Balkan 2012 shortlist

Since the quadrilateral $ODCE$ is cyclic, we have $\angle O_1=\angle C$. (The angles are as shown in the figure.) Since the quadrilateral $ODBZ$ is cyclic, we have $\angle O_2=\angle B$. Adding these we obtain $\angle EOUZ=\angle B+\angle C=180^{\circ }-\angle A$. Therefore the points $A$, $Z$, $O$, $E$ are concyclic and $c_3$ passes through $O$. We also have $\angle Z_2=\angle D_2=\angle E_1$. Since these three angles are subtended by the chords $AO$, $BO$, $CO$ of equal length in the circles $c_3$, $c_1$, $c_2$, respectively, the radii of these circles are equal. Therefore the angles $\angle Z_1$ and $\angle D_1$ are equal as they are subtended by chords $OK$ and $OM$ of the same length. It follows that the points $K$, $D$, $M$ are collinear. Similarly, $M$, $E$, $N$ are collinear and $Z$, $Z$, $K$ are collinear. Then $\angle BDK=\angle CDM$ and $\angle CEM=\angle AEN$, and $BK=CM=AN$. Therefore the triangle $NKM$ is obtained by rotating the triangle $ABC$ by a rotation with center $O$. Hence $ABC$ and $NKM$ are congruent.