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In $\Delta ABC$, $AD, BE, CF$ are the altitudes and $\Delta A'B'C'$ is the medial triangle. $K, L, M$ are the midpoints of $AH, CH, BH$. Consider the nine-point circle with centre $G$ (not to be confused with the centroid) and diameters shaded in yellow. Prove that $G$ is the midpoint of the Euler line $HO$.

This, frankly, is an amazing result. The fact that the nine-point circle exists at all is amazing in itself. But, I haven't seen a convincing proof for this fact yet.

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your looking for an analytic proof or a different proof? –  ulead86 Jul 25 '13 at 15:54
    
I'm looking for a geometrical proof –  Gerard Jul 25 '13 at 16:04
1  
How about this? If you squint at your diagram, the vertices of hexagon $A^\prime M B^\prime K C^\prime L$, along with points $H$ and $O$ look like the vertices of an orthogonal projection of a shoebox. ($O$ is the closest corner to us; $H$ the farthest from us. Or vice-versa.) The yellow segments $A^\prime K$, $B^\prime L$, $C^\prime M$, and segment $HO$, are projections of the long diagonals of the box; since a box's diagonals obviously meet at the box's center, the projected diagonals must meet at the projected center: a point (here, $G$) that bisects each of the projections. –  Blue Jul 25 '13 at 16:17

2 Answers 2

up vote 2 down vote accepted

Behold!

Nine-point Circle

Shoebox

(MSE doesn't like such concise answers, so here are some extra characters.)

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The orthocenter of $A'B'C'$ is $O$, while $H$ is also the orthocenter of $KLM$. $A'B'C'$ and $KLM$ have parallel sides and are congruent so you can map them by a central symmetry to each other. Thus $H$ and $O$ would be mapped under to each other under this transform. So $KA'$, $LB'$, and $MC'$ all pass through the midpoint of $OH$. Then it suffices to show that for instance $KB'$ is orthogonal to $A'B'$ (which is obvious).

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