In a right-angled triangle, the segment of the perpendicular bisector of the hypotenuse that falls between the lines of the legs is as long as the hypotenuse. What are the angles of the triangle?
Solution. Let the right triangle be $ABC$, where $\angle CAB = 90^\circ$. Let the intersection points of the perpendicular bisector with segment $CA$ and line $AB$ be $D$ and $E$, respectively, as shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_02_6f3bf1a4f5a229e8503cg-1.jpg?height=418&width=336&top_left_y=230&top_left_x=884)

Clearly, $\angle ACB = \angle AED$, since they are right angles, and similarly, $\angle ADE = \angle ABC$. Therefore, triangles $ADE$ and $ABC$ are similar. Since $ED = BC$, the two triangles are congruent. Hence, $AE = AC$, making $\triangle EAC$ an isosceles right triangle, with $\angle AEC = \angle ACE = 45^\circ$. The isosceles triangle $BEC$ has its $\angle BEC = \angle AEC$ angle bisected by the perpendicular bisector of $BC$, so

$$
\angle AED = \frac{45^\circ}{2} = 22.5^\circ, \quad \text{thus} \quad \angle ACB = \angle AED = 22.5^\circ.
$$

Therefore, the acute angles of the right triangle $ABC$ are 22.5 and 67.5 degrees.