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Homework Help: Prove that ABCD is a parallelogram

  1. Sep 21, 2010 #1
    1. The problem statement, all variables and given/known data
    Let [tex]z_1[/tex], [tex]z_2[/tex], [tex]z_3[/tex] and [tex]z_4[/tex] be te position vectors of the vertices of quadrilateral AMCD. Prove that ABCD is a parallelogram if and only if [tex]z_1-z_2-z_3+z_4=0[/tex].

    2. Relevant equations

    3. The attempt at a solution
    The solution obviously uses the fact that collinear vectors of equal magnitude are equal, but I get [tex]z_1-z_2+z_3-z_4=0[/tex]. Am I missing something obvious or is it just notations issue with the vertices of the quadrilateral.
     
  2. jcsd
  3. Sep 21, 2010 #2

    HallsofIvy

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    Re: Parallelogram

    [itex]z_1- z_2[/itex] is the side from "[itex]z_2[/itex]" to "[itex]z_1[/itex]" and [itex]z_3- z_4[/itex] is the side from "[itex]z_4[/itex]" to "[itex]z_3[/itex]". Since this is a parallelogram those sides are parallel and equal in length. In other words, the vectors are equal: [itex]z_1- z_2= z_3- z_4[/itex] whence [itex]z_1- z_2- z_3+ z_4= 0[/itex]
     
  4. Sep 21, 2010 #3
    Re: Parallelogram

    That is true only if you haven't numbered the vertics clockwise, but in the form of a 'z' (if you get what I mean). In my picture, [tex]z_1[/tex] is adjacent to [tex]z_2[/tex] and [tex]z_4[/tex], and [tex]z_2[/tex] is adjacent to [tex]z_1[/tex] and [tex]z_3[/tex], so in my case the vector [tex]z_1-z_2[/tex] is opposite to the vector [tex]z_3-z_4[/tex].
     
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