Belarus competition problems

Let $H_1$ and $H_2$ be the intersection points of the altitudes of the triangles $ADF$ and $BDE$. Note that the point $O$ is the circumcenter of both triangles $ADF$ and $BDE$. Therefore the coinciding lines $H_1O$ and $H_2O$ are the Euler lines of these triangles and they contain the intersection points $M_1$ and $M_2$ of the medians of the triangles $ADF$ and $BDE$ respectively.
Let $K$ and $N$ be the midpoints of the segments $AD$ and $BD$ respectively. Then $M_1 \in FK$, $M_2 \in EN$ and $KM_1 : M_1F = NM_2 : M_2E = 1:2$. Denote by $R$ the radius of the semicircle given in the condition.
Let $L_1$ be the point on the extension of the ray $OM_1$ beyond the point $M_1$, such that $M_1L_1 = 2OM_1$. Since $KM_1 : M_1F = OM_1 : M_1L = 1:2$, the triangles $KM_1O$ and $FM_1L_1$ are similar with the coefficient $1/2$. Hence
$$\angle FL_1O = \angle FL_1M = \angle KOM_1 = \angle EOL_2 \quad \text{and} \quad L_1F = 2OK = BD. \quad (1)$$
Let $L_2$ be the point on the extension of the ray $OM_2$ beyond the point $M_2$, such that $M_2L_2 = 2OM_2$. Similarly, we obtain the equalities
$$\angle EL_2O = \angle FOM_1 \text{ and } EL_2 = AD. \quad (2)$$
The equalities (1) and (2) implies that the triangles $OEL_2$ and $L_1FO$ are similar, therefore
$$\frac{EL_2}{EO} = \frac{OF}{FL_1} \iff \frac{AD}{R} = \frac{R}{BD} \iff AD \cdot BD = R^2.$$
Draw the altitude $DH$ of the triangle $ABD$. According to the formula for the length of the altitude drawn to the hypotenuse, $DH = \frac{AD \cdot BD}{AB} = \frac{R^2}{2R} = \frac{DO}{2}$. Hence in the right-angled triangle $DOH$, the angle $DOH$ is equal to $30^\circ$. If the point $H$ lies on the segment $OB$ then $\angle DOB = \angle DOH = 30^\circ$, and if it lies on the segment $OA$ then $\angle DOB = 180^\circ - \angle DOH = 150^\circ$.