20. In the figure below, $A B C$ is a triangle with $A B=10 \mathrm{~cm}$ and $B C=40 \mathrm{~cm}$. Points $D$ and $E$ lie on side $A C$ and point $F$ on side $B C$ such that $E F$ is parallel to $A B$ and $D F$ is parallel to $E B$. Given that $B E$ is an angle bisector of $\angle A B C$ and that $A D=13.5 \mathrm{~cm}$, fmd the length of $C D$ in $\mathrm{cm}$.
20. Answer: 24

Since $B E$ bisects $\angle A B C$, we have $A E: E C=A B: B C=1: 4$. Furthemore, since $E F / A B$ and $D F \not E B$, we see that $D F$ bisects $\angle E F C$. Hence $D E: D C=1: 4$. Let $A E=x$ and $D E=y$. Then we have $x+y=13.5$ and $4 x=5 y$. Solving the equations yields $x=7.5$ and $y=6$. It follows that $C D=4 y=24$.