contest-mathgeometry
$G$ 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.
Let $\Delta ABC$ be a triangle, denote by $M,N,P$ the mid points of the sides opposite to $A,B,C$. Let $D$ be a point on $BC$. The parallel to $AD$ through $M$ intersects $NP$ in a point denoted by $D'$.
Construct $E',F'$ in a similar way on the sides of $\Delta MNP$.
Because $MN$, $ND'$, $D'M$ are parallel to (respectively) $AB$, $BD$, $DA$, we obtain a triangle similarity, so $D'N:DB=MN:AB=1:2$. This gives $D'N:D'P=DB:DC$.
We finally apply Ceva (and the reciprocal) in the two triangles $\Delta ABC$, $\Delta MNP$: $$ \frac{D'N}{D'P}\cdot \frac{E'P}{E'M}\cdot \frac{F'M}{F'N} = \frac{DB}{DC}\cdot \frac{EC}{EA}\cdot \frac{FA}{FB} = -1\ . $$ $\square$
Later edit: Quick picture inserted:
Look at the drawing here.
What do we have?
$AD,BE,CF$ - they intersect in a single/common point - point $O$
$A'$ - midpoint of $BC$
$B'$ - midpoint of $CA$
$C'$ - midpoint of $AB$
$D'$ - midpoint of $AD$
$E'$ - midpoint of $BE$
$F'$ - midpoint of $CF$
From Ceva's theorem for triangle $ABC$ we get: $$\frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1 \tag{1}$$
Now the trick is to realize that:
$$\frac{B'F'}{F'A'} = \frac{AF}{FB} \tag{2}$$
$$\frac{C'D'}{D'B'} = \frac{BD}{DC} \tag{3}$$
$$\frac{A'E'}{E'C'} = \frac{CE}{EA} \tag{4}$$
Why is this so?
Because $B'C' || BC$ , $C'A' || CA$ and $A'B' || AB$
so these relations follow from the Intercept theorem.
Multiplying the last 3 equations and using $(1)$ we get:
$$\frac{B'F'}{F'A'}\frac{C'D'}{D'B'}\frac{A'E'}{E'C'} = \frac{AF}{FB}\frac{BD}{DC}\frac{CE}{EA} = 1$$
Thus:
$$\frac{B'F'}{F'A'}\frac{A'E'}{E'C'}\frac{C'D'}{D'B'} = 1 \tag{5}$$
Now using the inverse Ceva's theorem (for the triangle $A'B'C'$ and for the points $D', E', F'$), we can conclude from $(5)$ that the three lines $A'D', B'E', C'F'$ intersect at a single/common point. This is what we had to prove hence the problem is solved.
Best Answer
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.