## Orthocentric anti-inversion

Given triangle ABC, the orthic triangle DEF is formed by the feet of the altitudes of ABC. Some properties of this configuration have been studied in Orthocenter.html . There it was proved that the orthocenter (intersection point of the altitudes) satisfies: AH*HD = BH*HE = CH*HF = rh2. Thus we can define circle (h) centered at H with radius rh and consequently also the anti-inversion (f) with respect to this circle.
[1] f maps the circles kA, kB, kC with diameters respectively AH, BH, CH to the sides BC, CA, AB.
[2] f maps the euler circle (passing through D, E, F and the middles A', B', C' of HA, HB, HC respectively) to the circumcircle of ABC.
[3] f maps the incircle (g) of ABC to circle (d) which is simultaneously tangent to kA, kB, kC and the circumcircle of ABC.
[4] f maps the Feuerbach point X(11) of ABC to the point of tangency of (d) with the circumcircle. This coincides with triangle center X(108) in the list of Kimberling.

[1] Follows immediately from elementary properties of (anti)-inversions. In fact, f maps circle kA to a line and since A, D are (anti) inverse kA must map to a line orthogonal to AD at D i.e. to BC.
[2] The relation FH*HC = rh2 <=> (2FH)*(HC/2) = rh2, which shows that C' and C'' are anti-invese. Analogous relations hold for A', A'' and B', B''. This shows that f maps the euler circle (e) to the circumcircle (c).
The rest follows from the fact that the incircle is simultaneously tangent to the sides and the Euler-circle (Feuerbach's theorem), hence by the anti-inversion (f) the incircle (g) maps to a circle (d) simultaneously tangent to kA, kB, kC and (c). A consequence of this is that the center Q of (d) the incenter I and the orthocenter H are on a line. The identification of f(X(11)) with X(108) follows by a calculation involving the trilinears of X(11) (see Kimberling below).
Notice that circle (h) was met in the discussion of Autopolar triangles (see Autopolar.html ) and called there the conjugate circle of triangle ABC (later assumed there to be obtuse-angled).