National Math Olympiad in Slovenia

First, we will show that the points $E$, $D$, $M$, $F$ and $N$ are concyclic.

Let the circumcircle of the triangle $EDF$ intersect the line $AB$ at $M_1$. The quadrilaterals $BCED$ and $AFDC$ are cyclic, so $\angle M_1EF = \angle M_1DF = \angle ACB = \angle EDA = \angle EFM_1$. The triangle $EM_1F$ is isosceles with the apex at $M_1$, so $M_1$ lies on the bisector of the segment $EF$. This implies that $M_1 = M$ and the points $E$, $D$, $M$ and $F$ are concyclic.

The bisector of the chord $EF$ is the diameter of the circle passing through the points $E$, $D$, $M$ and $F$. We have $\angle MDN = \pi/2$ and $N$ lies on the bisector, so by Thales' theorem $N$ must also lie on the circle. This means that $E$, $D$, $M$, $F$ and $N$ are concyclic.

Now, $\angle ENF = \pi - \angle FME = \pi - \angle FDE = \pi - \angle FDA - \angle BDE + \pi = 2 \angle ECF$. Since $N$ lies on the bisector of the segment $EF$ and $\angle ENF = 2 \angle ECF$, we see that $N$ is the centre of the circumcircle of the triangle $EFC$.

The quadrilaterals $EDMF$ and $UDMC$ are cyclic and by the power of a point theorem we have $\overrightarrow{TE} \cdot \overrightarrow{TF} = \overrightarrow{TD} \cdot \overrightarrow{TM} = \overrightarrow{TU} \cdot \overrightarrow{TC}$. We conclude that $UEFC$ is also a cyclic quadrilateral. Thus, the points $C$ and $U$ both lie on the circle centred at $N$ and $|NC| = |NU|$.