Estonian Math Competitions

(a) Let $\angle BCD=\alpha$, then the equal radii $BC=BD$ give us $\angle BDC=\alpha$ (Fig. 2) and $\angle CBD=180^{\circ }-2\alpha$. Then $\angle OBF=\angle CBD=180^{\circ }-2\alpha$. But on the other hand $\angle BOE=2\angle BCE=2\angle BCD=2\alpha$ and therefore $\angle BOF=180^{\circ }-2\alpha$. Since $\angle OBF=\angle BOF$, the triangle $OBF$ is isosceles.





(b) Due to equal radii the triangle $CBO$ is isosceles (Fig. 3). Furthermore, it is similar to $OBF$ as they share the angle at $B$. We have $OB=2BD=2BC$, so by the similarity also $FB=2OB$. Therefore $FB=4BD$ and $\frac{FB}{BD}=4$.