Prove that the centre of the nine-point circle lies on the midpoint of the Euler line 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.

• 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
• 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

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

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).

The orthocentre of KLM is orthocentre of ABC and that of medial triangle is circumcentre of ABC and note that vertices of KLM and medial triangle are diametrically opposite so if we rotate KLM by 180 degrees its orthocentre will coincide circumcentre of ABC so nine point centre is midway between orthocentre and circumcentre

In your figure, the distance HK is equal to OA'. This can be proved very easily; the distance of orthocentre from vertex A is equal to 2RcosA and as HK=AH/2 ,HK=RcosA. Now in right triangle OA'B; OA'=OBcosA=RcosA... Hence, we proved that HK=OA'. Now connect A' and X, now as HK||OA' and HK=OA', we can conclude triangles OA'I and HKI are similar;giving us "OI=HI".