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6262Re: Some theorems on Miquel points

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  • jpehrmfr <jean-pierre.ehrmann@wanadoo.fr>
    Jan 4, 2003
      Dear Darij,
      I've tried to find an easy synthetic proof of Braun theorem
      > The Miquel point configuration is defined as follows:
      >
      > On the sidelines BC, CA, AB of a triangle ABC, points A', B', C'
      are
      > chosen and the circles AB'C', BC'A', CA'B' are drawn. Then, as we
      > know, these circles have a common point P (the Miquel point).
      >
      > Now the two theorems:
      >
      > 2. Let Ma, Mb, Mc be the centers of the circles AB'C', BC'A',
      CA'B'
      > [it is well-known that triangles MaMbMc and ABC are similar], let
      M'
      > be the circumcenter of triangle MaMbMc, and M the circumcenter of
      > triangle ABC. Then M'M = M'P.
      >
      > [This is a theorem by Peter Baum; proposed in the little German
      > periodical "Die Wurzel" (see the website wurzel.org), in the 12/98
      > issue. Solved by Sefket Arslanagic in the 5/99 issue; the solution
      > was quite involved.]

      PaPbPc is the pedal triangle of P.
      A direct similitude with center P, angle T, ratio 1/cos(T) maps
      PaPbPc to A'B'C'.
      As the circumcenter of PPbPc is the midpoint A'' of AP, the
      similitude maps this midpoint to Ma; hence the similitude maps the
      midpoint I of PM (circumcenter of A''B''C'') to M'(circumcenter of
      MaMbMc).
      As PM'/PI = cos(T) and <IPM' = T, it follows that M' lies on the
      perpendicular bisector of PM.
      Friendly. Jean-Pierre
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