AN ORTHOPOLAR LINE

THEOREM (proof?):

Let ABC, A'B'C' be two triangles inscribed in two concentric circles such that A,B,A',B' are collinear. [the triangles share the same circumcenter and a midpoint of a side]

Let L be a line passing through O, the common circumcenter of the circles.

Denote:

1 = the orthopole of L wrt ABC

2 = the orthopole of L wrt A'B'C'

The line 12 passes through the intersection point of the NPCs (N),(N') of ABC, A'B'C', resp. other than the common midpoint M of AB, A'B'.

COROLLARY.

Let ABC be a triangle and A'B'C' the orthic triangle.

Denote:

Ab, Ac = the reflections of A in C',B', resp.

Bc, Ba = the reflections of B in A',C', resp.

Ca, Cb = the reflections of C in B',A', resp.

The NPCs of AAbAc, BBcBa, CCaCb are concurrent.

The point of concurrence is X1986 in ETC.

Let L be a line passing through H.

Denote:

1 = the orthopole of L wrt AAbAc

2 = the orthopole of L wrt BBcBa

3 = the orthopole of L wrt CCaCb

The points 1,2,3, X1986 are collinear.

We have:

The three triangles AAbAc, BBcBa, CCaCb share the same circumcenter, the H of ABC

A' is the midpoint of BBc of BBcBa and CCb of CCaCb ==> X1986, 2, 3 are collinear.

B' is the midpoint of CCa of CCaCb and AAc of AAbAc ==> X1986, 3, 1 are collinear.

C' is the midpoint of AAb of AAbAc and BBa of BBcBa ==> X1986, 1, 2 are collinear.

Therefore 1,2,3, X1986 are collinear

Antreas P. Hatzipolakis, 4 April 2013