+5 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 $AF = 9$ and $DF = 3$, then what is $BD$?
+129771 (I cheated a little on this one !!! )
C
E
A 9 F 3 D x B
In triangle ABC, DE is paralllel to BC
So....AE / AC = AD / AB (1)
In triangle ACD, EF is parallel to CD
So.... AE/ AC = AF / AD (2)
So using (1) and (2)
AD/ AB = AF / AD
So
AD^2 = AB * AF
(9+ 3)^2 = (12 + x) (9)
12^2 = 108 + 9x
144 = 108 + 9x
144-108 = 9x
36 = 9x
36/9 = x = 4 = BD
