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:
$\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad \textbf{(B)}\ \frac{1}{(n+1)^2}\qquad \textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qquad \textbf{(D)}\ \frac{n^2}{(n+1)^2}\qquad \textbf{(E)}\ \frac{n(n-1)}{n+1}$
WLOG, let's assume that $\triangle ABC$ is equilateral. Therefore, $[ABC]=\frac{(1+n)^2\sqrt3}{4}$ and $[DBE]=[ADF]=[EFC]=n \cdot sin(60)/2$. Then $[DEF]=\frac{(n^2-n+1)\sqrt3}{4}$. Finding the ratio yields $\fbox{A}$.  -Dark_Lord