In right triangle $ACE$ (right angle at $C$), $AC=9$, $CE=12$, $EA=15$. Points $B,D,F$ lie on $AC,CE,EA$ respectively with $AB=3$, $CD=4$, $EF=5$. What is the ratio of the area of $\triangle DBF$ to that of $\triangle ACE$?

1. Each cut is exactly $\frac13$ of its side: $AB/AC=CD/CE=EF/EA=\frac13$.
2. If the uniform fraction is $k$, one computes $[DBF]/[ACE]=1-3k(1-k)$ (each corner triangle contributes a $k(1-k)$ share).
3. With $k=\frac13$: ratio $=1-3\cdot\frac13\cdot\frac23=1-\frac23=\boxed{\dfrac13}$.