Selection Examinations for the IMO

The segments $BF$ and $CE$ are parallel, so the quadrilateral $FBCE$ is an isosceles trapezoid and $|BC| = |EF|$. The inscribed angles over the chords $BC$ and $EF$ are equal, $\angle EDF = \angle BDC$. Let $\angle EDF = \alpha$. The segment $AD$ is the diameter of a circle, so by Thales' theorem we have $\angle AED = \frac{\pi}{2} = \angle DCA$. From here we get $\angle ELD = \frac{\pi}{2} - \angle EDL = \frac{\pi}{2} - \alpha$, so $\angle ELF = \frac{\pi}{2} + \alpha$. Similarly, $\angle DKC = \frac{\pi}{2} - \angle KDC = \frac{\pi}{2} - \alpha$, so $\angle CKB = \frac{\pi}{2} + \alpha$.

We have shown that $\angle ELF = \angle CKB$ and $|BC| = |EF|$. The angles $\angle ELF$ and $\angle CKB$ over the two chords $EF$ and $BC$ of equal length are equal. If we somehow map the triangle $ELF$ so that $E$ gets mapped to $C$, $F$ gets mapped to $B$ and the image of $L$ lies on the same side of the line $BC$ as $K$, then we will get a cyclic quadrilateral. The mapping with these properties is the reflection over the line of symmetry of the trapezoid $FBCE$. Denote this line by $\ell$.

Let $K'$ be the image of the point $K$ and let $L'$ be the image of the point $L$ under the reflection over $\ell$. Obviously, $\ell$ is the line of symmetry of the quadrilateral $LL'KK'$. This quadrilateral is an isosceles trapezoid (because $LL'$ and $KK'$ are parallel). Its diagonals $KL$ and $K'L'$ intersect on $\ell$. We wish to show that the intersection of the diagonals is also the midpoint of each of them, which means we wish to show that $LL'KK'$ is a rectangle. It suffices to show that the segment $L'K$ is parallel to $\ell$.

Let $A'$ be the image of $A$ and let $D'$ be the image of $D$ under the reflection over $\ell$. The point $L'$ is the intersection of the lines $BD'$ and $A'C$. Since $A'BCD$ is a cyclic quadrilateral, we have $\angle A'DB = \angle A'CB$. The points $K, L', B$ and $C$ are concyclic, so $\angle L'KB = \angle L'CB = \angle A'CB$. Hence, $\angle A'DB = \angle L'KB$ and the line $KL'$ is parallel to the line $A'D$. The quadrilateral $AA'DD'$ is a rectangle (the reflection makes it so that the segment $AA'$ is parallel to the segment $DD'$, but $AD$ and $A'D'$ are the diameters, so $\angle AD'D = \pi/2$ and $\angle A'DD' = \pi/2$). So, the segment $A'D$ is parallel to $\ell$, which implies that $KL' \parallel \ell$. So, $LL'KK'$ is a rectangle and its diagonals bisect each other. Their intersection lies on $\ell$. Since $M$ lies on $\ell$ the distances of $M$ to the points $C$ and $E$ are the same.