, so $Y,Z$ have degree $1$ and thus $T=NY\cap PZ$ has degree $2$. Thus we need to check $3$ cases of $\ell$. First if $\ell=AI$ and $E=F=I$ then $T=M$ by fact $5$. Now let $x$ meet $\Omega$ at $K$. If $K\in\ell$ then \[\measuredangle BEK+\measuredangle KDA=\measuredangle BIA=\measuredangle BNM+\measuredangle NMA=\measuredangle MAC+\measuredangle CBN=\measuredangle MKN=\measuredangle MKK+\measuredangle KKN=\measuredangle MAK+\measuredangle KBN=\measuredangle KBE+\measuredangle DAK.\]Thus $K\in x\implies\measuredangle KDA=\measuredangle DAK\implies\measuredangle BEK=\measuredangle KBE\implies K\in y$ Similarly $K\in z$, so $X,Y,Z=K$ and $T=K$. This works for both possible positions of $K$, so we have shown $3$ cases so we are done.