### Orthologic triangles

Hyacinthos #24738 (Antreas Hatzipolakis)

Let ABC be a triangle and A'B'C' the pedal triangle of I.

Denote:

A"B"C" = the orthic triangle of A'B'C'

Aa, Ab, Ac = the reflections of A" in IA', IB', IC', resp.

Pa, Pb, Pc = same points of AaAbAc, BaBbBc, CaCbCc, resp.

**
ABC, PaPbPc are orthologic.**

The orthologic center (ABC, PaPbPc) lies on the circumcircle of ABC.

It is the isogonal conjugate of the infinity point of the OP line, where P is the same to Pa,Pb,Pc point wrt triangle ABC.

The locus of the orthologic center (PaPbPc, ABC) as Pa, Pb, Pc move on the Euler lines of AaAbAc, BaBbBc, CaCbCc, being same, is the line X(1)X(21).

The triangle PaPbPc is inversely similar to ABC. The similtide centers lie on the hyperbola passing through X(1), X(40), and of asymptotes parallels at X(3) to the asymptotes of Jerabek hyperbola.

If V is the orthologic center (PaPbPc, ABC) the envelope of line VP as P move on the Euler ABC is the parabola, tangent to Euler line, to line X(1)X(21), and to line X(1)X(3) at X(999).

(To construct this parabola, see Paris Pamfilos.- A Gallery of Conics by Five Elements. Forum Geometricorum Volume 14 (2014) 295–348. Sect. 13.2. page 246.)

http://amontes.webs.ull.es/otrashtm/HGT2016.htm

Angel Montesdeoca. Nov, 2016