I have more ideas.
Let E (euler) midpoint of OH, Hc midpoint of HC, r is radius of circumcircle (OC=OD=OH'=r).
We know: HMc=McD (post #1) , HC'=C'H' (post #2).
Conclusions:
EHc=OC/2=r/2 ( as midline of triangle OHC ).
EMc=OD/2=r/2 ( as midline of triangle OHD ).
EC'=OH'/2=r/2( as midline of triangle OHH').
According to that we have:
Points Hc, Mc and C' have same distance with E (half of circumcircle radius).

Same for other six points, if we look at the side of point A and point B.

In the Attachment: Picture for post #1 without my comments.
If you need some corrections of the pictures(name of points),let me know.
Another thanks and god bles you, Alex.

https://www.cut-the-knot.org/Curriculum/Geometry/NinePointCircleSimple.shtml
https://www.cut-the-knot.org/Curriculum/Geometry/EulerLineSimple.shtml

Now, if you feel like you may want to make further contributions, I suggest you have a look at the wiki part of the site:

https://www.cut-the-knot.org/wiki-math/index.php

It's not difficult to learn. If you wish to try I'll send you the password by private email.

All the best,
Alex