Given a triangle $A B C$. On the extension of side $A C$ beyond point $C$, a point $N$ is taken such that $C N=2 / 3 A C$. Point $K$ 

is on side $A B$, and $A K: K B=3: 2$.

In what ratio does the line $K N$ divide side $B C$?
The first method. Through point $B$, draw a line parallel to $A C$. Let line $K N$ intersect it at point $P$, and line $B C$ at point $M$.

$C N=2 / 3 A C, A N=5 / 3 A C$. From the similarity of triangles $P K B$ and $N K A$ (coefficient $2 / 3$), we find that $P B=2 / 3 A N = 10 / 9 A C$, and from the similarity of triangles $P B M$ and $N C M$, we have $B M: M C = P B: C N = 5: 3$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-25.jpg?height=315&width=646&top_left_y=1486&top_left_x=709)

The second method. Through point $K$, draw a line parallel to $A C$, until it intersects $B C$ at point $L$.

From the similarity of triangles $K B L$ and $A B C$, we find that $K L=2 / 5 A C$ and $B L=2 / 5 B C$, and from the similarity of triangles $K L M$ and $N C M$, we have $L M: M C = N C: K L = 3: 5$. Therefore, $M C = 5 / 8 \cdot 3 / 5 B C = 3 / 8 B C$.

## Answer

$5: 3$, measured from point $B$.