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Re: {Disarmed} [EMHL] Re: A condition for a quadrilateral to be orthodiagonal

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  • Eisso J. Atzema
    Dear Cosmin, I like your argument too, although I would like to observe that your Lemma 1 (or at least the part that you need for your final statement) is
    Message 1 of 6 , Jul 1, 2009
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      Dear Cosmin,

      I like your argument too, although I would like to observe that your
      Lemma 1 (or at least the part that you need for your final statement) is
      again more easily phrased in terms of the isoptic cubic. Indeed, the
      locus of all points P in the plane of a quadrilateral ABCD such that
      its (perpendicular) projections onto the sides of ABCD are co-cyclic is
      again the isoptic curve. This actually follows from some simple ``angle
      chasing''. Consequently, by the same argument as in in my first e-mail,
      the projections of the diagonal point AC \cap BD of a quadrilateral
      ABCD onto its sides can be co-cyclic iff ABCD is orthodiagonal.

      Eisso

      Cosmin Pohoata wrote:
      > Dear Eisso,
      >
      > Your idea is great. As I don't usually work with cubics, it is always
      > a joy for me to see their appearance in such rather easy configurations.
      >
      > Here is my proof of:
      >
      > > > Problem: Let ABCD be a convex quadrilateral and let P = AC /\ BD, E =
      > > > AB /\ CD and F = AD /\BC. Prove that AC \perp BC if and only if
      > > > isog_{ABF}(P)=isog_{CDF}(P)=isog_{ADE}(P)=isog_{BCE}(P).
      >
      > We use the following easy lemma:
      > Lemma 1(some 8-pt circle theorem): Let ABCD be a quadrilateral, and
      > let P be the common point of its diagonals. Denote by U1, V1, W1, T1
      > the projections of P on AB, BC, CD, DA and by U2, V2, W2, T2 the
      > intersections of PU1, PV1, PW1, PT1 with CD, DA, AB, BC, respectively.
      > Then, ABCD is orthodiagonal if and only if U1, V1, W1, T1, U2, V2, W2,
      > T2 are all concyclic.
      >
      > This can be proved with a quick angle-chase (if I remember correctly):
      >
      > Now we also have this:
      >
      > Lemma 2(well-known). If ABC is a triangle and P, Q two (distinct)
      > points in plane. P and Q are isogonal if and only if their pedal
      > circles coincide.
      >
      > Combining these two we get our result.
      >
      > Regards,
      > Cosmin
      >
      ========================================
      Eisso J. Atzema, Ph.D.
      Department of Mathematics & Statistics
      University of Maine
      Orono, ME 04469
      Tel.: (207) 581-3928 (office)
      (207) 866-3871 (home)
      Fax.: (207) 581-3902
      E-mail: atzema@...
      ========================================
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