Let $G$ be the centroid of triangle $ABC$. Let $D$ be the midpoint of $BC$. A line through $G$ parallel to $BC$ meet $AB$ at $M$ and $AC$ at $N$. $MC$ meets $BG$ at $P$ and $NB$ meets $CG$ at $Q$. Prove that triangle $DQP$ is similar to triangle $ABC$.
I observed, by drawing an accurate diagram, that $E, P, D$ seem to be collinear and so are $D, Q, F$. It is obvious that the triangle $DEF$ is similar to $ABC$, and hence I was thinking of proving that the points mentioned above are collinear, AND $PQ//EF$ hence triangle $DPQ$ is similar to $DEF$ which is then similar to $ABC$. How should I go about proving this?