8. Variant 1.

A line parallel to the leg $A C$ of the right triangle $A B C$ intersects the leg $B C$ at point $K$, and the hypotenuse $A B$ at point $N$. On the leg $A C$, a point $M$ is chosen such that $M K=M N$. Find the ratio $\frac{A M}{M C}$, if $\frac{B K}{B C}=14$.
Answer: 7.

Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_0d3dbdd740b551c86dd8g-7.jpg?height=821&width=634&top_left_y=164&top_left_x=694)

Drop the altitude $M H$ from point $M$ in the isosceles triangle $M N K$. Then $M H K C$ is a rectangle and $M C=K H=H N$. Let $M C=K H=H N=y$. Let $K B=x$, then $C K=C B-K B=4 x-x=3 x$. From the similarity of triangles $N K B$ and $A C B$, it follows that $\frac{A C}{C B}=\frac{N K}{K B}$. Therefore, $\frac{A C}{4 x}=\frac{2 y}{x}$. Hence, $A C=8 y$. We get that $\frac{A M}{M C}=\frac{8 x-x}{x}=7$.