A right-angled triangle is cut into a triangle and a quadrilateral by the perpendicular bisector of its hypotenuse. The ratio of the diagonals of the quadrilateral is

$$
(1+\sqrt{3}): 2 \sqrt{2}
$$

What are the acute angles of the right-angled triangle?
Solution. Let the midpoint of the hypotenuse $AB$ of triangle $ABC$ be $O$, and let the perpendicular bisector of the hypotenuse intersect the leg $AC$ at point $D$. The diagonals of the resulting quadrilateral $OBDC$ are $f$ and $g$ as shown in the figure. Since $D$ lies on the perpendicular bisector of segment $AB$, it is equidistant from points $A$ and $B$: $AD = BD = f$. By the converse of Thales' theorem, $OA = OB = OC = g$.

![](https://cdn.mathpix.com/cropped/2024_05_02_bfb5a4851d3df0340f90g-1.jpg?height=255&width=588&top_left_y=344&top_left_x=758)

Since $\frac{1+\sqrt{3}}{2 \sqrt{2}} < 1$, we have

$$
\cos \angle OAD = \frac{AO}{AD} = \frac{g}{f} = \frac{1+\sqrt{3}}{2 \sqrt{2}}.
$$

From this,

$$
\angle BAC = \angle OAD = 15^\circ \quad \text{and} \quad \angle ABC = 75^\circ.
$$

Remark. The obtained result is exact, since

$$
\cos 15^\circ = \cos (45^\circ - 30^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{1+\sqrt{3}}{2 \sqrt{2}}.
$$