, 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 $\triangle ADO$ is right-angled triangle with an angle of $30^\circ$. From this we find that $\frac{DO}{OB} = \frac{1}{2}$.