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Let $ABC$ be an acute angled triangle and suppose $X$ is a point on the circumcircle of $\Delta ABC$ with $AX||BC$ and $X\neq A$. Denote by $G$ the centroid of triangle $ABC$, and by $K$ the foot of the altitude from $A$ to $BC$. Prove that $K,G,X$ are collinear.

I tried to apply Menelaus theorem but couldn't find a triangle to apply. I found a homothety centred at $G$ which maps the medial triangle to the main triangle. I guess the homothety maps $K$ to $X$ i.e. $K$ a point on the circumcircle of the medial triangle but i failed to proof this. Please help me.

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I usually enjoy overkills but Menelaus' theorem is too much even for my taste, here. enter image description here

Add a couple of points: $M$ as the midpoint of $BC$ and $J$ as the projection of $X$ on $BC$. In the rectangle $AKJX$ the centroid $G$ lies at $\frac{2}{3}$ of the segment joining $A$ with the midpoint of $KJ$, hence it trivially lies on the diagonal $KX$.

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  • $\begingroup$ Please try to solve the problem which I post yesterday.I didnot find any good answer. $\endgroup$ – Sufaid Saleel Apr 4 '18 at 3:48
  • $\begingroup$ @SufaidSaleel: done. $\endgroup$ – Jack D'Aurizio Apr 4 '18 at 4:08
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Let the line $AK$ cuts circumcircle of $ABC$ at $H'$ and let $H$ be an orthocenter of $ABC$. Say circumcenter of $ABC$ is origin of position vectors. Now we have $$\vec{H}=\vec{A}+\vec{B}+\vec{C} = 3\vec{G}$$ and $$ \vec{K} = {1\over 2}(\vec{H}+\vec{H'})$$ Since $\angle H'AX = 90^{\circ}$ we have $\vec{H'}=-\vec{X}$. If we put this all together we get: $$ \vec{K} = {1\over 2}(3\vec{G}-\vec{X})\;\;\;\Longrightarrow \;\;\; \vec{GX} = -2\vec{GK}$$ So $G,X,K$ are collinear and $G$ is a center of homothety with facotr $k=-2$ which takes $K$ to $X$.

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