$G$ 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.

  • 1
    $\begingroup$ I don't understand what "$AG$ is produced to $X$" means. Did you draw a picture? Can you share it with us? $\endgroup$
    – dfnu
    Mar 27, 2019 at 16:01
  • 3
    $\begingroup$ Are you sure that $GX = AC$? I've calculated that you need $GX=AG$ for this theorem to be true. $\endgroup$ Mar 27, 2019 at 16:06
  • $\begingroup$ Could you provide a picture?? $\endgroup$
    – Dr. Mathva
    Mar 27, 2019 at 17:37
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    $\begingroup$ @Matteo "Produce" is a slightly old-fashioned term in geometry meaning to extend or lengthen. You can find the definition in Wiktionary. $\endgroup$
    – FredH
    Mar 27, 2019 at 18:46
  • $\begingroup$ 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 $\endgroup$
    – Stomp
    Mar 28, 2019 at 1:18

1 Answer 1


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.

  • $\begingroup$ Thank you, if AG=GX then the problem becomes solvable with similarity.Can't thank you enough for that. $\endgroup$
    – Stomp
    Mar 28, 2019 at 1:20

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