H - Orthologic

Let ABC be a triangle, P = H = X(4) and Q a point on the Euler line.

Denote:

Bc, Cb = the orthogonal projections of B, C on HC, HB, resp.

Qa = same to Q point of the triangle ABcCb.
Similarly Qb, Qc.

ABC, QaQbQc are orthologic.

For Q = X(3) = O:
Orthologic centers = X(4) = H

For Q = G = X(2)
Orthologic center (ABC, GaGbGc) = ?
Orthologic center (GaGbGc, ABC) = G of orthic = X(51)

Q = H = X(4)
Orthologic center (ABC, HaHbHc)= X(3)= O
Orthologic center (HaHbHc, ABC) = ?

For Q = N = X(5)
Orthologic center (ABC, NaNbNc) = ?
Orthologic center (NaNbNc, ABC) = ?