G - Orthologic

Let ABC be a triangle, P = G = X(2) and Q a point on the Euler line.

Denote:

Bc, Cb = the orthogonal projections of B, C on GC, GB, resp.

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

ABC, QaQbQc are orthologic.

For Q = X(3) = O:
Orthologic center (ABC, GaGbGc)= X(36889)
Orthologic center (GaGbGc, ABC)= X(1352)
Euclid 9541

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

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

Orthologic center (ABC, NaNbNc)= ?
Orthologic center (NaNbNc, ABC)= ?