5. Let $S$ be the midpoint of the hypotenuse $AB$ of a right-angled triangle $ABC$, which is not isosceles. Let $D$ be the foot of the altitude $z$ from vertex $C$ and $R$ the intersection of the angle bisector at vertex $C$ with the hypotenuse $AB$. Determine the sizes of the interior angles of this triangle, given that $|SR|=2|DR|$.
SOLUTION. For the given right triangle $ABC$, we will assume that of its legs $AC$ and $BC$, the first one is longer, and thus the angle $BAC$ (marked with two arcs in Fig. 2) is less than $45^{\circ}$. In the opposite case, it is sufficient to interchange vertices $A$ and $B$ throughout the solution, including the final answer.

![](https://cdn.mathpix.com/cropped/2024_04_17_7921a9481992c0fe6541g-5.jpg?height=434&width=1048&top_left_y=1779&top_left_x=504)

Fig. 2

Since triangle $ASC$ is isosceles (point $S$ is the center of the Thales circle circumscribed around the right triangle $ABC$), we have $|\angle ACS| = |\angle BAC|$. Right triangles $ABC$ and $CBD$ are similar because they share an internal angle at vertex $B$, which implies the congruence of angles $BAC$ and $BCD$. Therefore, angles $ACS$ and $BCD$ are congruent and less than $45^{\circ}$, so they complement the right angle $ACB$ with a non-zero angle $SCD$, whose bisector is also the bisector of the entire angle $ACB$, which is the ray $CR$. This also implies the congruence of angles $SCR$ and $DCR$ (and that point $R$ lies between points $S$ and $D$).

Let $P$ be the midpoint of segment $SR$ and $Q$ the foot of the perpendicular from point $R$ to line $SC$. Right triangles $CQR$ and $CDR$ with right angles at vertices $Q$ and $D$ are congruent in the size of the internal angle at vertex $C$ and in the length of the (common) hypotenuse $CR$, so they are congruent. According to the problem's condition, $|QR| = |DR| = \frac{1}{2}|SR| = |PR|$. This means that triangle $PRQ$ is equilateral, so $|\angle PRQ| = 60^{\circ}$, $|\angle RSQ| = 30^{\circ}$, and $|\angle SCD| = 60^{\circ}$. Since the angle at vertex $C$ is a right angle, it follows that $|\angle BAC| = |\angle ACS| = 15^{\circ}$ and $|\angle ABC| = 75^{\circ}$.