Japan Mathematical Olympiad

$\frac{7}{9}$

For the triangles $ABD$, $DCE$, and $EAF$, we have $$\angle EDC=\angle ADC-\angle ADE=(\angle DAB+\angle ABD)-60^{\circ }=\angle DAB$$ and similarly, $$\angle FEA=\angle DEA-\angle DEF=(\angle EDC+\angle DCE)-60^{\circ }=\angle EDC$$

Also, since $\angle ABD=\angle DCE=\angle EAF=60^{\circ }$, the triangles $ABD$, $DCE$, and $EAF$ are similar. Therefore, we have $\frac{EA}{AF}=\frac{DC}{CE}=\frac{AB}{BD}=3$, and $$DC=BC-BD=2,CE=\frac{1}{3}DC=\frac{2}{3},EA=CA-CE=\frac{7}{3},AF=\frac{1}{3}EA=\frac{7}{9}$$ follows. Thus, the answer is $\frac{7}{9}$.