AHSME 1952 Problem 49

In the figure, $\overline{CD}$, $\overline{AE}$ and $\overline{BF}$ are one-third of their respective sides. It follows that $\overline{AN_2}: \overline{N_2N_1}: \overline{N_1D} = 3: 3: 1$, and similarly for lines BE and CF. Then the area of triangle $N_1N_2N_3$ is: $\text{(A) } \frac {1}{10} \triangle ABC \text{(B) } \frac {1}{9} \triangle ABC \text{(C) } \frac{1}{7}\triangle ABC \text{(D) } \frac{1}{6}\triangle ABC \text{(E) } \text{none of these}$