In the right-angled triangle $A B C$, the hypotenuse $A B$ is one unit, the angle at $A$ is $30^{\circ}$, and the centroid is $S$. Into what parts does the angle bisector of $\angle B S C$ divide the leg $B C$?
In the right-angled triangle $ABC$, it is exactly half of an equilateral triangle with side length $AB=1$, so $BC=\frac{1}{2}$, $AC=\frac{\sqrt{3}}{2}$. The length of the median $CE$ is $\frac{AB}{2}=\frac{1}{2}$, from which $SC=\frac{2}{3} \cdot \frac{1}{2}=\frac{1}{3}$. The length of the median $BD$ can be calculated using the Pythagorean theorem in the right-angled triangle $BCD$.

$$
BD=\sqrt{BC^{2}+CD^{2}}=\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{4}\right)^{2}}=\frac{\sqrt{7}}{4}
$$

From this, $BS=\frac{2}{3} \cdot \frac{\sqrt{7}}{4}=\frac{\sqrt{7}}{6}$.

In the triangle $BSC$, the angle bisector $SF$ divides the opposite side in the ratio of the adjacent sides, that is:

$$
\frac{BF}{FC}=\frac{BS}{SC}=\frac{\sqrt{7}}{2}
$$

The lengths of the segments in question are:

$$
BF=\frac{\sqrt{7}}{2} \cdot FC \quad \text{and} \quad \frac{\sqrt{7}}{2} FC+FC=\frac{1}{2}
$$

from which $FC=\frac{\sqrt{7}-2}{3}$ and $BF=\frac{7-2\sqrt{7}}{6}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_8e83875cc30c6f305ebfg-1.jpg?height=304&width=496&top_left_y=1103&top_left_x=801)