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Finding the diagonal of an irregular quadrilateral

  1. Jul 15, 2011 #1
    1. The problem statement, all variables and given/known data
    In an irregular quadrilateral ABCD, the length of all sides are AB=a BC=b CD=c DA=d and the length of the diagonal AC is x. Angle ABC + angle ADC = 180

    prove that (ab+cd)x2=(ac+bd)(ad+bc)


    2. Relevant equations

    Cosine formula c2 = a2 + b2 - 2abcosθ

    3. The attempt at a solution
    I really have no idea how to start
     
  2. jcsd
  3. Jul 15, 2011 #2

    HallsofIvy

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    Use the cosine formula, [itex]c^2= a^2+ b^2- 2abcos(C)[/itex], with x as "c", sides AB and BC as "a" and "b", and angle ABC as angle "C". Then do the same with x as "c" again but AD and BD as "a" and "b", and angle ADC as angle "C". Of course, the "x" in both of those is the same so they can be set equal.
     
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