In triangle $ABC$, points $D$ and $F$ are on $\overline{AB},$ and $E$ is on $\overline{AC}$ such that $\overline{DE}\parallel \overline{BC}$ and $\overline{EF}\parallel \overline{CD}$. If $AF = 1$ and $DF = 2$, then what is $BD$?
The answer is BD = 15*sqrt(2) + 8.
+80 
