Here's the pedestrian solution: Assume for simplicity $AB < AC$. Let $K$ be the contact point of the $A$-excircle on $BC$; also let ray $TD$ meet $\Omega$ again at $L$. From the fact that $\angle FTL = \angle FTD = 180^{\circ} - \angle FED$, we can deduce that $\angle FTL = \angle ACF$, meaning that $F$ is the reflection of $A$ across the perpendicular bisector $\ell$ of $BC$. If we reflect $T$, $D$, $L$ over $\ell$, we deduce $A$, $K$ and the reflection of $T$ across $\ell$ are collinear, which implies that $\angle BAT = \angle CAK$. Now, consider the reflection point $K$ across line $AI$, say $S$. Since ray $AI$ passes through the $A$-excenter, $S$ lies on the $A$-excircle. Since $\angle BAT = \angle CAK$, $S$ also lies on ray $AT$. But the circumcircles of triangles $DEF$ and $EFK$ are congruent (from $DF = KF$), so $S$ lies on the circumcircle of $\triangle DEF$ too. Hence $S$ is the desired intersection point. Of course, $T$ is actually the contact point of the mixtilinear incircle.

Found a pretty convoluted solution during the TST...

I heard that our friend ksun48 has a beautifully brilliant solution, would our friend post here?

trumpeter wrote: I heard that our friend ksun48 has a beautifully brilliant solution, would our friend post here? I thought ksun was the one with the four-page complex bash?

Here's a solution using $\sqrt{bc}-$ inversion. Let $K$ be the point where the $A-$ excircle is tangent to $BC$. This proof uses well known properties of $\sqrt{bc}-$ inversion.

Should the OP say circumcircle instead of circumcenter? Fixed ~dj

My solution: Let $\omega$ the incircle of $\triangle ABC$ and let $\Gamma$ the $A$-excircle. Let $D'=\Gamma\cap BC$ $\Longrightarrow $ $BD=CD'$, let $X$ the reflection of $D'$ respect to $AI$ such that $I$ is the incenter of $\triangle ABC$ $\Longrightarrow $ $X\in \Gamma$ and let $Y$ be a point such that $Y\in \omega$ and $EY$ is tangent to $\omega$ $\Longrightarrow $ $EYD'F$ is cyclic since $FD=FD'=FY$ and $EA$ is bisector of $\angle DEY$ $\Longrightarrow $ by reflection in $AI$ we get $EDFX$ is cyclic since $EYD'F$ is cyclic and $\angle XAB=\angle D'AC$ and let $Z=AD'\cap \Omega$ and let $T'$ be a point in $\Omega$ such that $T'Z\parallel BC$ $\Longrightarrow $ $\angle T'AB=\angle D'AC$. By angle-chasing we get $\angle FZD'=\angle FT'D=\angle DEA$ $\Longrightarrow $ $DEFT'$ is cyclic $\Longrightarrow $ $T=T'$ and $A,X,T'$ are collinear since $\angle XAB=\angle T'AB=\angle ZAC$ $\Longrightarrow $ $A,X,T$ are collinear, $X\in \Gamma$ and $X\in \odot (DEF)$ $\Longrightarrow $ $X=S_1$ or $X=S_2$ $\Longrightarrow $ $A,T,S_1$ are collinear or $A,T,S_2$ are collinear and $EX$ is tangent to $\Gamma$...

Here is a sketch of my solution: We will denote by $I$ the incenter of $\triangle{ABC}$ and by $J$ we will denote the point where the $A$-excircle touches $BC$.Also let $TD\cap\Omega\equiv L$. Easy angle-chasing yields $\angle{BTD}=\angle{ABC}$,so $\angle{LAB}=180^0-\angle{ABC}$,which means that $AL\parallel BC$.Hence $\triangle{ABC}\equiv\triangle{LCB}$(note that $AL\parallel BC$ and $L\in\Omega$),so $\angle{BDT}=\angle{LDC}=\angle{AJB}$. Hence $$\angle{ABT}=\angle{ABC}+\angle{DBT}=\angle{BTD}+\angle{DBT}=180^0-\angle{BDT}=180^0-\angle{AJB}=\angle{AJC}.$$But $\angle{ATB}=\angle{ACJ}$,so $\triangle{BAT}\sim\triangle{JAC}$.From here we obtain that lines $AT,AJ$ are isogonal in $\triangle{BAC}$.Thus $\angle{TAI}=\angle{I_aAJ}$,where $I_a$ is the center of the $A$-excircle By applying law of sines in $\triangle{BAI},\triangle{ATB}$ and $\triangle{AJI_a}$ we easily obtain $\frac{AT}{AI}=\frac{AI_a}{AJ}$. But $\angle{TAI}=\angle{I_aAJ}$,so $\triangle{TAI}\sim\triangle{I_aAJ}$. Now let $S\equiv AT\cap\odot(DEF)$.PoP yields $AT\cdot AS=AE\cdot AF\stackrel{\text{well known}}{=}AI\cdot AI_a$,so points $T,S,I_a,I$ are concyclic.Hence $\angle{AI_aS}=\angle{ATI}\stackrel{\triangle{TAI}\sim\triangle{I_aAJ}}{=}\angle{AI_aJ}$,but lines $AT,AJ$ are symmetric wrt $AI$,so $\angle{AI_aS}=\angle{AI_aJ}$ implies $I_aS=I_aJ$.Therefore $S$ lies on the $A$-excircle of $\triangle{ABC}$.This ends our proof,because $AT\cap\odot(DEF)\equiv S$(so $S\in\odot(DEF)$) and $S$ lies on the $A$-excircle of $\triangle{ABC}$,meaning that $S_1\in AT$ or $S_2\in AT$.

Here is an extension Let $ABC$ be a triangle inscribed in circle $(O)$ with bisector $AD$. $P$ is a point on $AD$. $AD$ cuts $(O)$ again at $E$ and cuts $(PBC)$ again at $J$. $U$ is projection of $J$ on $BC$. $V$ is symmetric of $U$ through midpoint of $BC$. $(VDE)$ cuts $(O)$ again at $Q$. Prove that $AQ$ passes through one of intersections of $(VDE)$ and $(J,JU)$.

v_Enhance wrote: Proposed by Evan Chen A certain person remarked that the author was most likely going to be Evan Chen...

This was cute and nice to try, but a bit easy may be for its position as I found #1 a lot more harder (probably because I suck at combo and/or Mixitilinear circles are my favorites all thanks to the "guessing" game handout ) Assume w.l.o.g. that $AB<AC$ and see that $T$ is the touch-point of the $A$ mixitlinear incircle with the circumcircle of triangle $ABC$. Now, let $J$ be the $A$-excenter and $S$ be the point on side $BC$ such that $BS=CD$. Then, we let $AT$ meet $(DEF)$ again at $X$ and see that $AX.AT=AE.AF=AB.AC=AT.AS$ and so $AX=AS$ and since $\angle EAX=\angle EAS$ we get that $AE$ is the perpendicular bisector of $SX$ and so $JS=JX$ and $X$ lies on the $A$-excircle thus, $X$ is one of $S_1$ or $S_2$ and we are done. $\square$ By the way, what about the other point since only one of $S_1,S_2$ lies on $AT$ so what happens to the other one? Is it also special? Also, the problem seems not-so easy to come up with, some motivation would be helpful On a side note, (trivial one) if we replace everywhere in the above solution "in" by "ex", it doesn't really matter This seems a rich configuration as EulerMacaroni said about line $ES$.Let's try to find more co-incidences then.

buratinogigle wrote: Here is an extension Let $ABC$ be a triangle inscribed in circle $(O)$ with bisector $AD$. $P$ is a point on $AD$. $AD$ cuts $(O)$ again at $E$ and cuts $(PBC)$ again at $J$. $U$ is projection of $J$ on $BC$. $V$ is symmetric of $U$ through midpoint of $BC$. $(VDE)$ cuts $(O)$ again at $Q$. Prove that $AQ$ passes through one of intersections of $(VDE)$ and $(J,JU)$. Let $ U'$ the symmetric of $ U$ wrt the $A$-angle bisector and $Q'$ the intersection of $AU'$ with the ircumcircle of $ABC$ .It 's clear that $ U' $ is on $(J,JU)$ . Besides $AU'\cdot AQ'=AU\cdot AQ' =AB \cdot AC =AD\cdot AE$ which means $Q'$ lies on the circle of $DEU'$ , in the other hand $\widehat{DVE}=\widehat{DUE} =\widehat{DU'E}$ thus $D,E,V,Q'$ and $U'$ are concyclic therefore $Q'=Q$ WCP

trumpeter wrote: A certain person remarked that the author was most likely going to be Evan Chen... Indeed, you can probably tell from the point $T$ alone anantmudgal09 wrote: This was cute and nice to try, but a bit easy may be for its position as I found #1 a lot more harder (probably because I suck at combo and/or Mixitilinear circles are my favorites all thanks to the "guessing" game handout ) In my opinion this problem is not so easy if you don't recognize $T$ as the mixtilinear touch point. In particular, it is IMO pretty tricky to realize that the desired $S_i$ is the reflection of the contact point of the excircle over the $\angle A$ bisector. Empirically (i.e. based on how people did), problems 2 and 3 seem to have been of comparable difficulty.

Yes, I agree with that it was pretty slippery without knowing about the Mixitilinear touch point. In that case, it's a floating diagrams where no piece fits nicely.And the reflection part is a bit tricky as in IMO 2015/4 and not easy to spot. Also, maybe you can show the motivation behind this in your blog post on posing geometry problems.

@ v_Enhance, how did you create this beautiful problem?

MathPanda1 wrote: @ v_Enhance, how did you create this beautiful problem? I think you should look at his handout "Writing Olympiad Geometry Problems" which can be found here.

I know. I'm just wondering how he came up with this problem.

anantmudgal09 wrote: Also, maybe you can show the motivation behind this in your blog post on posing geometry problems. MathPanda1 wrote: @ v_Enhance, how did you create this beautiful problem? Well, I was just playing around with the mixtilinear configuration in Geogebra. I had actually drawn both the mixtilinear incircle and the mixtilinear excircle (tangent to circumcircle at $T'$), as well as the $A$-Nagel line (because it is isogonal to $AT$); let me call it $AK$, where $K$ is the contact point of the excircle. Then by chance I noticed that the farther intersection of the $A$-Nagel line with the mixtilinear incircle was concyclic with $T'$, $E$, $F$, $K$ -- cyclic pentagon! Of course I already knew about $T'EFK$ from its mixtilinear analog, but the fifth point was initially a surprise to me. Now a $\sqrt{bc}$ inversion of this observation gives the TST problem.

We claim that $T=T'$, where $T'$ is the $A$-mixtilinear touchpoint. Since $\triangle ABT\sim\triangle CDT$, \[\measuredangle T'FE=\measuredangle T'FA=\measuredangle T'BA=\measuredangle T'DC=\measuredangle T'DE,\]so $T'$ lies on $(DEF)$, from which $T=T'$. Consider an inversion $\Phi$ about $A$ with radius $\sqrt{AB \cdot AC}$. Then, $\Phi$ swaps $E$ and $F$. Let $\Phi(T)=X$; then $(AX)(AT)=(AB)(AC)=(AE)(AF)$ as $\Phi$ swaps $B \Leftrightarrow C$, $E \Leftrightarrow F$, $T \Leftrightarrow X$. Note that converse of PoP yields that $DEFTX$ is cyclic. Since $\Phi$ maps the $A$-mixtilinear incircle to the $A$-excircle, $X$ lies on the $A$-excircle and is thus either $S_1$ or $S_2$, from which the conclusion follows.

Bhai, what a configuration!! Kudos to Evan Saar. Firstly, I claim that the point $T$ is the $A-$mixtilinear intouch point. For the proof, firstly $M_A$ denote the midpoint of the major arc $\widehat{BC}$, $P$ and $Q$ denote the $AC-$ and $AB-$ mixtilinear touch points respectively, $A_{SD}$ denote the $A-$sharky devil point and $\odot(AI)$ denote the circle with diameter $AI$. Now redefine $T$ as the $A-$mixtilinear intouch point. Now it is well known that $\overline{T-I-M_A}$ are collinear which gives that $\measuredangle MTI=90^\circ$ which means that $\odot(MTI)$ is tangent to $PQ$. Also $\odot(AI)$ becomes tangent to $PQ$. Now Radax on $\left\{\odot(AI),\odot(MI),\odot(ABC)\right\}$ gives that their Radax all concur say at $X$. Now $\odot(M)$ denote the circle with center $M$ and radius $MI$. Now apply Radax again on $\left\{\odot(M),\odot(AI),\odot(ABC)\right\}$ which with the previous Radax gives that $X=AA_{SD}\cap PQ\cap BC\cap MT$. Now from some angel chasing, we get,\[ \measuredangle AA_{SD}D=\measuredangle AA_{SD}M=\measuredangle ABM=\measuredangle AKC=\measuredangle AKD, \]which gives that $AA_{SD}DK$ is cyclic. Now finally we can use POP to get $XT\cdot XM=XA\cdot XA_{SD}=XD\cdot XK$ which finally proves that $DKMT$ is cyclic and we are done with our claim. Now perform an Inversion with center $A$ and radius $\sqrt{AB\cdot AC}$ followed by a reflection along the angle bisector of $\angle ABC$. So our current problem statement suffices to prove that if $E$ is the $A-$extouch point, then the $A-$nagel cevian passes through one of the intersections of the $A-$mixtilinear incircle and $\odot(MKE)$. Firstly note that as $\left\{K,M\right\}$ get swapped under the Inversion, we get that the circle gets fixed under this $\sqrt{bc}$ Inversion which means that the image of $\odot(KTM)$ is just the reflection of $\odot(KTM)$ over $AI$. Now the $A-$mixtilinear incircle is invariant under reflection across the angle bisector. This upon reflection across the angle bisector of $\angle ABC$ gives that the intersection of $T=A-\text{mixtilinear incircle}\cap\odot(KTM)\leftrightarrow A-\text{mixtilinear incircle}\cap\odot(KEM)\overset{\mathrm{def}}=S$ which indeed lies on the reflection of $AT$ across the $A-$angle bisector which is indeed the $A-$nagel cevian and we are done .

took me a while to see the mx-incircle touchpoint Let $S$ be the (unique) intersection of $(DEF)$ and $AT$; we wish to show that $S$ lies on the $A$-excircle. The main claim of the problem is the following: Claim: $T$ is the $A$-mixtilinear incircle touchpoint in $ABC$. Proof. This is just angle chasing; letting the touchpoint be $T_1$, we have that \[\measuredangle T_1FE = \measuredangle T_1FA = \measuredangle T_1BA = -(\measuredangle BAT_1 + \measuredangle AT_1B) = -(\measuredangle DT_1C + \measuredangle DCT_1) = \measuredangle T_1DE\](might have messed up the directed angles here oops). Hence $DEFT_1$ cyclic $\implies T_1 = T$, as desired. $\square$ Now let $D'$ be the $A$-excircle touchpoint - it is well known that $AT$ and $AD'$ are isogonal in $ABC$. Consider the $\sqrt{bc}$ inversion followed by a reflection about the $A$-angle bisector - denote all images of this transformation with an asterisk $^*$. Note that $E^* = F$ and $F* = E$, and $T^* = D'$ and $D'^* = T$. In particular, we see that $S*$ is the intersection of $(D'EF)$ and line $A'D$. However, \[AS\cdot AT=AE\cdot AF=AB\cdot AC\]so $S^*$ is the reflection of $T$ over $EF$; since the mixtilinear incircle is symmetric w.r.t. the $A$-angle bisector $EF$, it follows that $S^*$ lies on the mixtilinear incircle. Inverting back (and noting that mixtilinear incircle and excircle swap under $\sqrt{bc}$ inversion) gives the desired. $\blacksquare$

Let $T'\in\Omega$ be the mixtilinear intouch point and $G$ the excontact point on $BC$ so that $AT$ and $AG$ are isogonal. Since $AG$ intersects $\Omega$ at the reflection of $T$ over the perpendicular bisector of $BC$, $TD$ intersects $\Omega$ at the reflection $A'$ of $A$ over the perpendicular bisector of $BC$. Then we have $$\measuredangle FTD = \measuredangle FTA' = \measuredangle BEA = \measuredangle FED$$ so $T$ lies on $(DEF)$. Then if $S$ is the reflection of $G$ over the $A-$ internal angle bisector, we clearly have $A,S,T$ collinear and $S\in A-$ excircle and from $$AS\cdot AT = AS'\cdot AT = \sqrt{AB\cdot AC} = AE\cdot AF$$ it follows that $S$ lies on $(TEF)$, as desired.

After panicking on discord as to why the conjecture of iverting takes mixtilinear circle to excircle, i figured it out after like 2 minutes oops also the otis version has different point labels, just redefine E as K and F as M (ignore the extra F, that's the center of mixtilinear circle) angles are taken mod 180 and plus minus switched without any regards to order of directed angles because i do not want to label them correctly One of my favorite problems Note that $TDC=TMK=TBA$ so T is the A-mixtilinear intouchpoint. Take a sqrtbc inversion (without reflection over angle bisector), s.t. $K\leftrightarrow M,T\leftrightarrow S\impliedby AB\cdot AC=AK\cdot AM=AT\cdot AS$. And by the diagram you can see $AG\cdot AM=AI\cdot AJ=AB\cdot AC$ which is the inversion ratio, so indeed the mixtilinear circle is mapped to the excircle through this (it's well known GH passes through I and JL passes through M). Hence S lies on the mixtilinear circle as well, and consequently it's either $S_1$ or $S_2$.

somewhat silly Claim: $T$ is the mixtilinear intouch point. Proof: We have $\measuredangle FTD=\measuredangle FED=\measuredangle AEB=\measuredangle ACF$. Thus if $\overline{TD}$ intersects $(ABC)$ again at $X$, $\widehat{FX}=\widehat{AF}$, so $ABCX$ is an isosceles trapezoid and $\overline{TA}$ and $\overline{TD}$ are isogonal in $\triangle BTC$. By a well-known mixtilinear incircle property the result follows. Now let $S'=(DEFT) \cap \overline{AT} \neq T$ and let $D'$ be the $A$-extouch point. By power of a point, $AS'\cdot AT=AE\cdot AF$. By $\sqrt{bc}$ inversion and reflection about the $\angle A$-bisector, $AB\cdot AC=AE\cdot AF=AT\cdot AD'$, since this swaps the $A$-excircle and $A$-mixtilinear incircle. Therefore $AS'=AD'$, and since $\overline{AS'}$ and $\overline{AD'}$ are isogonal it follows that the perpendicular bisector of $\overline{S'D'}$ is the $\angle A$-bisector, hence if $I_A$ is the $A$-excenter we also have $I_AS'=I_AD'$, so $S'$ lies on the $A$-excircle as well. This implies the desired result. $\blacksquare$

Claim: $T$, the intouchpoint to $BC$, and the reflection of $A$ across the perpendicular bisector of $BC$ are collinear. Let $TD$ intersect the circumcircle again at $Q$. Then, $$\angle QBM=\angle QTM=\angle AKB=180-\alpha/2-\beta=\gamma+\beta/2=\angle ACM,$$hence $AQCB$ is an isosceles trapezoid which shows the claim. Now, let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. Let $D'$ be the extouchpoint to $BC$, and let $P$ be the reflection of $D'$ across $AI$. We claim that $P$ lies on $AT$, $(DKM)$, and the A-excircle, which would of course solve the problem. $P$ lies on A-excircle: This is simply because $D'$, a point on the excircle, was reflected over line that passes through the center of the excircle. $P$ lies on $(DKM)$: Note that the circumcircle of $(PKM)$ is just the circumcircle of $\triangle D'KM$ reflected over $KM$. We claim that the circumcircle of $(DKM)$ is also the reflection of the circumcircle of $(D'KM)$ over $KM$. Note that $DM=D'M$, so in the two triangles, $\angle DKM$ corresponds to $DM$ while $\angle D'KM$ corresponds to $D'M$, and since the angles add to 180 so they have the same sine, the have the same circumradius. Since their radical axis is $KM$, they are just reflections of each other across $KM$. Hence, $(DKM)$ and $(PKM)$ are the same circle, which shows that $P$ lies on $(DKM)$. $P$ lies on $AT$: This is the same as saying that $AT$ is isogonal to $AD'$. However, we have $$\angle BAT=\angle BA'T=\angle BA'D=\angle CAD'$$where the last step follows by reflection across the perpendicular bisector of $BC$. Thus, we are done.

By a well-known lemma $T$ is the $A$-mixtilinear intouch point. Thus, take a $\sqrt{bc}$ inversion at $A$ without the reflection. Since this takes the $A$-mixtilinear incircle to the excircle, the inverse of $T$ must be on the excircle. Let this inverse be $S.$ However, since $K$ and $M$ are inverses, we have $AK \cdot AM = AB \cdot AC = AS \cdot AT,$ so $S$ lies on circle $KMT.$ Thus, $S$ must be either $S_1$ or $S_2,$ so we are done,

Claim. $T$ is the $A$-mixtilinear touchpoint. Proof. This is well-known, but for completeness, this follows as $$\angle DTM = \angle DTC + \angle CTM = C+\frac A2 = \angle DKA. \ \ \blacksquare$$ Denote $E$ to be the $A$-excentral touchpoint, and set $S = \overline{AT} \cap (DKM)$. Note that $(DKM)$ and $(EKM)$ are symmetric about the $\angle A$-bisector as $\angle KDM = \angle KEM$. So the reflection of $E$ over $\overline{AI}$ lies both on $\overline{AT}$ and $(DKM)$, hence it is precisely $S$. On the other hand the previously described point must also lie on the $A$-excircle, which finishes the problem.

We begin with the following: Claim 1: $T$ is the $A$-mixtilinear incircle touch point. If we let $T'$ be this touch point, note \[\measuredangle DKM = \measuredangle CAK + \measuredangle KCA = \measuredangle CTM + \measuredangle DTC = \angle DTM. \quad {\color{blue} \Box}\] Claim 2: $S'$ lies on the $A$-excircle, where $S' = (MDK) \cap AT$. Consider the outcome of force overlaying this diagram. Because \[AS \cdot AT = AK \cdot AM = AB \cdot BC,\]we know that $T^*$ is simply the reflection of $S'$ over the angle bisector, but it is also well known to be the $A$-excircle touch point. Thus $S'$ and $T^*$ are equidistant from the $A$-excenter, concluding the proof. $\blacksquare$

Let $S$ be the second intersection of $(DEF)$ with $\overline{AT}$; we want to show that $S$ lies on the $A$-excircle. Claim: $T$ is the mixtillinear touch point. Proof: Consider $\sqrt{bc}$ inversion at $A$: since $AE \cdot AF = AB \cdot AC$, the radius of circle $(DEF)$ is not affected by this inversion, and its image is merely the reflection of $(DEF)$ about line $EF$. The intersection of this reflected circle with $\overline{BC}$ must be the point $D'$ on $\overline{BC}$ for which $\angle ED'F = \angle EDF$, otherwise known as the extouch point. So, $T$ gets sent to the $D'$, implying that $T$ is the mixtillinear touch point. Two more consequences follow from the $\sqrt{bc}$ inversion: one, $AS = AD'$, and two, line $AI$ bisects $\angle TAD'$. So, $S$ is the reflection of $D'$ over line $AI$, so the result follows.

Let $A'$ be the reflection of $A$ across the perpendicular bisector of $BC$. $\measuredangle FTD = \measuredangle FED = \measuredangle AEB = \measuredangle ABE + \measuredangle EAB = \measuredangle AFC + \measuredangle CAF = \measuredangle ACF = \measuredangle FBA' = \measuredangle FTA'$, so $T,D,A'$ are collinear. It follows that $T$ is the $A$-mixtillinear intouch point. Let $S$ be the second intersection between $AT$ and $(DEFT)$. It suffices to show that $S$ lies on the $A$-excircle. Let $T'$ be the $A$-extouch point, which is also the image of $T$ under $\sqrt{bc}$ inversion. Note that $E,F$ are also images of each other under $\sqrt{bc}$ inversion, so $AE \cdot AF = AB \cdot AC = AT \cdot AT' $. By pop, $AS \cdot AT' = AE \cdot AF$, so $AS=AT'$. Note that $AS$ and $AT'$ are reflections across the $A$-bisector, so the $A$-bisector is the perpendicular bisector of $ST'$. Let $I_A$ be the $A$-excenter. Then $I_AS=I_AT'$, so $S$ lies on the $A$-excircle as desired.

My diagram has like 10 circles

A solution with Reim's and only angle chasing Using Reim's on circles $(DEFT)$ and $\Omega$ with points $S$ and $E$. The intersections of the circles $(DEFT)$ and $\Omega$ are $T,F$. So $TS$ passes through $A $ if and only if the lines $AA$ (which is the tangent) and $FS$ are parallel. Let the tangent line to $A$-excircle from $E$ (other than $BC$) touches $A$-excircle at $S$. We know that $\measuredangle SEF = \measuredangle FEC = \measuredangle AAF$. Hence, tangent line from $A$ and the line $SE$ are parallel. Proving that $S$ is on the circle $(DEF)$ is ample now. Let $M$ be the midpoint of $BC$ and $D'$ be the touch point of $A$-excircle on $BC$. $\measuredangle SFD = \measuredangle SFE - \measuredangle DFE = \measuredangle EFD'-\measuredangle DFE = 2\measuredangle EFM = 2(90°-\measuredangle MEF) = 180°- \measuredangle MES= \measuredangle DES \blacksquare$

A solution that I found after inverting with $\sqrt{bc}$ inversion, which is really interesting! W.L.O.G. $AT$ passes through $S_1$. Denote the incircle and the $A$-excircle of $\triangle{ABC}$ by $\omega$ and $\Gamma$, respectively. Let their centers.be $I$ and $I_A$, respectively. Let $M,N$ be the contact points of $\omega$ with $AC,AB$, respectively. Let $M,N,G$ be the contact points of $\Gamma$ with $AC,AB,BC$, respectively. We claim that $T$ is the contact point of $A$-mixtilinear incircle (denote it as $\gamma$) with $\Omega$. To do so, let $T'$ be the desired point. Also, let $O_A$ be the center of the circle and $K,J$ be the contact points of the circle with $AC,AB$, respectively. It suffices to prove that $T'=(DEF)\cap\Omega$. Invert with center $A$ and radius $\sqrt{AB\times AC}$, followed by the reflection over $A$-angle bisector (also called by $\sqrt{bc}$ inversion), denote it as $\varphi$. Well known that $\gamma$ will be mapped to $\Gamma$ and $\Omega$ will be mapped to $BC$ under this inversion. Hence, $\varphi: \gamma\cap\Omega\leftrightarrow\Gamma\cap BC$, or equivalently $\varphi: T'\leftrightarrow G$. Thus, lines $AT'$ and $AG$ are isogonal w.r.t. $\angle{A}$, and so it's easy to prove that $\triangle{ABT'}\sim\triangle{AGC}$. It's well-known that $BD=CG$, so we have $\frac{BT'}{CG}=\frac{AT'}{AC}\Longleftrightarrow\frac{BT'}{BD}=\frac{AT'}{AC}$. But $\angle{T'BD}=\angle{T'BC}=\angle{T'AC}$, so by $SAS$ similarity we may conclude that $\triangle{BT'D}\sim\triangle{AT'C}$, so $\angle{T'FE}=\angle{T'FA}=\angle{T'CA}=\angle{BDT'}$, meaning that $T'$ lies on both $\Omega$ and $\gamma$, as desired. There exists a homothety centered at $A$ that brings $\gamma$ to $\omega$. Now, see that $JI\perp AI_A$ and $AN'\perp I_AN'$, so $IJN'I_A$ is cyclic. And so, $bc=AI\times AI_A=AJ\times AN'=AJ\times s\Longleftrightarrow AJ=\frac{bc}{s}$. So $\frac{AJ}{AN}=\frac{bc}{s(s-a)}$ is the ratio of this homothety. By that, we have $AP=AD'\cdot \frac{bc}{s(s-a)}$. Let the antipode of $D$ w.r.t. $\omega$ be $D'$. By homothety, it's clear that $\overline{A,D',G}$ and $AG=\frac{s}{s-a}\cdot AD'$. Let $AG$ hits $\gamma$ at $P,S_1^*$, respectively, with $P$ is higher than $S_1^*$. We claim that $S_1^*$ lies on $(EFG)$. By Power of Point's Theorem, we have \begin{align*} AJ^2&=AP\times AS_1^*\\ \frac{bc^2}{s^2}&=AD'\cdot \frac{bc}{s(s-a)}\times AS_1^*\\ AS_1^*&=\frac{bc(s-a)}{s\cdot AD'} \end{align*}So we have $AG\times AS_1^*=\frac{s}{s-a}\cdot AD'\times\frac{bc(s-a)}{s\cdot AD'}=bc=AE\times AF$, so our claim is true by Power of Point's Theorem. Hence, $S_1^*=(EFG)\cap\gamma$. Now by using the previous inversion, we have $\varphi:(DET)\cap\Gamma\leftrightarrow(EFG)\cap\gamma$, or equivalently $\varphi:S_1\leftrightarrow S_1^*$. But $\varphi: T\leftrightarrow G$ and $S_1^*\in AG$, so inverting back yields $\overline{A,S_1,T}$, as desired. $\blacksquare$