(Cristian Mangra)

Lemma: Let $ABCD$ be an isosceles trapezoid ($AB \parallel CD$) inscribed in a circle $C$ of center $O$, $M$ is the intersection point of the diagonals $AC$ and $BD$, and $T$ is the intersection point of lines $AD$ and $BC$. If the line through $M$ parallel to $AB$ meets the circle $C$ at $E$ and $F$, then $TE$ and $TF$ are tangent to the circle.

Proof of the Lemma: Assume $AB < CD$. It is clear that points $T$, $M$ and $O$ are collinear. Also, $\angle AOT = \frac{1}{2} \angle AOB = \angle ACT$, which means that the quadrilateral $TAOC$ is cyclic. From the power of point $M$ with respect to the circumcircle of triangle $AOC$ we get $MA \cdot MC = MO \cdot MT$. But $MA \cdot MC = ME \cdot MF$ from the power of $M$ with respect to $C$, so $MO \cdot MT = ME \cdot MF$. This means that $TEOF$ is cyclic, so $\angle TEO + \angle TFO = 180^\circ$. But $\angle TEO = \angle TFO$, hence $\angle TEO = 90^\circ$, i.e., $TE$ is tangent to $C$. $\square$

Let us now return to the problem. Denote by $E$ and $F$ the second intersection points with the circle $C$ of lines $BE$, and $DE$, respectively. Because $M$ is the midpoint of the diagonal $AC$, and angles $AMB$ and $AMD$ are equal, it follows that $BDEF$ is an isosceles trapezoid (it is symmetric with respect to the perpendicular bisector of $AC$). Moreover, $BF \parallel DE \parallel AC$. If lines $BD$ and $EF$ meet at $T$, from the lemma it follows that $TA$ and $TC$ are tangent to the circumcircle of $ABCD$. If $O$ is the center of this circle, it follows that $OA \perp TA$ and $OC \perp TC$. As $N$ is the

midpoint of $BD$, we have $ON \perp BD$. Thus, points $A, N$, and $C$ all lie on the circle of diameter $OT$. Assume $\angle B > \angle D$ (if $\angle B < \angle D$, simply swap $B$ and $D$; the case $\angle B = \angle D$ is easy). In this case, $\angle ANB = \angle AOT$, and $\angle TNC = \angle TOC$. But $\angle AOT = \angle COT$ means $\angle ANB = \angle CNT = \angle CNB$.

Alternative Solution. (Given in the contest by David-Andrei Anghel.)

We only treat the case in the figure below; the other cases are similar.

Let $O$ be the circumcenter of triangle $ABC$. $O$ is on the perpendicular bisector of $BD$ but also on the external bisector of angle $BMD$ (it is perpendicular to $MA$, the internal bisector of that angle). It is known that these two lines meet on the circumcircle of triangle $BMD$ (in the midpoint of the arc $BMD$), therefore $BDOM$ is cyclic.

We obtain that angles $BMD$ and $BOD$ are equal. It follows that
$$\angle BCD = \frac{1}{2} \angle BOD = \frac{1}{2} \angle BMD = \angle BMA.$$
As $\angle BAM = \angle BDC$, triangles $BAM$ and $BDC$ are similar. It follows that $\frac{AM}{AB} = \frac{CD}{BD}$, which leads to $AC \cdot BD = 2AB \cdot CD$. But according to Ptolemy's Theorem,
$$AC \cdot BD = AB \cdot CD + BC \cdot AD,$$
therefore $AB \cdot CD = BC \cdot AD$, i.e., the quadrilateral $ABCD$ is harmonic.

This leads to the similarity of triangles $ANB$ and $ADC$. Also, triangles $CNB$ and $CDA$ are similar. The two similarities imply
$$\angle ANB = \angle ADC = \angle BNC.$$