Can you solve this?
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 $BD$?
A
1
F
2
D E
B C
FE patallel to DC so triangles AFE and ADC are similar
AF / FE = AD / DC
1 / FE = 3 / DC
FE = (1/3)DC
3FE = DC
FE parallel to DC and DE parallel to BC so triangle FDE is similar to triangle DBC
FD / FE = BD / DC
FD / FE = BD / (3FE)
FD = BD / 3
3FD = BD
3*2 = BD = 6