Let $N$ be the midpoint of $C D$. We will prove that the circumcircles of the triangles $M K T$ and $M L S$ pass through $N$. (1)
First will prove that the circumcircle of $M L S$ passes through $N$.
Let $Q$ be the midpoint of $E C$. Note that the circumcircle of $M L S$ is the Euler circle (2) of the triangle $E F C$, so it passes also through $Q .\left({ }^{*}\right)(3)$
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We will prove that

$$\angle S L Q=\angle Q N S \quad \text { or } \quad \angle S L Q+\angle Q N S=180^{\circ}$$

Indeed, since $F L C$ is right-angled and $L S$ is its median, we have that $S L=S C$ and

$$\angle S L C=\angle S C L=\angle A C B$$

In addition, since $N$ and $S$ are the midpoints of $D C$ and $F C$ we have that $S N \| F D$ and similarly, since $Q$ and $N$ are the midpoints of $E C$ and $C D$, so $Q N \| E D$.
It follows that the angles $\angle E D B$ and $\angle Q N S$ have parallel sides, and since $A B<C D$, they are acute, and as a result we have that

$$\angle E D B=\angle Q N S \quad \text { or } \quad \angle E D B+\angle Q N S=180^{\circ}$$

But, from the cyclic quadrilateral $A B C D$, we get that

$$\angle E D B=\angle A C B$$

Now, from (2),(3) and (4) we obtain immediately (1), so the quadrilateral $L N S Q$ is cyclic. Since from $\left(^{*}\right)$, its circumcircle passes also through $M$, we get that the points $M, L, Q, S, N$ are cocylic and this means that the circumcircle of $M L S$ passes through $N$.
Similarly, the circumcircle of $M K T$ passes also through $N$ and we have the desired.