In this write up I construct the orthocenter H of triangle ABC, orthocenters of triangles HBC, HAB, and HAC...the circumcircles of triangles ABC, HBC, HAB, and HAC and explore the resulting figure for make conjectures...finally resulting in some proofs.

My first conjecture is that the orthocenters for HBC, HAB, and HAC will be, respectively, A,C, and B. Here is a picture of the situation.

Here is a proof for...the orthocenter for triangle HAC will be B:

By construction, line HB is perpendicular to AC and hence passes through the orthocenter for triangle HAC (this is how H was using the intersection of altitudes in triangle ABC). The orthocenter for HAC is also on the perpendicular from C to line AH. But H is on perpendicular AH (AH perpendicular to BC) again by construction of H. Thus, the orthocenter for AHC is on lines HB and CB. As long as HB and CB are distinct lines then they intersect in only one point, B. It now follows that B must be the orthocenter for HAC as desired.(The only problems occur when ABC is degenerate.) QED