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