![](https://cdn.mathpix.com/cropped/2024_05_06_1b324abf723c207e3278g-11.jpg?height=88&width=1129&top_left_y=1827&top_left_x=17)

On the sides $AB$ and $AC$ of triangle $ABC$, points $K$ and $L$ are located such that $AK: KB = 4: 7$ and $AL: LC = 3: 2$. The line $KL$ intersects the extension of side $BC$ at point $M$. Find the ratio $CM: BC$.
Through point $A$, draw a line parallel to $B C$. Let $T$ be the point of its intersection with line $K L$. From the similarity of triangles $A L T$ and $C L M$, we find that $A T = \frac{3}{2} C M$, and from the similarity of triangles $A K T$ and $B K M - B M = \frac{7}{4} A T = \frac{7}{4} \cdot \frac{3}{2} C M = \frac{21}{8} C M$.

Thus, $B C = \frac{21}{8} C M - C M = \frac{13}{8} C M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_1b324abf723c207e3278g-11.jpg?height=323&width=414&top_left_y=2411&top_left_x=822)

## Answer

$8: 13$.

Problem