5. The median to the hypotenuse of a right triangle cuts off a triangle whose area is three times smaller than the area of the original triangle. Determine the acute angles of the observed triangle.
First solution.

With the labels as in the figure, the condition of the problem is

$$
P(\triangle A D E)=\frac{1}{3} P(\triangle A B C)
$$

which means

$$
\frac{1}{2} \cdot \frac{|A B|}{2} \cdot|E D|=\frac{1}{3} \cdot \frac{1}{2} \cdot|A B| \cdot|C P|
$$

from which it follows that

$$
\frac{|C P|}{|E D|}=\frac{3}{2}
$$

Triangles $\triangle A D E$ and $\triangle A P C$ are similar because lines $E D$ and $C P$ are parallel. Therefore,

$$
\frac{|A P|}{|A D|}=\frac{|C P|}{|E D|}=\frac{3}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_30_20b7f5a96a45568d6d08g-6.jpg?height=414&width=526&top_left_y=1669&top_left_x=738)

The midpoint of the hypotenuse $D$ of the right triangle is the center of the circumscribed circle, so $|D C|=$ $|D B|=|D A|$, and due to the above ratio, $P$ is the midpoint of segment $D B$. $30^{\circ}$.