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22477Euler line - Pedal triangle - Orthologic

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  • Antreas Hatzipolakis
    Jun 29, 2014
    • 0 Attachment
      Let ABC be a triangle, HaHbHc the pedal triangle of H (orthic tr.),
      and H1,H2,H3 the midpoints of AH,BH,CH, resp.


      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.

      Let A*, B*, C* be points on OpA', OpB', OpC' such that
      OpA* / OpA' = OpB* / OpB' = OpC* / OpC' = t

      Let Ma, Mb, Mc be the midpoints of of P1A*, P2B*, P3C*, resp.

      The triangles MaMbMc, ABC are orthologic.
      The orthologic center (MaMbMc, ABC) is lying on the
      line POp.