8. Let the isosceles triangle $A B C$ be such that $|A C|=|B C|=12 \mathrm{~cm}$. A line parallel to

(20) the leg passes through the midpoint of the height to the base of the triangle and intersects the base at point $M$ and the other leg at point $N$. What is the length of the segment $\overline{M N}$?
First solution. The length $\overline{M P}$ is the median of triangle $C_{1} B C$ (where $C_{1}$ is the foot of the altitude on the base). Therefore, $M$ is the midpoint of $\overline{C_{1} B}$.

(10 points)

![](https://cdn.mathpix.com/cropped/2024_05_30_fac652c8a845a57c6fa7g-11.jpg?height=512&width=394&top_left_y=555&top_left_x=840)

Triangles $A M N$ and $A B C$ are similar (equal angles). The similarity coefficient is $k=\frac{|A M|}{|A B|}=\frac{3}{4}$.

(5 points)

From this, it follows that $|M N|=\frac{3}{4}|B C|=9 \mathrm{~cm}$.

(5 points)