Find the length of $EF$. 


$AB = 60\text{ cm}$
$CD:DB = 3:5$
$AF:AD = 4:5$
$E$ is the midpoint of $AC$
Find $EF$.


I have this following problem. I tried solving it, but it doesn't make sense alot. Can I have a hint or a guide for solving this question. Thanks!

First we notice that $CD=\frac{3CB}{8}$ and $DB=\frac{5CB}{8}$
Draw line $l$ parallel to AB from F.Let G be the point of intersection of $l$ and CB, then we have $DG=\frac{CB}{8}$ But then $CG=CD+DG=\frac{4CB}{8}$ and thus G is the mid point of CB and thus line $l$ passes through E.Now cause of similar triangles we have that $FG=\frac{AB}{5}=12$cm and $EG=\frac{AB}{2}=30$cm ,so $EF=EG-FG=18$cm 
