Given a cyclic quadrilateral $A B C D$. The rays $A B$ and $D C$ intersect at point $K$. It turns out that points $B, D$, as well as 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?
$M N$ is the midline in triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$ because quadrilateral $A B C D$ is cyclic.

Let points $M$ and $N$ lie on the same side of line $B D$. Then $M$ lies inside triangle $B C D$ and, therefore, inside triangle $B N D$, and thus inside its circumcircle. But then points $B, N, D$, and $M$ cannot lie on the same circle. Therefore, $N$ and $M$ lie on opposite sides of $B D$, and $\angle B D C = \angle B M N$. From the parallelism of $M N$ and $A K$, it follows that $\angle B M N = \angle A B M$, from which $\angle B A C = \angle B D C = \angle A B M$. Hence, $A M = M B$, meaning that in triangle $A B C$, the median $B M$ is equal to half the side $A C$. Therefore, $\angle A B C = 90^{\circ}$, and thus $\angle A D C = 90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-34.jpg?height=494&width=564&top_left_y=-1&top_left_x=747)

## Answer

$90^{\circ}$.

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