proof incenter of orthic

Let T=ABC be a triangle and T_h=A'B'C' its orthic. Let X_h(1) be the orthic's incenter.

1) T acute

- Th vertices are on the sides of T.

Lemma 1: [Rozikov] the altitudes of T are bisectors of Th

Corollary: since the altitudes meet at X_4, X_h(1) = X_4

2) T obtuse

- Th has one vertex on T's longest side, and two others on the extensions of its two sides, i.e., the latter two vertices are exterior to T.

Lemma: the triangle Te = A X4 C is the (acute) Excentral Triangle of Th.

- since AC' with CA' are altitudes of ABC, their intersection is at X_4.

- CC', AA', X4 B' are altitudes of Te, i.e., Th=A'B'C' is the orthic of A X4 C, i.e., Te is Th's excentral.

Theorem: the X_h(1) is B

By Lemma 1, altitudes CC', AA', X4 B' are bisectors of C', A', and B'. These meet at B, which completes the proof.

Corollary: X_4 is an excenter of Th {coxeter}

Since Th is the orthic of A X4 C, X_4 is an excenter of Th.

Observation: for T obtuse both ABC and AX4C have the same orthic. Under the orthic map, the pre-image of Th contains both T and the excentral of Th.  

When acute both preimages of Th are Te.

Given an orthic Th of an obtuse, to get its two pre-images:

a) acute Te: excentral of Th
b) obtuse Th: take X_h(1) and the endpoints of the longest side of Te or the side of Th which passes through the vertex of Th with the smallest angle (B' in the drawing).