3. In triangle $A B C$, points $D$ and $E$ lie on the interior of segments $A B$ and $A C$, respectively, such that $A D=1, D B=2, B C=4, C E=2$ and $E A=3$. Let $D E$ intersect $B C$ at $F$. Determine the length of $C F$.
Solution: First notice that the sidelengths of $\triangle A B C$ are 3,4 and 5. By Pythagoras this implies that triangle $A B C$ is right-angled at $B$. Now we can put the diagram on coordinate axes such that $B=(0,0)$ and $A=(0,3)$ and $C=(4,0)$. Furthermore we get $D=(0,2)$ and since $E$ divides $C A$ into the ratio $2: 3$ we get $E=(2.4,1.2)$, as shown in the diagram.

Now we can calculate the slope of the line $D E$ to be $\frac{-0.8}{2.4}=-\frac{1}{3}$. This means that the equation of line $D E$ is given by $y=-\frac{x}{3}+2$. Therefore the $x$-intercept of this line is the solution to $0=-\frac{x}{3}+2$. The solution is when $x=6$, and thus $F=(6,0)$. Hence $C F=2$.