Solution 1. For any interesting pair $(E, F)$, we will say that the corresponding triangle $EFK$ is also interesting.
Let $EFK$ be an interesting triangle. Firstly, we prove that $\angle KEF = \angle KFE = \angle A$, which also means that the circumcircle $\omega_1$ of the triangle $AEF$ is tangent to the lines $KE$ and $KF$.
Denote by $\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $AM$ intersect the line $ST$ and the circle $\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).
Since $EF \parallel TS$ and $M$ is the midpoint of $EF$, $N$ is the midpoint of $ST$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $ST$, we have $\angle KNS = \angle MNS = \angle LNT$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $ST$. Therefore $KL \parallel ST$.
Let $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $EF$, and we may assume that it lies on the ray $MF$. One has
$$\angle KGE = \angle KNS = \angle SNM = \angle KLA = 180^\circ - \angle KSA$$
(if $K = L$, then the angle $KLA$ is understood to be the angle between $AL$ and the tangent to $\omega$ at $L$). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $KSGT$ is a parallelogram, we obtain $\angle KEF = \angle KSG = 180^\circ - \angle TKS = \angle A$. Since $KE = KF$, we also have $\angle KFE = \angle KEF = \angle A$.

First method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.
Lemma. Let $ABC$ be an arbitrary triangle. Choose two points $E_1$ and $E_2$ on the side $AC$, two points $F_1$ and $F_2$ on the side $AB$, and two points $K_1$ and $K_2$ on the side $BC$, in a way that the triangles $E_1F_1K_1$ and $E_2F_2K_2$ are similar. Then the six circumcircles of the triangles $AE_iF_i$, $BF_iK_i$, and $CE_iK_i$ $(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_1F_1K_1$ to the triangle $E_2F_2K_2$.
Proof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $AE_iF_i$, $BF_iK_i$, and $CK_iE_i$ have a common point $Z_i$ by Miquel's theorem. Moreover, we have
$\text{(Z iF i, Z iE i) = (AB, CA), (Z iK i, Z iF i) = (BC, AB), (Z iE i, Z iK i) = (CA, BC)}$.
This yields that the points $Z_1$ and $Z_2$ correspond to each other in similar triangles $E_1F_1K_1$ and $E_2F_2K_2$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.
Finally, in order to show that $Z_1 = Z_2$, one may notice that $\text{(AB, AZ 1) = (E 1F 1, E 1Z 1) = (E 2F 2, E 2Z 2) = (AB, AZ 2)}$ (see Figure 2). Similarly, one has $\text{(BC, BZ 1) = (BC, BZ 2)}$ and $\text{(CA, CZ 1) = (CA, CZ 2)}$. This yields $Z_1 = Z_2$.
Now, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $AC$ and $AB$, respectively, and let $R$ be the midpoint of $BC$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $BCPQ$. Thus we obtain $\angle APQ = \angle B$ and $\angle RPC = \angle C$, which yields $\angle QPR = \angle A$. Similarly, we show that $\angle PQR = \angle A$. Thus, all interesting triangles are similar to the triangle $PQR$.

Figure 3

Figure 4
Denote now by $Z$ the common point of the circumcircles of $APQ$, $BQR$, and $CPR$. Let $E_1F_1K_1$ and $E_2F_2K_2$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_1F_1K_1$, $E_2F_2K_2$, and $PQR$ to some other of them. Therefore the triangles $ZE_1E_2$ and $ZF_1F_2$ are similar, as well as the triangles $ZE_1F_1$ and $ZPQ$. Hence
$$\frac{E_1E_2}{F_1F_2} = \frac{ZE_1}{ZF_1} = \frac{ZP}{ZQ}$$
Moreover, the equalities $\angle AZQ = \angle APQ = \angle ABC = 180^\circ - \angle QZR$ show that the point $Z$ lies on the line $AR$ (see Figure 4). Therefore the triangles $AZP$ and $ACR$ are similar, as well as the triangles $AZQ$ and $ABR$. This yields
$$\frac{ZP}{ZQ} = \frac{ZP}{RC} \cdot \frac{RB}{ZQ} = \frac{AZ}{AC} \cdot \frac{AB}{AZ} = \frac{AB}{AC}$$
which completes the solution.

Second method. Now we will start from the fact that $\omega_1$ is tangent to the lines $KE$ and $KF$ (see Figure 5). We prove that if $(E, F)$ is an interesting pair, then
$$\begin{equation*} \frac{AE}{AB} + \frac{AF}{AC} = 2 \cos \angle A. \tag{1} \end{equation*}$$
Let $Y$ be the intersection point of the segments $BE$ and $CF$. The points $B, K$, and $C$ are collinear, hence applying Pascal's theorem to the degenerated hexagon $AFFYEE$, we infer that $Y$ lies on the circle $\omega_1$.
Denote by $Z$ the second intersection point of the circumcircle of the triangle $BFY$ with the line $BC$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain
$$BF \cdot AB + CE \cdot AC = BY \cdot BE + CY \cdot CF = BZ \cdot BC + CZ \cdot BC = BC^2.$$
On the other hand, $BC^2 = AB^2 + AC^2 - 2AB \cdot AC \cos \angle A$, by the cosine law. Hence
$$(AB - AF) \cdot AB + (AC - AE) \cdot AC = AB^2 + AC^2 - 2AB \cdot AC \cos \angle A,$$
which simplifies to the desired equality (1).
Let now $(E_1, F_1)$ and $(E_2, F_2)$ be two interesting pairs of points. Then we get
$$\frac{AE_1}{AB} + \frac{AF_1}{AC} = \frac{AE_2}{AB} + \frac{AF_2}{AC},$$
which gives the desired result.

Third method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.
Let $EFK$ be any interesting triangle. The equalities $\angle KEF = \angle KFE = \angle A$ yield that the ratio $\nu = \frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation
$$\begin{equation*} K = \mu E + \nu F, \quad \text{ where } \quad \mu = 1 - \nu. \tag{2} \end{equation*}$$
Now let us choose the points $X$ and $Y$ on the rays $AB$ and $AC$, respectively, so that $\angle CXA = \angle AYB = \angle A = \angle KEF$ (see Figure 7). Then each of the triangles $AXC$ and $YAB$ is similar to any interesting triangle, which also means that
$$\begin{equation*} C = \mu A + \nu X = \nu X \quad \text{ and } \quad B = \mu Y + \nu A = \mu Y. \tag{3} \end{equation*}$$
Moreover, one has $X / Y = \overline{C / B}$.
Since the points $E, F$, and $K$ lie on $AC, AB$, and $BC$, respectively, one gets
$$E = \rho Y, \quad F = \sigma X, \quad \text{ and } \quad K = \lambda B + (1 - \lambda) C$$
for some real $\rho, \sigma$, and $\lambda$. In view of (3), the equation (2) now reads $\lambda B + (1 - \lambda) C = K = \mu E + \nu F = \rho B + \sigma C$, or
$$(\lambda - \rho) B = (\sigma + \lambda - 1) C.$$
Since the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\rho = \lambda$ and $\sigma = 1 - \lambda$. Therefore,
$$\begin{equation*} \frac{E}{Y} + \frac{F}{X} = \rho + \sigma = 1. \tag{4} \end{equation*}$$
Now, if $(E_1, F_1)$ and $(E_2, F_2)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get
$$\frac{E_1 - E_2}{Y} = \frac{F_2 - F_1}{X}, \quad \text{ so } \quad \frac{E_1 - E_2}{F_2 - F_1} = \frac{Y}{X} = \frac{\bar{B}}{\bar{C}}.$$
Taking absolute values provides the required result.

Solution 2. Let $(E, F)$ be an interesting pair. This time we prove that
$$\begin{equation*} \frac{AM}{AK} = \cos \angle A. \tag{5} \end{equation*}$$
As in Solution 1, we introduce the circle $\omega$ passing through the points $K, S, A$, and $T$, together with the points $N$ and $L$ at which the line $AM$ intersect the line $ST$ and the circle $\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\omega$ (see Figures 8 and 9). As in Solution 1, we note that $N$ is the midpoint of $ST$ and show that $KL \parallel ST$, which implies $\angle FAM = \angle EAK$.

Figure 8

Figure 9
Suppose now that $K \neq L$ (see Figure 8). Then $KL \parallel ST$, and consequently the lines $KM$ and $KL$ are perpendicular. It implies that the lines $LO$ and $KM$ meet at a point $X$ lying on the circle $\omega$. Since the lines $ON$ and $XM$ are both perpendicular to the line $ST$, they are parallel to each other, and hence $\angle LON = \angle LXK = \angle MAK$. On the other hand, $\angle OLN = \angle MKA$, so we infer that triangles $NOL$ and $MAK$ are similar. This yields
$$\frac{AM}{AK} = \frac{ON}{OL} = \frac{ON}{OT} = \cos \angle TON = \cos \angle A.$$
If, on the other hand, $K = L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $ST$ (see Figure 9). This implies that $AK$ is a diameter of $\omega$, which yields $AM = 2OK - 2NK = 2ON$. So also in this case we obtain
$$\frac{AM}{AK} = \frac{2ON}{2OT} = \cos \angle TON = \cos \angle A.$$

Thus (5) is proved.
Let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $AC$ and $AB$, respectively (see Figure 10). We claim that the point $M$ lies on the line $PQ$. Consider now the composition of the dilatation with factor $\cos \angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\angle BAC$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $BC$, the point $M$ lies on the line $PQ$.

Figure 10