jmc

In the figure, $\overline{CD}$, $\overline{AE}$ and $\overline{BF}$ are one-third of their respective sides. It follows that $\overline{A{N}_2}:\overline{N_2{N}_1}:\overline{N_{1}D}=3:3:1$, and similarly for lines BE and CF. Then the area of triangle $N_1{N}_2{N}_3$ is: $\text{(A) }\frac{1}{10}△ABC\text{(B) }\frac{1}{9}△ABC\text{(C) }\frac{1}{7}△ABC\text{(D) }\frac{1}{6}△ABC\text{(E) }\text{none of these}$