Question 49In the figure, $CD,$ $AE,$ and $BF$ are one-third of their respective sides. It follows that $AN_2:N_2N_1:N_1D=3:3:1,$ and similarly for lines $BE$ and $CF.$ Then the area of triangle $N_1N_2N_3$ is: $\dfrac1{10}\triangle ABC$$\dfrac19\triangle ABC$$\dfrac17\triangle ABC$$\dfrac16\triangle ABC$None of these
