Denote by $\Omega$ the circle internally tangent to $\omega$ at $T$ and tangent to $BC$ at $D.$ $TB,TC$ cut $\Omega$ again at $B',C'.$ Since $T$ is exsimilicenter of $\Omega \sim \omega,$ then $BC \parallel B'C'$ $\Longrightarrow$ arcs $DB'$ and $DC'$ of $\Omega$ are equal, i.e. $TD$ bisects $\angle BTC$ $\Longrightarrow$ $M \in TD.$ $ME$ cuts $\omega$ again at $F$ and $FI_A$ cuts $\omega$ again at $S^*.$ It's well known that $I,I_A$ lie on the circle $(M)$ with center $M$ and radius $MB=MC,$ thus inversion WRT $(M)$ takes $D,E$ into $T,F$ and $I,I_A$ into themselves $\Longrightarrow$ $\odot(IDT)$ and $\odot(I_AEF)$ are orthogonal to $(M)$ $\Longrightarrow$ $\angle MFS^*=\angle EI_AM=\angle DIM=\angle MTS$ $\Longrightarrow$ $S \equiv S^*.$ Lines $ME$ and $SI_A$ intersect at $F \in \omega.$

My solution: Let $ T'=ME \cap (ABC) $ . From homothety we get $ T, D, M $ are collinear . Since $ BD=CE $ , so $ T' $ is the reflection of $ T $ in the bisector of $ BC $ . Since $ \triangle MBD \sim \triangle MTB $ , so we get $ MI^2=MB^2=MD \cdot MT $ . ie. $ \triangle MID \sim \triangle MTI $ Since $ \text{Rt} \triangle BDI \sim \text{Rt} \triangle I_AEB $ , so we get $ DM \cdot ET'=DM \cdot DT=DB \cdot DC=DB \cdot EB=DI \cdot EI_A $ , hence combine with $ \angle I_AET'=\angle MDI $ we get $ \triangle I_AET' \sim \triangle MDI $ , so from $ \angle ET'I_A=\angle DIM=\angle ITM=\angle ET'S $ we get $ T', S, I_A $ are collinear . ie. $ ME \cap I_AS \in (ABC) $ Q.E.D

Telv Cohl thank you for nice proof..My solution similar to Luiz Gonzalez's solution(but without invesion).

Other solution: Claim 1: Are given two circle internally tangent at a point $T$, $AB$ a chord of the bigger circle tangent to the smaller circle in a point $C$. So $TC$ is the internal bisector of $\angle ATB$. Prove: A simple homothety with center $T$ carring one circle in the other. Applying Claim 1 in the problem, we have that $T, D, E$ are collinear. Also is well known that $CE = BD$, and since $MB = MC$, follows that $MD = ME$. Denote $ME \cap \omega = T'$, and since $MD = ME$, and the tangent to $\omega$ at $M$ is parallel to $DE$, follows that $MT' = MT$. Now we gonna prove that $T', S$ and $I_A$ are collinear. Note that $(A, M; I, I_A) = (A, M; T, SI_A \cap \omega)$, where $(A, M; I, I_A) = \dfrac{AI}{AI_A} = \dfrac{r}{r_A}$, since $MI = MI_A = MB = MC$. But also note that $(A, M; T, T') = \dfrac{AT}{AT'}$, since $MT = MT'$, where $\dfrac{AT}{AT'} = \dfrac{\text{sen}(\angle TMA)}{\text{sen}(\angle AMT')} = \dfrac{DR}{RE}$, since $MD = ME$, where $R = IM \cap DE$. But $\triangle IDR$ and $\triangle I_AER$ are similar, thus $\dfrac{DR}{RE} = \dfrac{r}{r_A}$. Therefore, $(A, M; T, SI_A \cap \omega) = (A, M; T, T')$, in other words $T' = SI_A \cap \omega$, as desired.

Similar to Telv Cohl's solution. Assume without loss of generality, $AB <AC$.(The case $AB=AC$ being trivial.) Let $\omega$ denote the circumcircle of $\triangle ABC$ and let $ME$ meet $\omega$ again at $X$. Now, let $AM$ meet $BC$ at $Y$. Clearly, $MI^2= MY.MA=ME.MX=MD.MT$ and so $A,Y,E,X$ are concyclic. Now applying $\sqrt {bc}$ inversion we get that $X \longrightarrow X'$ where $X'$ is the point of intersection of $BC$ with $T_AM$ where $T_A$ is the touch-point of the $A$-mixitilinear in-circle with $\omega$. Now it is well-known that $X'$ lies on the mixitilinear touch-chord of the $A$-mixitilinear in-circle and so we conclude that $\angle AXI_A=90$. Now let $AI_A$ meet $\omega$ again at $S'$. Then clearly, $S'$ is the antipodal point of $A$ w.r.t $\omega$. So it suffices to showing that $T,I,S'$ are col-linear. This is equivalent to the fact that $\angle MTI=\angle MAS'$. But, $\angle MTI=\angle MID=\frac{\angle B-\angle C}{2}=\angle MAS'$. Hence the conclusion follows.

I think this is the simplest solution yet... Invert about $M$ with radius $MB$. The problem becomes to show that the circumcircles of $(MDI)$ and $(MEI_a)$ intersect on $BC$. Let $K$ be the intersection of $(MEI_a)$ with $BC$. Then since $I_aE\perp EK$, $\angle I_aEK=\angle I_aMK=\angle AMK=\angle AMS=90^{\circ}$, where $S$ is the intersection of $MK$ with $\omega$, so $S$ is the antipode of $A$ in $\omega$. Now, let $K'$ be the intersection of $(MDI)$ with $BC$, and $S'$ be the intersection of $MK'$ with $\omega$. Then $\angle IDK'=\angle IMK'=\angle AMS'=90^{\circ}$, so $S\equiv S'$, hence $K\equiv K'$ and we're done.

Every problem needs a nice BARY-BASH! The following computation is very instructive in the ways of bashing. Let $A=(1,0,0),B=(0,1,0),C=(0,0,1)$. Then \[D=(0:s-c:s-b), E=(0:s-b:s-c),I=(a:b:c),I_a=(-a:b:c),M=(-a^2:b^2+bc:c^2+bc).\]All fine and good, but now how do we proceed? We need to think very carefully about how to make this barybash feasible. In particular, the definition of $T$ is nearly impossible to use as is. We need the synthetic observation that $T,D,M$ are collinear by homothety. This makes $T$ a workable point. So now, we are going to intersect a bunch of lines with the circumcircle. The only way this is remotely possible is by using Vieta. Indeed, this bash will show just how ridiculously overpowered Vieta is... Let $X=ME\cap \omega$. Our solution path is as follows: compute $T=MD\cap \omega$ and $X$ by intersecting each line with the circle. Then we'll find $S=XI_a\cap \omega$ by intersecting this line with the circle, and finally check that $S$ lies on $XI$. Let's get started! First, the equation of $MD$ is \[\begin{vmatrix} x & y & z \\ -a^2 & b^2+bc & c^2+bc \\ 0 & s-c & s-b \end{vmatrix}=0\iff\]\[\iff x(b+c)(b(s-b)-c(s-c))+y(a^2)(s-b)-z(a^2)(s-c)=0\]\[\iff x(c^2-b^2)(s-a)+y(a^2)(s-b)-z(a^2)(s-c)=0\]\[\iff z=x\cdot \frac{(c^2-b^2)(s-a)}{a^2(s-c)}+y\cdot \frac{(s-b)}{(s-c)}.\]The circumcircle has equation $a^2yz+b^2zx+c^2xy=0$, so now we'll plug $z$ into the equation: \[y^2\cdot \frac{a^2(s-b)}{(s-c)}+x^2\cdot \frac{b^2(c^2-b^2)(s-a)}{a^2(s-c)}+Qxy=0,\]for some function $Q$ of $a,b,c$. Note that Vieta allows us to not care what $Q$ is! Now if we normalize to $x=1$ and observe that one solution for $y$ is when $M$ is the considered point, i.e. $y=\frac{-b(b+c)}{a^2}$, we have \[\left( \frac{-b(b+c)}{a^2}\right) \left(\frac{a^2(s-b)}{(s-c)}\right)y=\frac{(c^2-b^2)(s-a)}{a^2(s-c)}. \]This reduces to \[y=\frac{b(b-c)(s-a)}{a^2(s-b)}.\]Now since $x=1$ we can easily calculate $z$ as \[z=\frac{c(c-b)(s-a)}{a^2(s-c)},\]which is acceptable since it has to agree (symmetrically) with $y$. Now we re-normalize to the cleaner form \[T=\left(a^2(s-b)(s-c):b(b-c)(s-a)(s-c):c(c-b)(s-a)(s-b)\right).\]The computation of $X$ is virtually identical; we find \[X=\left( a^2(s-b)(s-c):b(c-b)s(s-b):c(b-c)s(s-c) \right).\](Note to reader: We can also synthetically observe that $T,X$ have the same x-coordinate [when normalized, although in the above forms they still do.] This could be used to find $X$. The time expenditure of setting up such a bash instead of copying the one we already have for $T$ probably makes it counter-productive.) Now we need to compute the equations of line $IT,I_aX$. First, line $IT$ has equation \[\begin{vmatrix} x & y & z \\ a & b & c \\ a^2(s-b)(s-c) & b(b-c)(s-a)(s-c) & c(c-b)(s-a)(s-b) \end{vmatrix}=0\]\[\iff x(bc)(s-a)(c-b)a+y(ac)(s-b)(a(s-c)+(b-c)(s-a))-z(ab)(s-c)(a(s-b)+(c-b)(s-a))=0.\]Now, we divide through by $a$ and, more importantly, observe that \[a(s-c)+(b-c)(s-a)=S_c, a(s-b)+(c-b)(s-a)=S_b.\]Thus line $IT$ has equation \[x(bc)(s-a)(c-b)+y(c)(s-b)S_c-z(b)(s-c)S_b=0.\]Again computing $I_aX$ is done in nearly the same way; somewhat surprisingly, we again find a nice simplification, this time \[a(s-b)+(b-c)s=S_c, a(s-c)+(c-b)s=S_b,\]which allows us to reduce the equation of $I_aX$ to \[x(bcs)(b-c)+y(c)(s-c)S_c-z(b)(s-b)S_b=0.\]Now we will compute $S$ as the intersection of $I_aX$ and $\omega$. Perhaps it seems unintuitive to not intersect the two lines with each other; however, there are good reasons for intersecting with the circle: (1) There isn't a terribly easy way to intersect two lines, (2) we have already seen how powerful Vieta is, and (3) if $S$ turns out to be nasty, it will be much easier to check if it lies on a line rather than if it lies on a circle. (The choice to use $I_aX$ here instead of $IT$ is essentially arbitrary; their equations are very similar.) So let's do this! From the equation for $I_aX$ we obtain \[z=x\cdot \frac{cs(b-c)}{(s-b)S_b}+y\cdot \frac{c(s-c)S_c}{b(s-b)S_b}.\]Plugging this into the circumcircle equation we find \[y^2\cdot \left(\frac{a^2c(s-c)S_c}{b(s-b)S_b}\right)+x^2\cdot \left( \frac{b^2cs(b-c)}{(s-b)S_b}\right)+Q\cdot xy=0,\]for some function $Q$ of $a,b,c$. Normalize to $x=1$, then one solution of this is \[y=\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}.\]Therefore the one we care about satisfies \[\frac{b(c-b)s(s-b)}{a^2(s-b)(s-c)}\cdot \frac{a^2c(s-c)S_c}{b(s-b)S_b}\cdot y=\frac{b^2cs(b-c)}{(s-b)S_b,}\]which miraculously simplifies to $y=\frac{-b^2}{S_c}$. Back substituting, we find $z=\frac{-c^2}{S_b}$, which is good since $y$ and $z$ are symmetrical. Re-normalizing, we find the cleaner form \[S=(-S_bS_c:b^2S_b:c^2S_c).\]We're almost there! We just need to check that this lies on $IT$, whose equation we already have. Plugging in, we want \[-S_bS_cbc(s-a)(c-b)+b^2c(s-b)S_bS_c-c^2b(s-c)S_bS_c=0.\]We can cancel the $6$th degree $bcS_bS_c$ from everything, leaving only \[-(s-a)(c-b)+b(s-b)-c(s-c)=0.\]Now factoring $(b-c)$ from this, it becomes \[(b-c)(2s-a-b-c)=0,\]which is clearly true. Hooray! Most important thing to note from this: Vieta is incredible, making intersecting a line with a circle very very easy (look at how ridiculously easily we compute $S$.) Also! I have seen this point $S$ before when bary bashing another problem... bash configuration recognition? FWIW: This only took about 1.5 pages to solve, and is probably doable in an hour or so (I did it while watching a movie, so it took a while )

My solution: Define $E'$ as the intersection of $ME$ with $\omega$ and it suffices to prove that $E',S,I_A$ are collinear. By the homothety of center $T$ which sends the circle tangent to $BC$ and $\omega$ to $\omega$ we easily deduce that $T,D,M$ are collinear. Now we have that $\angle BTM=\angle BAM=\angle MBC$ so $MB$ is tangent to the circumcircle of $TBD$ so $MD \cdot MT=MB^2=MI^2$ so $MI$ is tangent to the circumcircle of $TDI$ so $\angle ITD=\angle DIM$. Now invert about the circle $(IBI_AC)$ with center $M$ and radius $MI^2$.This inversion sends ($BC$ common chord of $\omega$ and $M$) $\omega$ to $BC$,$BC$ to $\omega$.Then obviously $T$ goes to $D$,$E$ goes to $E'$ and $S$ goes to the intersection of $MS$ with $BC$, let it be $S'$. Then since $MS \cdot MS'=MI^2=MT \cdot MD$, $S'SDT$ is cyclic so $\angle MS'D=\angle DTS= \angle DIM$ and thus $S'MDI$ is also cyclic so $\angle IMS'=IDS'=90$ Now since $\angle S'MI_A=\angle I_AES'=90$ we obtain that $MES'I_A$ is cyclic. Thus the circle that passes through these 4 points (and $M$ the center of the inversion) after the inversion will become a line passing through the images of $I_A,E,S'$.But since $I_A$ lies on $M$ it will map to itself and as we mentioned,$S'$ maps to $S$ and $E$ maps to $E'$.So $I_A,S,E'$ are collinear and we are done.

my solution : lemma for ABC is a triangle have (I) is its incircle and (O) is its circumcircle. ID perpendicular BC . OM is a diameter of (O) MI cut (O) at S.prove SD through the midpoint of BC chord not contain A. (easy). from lemma we have AS is a diameter. AI*Aia =bc=AD*AS(AD is perpendicular BC) => DIM+ASIa=180 we only need prove ID//ME (easy by trigo)

$ME\cap \odot ABC=\{R\}$,$ME\cdot MR=MB^2=MD\cdot MT$ ,$ME=MD$ $\implies$ $MR=MT$ $\implies$ $BR=CT$.Let $\odot I$ touch $AB,AC$ in $P,Q$.$M-D-T$ $\implies$ $\tfrac{TP}{TQ}=\tfrac{TB}{TC}=\tfrac{BD}{CD}=\tfrac{BP}{CQ}$ $\implies$ $T\equiv \odot ABC\cap \odot APQ$ $\implies$ $\angle ATI=90^{\circ}$ and so $S$ is the antipode of $A$.$SM\perp AM$,$IM=I_AM$ $\implies$ $\angle ISM=\angle MSI_A$ .Let $SI_A\cap \odot =\{R'\}$ $\implies$ $\angle R'ST=\pi-2\angle TSM$. On the other hand $\angle RST=\angle RMT=\pi-2\angle MTR'$ but as $MR=MT$ $\angle TSM=\angle MTR'$ $\implies$ $R\equiv R'$.$\blacksquare$

nikolapavlovic wrote: $M-D-T$ $\implies$ $\tfrac{TP}{TQ}=\tfrac{TB}{TC}=\tfrac{BD}{CD}=\tfrac{BP}{CQ}$ Could you please explain why this is true?

$M$ is the midpoint of $\widehat{BC}$ and so by angle bisector theorem$\tfrac{TB}{TC}=\tfrac{BD}{CD}$ but as $BP=BD,CD=CQ$ we have that $T$ is the center of spiral similarity carrying $PQ\mapsto BC$ as $T\in \odot ABC$.That's why the first eqality holds (it was there so that i can explain why the next paragraph would follow)

nikolapavlovic wrote: $M$ is the midpoint of $\widehat{BC}$ and so by angle bisector theorem$\tfrac{TB}{TC}=\tfrac{BD}{CD}$ but as $BP=BD,CD=CQ$ we have that $T$ is the center of spiral similarity carrying $PQ\mapsto BC$ as $T\in \odot ABC$.That's why the first eqality holds (it was there so that i can explain why the next paragraph would follow) Thank you!

TelvCohl wrote: My solution: Let $ T'=ME \cap (ABC) $ . From homothety we get $ T, D, M $ are collinear . Since $ BD=CE $ , so $ T' $ is the reflection of $ T $ in the bisector of $ BC $ . Since $ \triangle MBD \sim \triangle MTB $ , so we get $ MI^2=MB^2=MD \cdot MT $ . ie. $ \triangle MID \sim \triangle MTI $ Since $ \text{Rt} \triangle BDI \sim \text{Rt} \triangle I_AEB $ , so we get $ DM \cdot ET'=DM \cdot DT=DB \cdot DC=DB \cdot EB=DI \cdot EI_A $ , hence combine with $ \angle I_AET'=\angle MDI $ we get $ \triangle I_AET' \sim \triangle MDI $ , so from $ \angle ET'I_A=\angle DIM=\angle ITM=\angle ET'S $ we get $ T', S, I_A $ are collinear . ie. $ ME \cap I_AS \in (ABC) $ Q.E.D Which app you use to draw this picture Thx

By Archimedes' Lemma, it follows that $M,D,T$ are collinear. Rename $E$ to be $D',$ and let $\triangle DEF$ be the intouch triangle of $\triangle ABC.$ Lemma: $T$ is the spiral center that sends $FB\to EC.$ Proof: Evidently $\angle TBF\angle TBA=\angle TCA=\angle TCE.$ Now $$\dfrac{TB}{TC}=\frac{DB}{DC}=\frac{BF}{EC},$$so by SAS the claim follows. This lemma implies that $\angle T\in \odot(AFIE),$ so $\angle ATI=90^{\circ}$ and $S$ is actually the antipode of $A$ in $\odot (ABC).$ Now that we have removed disguises, we may quickly finish. Inverting at $M$ with radius $MB^2$ fixes $B,I,C,I_A$ by Fact 5. Moreover, $\odot (ABC)\to BC$ and $T\to D$ by Shooting Lemma. Then if $S'=MS\cap BC$ is the inverse of $S$ in this inversion, it suffices to show that $\odot (S'MI_A)$ meets $BC$ at $D'.$ But this is obvious, as $\angle I_AD'S'=90^{\circ}=I_AMS'.$ We may conclude.

Clearly $T=MD \cap \omega$. Let $P=ME \cap \omega$. It suffices to show that $TI, PI_A$ concur on $\omega$, which is equivalent to showing $\angle MTI=\angle MPI_A$. Note that $MI_A^2=MI^2=MD \cdot MT=ME \cdot MP$ which gives $\triangle MDI \sim \triangle MIT$ and $\triangle MEI_A \sim \triangle MI_AP$. Hence, $I_AE \parallel DI$ gives $\angle MPI_A=\angle MI_AE=\angle MID=\angle MTI$. $\blacksquare$. 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Let $ME \cap \omega =X, AI \cap BC =K$. By an easy homothety argument, we get that $M,D,T$ are collinear. We have to show that $TI$ and $XI_A$ meet at $S$, i.e. on $\omega$, which is equivalent to showing that $\angle I_AXE=\angle ITD$. Now, By Shooting Lemma and Fact 5, we get that $MI^2=MD \cdot MT$ and $MI_A^2=ME \cdot MX$. Thus, $MI$ and $MI_A$ are tangent to $\odot (ITD)$ and $\odot (I_AXE)$. Therefore, $\angle I_AXE =\angle MI_AE =90^{\circ}-\angle I_AKE=90^{\circ}-\angle IKD=\angle MID =\angle ITD$. Hence, done.

Can you explain what the shooting lemma is please?

See here.

I didn't want to invert this one, so I found solution without inversion. It's well-known that $M, D, T$ are collinear. Claim: $AS$ is diameter of $\omega$. Proof: Since $MI^2 = MD \cdot MT$, so $\angle{MDT} = \angle{MIT}$. Note that $\angle{MDT} = \angle{MDC} + 90^{\circ} = \angle{TAI} + 90^{\circ}$. Hence $\angle{ATI} = 90^{\circ}$, so $AS$ is diameter of $\omega$. $\blacksquare$ Since $\angle{ATI} = 90^{\circ}$, thus $(ATI)$ and $(BICI_A)$ are tangent to each other. Let $l$ be line passing through $I$ and tangent to $(BICI_A)$. Then three radical axis of $(ATI), (BICI_A), \omega$ are concurrent, so $l, BC, AT$ are concurrent. Let $R = l \cap BC \cap AT$. Let $X = ME \cap \omega$. We'll prove that $X, S, I_A$ are collinear. Since $\angle{AXS} = 90^{\circ}$, so it's enough to show that $\angle{AXI_A} = 90^{\circ}$. Claim: $A$ is the center of spiral similarity sending $RI$ to $I_AX$. Proof: Since $TM, MX$ are symmetric about perpendicular bisector of $BC$, so $\angle{RAI} = \angle{XAI}$, so $\angle{RAB} = \angle{CAX}$. Since $\angle{ABR} = \angle{AXC}$, so $A$ is the center of spiral similarity sending $RB$ to $CX$. Thus $AR \cdot AX = AB \cdot AC$. It's not hard to check that $AI \cdot AI_A = AB \cdot AC$, so $\frac{AR}{AI} = \frac{AI_A}{AX}$ and since $\angle{RAI} = \angle{I_AAX}$, therefore $A$ is the center of spiral similarity sending $RI$ to $I_AX$. $\blacksquare$ Hence $\angle{AXI_A} = \angle{AIR} = 90^{\circ}$, as needed. $\blacksquare$

Well I did read something related to fact 5 inversion in this thread but to my defence I can say that atleast the idea came to me naturally xD And of course if I don't solve a problem normally I always invert , also I am so out of touch that I didn't even recognize the sharky devil point It is obvious that $T=MD \cap \omega$ because of Curvilinear circle lemma from EGMO (or homothety or inversion). Now invert with respect to $\odot(BICI_A)$ and let the image of $S$ be $S'$ and it becomes obvious that $S$ is actually just the antipode of $A$ (because $MS' \perp AM$ which would be obvious if I had recognized the point ). Due to the inversion we also obtain that $S'E \perp EI_A$ as $S' \in \overline{BC}$ and hence we obtain that $ES'I_AM$ is cyclic. By inverting back we get that this concyclicity directly implies the question as the intersection point is the inverted image of $E$ $\blacksquare$

Very beautiful configuration Here is a sketch Observe - TDM collinear - S is antipode of A - Let Excentre of A and S meet the circumcircle at X Angle EXS=DTH' (Here H' is reflection of orthocenter on the circumcircle) This is due to BD=CE

We note that $T$ is simply $MD \cap (ABC)$, or the Sharkydevil point. Thus $S$ is the antipode of $A$, so if we define $K = ME \cap (ABC)$, it suffices to prove $\angle AKI_A = 90$. Shooting Lemma along with Incenter-Excenter tells us \[ME \cdot MK = MC^2 = MI_A^2 \implies \triangle MEI_A \sim \triangle MI_AK,\]which indeed gives \[\measuredangle AKI_A = \measuredangle AKM + \measuredangle MKI_A = \measuredangle (AM, BC) + \measuredangle EI_AM = 90. \quad \blacksquare\]

Note that by homothety $T$, $D$, $M$ are collinear, which means that $T$ is the $A$-Sharkydevil point. This also means that $T$ lies on $(AI)$, so $S$ is the antipode of $A$ on $(ABC)$. Furthermore, $MD$ and $ME$ are reflections across the perpendicular bisector of $BC$, so if $ME$ intersects the circumcircle at $K$, then $K$ is the reflection of the $A$-sharkydevil point across the perpendicular bisector of $BC$. It suffices to show that $K$, $S$, and $I_a$ are collinear. To do this, we perform a $\sqrt{bc}$ inversion followed by a reflection across the bisector. The image of $I_a$ is $I$. The image of $S$ is the foot from $A$ to $BC$, which we call $X$. Since $TBCK'$ is an isosceles trapezoid, $AT$ is isogonal to $AK$, which means that the image of $K$ lies on $AT$, which means it is the intersection of $AT$ with $BC$. However, by Radical Axis Theorem on $(ABC)$, $(AI)$, and $(BIC)$, we have that $AT$, $II$, and $BC$ concur, so $K'$ lies on $II$. Since $(AI)$ and $(BIC)$ are tangent, we have $\angle K'IA=90$. Since $\angle K'XA=90$ as well, $K'XIA$ is cyclic, so we are done. remark: After the antipode remark, an alternative way to finish would be to note that $ME\cdot MK=MD\cdot MT=MB^2=MI_a^2$ by the tangent circles configuration, after which angle chasing suffices as we can show $\angle AKI_a=90$. The idea here is that this angle that we want to equal 90 is composed of an angle that we know very well (since it is inscribed in $AM$) and a more strange angle, suggesting that the strange angle has a nice value as well. We can then actually work out what that angle must be, and from our angle chasing realize that it's equal to $\angle MI_aE$ and we immediately know what to do from here.

Note that $T,D,M$ are collinear, so $T$ is the $A$-Sharky-Devil point. It follows that $S$ is the antipode of $A$ in $\omega$. Let $ME$ intersect $\omega$ again at $X$, then $X$ is the reflection of $T$ across the angle bisector of $BC$. Let $S'=MS \cap BC$. Inverting about $(BIC)$, $X$ goes to $E$ and $S$ goes to $S'$. It suffices to show that $M,E,S',I_A$ are concyclic. But $\measuredangle I_AMS' = \measuredangle AMS = 90^{\circ} = \measuredangle I_AES'$, so we are done. $\square$

Let $X = \overline{AI} \cap \overline{BC}$. Perform a $\sqrt{bc}$ inversion followed by reflection across $\overline{AI}$ and note, $E$ is sent to the mixintillinear touch point, which we call $K$. $I$ and $I_a$ swap positions. $X$ and $M$ swap positions. $S$ maps to the foot from $A$ to $\overline{BC}$, say $F$. Then it suffices to show that $(AFI)$ and $(AKX)$ concur on $\overline{BC}$. Now let $\overline{AT}$ meet $\overline{BC}$ at $Y$. However as $YT \cdot YA = YB \cdot YC$ as $ABTC$ cyclic, we find $\overline{IY} \perp \overline{AI}$. Then inverting about $(BIC)$ note that $Y$ and $K$ swap positions and hence $M$, $K$ and $Y$ collinear. Now note that $A$ and $D$ swap under this transformation and hence $AM \cdot MD = MY \cdot MK$ which gives our first desired concylicity. The other follows as $\angle AFY = \angle AIY = 90$. Thus we are done.

Finally got around to writing solutions.

I don't know why this is in spi sim when its trivial by inversion. Invert at $M$ with radius $MB$. Under this \begin{align*} & I \leftrightarrow I\\ & I_a \leftrightarrow I_a\\ & T \leftrightarrow D\\ & S \leftrightarrow MS\cap BC\\ & ME \leftrightarrow ME\\ & I_aS \leftrightarrow (I_aMS^{*})\\ \end{align*}The problem reduces to showing that $ME \cap (I_aMS^{*}) \in BC$ Then observe that $M,D,I,S^{*}$ is cyclic(due to inversion). Hence $$\angle S^{*}MI_a = \angle S^{*}MD = 90 = \angle S^{*}EI_a$$So $E \in (I_aMS^{*})$ hence we are done as $E\in BC$.

Let $P$ and $Q$ be the feet of the perpendiculars from $I$ to $AB$ and $AC$, and let $N$ be the antipode of $M$. Let $T'$ be the reflection of $T$ over the perpendicular bisector of $BC$. We claim that this is the desired point. First, note that by the shooting lemma we get $TDM$ collinear. Reflecting this over the perpendicular bisector of $BC$ shows $T'EM$ collinear. So it suffices to show that $I_AST'$ collinear. Then we want \[-1=(TT';NM)\stackrel{S}{=}(II_A;(NS\cap AM)M).\]Thus we want $NS\parallel AM$. This is equivalent to $S$ being the $A$-antipode, which is equivalent to $\angle ATI=90^{\circ}$. By the shooting lemma, \[MI^2=MB^2=(MD)(MT)\Longrightarrow \angle ITM=\angle DIM=\angle AMN=\angle ATN.\]So $\angle NTM=\angle ATI$ too. Yay. $\blacksquare$

Denote by $AT\cap BC=R,ME\cap (ABC)=K$. By shooting lemma $T$ is $A-$sharky devil point hence $S$ is the antipode of $A$ on $(ABC)$. $A$ Under $\sqrt{bc}$ inversion and reflection over the angle bisector of $\measuredangle CAB$, $I_A,K$ swap with $I,R$ and $S$ swaps with the foot of the altitude from $A$ to $BC$ which all lie on the circle with diameter $AR$ as desired.$\blacksquare$

Invert >>>> spiral

Alternate spiral solution

easy one

Solution using projective. By shooting lemma $T-D-M$ collinear, so $T$ is the $A$-Sharky-Devil point, so $S$ is the $A$-antipode in $(ABC)$. Now let $N$ be the midpoint of arc $BAC$, then clearly $AM//SN$. We have $-1=(I,I_A;M,\infty_{AM}) \stackrel{S}{=} (T,SI_A \cap (ABC);M,N)$, so $SI_A$ intersects $(ABC)$ at the reflection of $T$ across $\perp_{BC}$, which lies on $ME$ since $BD=CE$. $\square$