Task 2. (10 points) In triangle $ABC$ with sides $AB=9$, $AC=3$, $BC=8$, 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 $KN$.

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# 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-17.jpg?height=366&width=1287&top_left_y=1462&top_left_x=453)

Fig. 1.

By the property of the angle bisector $\frac{B K}{C K}=\frac{A B}{A C}=\frac{9}{3}=\frac{3}{1} \Rightarrow \frac{B K}{8-B K}=\frac{3}{1} \Rightarrow B K=6$

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

Therefore, $\frac{A L}{B C}=3, A L=3 B C=24$.

Consider similar triangles $L A N$ and $B K N: \frac{A N}{N K}=\frac{A L}{B K}=\frac{24}{6}=\frac{4}{1}$.

The length of the angle bisector is

$A K=\sqrt{A B \cdot A C-B K \cdot C K}=\sqrt{9 \cdot 3-6 \cdot 2}=\sqrt{15}$.

$K N=\frac{1}{5} A K=\frac{\sqrt{15}}{5}$

Answer. $\frac{\sqrt{15}}{5}$.