Taiwan IMO Selection Camp

Let $\omega_1$ be the circumcircle of the quadrilateral $KSAT$. Let $N$ be the intersection of line $AM$ with line $ST$, and let $L$ be the second intersection point of $AM$ with $\omega_1$, as shown in the figure below.

Since $EF \parallel TS$ and $M$ is the midpoint of $EF$, $N$ is also the midpoint of $ST$. Furthermore, since $K$ and $M$ are symmetric with respect to line $ST$, we have $\angle KNS = \angle MNS = \angle LNT$. Therefore $K$ and $L$ are symmetric with respect to the perpendicular bisector of $ST$, so $KL \parallel ST$.
Let $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on line $EF$. Without loss of generality, assume $G$ lies on ray $MF$. We obtain
$$\angle KGE = \angle KNS = \angle SNM = \angle KLA = 180^\circ - \angle KSA$$
(when $K=L$, $\angle KLA$ refers to the angle between $AL$ and the tangent to $\omega$ at $L$). Hence $K,G,E,S$ form a cyclic quadrilateral. Because $KSGT$ is a parallelogram, we know $\angle KEF = \angle KSG = 180^\circ - \angle TKS = \angle A$. Also, since $KE = KF$, by symmetry we get $\angle KFE = \angle KEF = \angle A$.