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