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 $K^{\prime }$ be the image of the point $K$ and let $L^{\prime }$ be the image of the point $L$ under the reflection over $\ell$. Obviously, $\ell$ is the line of symmetry of the quadrilateral $L{L}^{\prime }K{K}^{\prime }$. This quadrilateral is an isosceles trapezoid (because $L{L}^{\prime }$ and $K{K}^{\prime }$ are parallel). Its diagonals $KL$ and $K^{\prime }{L}^{\prime }$ 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 $L{L}^{\prime }K{K}^{\prime }$ is a rectangle. It suffices to show that the segment $L^{\prime }K$ is parallel to $\ell$.

Let $A^{\prime }$ be the image of $A$ and let $D^{\prime }$ be the image of $D$ under the reflection over $\ell$. The point $L^{\prime }$ is the intersection of the lines $B{D}^{\prime }$ and $A^{\prime }C$. Since $A^{\prime }BCD$ is a cyclic quadrilateral, we have $\angle A^{\prime }DB=\angle A^{\prime }CB$. The points $K,L^{\prime },B$ and $C$ are concyclic, so $\angle L^{\prime }KB=\angle L^{\prime }CB=\angle A^{\prime }CB$. Hence, $\angle A^{\prime }DB=\angle L^{\prime }KB$ and the line $K{L}^{\prime }$ is parallel to the line $A^{\prime }D$. The quadrilateral $A{A}^{\prime }D{D}^{\prime }$ is a rectangle (the reflection makes it so that the segment $A{A}^{\prime }$ is parallel to the segment $D{D}^{\prime }$, but $AD$ and $A^{\prime }{D}^{\prime }$ are the diameters, so $\angle A{D}^{\prime }D=\pi /2$ and $\angle A^{\prime }D{D}^{\prime }=\pi /2$). So, the segment $A^{\prime }D$ is parallel to $\ell$, which implies that $K{L}^{\prime }\parallel \ell$. So, $L{L}^{\prime }K{K}^{\prime }$ 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.