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
Because EF parallel to CD
AF / AE = AD / AC
1 / AE = 3 / AC
AC / AE = 3 /1 = 3
And because DE parallel to BC
AB / AD = AC / AE
AB/ AD = 3
AB / 3 = 3
AB = 3 * 3 = 9
So
BD = AB - AD = 9 - 3 = 6
+842 
