Any ideas?

First consider $K=FE\cap AL$, after inversion centered at $A$ with radius $AF$ it is easy to check that $K$ is the image of $M$ and that $AKDF$ is clyclic. Angle chasing to show that $\angle LAE=\angle ADL$ and $\angle LKE=\angle KDL$, these imply that $L$ is the midpoint of $AK$ and then $L,Q,T$ are collinear and $LT\parallel FE$. Now, $ALDT$ is cyclic then $\angle TDA=\angle TLA$, but $\angle TLA=\angle FKA=\angle FDA$, and then $T,F,D$ are collinear. $TL$ meets $AF$ at the midpoint, then $S\in TL$, moreover $\angle ATL=\angle ADL=\angle XFE$ then $F,X,S$ are collinear. $AS\parallel TD$ implies $\angle XAS=\angle FDA=\angle FKA=\angle SLA$ then $XA$ is tangent to the circumcircle of $\triangle ASL$ at $A$. $ASXN$ cyclic and $XS\parallel TA$ imply $\angle ATS=\angle XST=\angle XAN$ then $XA$ is tangent to the circumcircle of $\triangle ATN$ at $A$ and we are done.

It is easy to get that $AM \mid\mid EX$ , now let $T'$ be a point of $DF$ such that $AT' \mid\mid FX$, note that quadrilaterals $AT'DL$ and $XFDE$ are similar, thus $AT'DL$ is harmonic.Now let $P$ be the point at infinity of line $BC$ and because the tangent at $L$ to circle $ALDT'$ is parallel to $BC$, we have that $MT'$ bisects $AE$, because $\ -1=L(A,E;T',L)=(A,E;MT'\cap AE,P)$. Now the tangency of the two circles can be easily established, by proving that $AD$ is tangent to both of them.

My solution: At first using cyclic quadrilaterals note that $\angle{XEA}=\angle{XFM}=\angle{EAM}$ so $AM \parallel XE$. CLAIM 1: $T,F,D$ are collinear. Proof:Note that since $\angle{DFE}=\angle{DXE}=\angle{DAL}=\angle{LAD}=\angle{LTD}$, it suffices to show that $TL \parallel FE$. Now,since $\angle{XDE}=\angle{XFE}=\angle{EAM}$ deduce that $LA$ is tangent to the circumcircle of triangle $DEA$ so $LA^2=LE\cdot LD$(1).But since $Q$ is the midpoint of $AE$, $QA^2=QE^2$(2),and (1) and (2) imply that $L,Q$ have equal powers with respect to the degenerate circle with center $A$ and the incircle,so $TL$ is the radical axis of these circles and thus is perpendicular to $AI$ where $I$ is the incenter of $ABC$.But also $AI\perp EF$ so $TL\parallel FE$,whih proves claim1. CLAIM 2:$S\equiv TL\cap FX$ Proof:Let $S'\equiv TL\cap FX$ and it must be shown that $S\equiv S'$.Let $Y\equiv TL\cap AF$ and $R\equiv XD\cap FE$.Obviously,$Y$ is the midpoint of $AF$ and the pencil $(FA,FX,FR,FD)$ is harmonic.So since $S'\equiv TL\cap FX$ ,$T\equiv TL\cap FD$ and $TL \parallel FE$,by a well known lemma in harmonics,$Y\equiv TL\cap FA$ must be the midpoint of $S'T$.But it is also the midpoint of $AF$ so $ATFS'$ is a parallelogram so $S\equiv S'$ which proves claim 2. CLAIM 3:$X,E,N$ are collinear. Proof:Note at first that since $\angle {MAQ}=\angle {SFE}=\angle {QSF}$ and $\angle {QEM}=\angle {AFM}=\angle {ADF}=\angle {ALQ}$,quadrilaterals $AMQS,MQEL$ are cyclic.So by angle chasing,$\angle {AXN}=\angle {ASQ}=\angle {AMQ}=\angle {QEL}=\angle {DEC}=\angle {DXE}$ which proves claim 3 CLAIM 4:the circumcircles of triangles $TAN$ and $LSA$ are tangent to each other. Proof:It is easy to see that triangles $ATL,XFE$ are similar since they have parallel sides.Let $Ax$ be the tangent line to $A$ of the circumcircle of triangle $ATN$ (in fact this line is just $AX$,but this is not necessary to continue).Note that $\angle {SAx}=\angle {NAS}-\angle{NAX}=180-\angle{NXS}-\angle{ATN}=180-\angle {FXE}-\angle {XFE}=\angle {XEF}=\angle {ALS}$ which proves that line $Ax$ is also tangent to the circumcircle of triangle $LSA$,which meas that thes circles are mutually tangent,as desired $\blacksquare$

Nice problem... Let $V$ be the point of intersection of $FE$ and $XD$ First we prove that $AL \parallel XE$ $ \angle MAE = \angle MFE = \angle XDE $ $ \angle LAD = \angle LEA = \angle DEC = \angle EXD $ Now let prove that $LQ \parallel EF$. Let $K$ be the midpoint of $XD$. First note that triangle $ALE$ similar to triangle $DEX$ so $\angle QLE = \angle XEK = \angle FED$ since $FXED$ is harmonic. Now consider homothety with center $D$ and which sends $XE$ to $AL$ and $LT$ to $EF$. It is clear that it sends $(XFDE)$ to $(ATDL)$, $XF$ to $AT$ to so $XF \parallel AT$ and $D - F - T$ are collinear. Let prove that $S\equiv TL\cap FX$. Let $S'\equiv TL\cap FX$. We prove that $S'F = AT$ and since $AT \parallel XF$, $ATFS'$ is parallelogram. First note that $ \angle TS'F = \angle ATS' = \angle XFE = \angle XDE $ and $ \angle LTF = \angle EFD = \angle EXD $ So triangle $TS'F$ similar to triangle $XDE$ and we have $\frac{FS'}{DE} = \frac{TF}{XE}$ $FS' = \frac{DE*TF}{XE} = \frac{XV}{VD} *TF = \frac{AX}{AD} *TF$ Last equality holds because $(AV, XD) = -1$. Since $1-\frac{AX}{AD} = \frac {XD}{DA} = \frac {XF}{AT}$ It is easy to calculate $AT$ and conclude that $AT = S'F$ Now since $\angle SAD = \angle ADT = \angle XEF = \angle ALT$ And $\angle NAD= \angle NSF = \angle ATN$ We conclude that $AD$ is tangent to both of out circles so thet tangent to each other. Done))

This is a nice problem *$(ABC)$ refers to the circumcircle of $\triangle ABC$

Claim 1: (ADL) is tangent to BC

Claim 2: Line LQ contains the A-midline of \triangle AEF

Claim 3: AL is parallel to XE

Claim 4: Points D,F,T are collinear

Claim 5: S is the intersection of TL and FM

Claim 6: Points N,X,E are collinear

Now, it suffices to show that $$\angle ATN+\angle ALS=\angle NAS$$since it implies that we can draw a line $l$ through $A$ such that the angle formed by $l$ with $AN$ equals $\angle ATN$ and the angle formed by $l$ with $AS$ equals $\angle ALS$. This implies that $l$ is a common tangent to $(ATN)$ and $(ALS)$, which implies that the two circles are tangent. Now, we have $$\angle ATN+\angle ALS=180- \angle TAL =180 -\angle SML =180-\angle SXN =\angle NAS$$Which proves the problem $\blacksquare$

Bonus property

Let $K$ be the point where $EF$ and $AL$ meet. Note that $$\angle AMX=\angle AMF=\angle AEF=\angle EDF=\angle MXE \Rightarrow AL \parallel EX$$By the converse of Reim's Theorem, we get that $BC$ is tangent to $\odot (ALD)$ at point $D$. We have the following crucial claim:- CLAIM $L$ is the midpoint of $AK$. Proof of Claim Notice that $\angle DAK=\angle DXE=\angle DFE$ which gives that $AFDK$ is cyclic. Then we have $$\angle KDA=\angle KFA=\angle EFA=\angle EDF \Rightarrow \{DE,DA\} \text{ and } \{DK,DF\} \text{ are isogonal}$$This gives us $\angle KDE=\angle ADF=\angle AKF$, and so $AK$ is tangent to $\odot (DEK)$. Also, $\angle KAE=\angle AEX=\angle EDA$ gives that $AK$ is also tangent to $\odot (ADE)$. Thus, $L$ lies on the radical axis as well as the common tangent of $\odot (ADE)$ and $\odot (DEK)$, which directly gives $LA=LK$. $\Box$ Return to the problem at hand. By our Claim, we get that $LQ$ is parallel to $EF$; or equivalently that it bisects $AF$. But, as $ATFS$ is a parallelogram, so the midpoint of $AF$ must lie on line $ST$. Thus, we have $S \in LQ$. And, $$\angle DTL=\angle DAL=\angle DXE=\angle DFE \Rightarrow \text{Using the fact that } LT \parallel EF \text{, we get } T \in DF$$Then Reim's Theorem gives $XF \parallel AT$, which means that $S$ also lies on line $FM$. This gives us $$\angle XAN=\angle XSN=\angle ATN \Rightarrow AX \text{ is tangent to } \odot (TAN)$$Finally, since $FT \parallel AS$, so we get $$\angle XAS=\angle TDA=\angle ALS \Rightarrow AX \text{ is tangent to } \odot (LSA)$$Thus, $\odot (TAN)$ and $\odot (LSA)$ are tangent to each other at $A$. Hence, done. $\blacksquare$

This is such a great and rich configurational geometry problem. One of the best example to illustrate dissecting figure into pieces before assembling it into one. Let us first delete the point $T,S,N$ for convenience.

Some Motivation before starting the problem

Claim 01. $XE \parallel AM$. Proof. We have $\measuredangle FXE = \measuredangle FDE = \measuredangle FEA = \measuredangle FMA$. Claim 02. $LQ \parallel EF$. Proof. Notice that $(X,D;E,F) = -1$. Therefore, \[ -1 = (X,D;E,F) \overset{E}{=} (XE \cap AM,L;A, EF \cap DL)\]Let $EF \cap DL = G$. We then conclude that $L$ is the midpoint of $AG$. Claim 03. $AGDF$ is cyclic. Proof. We have $\measuredangle DFG \equiv \measuredangle DFE = \measuredangle DXE = \measuredangle DAM = \measuredangle DAG$. Now, we are ready to deal with other points $T,S$ and $N$. Claim 04. $D,F,T$ are collinear Proof. $\measuredangle ADT = \measuredangle ALT = \measuredangle AL'F = \measuredangle ADF$ Claim 05. $AD$ tangent $(ASL)$. Proof. We have \[ \measuredangle ALS = \measuredangle AGF = \measuredangle ADF \equiv \measuredangle ADT \overset{AS \parallel TF}{=} \measuredangle DAS \] Claim 06. $N,X,E$ are collinear. Proof. We have \[ \measuredangle NXA = \measuredangle TSA = \measuredangle STD = \measuredangle EFD = \measuredangle EXD \] Claim 07. $AN \parallel DE$. Proof. From the previous claim, we have $N,X,E$ are collinear. Therefore $AL \parallel NE$. Moreover, $Q$ is the midpoint of $AE$ and $L,Q,N$ are collinear by definition of $N$. Then, $ANEL$ is parallelogram, forcing $AN \parallel LE \equiv DE$. Claim 08. $AD$ tangent $(ANT)$. Proof. To prove this, \[ \measuredangle ATN \equiv \measuredangle ATL= \measuredangle ADL \overset{AN \parallel DE}{=} \measuredangle NAX \] To finish this, we notice that $AD$ is tangent to both $(ANT)$ and $(ASL)$, which conclude that $(ANT)$ and $(ASL)$ are tangent to each other at $A$.

Remark:Other Facts

What is EMC?

European Mathematical Cup

Nice probelm Let $\ell$ denote the $A$-midline of $\triangle AEF$. Note that $\ell$ is the radical axis of $\odot (DEF)$ and $\odot (A,0)$. Note that we have : $$\angle FXE = \pi- \angle FDE = \pi- \angle AEF = \pi - \angle AMF = \angle LMX$$ Hence $XE \parallel ML$. We now contend that $L,T \in \ell$. Again we angle chase : $ \angle LAE = \angle AEX =\angle XDE$, which implies that $LA$ is tangent to $\odot (ADE)$. Hence $LA^2=LD\cdot LE$. So $L$ lies on the radical axis of $\odot (DEF)$ and $\odot (A,0)$. Hence we have $LT \parallel EF$. Now consider the homothety at $D$ that sends $X\mapsto A$. Note that $E\mapsto X$ and since $EF \parallel LT$, hence $F \mapsto T$. So $D,F,T$ are collinear. We can prove as above that $XF \parallel AT$. Now note that if $P$ is the midpoint of $AF$, then $T,P,S$ are collinear, so $S \in \ell$. Also note that $SF\parallel AT$, so $S\in XF$. Redefine $N$ to be the intersection of $XE$ with $\ell$. We have $ALEN$ is a parallelogram. We show that $A,N,X,S$ lie on a circle. We have : $$ \angle NAX = \angle XDE = \angle XFE = \angle NSX$$So the claim follows. Finally we prove that $AX$ is tangent to both $\odot (TAN)$ and $\odot (LAS)$. Note that $\angle NAX = \angle ADL = \angle ATL = \angle ATN$. So $AX$ is tangent to $\odot (TAN)$. Similarly its tangent to $\odot (LAS)$. We're done. $\square$

Motivation

PS

Interesting diagram... Claim $: AL || XE$. Proof $:$ $\angle MAE = \angle MFE = \angle XFE = \angle AEX$. Let $FE$ meet $AL$ at $P$. Claim $: AFDP$ is cyclic. Proof $:$ $\angle APF = \angle XEF = \angle XDF = \angle ADF$. Claim $: AL = LP$. Proof $:$ Note that $\angle LAE = \angle AEX = \angle EDX = \angle LDA \implies LA^2 = LE.LD$ and $\angle PDA = \angle PFA \implies \angle PDL = \angle XFA = \angle XDF = \angle APF = \angle LPE \implies LP^2 = LE.LD$. so $LQ || EF$. Claim $: LMQE$ is cyclic. Proof $:$ $\angle MEQ = \angle EPL = \angle MLQ$. Claim $: T,F,D$ are collinear. Proof $:$ $\angle ADT = \angle ALT = \angle APF = \angle ADF$. Let $TL$ meet $MF$ at $S'$. Claim $: AMQS'$ is cyclic. Proof $:$ $\angle MAQ = \angle MAE = \angle MFE = \angle MS'Q$. Claim $: S'$ is $S$. Proof $:$ $\angle AS'M = \angle AQM = \angle ALE = \angle ALD = \angle ADT \implies AS' || FT$ and $\angle TAX = \angle TAD = \angle TLD = \angle LEP = \angle FED = \angle FXD \implies AT || XF$. Claim $: ASL$ is tangent to $AD$ at $A$. Proof $:$ Note that $\angle DAM = \angle DAP = \angle DFP = \angle DFE = \angle 180 - \angle AED = \angle 180 - \angle QML = \angle 180 - \angle QSA = \angle LSA$. Claim $: ANT$ is tangent to $AD$ at $A$. Proof $:$ Note that $\angle NTA = \angle NSX = \angle NAX$. so $ANT$ and $ASL$ are tangent to each other at $A$.

Wow, what a reach configuration it's! $\angle MAE=\angle MFE=\angle XEA \implies AM||XE$. Let $AL\cap EF=J$. $\angle EAJ=\angle XFE$ , $\angle AJE=\angle XEF$, and $\angle LEJ=\angle FED=\angle FXD$. Also $XD$ is $X-$symmedian in triangle $XEF$, so $EL$ is median in triangle $AEJ$. So $AL=LJ$, which implies $LQ||EF$. So $LQ$ bisects $AF$. Let $DF\cap LQ=T'$. $\angle ALT'=\angle AJE=\angle XEF=\angle ADT' \implies ALDT'$ is cyclic $\implies T'=T \implies T\in FD$. Since triangles $DEF$ and $DLT$ are homothetic, we get $(DLT)$ touches to $BC$ at $D$. Let $AB\cap LT=G$ and $FX\cap TL=S'$. $\angle ATL=\angle ADL=\angle XFE=\angle TS'F\implies AT||FS'$. Since $AG=GF$, we get $TFS'A$ is parallelogram. So $S'=S$. Let $EX\cap TL=N'$, Since $AS||DF$, we get $\angle SAD=\angle XDF=\angle XEF=\angle SN'X \implies ASXN'$ is cyclic $\implies N'=N$. $\angle NAX=\angle NSX=\angle XFE=\angle XDE=\angle ATN\implies (TAN)$ touches to $AD$ at $A$. $\angle XAS=\angle SNX=\angle XEF=\angle AJE=\angle ALS\implies (LSA)$ touches to $AD$ at $A$. So $(TAN)$ and $(LSA)$ touch each other at $A$. Done!

Nice geo, indeed. We start with a bit of angle chasing: $\angle MAE = \angle MFE =\angle XEA$, so we get that $XE \parallel AL$. In addition, $\angle LAE= \angle MFE = \angle ADL$, hence PoP gives that $LA^2=LE.LD$. Using the well-known trick with point circles and midlines, it is easy to see that $L$ lies on the radical axis of $A$ and $(DEF)$, so $LQ \parallel FE$ as a midline in $\triangle AFE$. Now, the homothety at $D$ taking $X$ to $A$, $E$ to $L$ and $(XED)$ to $(ALD)$ should take $F$ to $T$ since $EF \parallel LT$, so $XF \parallel AT$ and $D, T, F$ are collinear. Redefine $S=XF \cap LQ$. We proved that $AT \parallel SF$ and that $ST$ contains the midpoint of $AF$, which is enough to imply that $FTAS$ is a parallelogram. Now we are ready to finish: we will prove that both circles are tangent to $AD$ by angle chasing. Indeed, $\angle DAS = \angle ADT = \angle ALT$, so $AD$ is tangent to $(ASL)$. For the second tangency, note that $\angle NAX= \angle NSX = \angle ATN$, so we are done.

buh nice problem ig but boring Claim: $ALTD$ is harmonic. Proof. It suffices to show that the $L$ tangent to $(ALD)$ is parallel to $AC$. This is true as \[\measuredangle LDA = \measuredangle EDX = \measuredangle EFX = \measuredangle EFM = \measuredangle EAM = \measuredangle EAL\] Claim: $EX \parallel AL$ Proof. We show that $\triangle EFD \sim \triangle XAM$. To see why, $\angle AMX =\angle AEF = \angle EDF$ and $\angle MXA = \angle FXD = \angle FED$ as desired. Thus there is a homothety at $D$ sending $F,X,E$ to $T,A,L$. I claim that $S$ is the intersection of $XF$ and $TL$. Note that $S$ lies on $FX$ since $FX \parallel TA$, and similar work to the first claim gives $TL$ bisects $AF$. Thus we showed $S$ is $MF\cap TL$. To finish, I claim that $AD$ is tangent to the circles we want to show are tangent. This is just angle chasing: \[\measuredangle DAN = \measuredangle XAN = \measuredangle XSN = \measuredangle XFE = \measuredangle XDE = \measuredangle ADL = \measuredangle ATN\]and similarly \[\measuredangle DAS = \measuredangle XAS = \measuredangle XNS = \measuredangle XEF = \measuredangle XDF = \measuredangle ADT = \measuredangle ALS\]