Ukrajina 2008

Under the problem statement we can easily find the following angles (fig.1): $\angle ABD=30^{\circ }$, $\angle BDC=45^{\circ }$, $\angle DBC=75^{\circ }$. Let's draw rays $ADE$ and $ABF$. Then $\angle EDC=45^{\circ }$, $\angle FBC=75^{\circ }$. Therefore $BC$ is a bisector of $\angle DBF$ and $DC$ is a bisector of $\angle BDE$, which implies that $AC$ is a bisector of $\angle BAD$. The last statement is easily proved by the locus of the bisector. Thus $\angle BAO=30^{\circ }$ and $\angle AOB$ is an isosceles triangle. Therefore $AO=BO$ and $DO=\frac{1}{2}AO$ as $△ADO$ is right-angled triangle with an angle of $30^{\circ }$. From this we find that $\frac{DO}{OB}=\frac{1}{2}$.