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?
Refer to the following figure:
To prove $E,P$ and $D$'s collinearity, you may notice that $\triangle GPM$ and $\triangle BPC$ are similar. $\triangle GPM$ can be transformed into $\triangle BPC$ by rotating $180^\circ$ (and enlarging), and vice versa $\color{red}{\left(1\right)}$. Hence their medians from $P$ form a straight line. $MG\parallel BC$, and then $ED$ cuts $MG$ and $BC$ in half $\color{red}{\left(2\right)}$. The medians and $ED$ pass through the midpoints of $MG$ and $BC$, and only one straight line can pass through two points (on a plane). Therefore, the medians and $ED$ are the same straight line, and $E,P$ and $D$ are collinear.