AN ORTHOCENTER

Let ABC be a triangle and A'B'C' the orthic triangle

Denote

Ma, Mb, Mc = the midpoints of AA', BB', CC', resp.

Ha, Hb, Hc = the orthocenters of HMbMc, HMcMa, HMaMb, resp.

Then
Orthocenter of HaHbHc = Nine Point Circle Center N

Euclid 6368
Romantics of Geometry

PROOF

The perpendicular from Mb to HMc intersects B'C' at A". In the triangle BB'C' we have MbA" // BC' and Mb = midpoint of BB', therefore A" is the midpoint of B'C'.
Similarly in the triangle CC'B' the perpendicular from Mc to HMb passes through the midpoint of B'C'. Therefore A" is the orthocenter Ha of HMbMc.
Similarly the orthocenters Hb, Hc of HMcMa, HMaMb are the midpoints of C'A', A'B', resp., that is HaHbHc is the medial triangle of A'B'C'

The orthocenter of HaHbHc is the circumcenter of A'B'C'

The circumcenter of A'B'C' is the Nine Point Circle center N of ABC.
H of HaHbHc = O of A'B'C' = N of ABC
QED