Since $EF \parallel ST,$ then $AM$ cuts $ST$ at its midpoint $N.$ Since $NK$ is the reflection of $NA$ on $ST,$ then it follows by symmetry that $KN$ cuts $\odot(AST)$ again at $D$ forming the isosceles trapezoid $ATSD.$ Thus again, by symmetry, if $AN$ cuts $\odot(AST)$ again at $K',$ then $KK' \parallel ST$ $\Longrightarrow$ $AK$ and $AM \equiv AK'$ are isogonals WRT $\angle EAF$ $\Longrightarrow$ $AK$ is the A-symmedian of $\triangle AEF$ $\Longrightarrow$ $KE,KF$ are tangents of $\odot(AEF)$ $\Longrightarrow$ $\angle KEF=\angle KFE=\angle A.$

Let us relabel points $S, T, M$ as $B, C, X$, respectively (we can ignore the points $B, C$ specified in the problem statement). Now, let $H$ be the orthocenter of $\triangle ABC$ and let $M$ be the midpoint of $\overline{BC}.$ $\Gamma_1, \Gamma_2$ denote $\odot (ABC), \odot (BHC)$, respectively and $U, V$ are reflections of $A$ in $BC, M$, respectively. Since $EF \parallel BC$, it follows that $A, X, M$ are collinear. From $\triangle AFX \sim \triangle ABM$ we obtain $\tfrac{XF}{MB} = \tfrac{AX}{AM}.$ From $\triangle MXK \sim \triangle MAU$ we find $\tfrac{XK}{AU} = \tfrac{MX}{AM}.$ It follows that $\tfrac{XF}{XK} = \tfrac{AX}{MX} \cdot \tfrac{MB}{AU}.$ Now, because the reflection $X$ in $BC$ lies on $\Gamma_1$, it follows that $X$ lies on the reflection of $\Gamma_1$ in $BC$, which is just $\Gamma_2$ (well-known). Meanwhile, since $\Gamma_1$ and $\Gamma_2$ are symmetric about $BC$ and $M$, it follows that $U, V$ lie on $\Gamma_2.$ By Power of a Point, we obtain $AX \cdot AV = AH \cdot AU$ and $MX \cdot MV = MB^2.$ Therefore, \[\frac{AX}{MX} \cdot \frac{MB}{AU} = \frac{AH}{MB} \cdot \frac{MV}{AV} = \frac{AH}{2MB} = \frac{OM}{MB},\] where $O$ denotes the center of $\Gamma_1$, and we have used the well-known fact that $AH = 2OM.$ Then by side-angle-side similarity, we deduce that $\triangle KFX \sim \triangle BOM$, and hence $\angle KFE = \angle BOM = \angle A.$ $\square$

Suppose $GH\parallel EF$ and passes through $K$ with $G,H$ on $AE,AF$ respectively. Let $GF$ intersect $HE$ at $X$. Then by parallelism \[ \angle FXE=\angle GXH=\angle TKS=180-\angle FAE \] so $FAEX$ is cyclic. By Brokard's Theorem, $K'$, the intersection of the tangents from $E,F$ to $(AEXF)$, lies on $GH$. But of course $K'$ lies on the perpendicular bisector of $EF$, so $K'=K$, and we're done.

Dear pi37: What's $X$ mean?

Sorry, I edited it in.

Consider the circumcircle of $\triangle AST$. Obviously since $ST||EF$, $AM$ is the median of $\triangle AST$. Now we have that a point on the arc $ST$ not containing $A$whose reflection in $ST$ lie on the $A-$median. Let $X,Y$ be two such points on the arc $ST$ whose reflections in $ST$ lie on $A-$median. Denote these reflections by $X'Y'$ Thus since $S,T,X,Y$ are concyclic, $S,T,X'Y'$ are also concyclic which is not possible as they lie on the same side of $ST$. Thus there exists only one such point on the arc $ST$. It is well known and easy to show that this point ($K$ in the question) lies on the symmedian. Now the question is easy.

Never seen an uglier problem.

Actually, this question can be done by Bary. Just ignore B and C, and the rest is straightforward caculate.

For starters let's reformulate the problem so that $ST$ become $BC$: nicer formulation wrote: In $\triangle ABC$ points $E,F$ lay on $AC,AB$ so that $EF \parallel BC$ and reflection ($M \mapsto K$) of midpoint of $EF$ over $BC$ lays on $\odot ABC$. Prove that $\angle{KFE}=\angle{A}$. Let $AM$ cut ($ABC$) at $L$. $AL$ goes through midpoint of chord $BC$ and reflected over it. Due to this symmetry $KL \parallel BC$, $\angle KAB = \angle LAC$ $\implies AK$ - $A$-symmedian of $\triangle AEF$. With $MK$ being perp. bisector of $EF$ it means that $KF$ is tangent to $(AEF)$ and result follows.

it's clear that if $K$ is moving on the perpindacular bisector of $EF$ then there exists a unique point $K_0$ outside $\triangle AEF$ such that $ATSK$ is cyclic I'll claim that this point is the intersection of tangents at $E,F$ to $(AEF)$ proof: let $\omega $ the circle with radius zero centered at $K$ since $ST || EF$ we have that $TS$ goes from the modpoints of $KF,KE$ so $TS$ is the radical axis of $\omega$ and $(AEF)$ so $$TM^2=TK^2=TF.TA$$then $TM$ is tangent to$(AFM)$ so $\angle EAK=\angle FAM = \angle TMF=\angle STM =\angle KTS$ and we win

\begin{align*} \textbf{Let's review Humpty points} \end{align*}Notice it's enough to prove $AK$ is $A$ symmedian in $\triangle AST$ . Notice $M$ in triangle $AST$. Call orthocenter of $AST$ as $H$. Reflection of $M$ wrt $ST$ is on $\odot (AST)$ so $M \in \odot(SHT)$ . It's well known that In $\triangle ABC$ with orthocenter $H$ . Intersection of $A-$median and $\odot (BHC)$ is $A-$Humpty point. So $M$ is $A$-humpty point in $\triangle AST$. It's well known that $A$-humpty point is the reflection of intersection of $A$-symmedian with circumcircle wrt $ST$. So $AK$ is $A-$symmedian in $\triangle AST$ $\blacksquare$