Belarus2022

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 $H_{1}O$ and $H_{2}O$ 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 $K{M}_1:M_{1}F=N{M}_2:M_{2}E=1:2$. Denote by $R$ the radius of the semicircle given in the condition. Let $L_1$ be the point on the extension of the ray $O{M}_1$ beyond the point $M_1$, such that $M_1{L}_1=2O{M}_1$. Since $K{M}_1:M_{1}F=O{M}_1:M_{1}L=1:2$, the triangles $K{M}_{1}O$ and $F{M}_1{L}_1$ are similar with the coefficient $1/2$. Hence $$\angle F{L}_{1}O=\angle F{L}_{1}M=\angle KO{M}_1=\angle EO{L}_2\text{and}{L}_{1}F=2OK=BD.(1)$$ Let $L_2$ be the point on the extension of the ray $O{M}_2$ beyond the point $M_2$, such that $M_2{L}_2=2O{M}_2$. Similarly, we obtain the equalities $$\angle E{L}_{2}O=\angle FO{M}_1\text{ and }E{L}_2=AD.(2)$$ The equalities (1) and (2) implies that the triangles $OE{L}_2$ and $L_{1}FO$ are similar, therefore $$\frac{E{L}_2}{EO}=\frac{OF}{F{L}_1}⟺\frac{AD}{R}=\frac{R}{BD}⟺AD\cdot BD=R^2.$$ 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 $\angle DOB=180^{\circ }-\angle DOH=150^{\circ }$.