5. Triangle $A B C$ has side lengths $A B=2 \sqrt{5}, B C=1$, and $C A=5$. Point $D$ is on side $A C$ such that $C D=1$, and $F$ is a point such that $B F=2$ and $C F=3$. Let $E$ be the intersection of lines $A B$ and $D F$. Find the area of $C D E B$.
Answer: $\frac{22}{35}$
Solution: Draw segment $A F$. Then notice $A F=4$, and we have a right triangle. Now draw line $C E$, let it intersect $A F$ at $G$. By Ceva, $F G=\frac{4}{3}$ and $A G=\frac{8}{3}$. Using mass points we find that $\frac{A E}{E B}=6$ so $\frac{[A E F]}{[B E F]}=6$, and since $[A B F]=4,[B E F]=\frac{4}{7}$. It's easy to see that $[C D F]=\frac{1}{5}[A C F]=\frac{6}{5}$, so
$$
[B C D E]=[C D F]-[B E F]=\frac{6}{5}-\frac{4}{7}=\frac{22}{35} .
$$