$\Delta ABC$ has $AB = 10$ cm, $BC = 40$ cm. Points $D$ and $E$ lie on side $AC$ and point $F$ on side $BC$ such that $EF \parallel AB$, $DF \parallel EB$. Given that $BE$ is an angle bisector of $\angle ABC$ and $AD = 13.5$ cm, find $CD$ .
What I Tried: Here is a picture :-
A first idea to immediately come in my mind was similarity, because one will find a lot of similar triangles here. For example $\Delta CFE \sim \Delta CBA$ , $\Delta CFD \sim \Delta CBE$ , $\Delta DFE \sim \Delta EBA$ and so on. I have also marked the angles based on the equality of them, noting that the green angle is equal to the sum of $1$ blue and $1$ brown angle. I also got $BF = FE$ .
I am not sure, however, how to deal with the sides. I could assume $BF = x$ and continue with similarity, or I could use $DE = y$ and continue, and there are a lot of possibilities and also time consuming. So is there a way to quickly deduce the value of $CD$ with a decent method of similarity? Thank You.
$CD=x,DE=y,EA=z$ By similarity/thales theorem using$DF\parallel EB,EF\parallel AB$ $$\frac{x+y}{z}=\frac{CF}{FB}=\frac{x}{y}$$ also $$y+z=13.5$$ by angle bisector theorem $$\frac{z}{x+y}=\frac{10}{40}$$
Three variables three equations....