AHSME 1967 Problem 34

Points $D$, $E$, $F$ are taken respectively on sides $AB$, $BC$, and $CA$ of triangle $ABC$ so that $AD:DB=BE:CE=CF:FA=1:n$. The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is: $\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2} \textbf{(B)}\ \frac{1}{(n+1)^2} \textbf{(C)}\ \frac{2n^2}{(n+1)^2} \textbf{(D)}\ \frac{n^2}{(n+1)^2} \textbf{(E)}\ \frac{n(n-1)}{n+1}$