Hellenic Mathematical Olympiad

We have $DE\perp BC$ and $DF\perp AB$ (because ${\Gamma }_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$).

Figure 4

Figure 5

Hence the quadrilateral $BEDF$ is cyclic and let ${\Gamma }_3$ be its circumcircle. Moreover we have the equalities: $$BE=BF\text{ and }DE=DF.\text{\hspace{1em}(1)}$$

The $EF$ is the common chord of the circles ${\Gamma }_2$ and ${\Gamma }_3$ and $KG$ is the common chord of the circles ${\Gamma }_1$ and ${\Gamma }_2$. Since $EF$ and $KG$ meet at $M$, the common chord $BD$ of the circles ${\Gamma }_1$ and ${\Gamma }_3$ will pass through $M$ (radical center of the circles ${\Gamma }_1,{\Gamma }_2,{\Gamma }_3$). Since the orthogonal triangles $AFD$, $CED$ are equal ($FD=ED$, $\angle FAD=\angle ECD$), we have $DA=DC$. Hence $DO$ is the perpendicular bisector of $AC$, and since $DO\perp KG$, it follows that $AC\parallel KG$ and $\angle ACD=\angle MND$. Since $BD$ bisects the angle $\angle FBE$ and $BDCA$ is inscribed into the circle ${\Gamma }_1$ we have $\angle MBC=\angle FBM=\angle ACD=\angle MND$. Hence $BCNM$ is cyclic.

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Alternative solution.

We have $DE\perp BC$ and $DF\perp AB$ (because ${\Gamma }_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$). Hence the quadrilateral $BEDF$ is cyclic and let ${\Gamma }_3$ be its circumcircle. We consider the inversion with respect to the circle ${\Gamma }_2$ (with center $D$). Since circle ${\Gamma }_3$ is passing through $D$, its image is the line of common chord $FE$, and therefore: The image of $B$ belongs to the line $FE$. (5) Since circle ${\Gamma }_1$ is passing through $D$, its image is the line of common chord $KG$, and therefore: The image of $B$ belongs to the line $KG$. (6) From (5), (6) we conclude that the image of $B$ under the inversion we have considered is the point $M$. Also, $C$ has image the point $N$. Hence the quadrilateral $BMNC$ is cyclic.