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9156Re: [EMHL] Re: Cyclic quadrilateral problem

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  • Alexey.A.Zaslavsky
    Jan 30, 2004
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      Dear Eisso!
      >* Your properties 1 & 2 give rise to an easy and/or elegant
      >construction of a bi-centric quadrilateral, something I have been
      >looking for for a while. All you need to do is to take a circle and a
      >point P inside the circle. The four points of intersections of any pair
      >of perpendicular lines through with the circle will form a cyclic
      >quadrilateral and the KLMN that you construct will be bi-centric. When
      >you construct a bi-centric KLMN in this way, it turns out that both
      >``centers'' of KLMN and the center on the circumscribed circle of ABCD
      >are collinear. In fact, the circumcenter to KLMN lies exactly halfway
      >between the incenter of KLMN and the circumcenter of ABCD. Perhaps this
      >property can be used to give an alternative proof of Krafft's (?)
      >formula for the relation between the radii of incenter and excenter of
      >a bicentric quadrilateral and the distance between the centers of the
      You are right. This proof of Ponsele theorem for n=4 is typed in Sharygin's
      book "Problems of planymetry".

      Sincerely Alexey
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