Concurrence of diagonals on a particular hexagon

Question

Let $\triangle ABC$ a triangle, and $P$ a point inside $\triangle ABC$. Let $D$, $E$, $F$ mid points of $\overline{AP}$, $\overline{BP}$ and $\overline{CP}$ respectively and $L=\overline{BF}\cap\overline{CE}$, $M=\overline{CD}\cap\overline{AF}$, $N=\overline{AE}\cap\overline{BD}$.

1. Prove that the hexagon $DNELFM$ has one third the area of the $\triangle ABC$
2. Prove that the diagonals $\overline{DL}$, $\overline{FN}$, $\overline{ME}$ of the hexagon $DNELFM$ concur at one point.

I already proved part 1, but I'm having trouble with part 2 of this question, any advices?

Answer 1

This is pretty simple by exploiting vectors or affine maps.
$D$ is the midpoint of $AP$, hence $D=\frac{A+P}{2}$. $L$ is the centroid of $BCP$, hence $L=\frac{B+C+P}{3}$. $$2D+3L=A+B+C+2P = 3G+2P = 2E+3M = 2F+3N$$ hence the diagonals of $DMFLEN$ meet at $\frac{3}{5}G+\frac{2}{5}P$ with $G$ being the centroid of $ABC$.