4. Let quadrilateral $A B C D$ be inscribed. The rays $A B$ and $D C$ intersect at point K. It turns out that points $B, D$, and the midpoints $M$ and $N$ of segments $A C$ and $K C$ lie on the same circle. What values can the angle $A D C$ take?

![](https://cdn.mathpix.com/cropped/2024_05_06_cf77befabfd0e5d082b4g-1.jpg?height=388&width=457&top_left_y=1962&top_left_x=845)
Solution. Note that $MN$ is the midline in triangle $AKC$, so $\angle BAC = \angle NMC$. Moreover, $\angle BAC = \angle BDC$ since quadrilateral $ABCD$ is cyclic. Suppose points $M$ and $N$ lie on the same side of line $BD$. Then $M$ lies inside triangle $BCD$ and, therefore, inside triangle $BND$, and thus inside its circumcircle. However, this means that points $B, N, D$, and $M$ cannot lie on the same circle. Therefore, $N$ and $M$ lie on opposite sides of $BD$, and $\angle BDC = \angle BMN$.

From the parallelism of $MN$ and $AK$, it follows that $\angle BMN = \angle ABM$, hence $\angle BAC = \angle BDC = \angle ABM$. From this, we get $AM = MB$, meaning that in triangle $ABC$, the median $BM$ is equal to half the side $AC$. Therefore, $\angle ABC = 90^\circ$, and thus $\angle ADC = 90^\circ$. Answer: $90^\circ$.

Comment. If the solution is correct but lacks reasoning about the positions of points $N$ and $M$ relative to segment $BD$, deduct 2 points.