
Finding the length
Hi,
I've been having problems ith this question. Been trying to use ideas of area and similarity but nothing seems to work. One idea I have is to make the like go from F though the points E and F but will this make the question too specific?
Let triangle ABC have E on AC, D on AB an point F on the extended line BC. Given that AE=2EC and AD=3DB and CB=5, find the length of BF
Thanks!

Presumably D, E and F are supposed to lie on a straight line?
. . . . . . . . . . . . . . . . . $\displaystyle \setlength{\unitlength}{1.17mm}
\begin{picture}(70,45)
\thicklines
\put(3,3){\line(1,0){63}}
\put(3,3){\line(1,4){9}}
\put(24,3){\line(1,3){12}}
\put(66,3){\line(3,2){54}}
\put(24,3){\line(3,2){18}}
\put(66,3){\line(5,1){60}}
\put(11,40){$A$}
\put(2,0){$C$}
\put(23,0){$B$}
\put(21,13){$D'$}
\put(65,0){$F'$}
\put(3,14){$E$}
\put(13,0){\small$5$}
\end{picture}
$
This problem is much easier if you guess the answer and then work backwards. In the picture, D and F have been replaced by D' and F', which are defined as follows. The line AF' is parallel to EB, and meets the extended line CB at F'. The line EF' meets AB at D'.
The triangles CEB and CAF' are similar (equal angles). Since CA = 3CE, it follows that AF' = 3EB.
The triangles BD'E and AD'F' are also similar (equal angles). Since AF' = 3EB, it follows that AD' = 3D'B. But that means that D' = D, and consequently F' = F.
From the similar triangles CEB and CAF it then follows that CF = 3CB = 15, and thus BF = 10.