66th Belarusian Mathematical Olympiad

Since $R$ is the midpoint of the arc $\angle AB$, the diameter of the circumcircle passing through $R$ is perpendicular to the chord $AB$. So, $RD$ is a perpendicular bisector of the segment $AB$. Therefore, $AD=BD$ and $\angle ABD=\angle BAD$.

By condition, $KB=KN$, then $\angle KBN=\angle KNB$. Hence, $\angle KAD=\angle BAD=\angle ABD=\angle KBN=\angle KNB=\angle KND$.

The points $A$, $K$, $D$, and $N$ are concyclic since the angles $KAD$ and $KND$ subtend the same segment $KD$ and lie in the same half-plane with respect to the line $KD$.