2.2. $A L, B M, C N$ - medians of triangle $A B C$, intersecting at point $K$. It is known that quadrilateral $C L K M$ is cyclic, and $A B=2$. Find the length of median $C N$.
2.2. Answer: $\sqrt{3}$.

Since $L M$ is the midline of triangle $A B C$, then $L M \| A B$. (see fig. 10).

Thus, $\angle L M B = \angle A B M$. From the cyclic quadrilateral $C L K M$, we get that $\angle L M B = \angle L C N$. Therefore, $\angle L C N = \angle A B M$, so triangles $N B K$ and $N C B$ are similar (they share $\angle B N C$). From the similarity of these triangles, we can conclude that $\frac{C N}{N B} = \frac{N B}{N K}$. Notice now that $N B = \frac{A B}{2} = 1$, and $N K = \frac{C N}{3}$, since the centroid divides the median $C N$ in the ratio $2:1$. Thus, $\frac{C N^2}{3} = C N \cdot N K = N B^2 = 1$, so $C N = \sqrt{3}$.