# 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.

• I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us?
– dfnu
Mar 27, 2019 at 16:01
• Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true. Mar 27, 2019 at 16:06
• Could you provide a picture?? Mar 27, 2019 at 17:37
• @Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary. Mar 27, 2019 at 18:46
• Thanks you guys ,if AG is equal to GX then the problem becomes much easier.I think it was misprint that wasted a week of my time. thanks anyways Mar 28, 2019 at 1:18

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.