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

I have tried using Menelaus theorem and thus tried to multiply the corresponding ratios to obtain $-1$ but I can't do so. I don't seem to understand the relevance of $GX=AC$ and that's probably why I can't solve it. Any hint would be appreciated, thank you.
 A: I can prove that you'd need $X$ such that $|GX| = |AG|$ using vector algebra.
Let us describe the points of the plane using vectors from point A. Then $\vec{A} = \vec{0}$, and $\vec{B}$ and $\vec{C}$ are linearly independent. The centroid $G$ is given by a vector
$$ \vec{G} = \frac13(\vec{A}+\vec{B}+\vec{C}) = \frac13(\vec{B}+\vec{C}) $$
Point $X$ lies on the line $\overline{AG}$, which means that 
$$ \exists \lambda\in\mathbb{R} : \vec{X} = \lambda (\vec{B}+\vec{C}) $$
For point $L$ we have
\begin{align} \big(L\in \overline{BC}\big) &\Rightarrow \big(\exists \alpha_L\in\mathbb{R} : \vec{L} = \alpha_L \vec{B} + (1-\alpha_L) \vec{C} \big) \\
\big(\overline{XL} \parallel \overline{AC} \big) &\Rightarrow \big(\exists \beta_L\in\mathbb{R} : \vec{L} = \vec{X} + \beta_L \vec{C} = \lambda \vec{B} + (\lambda + \beta_L)\vec{C}\big) \end{align}
since vectors $\vec{B}=\vec{C}$ the only solution for these conditions is
$$ \vec{L} = \lambda \vec{B} + (1-\lambda)\vec{C}$$
For point $M$ we have
\begin{align} \big(M\in \overline{AC}\big) &\Rightarrow \big(\exists \alpha_M\in\mathbb{R} : \vec{M} = \alpha_M \vec{C} \big) \\
\big(\overline{XM}\parallel \overline{AB}\big) &\Rightarrow \big(\exists \beta_M\in\mathbb{R} : \vec{M} = \vec{X} + \beta_M \vec{B} = (\lambda + \beta_M) \vec{B} + \lambda \vec{C}\big) \end{align}
The only solution for these conditions is
$$ \vec{M} = \lambda \vec{C}$$
Finally, for point $N$ we have
\begin{align} \big(N\in \overline{AB}\big) &\Rightarrow \big(\exists \alpha_N\in\mathbb{R} : \vec{N} = \alpha_N \vec{B} \big) \\
\big(\overline{XN}\parallel \overline{BC}\big) &\Rightarrow \big(\exists \beta_N\in\mathbb{R} : \vec{N} = \vec{X} + \beta_N (\vec{B}-\vec{C}) = (\lambda + \beta_N) \vec{B} + (\lambda-\beta_N) \vec{C}\big) \end{align}
and the solution for these conditions is
$$ \vec{N} = 2\lambda \vec{B}$$
Now, if $L$, $M$, $N$ are supposed to be colinear that means that
\begin{align} \exists\gamma\in\mathbb{R} &: (\vec{L}-\vec{M}) = \gamma (\vec{N}-\vec{L}) \\ 
\exists\gamma\in\mathbb{R} &: \lambda\vec{B}+(1-2\lambda)\vec{C} = \gamma (\lambda\vec{B}+(\lambda-1)\vec{C})\end{align}
This can only be true if $\lambda=\frac23$, or $\lambda=0$. The second option would correspond to $X=A$ and it isn't the case we're interested in. That means that $\vec{X}=\frac23(\vec{B}+\vec{C}) = 2\vec{G}$
If you choose any other point $X$ on line $\overline{AG}$, points $L$, $M$, $N$ won't be colinear.
