5.17. In triangle $A B C$, point $N$ lies on side $A B$ and $A N=$ $=3 N B$; median $A M$ intersects $C N$ at point $O$. Find $A B$, if $A M=C N=7$ cm and $\angle N O M=60^{\circ}$.
5.17. By the condition, $A M$ is the median of $\triangle A B C$ (Fig. 5.10) and, therefore,

$$
\overline{A M}=\frac{1}{2}(\overline{A B}+\overline{A C})
$$

Further,

$$
\overline{A C}=\overline{A N}+\overline{N C}=\frac{3}{4} \overline{A B}-\overline{C N}
$$

Substituting (2) into (1), we get $\overline{A M}=\frac{7}{8} \overline{A B}-\frac{1}{2} \overline{C N}$, from which

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-605.jpg?height=506&width=355&top_left_y=893&top_left_x=61)

Fig. 5.10

$$
\begin{aligned}
& \overline{A B}=\frac{8}{7} \overline{A M}+\frac{4}{7} \overline{C N} . \text { Therefore, } \\
& A B^{2}=\frac{64}{49} A M^{2}+\frac{64}{49} \overline{A M} \cdot \overline{C N}+\frac{16}{49} C N^{2}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{64}{49} \cdot 49+\frac{64}{49} \cdot 7 \cdot 7 \cos 60^{\circ}+\frac{16}{49} \cdot 49=112 \\
& \text { i.e. }|\overrightarrow{A B}|=\sqrt{112}=4 \sqrt{7}(\mathrm{cm}) .
\end{aligned}
$$

Answer: $4 \sqrt{7}$ cm.