Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic tr.),

and H1,H2,H3 the midpoints of AH,BH,CH, resp.

Denote:

A*, B*, C* = the reflections of N in HHa,HHb,HHc, resp.

Ma, Mb, Mc = the midpoints of A*H1, B*B1, C*C1, resp.

The triangles ABC, MaMbMc are orthologic

(The othologic center (MaMbMc, ABC) is on the Euler line)

A generalization:

Let OaObOc be the pedal triangle of the isogonal

conjugate of H (=O) (medial triangle)

Observe that:

H1,H2,H3 are the antipodes of Oa,Ob,Oc in the common pedal circle

of HaHbHc, OaObOc (=NPC)

If A',B',C' are the orthogonal projections of N on OOa, OOb, OOc, resp.

then NA* / NA'= NB* / NB' = NC* / NC' = -2/1

In general:

Let ABC be a triangle, P a point, PaPbPc the pedal

triangle of P and Op the center of the pedal circle

of P (ie the circumcenter of the triangle PaPbPc)

Let P1, P2, P3 be the antipodes of Pa,Pb,Pc

(P1P2P3 is the reflection of PaPbPc in Op)

Let A',B',C' be the ortogonal projections of Op on

PPa,PPb, PPc, resp.