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. 
 A: 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).
A: Behold!


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