Prove that $L$, $M$, and $N$ are collinear. 
Let $G$ denote the centroid of $\triangle ABC$;  $AG$ is produced to $X$ such that $GX = AG$. If we draw parallels through $X$ to $CA$, $AB$, and $BC$ meeting $BC$, $CA$, and $AB$ at $L$, $M$, and $N$ respectively, prove that $L$, $M$, and $N$ are collinear.

This question is from an ebook (source: CTPCM).
I am unable to proceed in the question. I applied Ceva's Theorem and proportionality theorem, but I get stuck leading nowhere.


Let $D$ be the intersection of line $XL$ with $AB$, $F$ the intersection of $XM$ and $BC$, and $E$ the midpoint of $BC$ (so $A$, $G$, and $E$ are colinear).
Since $XL\parallel AC$, $\triangle XLE$ is similar to $\triangle ACE$, and we have
$$AE:EX=EC:EL=AC:XL=3:1.\tag{1}$$
Similarly, $\triangle XEF$ is similar to $\triangle ABE$ and
$$AE:EX=BE:EF=AB:FX=3:1.\tag{2}$$
From $EC:EL=BE:EF=3:1$ and $BE=EC$, we see that $BL=LF=FC=BC/3$.
Since $DL\parallel AC$, we have
$$AC:DL=BC:BL=3:1.$$
This and (2) imply that $DL=XL$. So $L$ is the midpoint of $XD$. Similarly, $F$ is the midpoint of $XM$. So $DM\parallel LF$ and both of them are parallel to $XN$.
It follows that $DMXN$ is a parallelogram (having parallel opposite edges). That means $MN$ contains the midpoint of $DX$, which is $L$.
