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Area of a quadrilateral II

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  • brozolo
    Once I ve known the right formula (Blaschke s formula) from Sergei Markelov, my proof has become much easier. I send it here hoping it could be of some
    Message 1 of 1 , Sep 30, 2002
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      Once I've known the right formula (Blaschke's formula) from Sergei
      Markelov, my proof has become much easier.
      I send it here hoping it could be of some interest for someone.
      This is the proof in the general case, non only for cyclic

      Given quadrilateral ABCD, construct quadrilateral SPQR drawing
      perpendiculars in the vertexes to their own bisectors (SP is
      perpendicular in A to the A-bisector, PQ is perpendicular in B to the
      With some angle considerations SPQR results to be cyclic.
      All quadrilaterals inscribed in SPQR having their sides parallel to
      those of ABCD have the same perimeter of ABCD (this can be seen with
      trigonometric considerations or in I. M. Yaglom's Geometric
      Transformations II, solution to problem n. 76).
      It is clear that the distance between the parallel sides of any two
      quadrilaterals A'B'C'D' and A''B''C''D'' so constructed is the same:
      I mean that the distance between A'B' and A''B'' is equal to the
      distance between B'C' and B''C'', …( I'm indebted with El Filibustero
      from it.scienza.matematica for this observation).
      In particular there is one quadrilateral of this family, let's call
      it A'B'C'D', that is circumscriptible (it can be constructed joining
      the projections from the point of intersection of the diagonals of
      SPQR to its sides; I learned this from P. Yiu's Euclidean Geometry
      notes, pg. 165) .
      Let's call r the radius of its incircle, O its center and h the
      common distance between sides of A'B'C'D' and those of ABCD.
      Let's suppose that, for example, BC+DA>=AB+CD (i.e. sides BC and DA
      are the sides of ABCD that intersect the sides of A'B'C'D'):

      A'B'=AB+ h tan(A/2)+h tan(B/2)
      B'C'=BC - h tan(B/2)-h tan(C/2)
      C'D'=CD + h tan(C/2)+h tan(D/2)
      D'A'=DC - h tan(D/2) –h tan(A/2)



      So we can calculate h:

      2h=(-AB+BC-CD+DA)/(tan (A/2)+ tan (C/2)+ tan (C/2)+ tan (D/2))

      moreover, with easy trigonometric considerations:

      2r=(A'B'+B'C'+C'D'+D'A')/(cot (A/2)+ cot (C/2)+ cot (C/2)+ cot (D/2))


      A'B'+B'C'+C'D'+D'A'= AB+BC+CD+DA


      2r=(AB+BC+CD+DA)/(cot (A/2)+ cot (C/2)+ cot (C/2)+ cot (D/2))

      The area of ABCD can


      after substitution of h and r one obtains the sought formula.
      I hope this all is correct.

      Best regards
      Marcello Tarquini
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