Japan Mathematical Olympiad Initial Round

$$\boxed{\frac{15}{8}}$$ Let $P$ be the point of intersection of lines $AD$, $BE$ and $CF$. Then, we have $\angle PBA=\angle PDE$, since they are subtended by the same arc $\text{overarc}EA$ of the circle at the points $B$ and $D$ on the circumference. We also have $\angle BPA=\angle DPE$ so that the triangles $BPA$ and $DPE$ are similar. Consequently, we get $$(1)PA:PE=BA:DE=1:4.$$ In the same way, we see that the triangles $CPB$ and $EPF$ are similar, and therefore, we get $$(2)PE:PC=EF:CB=5:2.$$ From (1) and (2) above, we get $PC:PA=8:5$. Also, from the similarity of the triangles $FPA$ and $DPC$, we obtain $PC:PA=DC:FA$, from which it follows that $FA=\frac{5}{8}\cdot DC=\frac{15}{8}$, which is the desired answer.