Let $I$ be the incenter and $L$ is the tangency point of $\omega$ with $BC.$ Now let $AL\cap \omega =S.$ Now it is easily seen that $FLES$ is harmonic quadrangle and so $TS$ is tangent to $\omega.$ (It follows easily from polus and polar) Let $AL\cap PQ=R.$ Now it is easily seen that a homothety $h$ with center $A$ which maps $\omega$ to $\omega_{a}$ also maps $PQ$ to $BC.$ So now we conclude that if $PQ\cap\omega =W$ we have $PR=WQ.$ Because of the fact that $PQ\parallel BC$ it follows that $WL$ is diameter of the incurcle and therefore $\angle WSR=\pi/2.$ But if $TS\cap PQ=M_{1}$ in the right-angle triangle $RSQ$ we have $M_{1}W=M_{1}S$ (because $SM_{1}$ and $M_{1}W$ are tangents to $\omega$) and from well-known problem we conclude that $M_{1}$ is the midpoint of $RW.$ Thus $M_{1}P=M_{1}R+RP=M_{1}W+WQ=M_{1}Q$ and so $M_{1}=M.$ QED

I have a solution with complex numbers! If the origin of complex plane is the incenter, and the incircle is the unit circle, we will have: $m = \frac{x(xz+xy-2yz)}{(x-z)(x-y)}$ and $t = \frac{x(xy+xz-2yz)}{x^{2}-yz}$, where y = E, z = F and x = K (K is the intersection of the incircle with BC)! To prove that TM is tangent to the incircle is easy!

Another "projective" solution, a little simpler from the one bilarev gave, should be by considering $N$ the midpoint of $(BC)$, and denote $V \in BQ \cap CP$. Because $BCQP$ is a trapezium, we have that $A,M,V,N$ are collinear and $(A,M,V,N)$ form a harmonical division. Now because $BCQP$ is a circumscribe quadrilateral from the Newton theorem, we have that $V \in EF$, thus $T(A,M,V,N)$ is a harmonical division, and now by intersecting it with $AD$ we have that $(AX_{1}YD)$ is a h.div, where $X \in TM \cap AD$ and $Y \in TV \cap AD$. But because $EF$ is the polar of $A$, we have that $(AX_{2}YD)$ is a h.div, where $X_{2}\in AD \cap w$. Thus $X_{1}\equiv X_{2}$, but now $TX_{2}$ is trivially tangent to $w$, because $AD$ is the polar of $T$, thus the point $X_{2}\in AD \cap w$, is the place where the second tangent cuts the circle, thus $TM$ is tangent to $w$.

Actually I improved my solution and now I think this is the shortest one: Just consider $ X \in AD \cap w$ and $ Y \in AD \cap EF$, because that $ AD$ is the polar of $ T$ we have that $ TX$ is tangent to $ w$, but we have that $ (AXYD) = - 1$, so $ B(AXYD) = - 1$, therefore by intersecting it with $ PQ$, we have that $ (P,M,Q, \infty) = - 1$, thus $ M$ is the midpoint of $ (PQ)$., and thus the problem is solved. (I used that $ BY$ cuts $ PQ$ in $ Q$ because of the Newton theorem).

PP (Iran TST 2007). Let $\omega$ be the incircle of the triangle $ABC$. Let $P\in AB$ and $Q\in AC$ be two points such that $PQ\parallel BC$ and $PQ$ is tangent to the circle $\omega$. The lines $AB$, $AC$ touch the circle $\omega$ at $F$, $E$ respectively. Denote the middlepoint $M$ of $[PQ]$ and $T\in EF\cap BC$. Prove that the line $TM$ is tangent to the circle $\omega$ (Ali Khezeli) Proof. Denote $D\in BC\cap w$ , $N\in IM\cap BC$ and the midpoints $S$ , $V$ of the segments $[AD]$ , $[BC]$ respectively. Observe that $M\in AV$ , $IM=IN$ . From the well-known properties $\boxed{\ S\in IV\ }\ (1)$ and $\boxed{\ TI\perp AD\ }\ (2)$ obtain immediatelly that $MN\parallel AD$ , i.e. $TI\perp MN$ and the line $TM$ is tangent to the circle $w$ because the lines $TM$ , $TN$ are symmetrically w.r.t the line $TI$ . $(1)\blacktriangleright$ Suppose w.l.o.g. $b\ne c$ . Denote the intersection $L\in AI\cap BC$ . Prove easily that $VL=\frac{a|b-c|}{2(b+c)}$ , $VD=\frac{|b-c|}{2}$ , $\frac{IA}{IL}=\frac{b+c}{a}$ . Apply the Menelaus' theorem to the transversal $\overline{VIS}$ and the triangle $ADL$ , i.e. $\frac{VD}{VL}\cdot\frac{IL}{IA}\cdot\frac{SA}{SD}=1$ $\Longleftrightarrow$ $S\in IV$ . $(2)\blacktriangleright$ The line $EF$ is the polar $\alpha$ of the point $A$ w.r.t. the circle $w$ and the point $T\in\alpha$ $\implies A$ belongs to the polar $\tau$ of the point $T$ w.r.t. the circle $w$ and $D\in\tau$ $\implies$ the polar of $T$ is $\tau=AD$ $\implies$ $TI\perp AD$ . Otherwise prove easily that $TA^{2}-TD^{2}=IA^{2}-ID^{2}$ . See PP14 from here

let $AD$ intersect incircle at $S$ and $PQ$ at $D'$ and consider incircle is tangent to $PQ$ at $R$ , we know that $MD'=MR$ and $\angle RSD=90^{\circ}$ hence $MD'=MR=MS$ also we have $\angle ADB=\angle MD'S=\angle D'SM$ hence $MS$ is tangent to incircle , let $MS\cap BC=T'$ , polar of $T'$ pass through $A$ so polar of $A$ (EF) pass through $T'$ hence $T'\equiv T$

E.lopes, how do you prove that the line TM is tangent to the unit circle?

Joao Guerreiro wrote: E.lopes, how do you prove that the line TM is tangent to the unit circle? if $z+t.z' = c$ is the equation of TM (where $z'$ is the conjugate of $z$), the foot of the perpendicular, trought the incenter, to TM, will have cordinates $\frac{c}{2}$. So, we can prove that $|\frac{c}{2}| = 1$!

hoch ghaste tohin be aghaye khezeli ro nadaram valiiiiiiii in soal akhe chie ke dadan bara TS.... mmidadan marhale 2 laaghal mishod ye karish kard:d

Another approach: By direct calculation we get $RM\cdot TD=IR^2$, i.e. $\triangle RIM \sim \triangle DTI$, done. Best regards, sunken rock

What is R? What is D? What is I? Your solution, sunken rock, is very unclear!

@CDP100: Sorry, $\{R\} \equiv \omega \cap PQ$, $I$ - the center of $\omega$, $\{D\} \equiv \omega \cap BC$. Further, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\cdot TD$, the reflection of $D$ in $IT$ coincides with the reflection of $R$ in $IM$, which solves the problem. Note: It has been a typo, I edited it. Sorry for any inconveniences caused! Best regards, sunken rock

Dear Mathlinkers, 1. let I the center of the incircle, D be the point of contact of the incircle and BC, D' the antipode of D, D" the second point of intersection of AD with the incircle and D+ the point of intersection of AD and PQ. 2. It is known that TD" is tangent to the incircle at D" 3. M is the midpoint of D'D+ 4. IM is parallel to AD and perpendicular to D'D" 5. By symmetry wrt MI, MD" is tangent to the incircle at D"… and we are done Sincerely Jean-Louis

sunken rock wrote: @CDP100: Sorry, $\{R\} \equiv \omega \cap PQ$, $I$ - the center of $\omega$, $\{D\} \equiv \omega \cap BC$. Further, after having seen, by compiling $TD$ and $RM$ that $IR^2=RM\cdot TD$, the reflection of $D$ in $ID$ coincides with the reflection of $R$ in $IM$, which solves the problem. Best regards, sunken rock Thank you very much!

Tangent from T intersect PQ at N. From Brianchon Theorem (we know that NTCAFP is a tangent hexagon) AN, FT and CP are concurrent. From above solution we know also AM ET and CP are concurrent. It means M=N. Solution over.

Let $D$ be the incircle's tangency point of $BC$, $H$ the tangency point of $PQ$, and $I$ the incenter. Let $AD\cap PQ=J$. We note that since $T$ lies on the polar of $A$ ($EF$), $AD$ is the polar of of $T$ and let $AD$ intersect the incircle at $G$. We note that since the incricle is the excircle of $\triangle APQ$, $B,I,H$ are collinear. By similar triangles, $J$ is the tanceny point of the inircle of $APQ$ with $PQ$, thus $HP=JQ$. Note that as $BD$ is a diamater, $\angle HGB=\angle HGJ=90$. Then the midpoint of $HJ$, $M$, has $MG$ tangent to to the incircle as $MG=MH$. Thus, $M,T,G$ are collinear and we are done.

Suppose in the complex plane $I=0,D=a,E=b,F=c,|a|=|b|=|c|=1$. Now so $A=\frac {2bc}{b+c},B=\frac {2ac}{a+c},C=\frac {2ab}{a+b}$.Now the point where $PQ$ touches in circle is $-a$.Thus $P=\frac {-2ac}{a+c},Q=\frac {-2ab}{a+b}$. And so $M=-a(\frac {c}{b-c}+\frac {b}{b-a})$.Suppose the line other than $PQ$ passes through $M$ tangent to circle be $x$. Then certainly $\frac {2x}{x-a}=(\frac {c}{c-b}+\frac {b}{a-b})$.Now suppose that line meets $BC$ at $T'=t$. Then we've $t=\frac {2ax}{a+x}$. So now eliminating $x$ we've $a=t(\frac {b}{b-c}+\frac{b}{b-a})$. Now by this we've $\frac {t-c}{t-b}$ real, and so that implies $E,G,T'$ are co linear,so $T=T'$ and that's what we were asked show, so done.

subham1729 wrote: Then certainly $\frac {2x}{x-a}=(\frac {c}{c-b}+\frac {b}{a-b})$.Now suppose that line meets $BC$ at $T'=t$. Then we've $t=\frac {2ax}{a+x}$. So now eliminating $x$ we've $a=t(\frac {b}{b-c}+\frac{b}{b-a})$. Now by this we've $\frac {t-c}{t-b}$ real, and so that implies $E,G,T'$ are co linear,so $T=T'$ and that's what we were asked show, so done. I haven't understood this last part Is x the complex number associated to the tangency point of T'M? Then why is that so obvious?

So easy problem... First, call the point that AD and incircle meets X. And call the point that AD and PQ meets Z. And call the point that PQ and incircle meets Y. Than YM=MZ, and angle YXZ=angle YXD=90 So YM=MX, therefore MX is tangent to incircle. And pole of XD is T. So XT is tangent to incircle. QED.

Let $BQ \cap CP = X$. $X \in EF$ (by Brianchon on $PQECBF$). Since $X \in AM$ we have $TM$ is tangent to $\omega$ (by converse of Brianchon on $TBAEQM$).

Let $AD$ intersect $TM$ at $U$, $D$ be the $A$-intouch point, and $DG$, where $G$ is the antipode of $D$ on $\omega$, intersect the line parallel to $BC$ at $Y$. Let the concurrency point of $EF, DG$, and $AN$ (where $N$ is the midpoint of $BC$) be $X$, and let $FE$ intersect $AD$ be $V$. Then observe $$(A, V; U, D) = (A, X; M, N) = (Y, X; G, D) = (P_{\infty}, N; AG \cap BC, D) = -1.$$As $V$ lies on the polar of $A$, $U \in \omega$, and as $AD$ is the polar of $T$, this is equivalent to $U$ being the other tangency point of $T$.

Harmonic cookie (also probably the most complicated solution with a million intersections) Let $K=\overline{AD} \cap \omega$, $(K,D;F,E) = -1 = (\overline{TT} \cap \omega,D;F,E)$ so $TK$ is tangent to $\omega$. Let $D'$ be the andipode of $D$ in $\omega$ which is the touchpoint of $\overline{PQ}$ with $\omega$, also let $N$ be the midpoint of $\overline{BC}$, $D'' = \overline{AD'} \cap \overline{BC}$, $R=\overline{DD'} \cap \overline{EF}$ and $S=\overline{KD} \cap \overline{EF}$. By incircle concurrency lemma $R$ lies on $\overline{AN}$. Key Claim: $(A,R;M,N) = -1$. Proof: $$(A,R;M,N) \stackrel{D'}{=} (D'' , D;\infty , N) = -1$$where $N$ is midpoint of $\overline{D' D''}$ because of a well known config. Then note that $$(A, R ; \overline{TK} \cap \overline{AN}, N) \stackrel{T}{=} (A,S; K,D) = -1 = (A,R;M,N)$$So $M \equiv \overline{TK} \cap \overline{AN} \implies T-K-M$ and thus $TM$ is tangent to $\omega$ at $K$. This is such a bruh solution

Let $N,N'$ be touch point of $\omega$ with $AB,PQ.$ Let $TR$ tangent to $\omega$ at $R\neq N,$ and redefine $M$ as $RT\cap PQ.$ Observe that $NFRE$ is harmonic, so by duality wrt $\omega$ $$(PQM\infty )=N'(EFRN)=-1,$$i.e. $RT$ bisects $PQ$ as desired.

[asy][asy] /* TAKEN FROM V_ENHANCE AND EDITED */ size(10cm); defaultpen(fontsize(9pt)); pair A = dir(130); pair B = dir(210); pair C = dir(330); draw(A--B--C--cycle); draw(incircle(A, B, C), blue); pair I = incenter(A, B, C); pair F = foot(I, A, B); pair E = foot(I, A, C); pair T = extension(E, F, B, C); draw(E--T--B); pair P = extension(A, B, I, I+rotate(-90)*(I-B)); pair Q = extension(A, C, I, I+rotate(90)*(I-C)); draw(P--Q); pair M = midpoint(P--Q); draw(T--M, red); pair D = foot(I, B, C); draw(A--D, dotted); pair X = extension(A, D, M, T); pair Y = 2*I-D; pair Sp = extension(A,D,P,Q); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(I)); dot("$F$", F, dir(150)); dot("$E$", E, dir(45)); dot("$T$", T, dir(T)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(20)); dot("$M$", M, dir(90)); dot("$D$", D, dir(D)); dot("$S$", X, dir(170)); dot("$R$", Y, dir(90)); dot("$S'$", Sp, dir(110)); /* Source generated by TSQ */ [/asy][/asy] Define $D$ the tangency of $\omega$ with $BC.$ Let line $AD$ intersect $\omega$ at $S$ and $PQ$ at $S'.$ Let $R$ be the tangency of $PQ$ with $\omega.$ Since $AF$ and $AE$ are tangents of $\omega$ at $E$ and $F,$ it follows $FDES$ is a harmonic quadrilateral. By well known properties, this implies $TS$ is tangent to $\omega$ at $S.$ It suffices to show $MS$ is tangent to $\omega.$ First, consider the homothety $\mathcal{H}$ centered at $A$ bringing $\omega$ to the $A$-excircle. It follows $\mathcal{H}(S')=D$ and $\mathcal{H}(R)=R',$ where $R'$ is the tangency point of $BC$ and the $A$-excircle. By excircle properties, it follows $BD=CR',$ which implies the midpoint of $DR'$ is also the midpoint of $BC,$ and therefore by the homothety the midpoint of $S'R$ is also the midpoint of $PQ,$ which happens to be $M.$ Thus, $MS'=MR.$ Now notice since $D,I,R$ are collinear it follows $R$ and $D$ are antipodes. Thus $\angle DSR=90^\circ.$ Note then $\angle RSS'=90^\circ,$ so the circumcircle of $SS'R$ is centered at $M,$ and thus $MS'=MR=MS,$ so $MS$ is tangent to $\omega,$ and therefore $TM$ is tangent to $\omega,$ as required.

Let $N$ the midpoint of $BC$, and let the tangents from $N$ to $\omega$ be $D,K$ where $D$ lies on $BC$, let $G$ the point where $PQ$ is tangent to $\omega$ (so $I,D,G$ are colinear), let $AM$ intersect $\omega$ at $X,Y$ and let $AD \cap \omega=L$. Now since its known that $A,G,K$ are colinear we project and take polars w.r.t. $\omega$ $$-1=(X,Y; D, K) \overset{A}{=} (Y, X; L, G) \implies \mathcal P_M=LG \implies M \in \mathcal P_L$$But since $\mathcal P_L=TL$ we have that $T,L,M$ are colinear hence $TM$ is tangent to $\omega$ as desired.

Let $D$ be the contact point of $\omega$ with $BC$, and let $R$ be the antipode of $D$ in $\omega$. Let $S\neq R$ in $\omega$ such that $MS$ tangent to $\omega$. Finally, let $D'=AD\cap PQ$. By homothety $D'$ is the contact point of the incircle of $\triangle APQ$ with side $PQ$, and so $MD'=MR=MS$, so $RS\perp SD'$. Also, $DR$ diameter in $\omega$, so $RS\perp SD$, and so $A-D'-S-D$ collinear. Finally, $T\in \mathcal{P}(A) \Longrightarrow AD=\mathcal{P}(T) \Longrightarrow S\in \mathcal{P}(T) \Longrightarrow T\in \mathcal{P}(S)$, and so $TM$ is tangent to $\omega$ at $S$.

Let $\omega$ touch $BC$ at $D$, $AD$ meet $\omega$ again at $R$ and intersect $PQ$ at $Y$, $PQ$ touch $\omega$ at $X$, and $AX \cap BC = X_1$. Since $DERF$ is harmonic, we know $TR$ is tangent to $\omega$. Hence, it suffices to show $MR$ is tangent to $\omega$. The Diameter of the Incircle Lemma implies $BD = CX_1$. Now, since $APQ$ and $ABC$ are homothetic, it follows that $MX = MY$. Because $DX$ is a diameter of $\omega$, we have $$\angle XRY = 180^{\circ} - \angle XRD = 90^{\circ}$$which means $M$ is the circumcenter of $XRY$. This yields $MR = MX$, as required. $\blacksquare$ Remark: The solution is quite simple. At first, however, I sort of overcomplicated this problem by adding $AM \cap DX \cap EF$ and attempting to use harmonic bundles.

Let the incircle be the unit circle, and set $d = -i$. Then, we have $$p = \frac{2fi}{f+i} \text{ and } q = \frac{2ei}{e+i},$$so we can compute$$m = \frac{2ief-e-f}{(e+i)(f+i)}.$$Next, using a Cartesian coordinate system, we can compute$$t = \frac{f+\overline f}2 - \frac{i(e+\overline e - f - \overline f)}{e-\overline e - f + \overline f}\left(\frac{f-\overline f}{2i} + 1\right) - i = \frac{-i(e+i)(f+i)}{ef+1} - i.$$Then, the reflection $d'$ of $D$ over $\overline{TO}$ is$$d' = \frac{t \cdot i}{\overline t} = \frac{2ef+i(e+f)}{e+f+2i}$$by the complex reflection formula; we have $\overline{TD"}$ is tangent to the unit circle. Then, it suffices to show that $M$ is the tangent intersection of the tangent at $i$ and the tangent at $d'$. In other words, we can simply check that$$\frac{2i\left(\frac{2ef+i(e+f)}{e+f+2i}\right)}{\frac{2ef+i(e+f)}{e+f+2i} + i} = \frac{2i ef-e-f}{(e+i)(f+i)} = m,$$so it follows that $\overline{TD'}$ passes through $M$, and thus $\overline{TM}$ is tangent to the incircle.

Let $D$ be the $A$-contact point, $D'$ be the tangency point of $\overline{PQ}$ with $\omega$, and $D'',X$ respectively be the intersections of $\overline{AD}$ with $\overline{PQ}$ and $\omega$. Then $D'$ and $D''$ are reflections of each other over $M$ and $D'$ and $D$ are reflections of each other over the incenter $I$. Because $T$ lies on the polar of $A$ wrt $\omega$, $A$ lies on the polar of $T$, which passes through $D$ as well. Hence it passes through $X$, so $\overline{TX}$ is tangent to $\omega$. It then suffices to show that $\overline{MX}$ is also tangent to $\omega$, i.e. $MX=MD'$. Since $\angle D''XD'=\angle DXD'=90^\circ$ and $M$ is the midpoint of $\overline{D'D''}$, it follows that it is the circumcenter of right triangle $D'XD''$, hence $MX=MD'=MD''$. $\blacksquare$

Take the circle the unit circle. Let the circle touch $BC$ at $D$ .We denote the complex co-ordinate of a point $X$ by $x$ . We set $d=1$ it is clear $PQ$ toiuches the unit circle at $-1$ .By standard formulae $b=\frac {2f}{1+f}$ and $c=\frac{2e}{1+e}$ and $p=\frac{-2f}{f-1}$ and $q=\frac{-2e}{e-1}$. Note that $ \Re{(t)} = 1$ and the equation of chord $EF$ is $z=e+f-ef\overline{z}$ and $t+\overline{t} = 2 \implies \overline t = 2-t$ and putting this in the equation of $EF$ we get $t = \frac{e+f-2ef}{1-ef}$ and $ m= \frac{e+f-2ef}{(e-1)(f-1)}$.let $o'$ be the complex co-ordinate of the reflection of origin wrt the line $MT$ $$\frac{\overline{m}}{\overline t - \overline m } = \frac{o' - m}{t-m}$$. let the mid-point of $O$ and $O'$ is $P$. then $p=\frac{o'+0}{2} = -\frac{e+f-2ef}{e+f-2}$(Stepped the calculation part ) Now note that $ \overline {p} = -\frac{e+f-2}{e+f-2ef}$.So , $p\overline p = 1$ and so $| p| = 1$ and so $P$ lies on the unit circle and so $OP\perp TM$ and hence we are done .

dkaoskoas hi because $I$ is the incenter $N=AD\cap \omega$ $Y$ is midpoint of $BC$ $D'=DI\cap \omega$ $X=DI\cap AM\cap EF$ (well-known to exist) $Z=AD\cap EF$ Since $T$ lies on the polar of $A$ so $A$ lies on the polar of $T$ hence $AD$ is the polar hence $TN$ is tangent to $\omega$ hence it suffices to show that $M,N,T$ are collinear. If we can show that $(A,X;M,Y)=-1$ we will be done since we also have $(A,Z;N,D)=-1$ and we could finish by Prism Lemma (aka phantom points). Now let $\infty_{BC}$ be the point at infinity along line $BC$ and project onto $\overline{DID'}$, suppose $A\to A'$. It suffices to show that $(A',X;D',D)=-1$ which is trivial: realize that $A'\in (AEIF)$ hence by Shooting Lemma \[IX\cdot IA'=IE^2=IF^2=ID^2=ID'^2\]done. Yay!

Let $\omega$ intersect $\overline{BC}$ and $\overline{PQ}$ at $D$ and $G$, and define $K = \overline{AD} \cap \omega, L = \overline{AD} \cap \overline{PQ}$. First, we note that $KDFE$ is harmonic, so $\overline{TK}$ is tangent to $\omega$. Thus, it suffices to show that $\overline{MK}$ is tangent to $\omega$ too. However, since a homothety at $A$ takes $\overline{PQ}$ to $\overline{BC}$ and $G$ to the $A$-extouch point, we have $LM = MG$. But $\overline{GD}$ is a diameter of $\omega$ because $\overline{PQ} \parallel \overline{BC}$, so $\overline{GK} \perp \overline{LKD}$, and $M$ is the circumcenter of right $\triangle LKG$. Therefore $MK = MG$, and since $MG$ is tangent to $\omega$, $MK$ is as well.

Let $D$ and $D'$ be the intersections of $\overline{BC}$ and $\overline{PQ}$ with $\omega$, respectively – let $D''$ be the intersection between $\overline{AD}$ and $\overline{PQ}$, and lastly, let $U$ be the second intersection of $\overline{AD}$ with $\omega$. Because $DD'$ is a diameter of $\omega$, $U$ is the foot from $D'$ to $\overline{AD}$. We claim that $\overline{MU}$ and $\overline{TU}$ are both tangent to $\omega$, which implies the problem. To prove that $\overline{MU}$ is tangent to $\omega$: a homothety of factor $\tfrac{1}{2}$ centered at $D'$ sends $D$ to $I$ and sends $D''$ to $M$. Thus, $MI$ is the perpendicular bisector of $\overline{UD'}$. Since $MD'$ is obviously tangent to $\omega$, it follows that $MU$ is tangent to $\omega$ as well. To prove that $\overline{TU}$ is tangent to $\omega$: let $G$ be the foot from $I$ to $\overline{AD}$. Note that $AFGIE$ is cyclic with diameter $AI$. Now, radical axis theorem on $(AFGIE)$, $(GID)$ and $\omega$ means that $T$, $G$ and $I$ are collinear. So, $TI$ is the perpendicular bisector of $\overline{UD}$, so $\overline{TU}$ is tangent to $\omega$.

Lemma: $DI,EF,AM$ are concurrent. Proof: Let $DI\cap EF=S,D'$ be the antipode of $D$ and $AD'\cap (I)=X,N$ be the midpoint of $BC$. Since $EF$ is the polar line of $A$ with respect to $(I)$, $K,S,X$ are collinear. Pascal at $XXKDDD'$ gives that $N,S,A$ are collinear. By homothety, $A,N,M$ are collinear hence $A,M,S$ are collinear.$\square$ Pascal at $KKDD'D'X$ yields $KK\cap D'D',A,S$ are collinear. Thus, $MK$ is tangent to $(I)$. By La Hire, since $AD$ is the polar line of $T$ with respect to $(I)$, $TK$ is tangent to $(I)$ hence $T,K,M$ are collinear as desired.$\blacksquare$