For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting. Let $E F K$ be an interesting triangle. Firstly, we prove that $\angle K E F=\angle K F E=\angle A$, which also means that the circumcircle $\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $K F$. Denote by $\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $A M$ intersect the line $S T$ and the circle $\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1). Since $E F \| T S$ and $M$ is the midpoint of $E F, N$ is the midpoint of $S T$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $S T$, we have $\angle K N S=\angle M N S=$ $\angle L N T$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $S T$. Therefore $K L \| S T$. Let $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $E F$, and we may assume that it lies on the ray $M F$. One has
$$\angle K G E=\angle K N S=\angle S N M=\angle K L A=180^{\circ}-\angle K S A$$
(if $K=L$, then the angle $K L A$ is understood to be the angle between $A L$ and the tangent to $\omega$ at $L$ ). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $K S G T$ is a parallelogram, we obtain $\angle K E F=\angle K S G=180^{\circ}-\angle T K S=\angle A$. Since $K E=K F$, we also have $\angle K F E=\angle K E F=\angle A$. After having proved this fact, one may finish the solution by different methods. ! Figure 1 ! Figure 2

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 $A B C$ be an arbitrary triangle. Choose two points $E_{1}$ and $E_{2}$ on the side $A C$, two points $F_{1}$ and $F_{2}$ on the side $A B$, and two points $K_{1}$ and $K_{2}$ on the side $B C$, in a way that the triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ are similar. Then the six circumcircles of the triangles $A E_{i} F_{i}$, $B F_{i} K_{i}$, and $C E_{i} K_{i}(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_{1} F_{1} K_{1}$ to the triangle $E_{2} F_{2} K_{2}$. Proof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $A E_{i} F_{i}, B F_{i} K_{i}$, and $C K_{i} E_{i}$ have a common point $Z_{i}$ by Miquel's theorem. Moreover, we have $\Varangle\left(Z_{i} F_{i}, Z_{i} E_{i}\right)=\Varangle(A B, C A), \quad \Varangle\left(Z_{i} K_{i}, Z_{i} F_{i}\right)=\Varangle(B C, A B), \quad \Varangle\left(Z_{i} E_{i}, Z_{i} K_{i}\right)=\Varangle(C A, B C)$. This yields that the points $Z_{1}$ and $Z_{2}$ correspond to each other in similar triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$. 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 $\Varangle\left(A B, A Z_{1}\right)=\Varangle\left(E_{1} F_{1}, E_{1} Z_{1}\right)=$ $\Varangle\left(E_{2} F_{2}, E_{2} Z_{2}\right)=\Varangle\left(A B, A Z_{2}\right)$ (see Figure 2). Similarly, one has $\Varangle\left(B C, B Z_{1}\right)=\Varangle\left(B C, B Z_{2}\right)$ and $\Varangle\left(C A, C Z_{1}\right)=\Varangle\left(C A, C Z_{2}\right)$. This yields $Z_{1}=Z_{2}$.

Now, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively, and let $R$ be the midpoint of $B C$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $B C P Q$. Thus we obtain $\angle A P Q=\angle B$ and $\angle R P C=\angle C$, which yields $\angle Q P R=\angle A$. Similarly, we show that $\angle P Q R=\angle A$. Thus, all interesting triangles are similar to the triangle $P Q R$. ! Figure 3 ! Figure 4

Denote now by $Z$ the common point of the circumcircles of $A P Q, B Q R$, and $C P R$. Let $E_{