5. In a right-angled triangle, the height drawn to the hypotenuse divides the hypotenuse into two segments whose difference is equal to the length of one of the legs. Determine the angles of this triangle.
Solution. Let $ABC$ be the given right triangle with a right angle at vertex $C$ and, for example, $CA < BC$. Further, let $D$ be the foot of the altitude drawn from vertex $C$ and $E$ be a point on segment $DB$ such that

![](https://cdn.mathpix.com/cropped/2024_06_05_3c43981f47282d5136d0g-2.jpg?height=339&width=544&top_left_y=823&top_left_x=920)
$DE = AD$. Then triangle $ACE$ is isosceles. From the condition of the problem, it follows that $BE = AC = CE$, which means that triangle $BCE$ is also isosceles, with $\alpha = \measuredangle AEC$ being its exterior angle, so $\alpha = 2\beta$. However, $\alpha + \beta = 90^\circ$, so $\beta = 30^\circ$ and $\alpha = 60^\circ$.

## Eighth Division