Task 2. (10 points) In triangle $ABC$ with sides $AB=8, AC=4, BC=6$, the bisector $AK$ is drawn, and a point $M$ is marked on side $AC$ such that $AM: CM=3: 1$. Point $N$ is the intersection of $AK$ and $BM$. Find $AN$.
# Solution.

Draw a line through point $A$ parallel to line $B C$ and intersecting line $B M$ at point $L$ (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_06_b2236406d6db6e8875c1g-12.jpg?height=380&width=1308&top_left_y=1495&top_left_x=451)

Fig. 1.

By the property of the angle bisector $\frac{B K}{C K}=\frac{A B}{A C}=\frac{8}{4}=\frac{2}{1} \Rightarrow \frac{B K}{6-B K}=\frac{2}{1} \Rightarrow B K=4$

Consider similar triangles $A M L$ and $C M B: \frac{A L}{B C}=\frac{A M}{M C}$.

Therefore, $\frac{A L}{B C}=3, A L=3 B C=18$.

Consider similar triangles $L A N$ and $B K N: \frac{A N}{N K}=\frac{A L}{B K}=\frac{18}{4}=\frac{9}{2}$.

The length of the angle bisector is

$$
A K=\sqrt{A B \cdot A C-B K \cdot C K}=\sqrt{8 \cdot 4-4 \cdot 2}=2 \sqrt{6}
$$

$A N=\frac{9}{11} A K=\frac{18 \sqrt{6}}{11}$