9.7. 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 of segments $A C$ and $K C$, lie on the same circle. What values can the angle $A D C$ take?

(G. Zhukov $)$
Answer: $90^{\circ}$.

Solution. Let $N$ and $M$ be the midpoints of segments $K C$ and $A C$, respectively. Then $M N$ is the midline of triangle $A K C$, so $\angle B A C = \angle N M C$. Additionally, $\angle B A C = \angle B D C$, since quadrilateral $A B C D$ is cyclic.

Suppose points $M$ and $N$ lie on the same side of line $B D$. Then $M$ lies inside triangle $B C D$; hence, it lies inside triangle $B N D$, and therefore inside its circumcircle. But then points $B, N, D$, and $M$ cannot lie on the same circle. Therefore, $N$ and $M$ must 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 $\angle B M N = \angle A B M$, hence $\angle B A C = \angle B D C = \angle A B M$. This implies $A M = M B$, meaning that in triangle $A B C$, the median $B M$ is equal to half the side $A C$, from which $\angle A B C = 90^{\circ}$, and therefore $\angle A D C = 90^{\circ}$.

Comment. Points are not deducted for the lack of justification for the placement of points $M$ and $N$.

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