12.157. In a right-angled triangle, find the angle between the median and the bisector drawn from the vertex of the acute angle equal to $\alpha$.
Solution.

Let in $\triangle A B C$ (Fig. 12.30) $\angle A=90^{\circ}, \angle A B C=\alpha$,

$B D$ - bisector, $B E$ - median. Since $\angle A>\angle C$, point $D$ is located between points $A$ and $E$. If $A B=x$, then $A C=x \operatorname{tg} \alpha$,

$$
\begin{aligned}
& A E=\frac{1}{2} x \operatorname{tg} \alpha . \mathrm{B} \triangle B A E: \operatorname{tg} \angle A B E=\frac{A E}{A B}=\frac{1}{2} \operatorname{tg} \alpha ; \angle A B E=\operatorname{arctg}\left(\frac{\operatorname{tg} \alpha}{2}\right) \\
& \angle D B E=\angle A B E-\angle A B D=\operatorname{arctg}\left(\frac{\operatorname{tg} \alpha}{2}\right)-\frac{\alpha}{2}
\end{aligned}
$$

Answer: $\operatorname{arctg}\left(\frac{\operatorname{tg} \alpha}{2}\right)-\frac{\alpha}{2}$.