On the circumference of a semicircle - whose diameter is $A B$ - determine the point $D$ such that $D A=D C$, where $C$ is the point where the tangent drawn at $D$ intersects the diameter $A B$.
The triangles $A O D$ and $A C D$ are isosceles, therefore $\alpha=\beta=\gamma$ and thus $3\alpha+\delta=180^{\circ}$. Since $\delta=90^{\circ}$, it follows that $\alpha=30^{\circ}$. Therefore, $\angle D O D_{1}=\angle D B D_{1}=60^{\circ}$ and $D B=D O=O B$. Therefore, the arc drawn from point $B$ with radius $O B$ intersects the semicircle drawn over $A B$ at the desired point $D$.

(Kiss József, Pápa.)

The problem was also solved by: Bánó L., Eckstein V., Friedländer H., Fuchs A., Füstös P., Harsányi Z., Heimlich P., Jánosy Gy., Lázár B., Miklóssy K., Paunz A., Pichler S., Rajz E., Sárközy E., Schuster Gy., Steiger J., Tandlich E., Tóth B., Wáhl V.