Dear Alex,

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