An isosceles triangle $ABC$ with base $AB$ of length $10 \, \text{cm}$ and legs of length $20 \, \text{cm}$ is given. Point $S$ is the midpoint of the base $AB$. Divide the triangle $ABC$ into five parts of equal area using four lines passing through point $S$. Determine the lengths of the segments cut by these lines on the legs of the triangle $ABC$.

(E. Trojáková)

Idea. The given construction is axially symmetric.
Triangle $ABC$ is symmetric with respect to the axis $CS$, so the angle bisectors must also be symmetric with respect to the same axis. Corresponding parts will then form two pairs of symmetric triangles and one (symmetric) quadrilateral with vertex $C$. Let the intersections of the two angle bisectors with one of the legs be $X$ and $Y$, see the figure.

![](https://cdn.mathpix.com/cropped/2024_04_17_39f221c1ba6e0b54eafbg-2.jpg?height=845&width=462&top_left_y=1442&top_left_x=797)

According to the problem, the areas of triangles $ASX$, $XSY$, and twice the area of triangle $YSC$ are equal. These three triangles, however, have the same height from the common vertex $S$, so the areas will be in the given ratio precisely when the opposite sides satisfy

$$
|AX|=|XY|=2|YC| \text{.}
$$

At the same time, we know that

$$
|AC|=|AX|+|XY|+|YC|=20 \text{ cm}.
$$

From the above, it follows that $5|YC|=20 \text{ cm}$, i.e., $|YC|=4 \text{ cm}$ and $|AX|=|XY|=8 \text{ cm}$. The angle bisectors cut segments of lengths 4 and $8 \text{ cm}$ on the legs of the triangle.