10.254. In triangle $ABC$, medians $AL$ and $BM$ are drawn, intersecting at point $K$. Vertex $C$ lies on the circle passing through points $K, L, M$. The length of side $AB$ is $a$. Find the length of median $CN$.
Solution.

Since $CN$ is the median, then $CK=\frac{2}{3} CN$ (Fig. 10.59). By connecting points $L$ and $M$, we obtain the midline $LM$; therefore, $\frac{LM}{AB}=\frac{CF}{CN}=\frac{1}{2}$, i.e., $CF=\frac{1}{2} CN, LM=\frac{a}{2}, LF=FM=\frac{1}{2} AN=\frac{a}{4}, FK=\frac{1}{6} CN$. We have

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0751.jpg?height=512&width=502&top_left_y=79&top_left_x=193)

Fig. 10.60

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0751.jpg?height=514&width=556&top_left_y=78&top_left_x=703)

Fig. 10.61

$LF \cdot FM = FK \cdot CF$ (the product of the segments of chords passing through point $F$).

Therefore, $\frac{a^2}{16} = \frac{1}{6} CN \cdot \frac{1}{2} CN$, from which $CN = \frac{a \sqrt{3}}{2}$.

Answer: $\frac{a \sqrt{3}}{2}$.