I suggest one another proof for:
https://www.cut-the-knot.org/Curriculum/Geometry/CirclesThroughOrthocenter.shtml

This proof based on one more generalized theorem. Detail as following:

Theorem:
If three circles concur at one Cevian point and their other three common points lie on three Cevian lines and these three circles cut triangle sidelines at six points then these six points are concyclic.

Proof:
Suppose P is a point on plane of triangle ABC; (Oa), (Ob), (Oc) are three circles concur at P and has three other common points: Ao on Cevian AP, Bo on Cevian BP, Co on Cevian CP. It means: (Oa) is circumcircle of PBoCo, (Ob) is circumcircle of PCoAo, (Oc) is circumcircle of PAoBo .Suppose (Oa), (Ob), (Oc) cut triangle sidelines BC, CA, AB at A1, A2, B1, B2, C1, C2 respectively. We proof these six points are concyclic.

We use power of circle theorem:

- Power of A wrt circles (Ob) and (Oc):
AC1*AC2 = AAo*AP = AB1*AB2 therefore B1, B2, C1, C2 are concyclic on one circle, suppose it is circle (K) and (K) cuts BC at D and E.

- Power of B wrt circle (K): BC1*BC2 = BD*BE
- Power of B wrt circles (Oc): BC1*BC2 = BBo*BP
Therefore BD*BE = BBo*BP so D, E, Bo, P are on circle say (K1)

- Power of C wrt circle (K): CB1*CB2 = CD*CE
- Power of C wrt circles (Ob): CB1*CB2 = CCo*CP
Therefore CD*CE = CCo*CP so D, E, Co, P are on circle say (K2)

Two circles (K1), (K2) have three common points D, E, P therefore they are the same one, say (K'). This circle (K') passes P, Bo, Co so it is the same circle (Oa) = (K') and D, E are intersections of (Oa) and sideline BC. It means D, E are A1, A2 and six points A1, A2, B1, B2, C1, C2 are concyclic on one circle (K).

Remark: Generally three circles (Oa), (Ob), (Oc) may be do not cut sidelines of ABC but in each special case of P we can choose proper circles (Oa), (Ob), (Oc) such that they always cut sidelines of ABC.

Now we apply this theorem for "Circles through the Orthocenter":

Line connected two circles centered at Mb, Ma should be perpendicular with MaMb or AB. Therefore two intersections of these circles are on line CA. Similar with other circles then we have: 3 circles centered at Ma, Mb, Mc follow our theorem conditions:
- They are concurrent at H
- Three their other common points are on HA, HB, HC
So they cut sidelines of ABC at six concyclic points.

Best regards,
Bui Quang Tuan

I've been working on that. The proof is actually the same. I think that the organizers chose to work with the orthocenter because the generalization is more suggestive of the method of solution.

We can use this fact to prove nine point circle. Detail as following:
Suppose Ha, Hb, Hc are altittude foots, Ma, Mb, Mc are midpoints of BC, CA, AB and H is orthocenter of triangle ABC.
We construct three circumcircles (Oa), (Ob), (Oc) of HHaMa, HHbMb, HHcMc. Of course three centers Oa, Ob, Oc are midpoints of HMa, HMb, HMc. Two circles (Ob), (Oc) share one common point at H. Their another common point is reflection of H in ObOc therefore it is on altitude AHa. Similarly we can show: three circles (Oa), (Ob), (Oc) share one common point at H and another common points are one Cevians (altitudes) AHa, BHb, CHc. By six concyclic theorem Ha, Hb, Hc, Ma, Mb, Mc are concyclic on one circle say (N)
This fact can be formulated as: in given triangle altitude foots and midpoints of sides are concyclic.
By applying this fact for triangle CAH: the circle (N) passes through altitude foots Ha, Hb, Hc therefore (N) must pass through midpoints of HA, HC and AC.
Similarly with triangles ABH, BCH, we can show that (N) passes through nine points: Ha, Hb, Hc, Ma, Mb, Mc and three midpoints of HA, HB, HC.

Best regards,
Bui Quang Tuan