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 $CE =3$ and $DF = 3$, then what is $BD$?
A
F
3
D E
x 3
B C
AF / AE = AD / AC ⇒ AE/AC = AF/ AD
AD/ AE = AB / AC ⇒ AE / AC = AD/ AB
AF / AD = AD / AB
AD^2 = AF * AB
We need to know AF to solve this rather than EC
+1301 
