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9889Re: [EMHL] The Brocard points of a quadrilateral (Geoff Millin)

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  • Darij Grinberg
    Jun 9, 2004
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      In Hyacinthos message #9843, I wrote:

      >> This is very interesting! At first, consider
      >> the Brocard points R and R' of the harmonic
      >> quadrilateral ABCD, and the symmedian point
      >> K (this is the point whose distances to the
      >> sides of the quadrilateral are proportional
      >> to the lengths of the sides). Also consider
      >> the circumcenter O of our quadrilateral.
      >> Then, the points O, K, R and R' lie on one
      >> circle, which is analogous to the "Brocard
      >> circle" of a triangle. The points O and K
      >> are the endpoints of a diameter of this
      >> circle. The points R and R' lie
      >> symmetrically with respect to this diameter.
      >> The circle through the points O, K, R and R'
      >> also passes through the midpoints of the
      >> diagonals AC and BD. The line joining these
      >> midpoints meets the diameter OK at the
      >> midpoint of the segment RR'. The angles
      >> < ROK and < R'OK are equal to the Brocard
      >> angle of the quadrilateral ABCD.

      These remarks started an interesting thread,
      which I could not follow since one half of the
      mathematics involved was new to me, but let me
      just make one remark: In a harmonic
      quadrilateral ABCD, the point K is the point
      of intersection of the diagonals AC and BD.
      [This is quite easy to prove.] Now, since the
      midpoints M and N of the diagonals AC and BD
      are the orthogonal projections of the
      circumcenter O on these diagonals, the angles
      < OMK and < ONK are right angles. Hence, M and
      N lie on the circle with diameter OK.

      Darij Grinberg
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