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Prove rhombus

  1. Apr 2, 2005 #1
    Hello

    How do you prove that the diagonals od a rhombus bisect each other??

    Thanks
     
  2. jcsd
  3. Apr 2, 2005 #2
    when u prove that they make right angles between them.
    then, you will get the answer.
    180/2 = 90!
     
  4. Apr 2, 2005 #3
    What I can figure out is that all sides are equal and the oposite corners angles = each other. Not sure what to do
     
  5. Apr 2, 2005 #4
    it is the theory which tells you that in a rombus the diagonals are perpendicular
    so far they bisect each other
    is it clear?
     
  6. Apr 2, 2005 #5

    Curious3141

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    Homework Helper

    You can actually prove a stronger result easily using vectors. The diagonals of a parallelogram bisect one another. The rhombus is just a special case of a parallelogram with all sides being equal.

    Let the parallelogram be drawn on a Cartesian plane as shown in the diagram, and the sides labelled as vectors as shown. You can see that one diagonal is [tex]\vec a + \vec b[/tex] and the other is [tex]\vec b - \vec a[/tex]

    Let [tex]\vec{WO}[/tex] be [tex]k_1(\vec a + \vec b)[/tex]

    and

    [tex]\vec{OZ}[/tex] be [tex]k_2(\vec b - \vec a)[/tex]

    where [tex]k_1, k_2[/tex] are some scalars (which we are to determine).

    In triangle WOZ, you can further see that

    [tex]\vec{WZ} = \vec{WO} + \vec{OZ}[/tex]

    hence

    [tex]\vec b = k_1(\vec a + \vec b) + k_2(\vec b - \vec a)[/tex]

    [tex]\vec b = (k_1 - k_2)\vec a + (k_1 + k_2)\vec b[/tex]

    giving [tex]k_1 - k_2 = 0[/tex] ---eqn (1)

    and [tex]k_1 + k_2 = 1[/tex] ---eqn(2)

    Solving those simple simultaneous equations yields [tex]k_1 = k_2 = \frac{1}{2}[/tex]

    so you know that the diagonals bisect each other. (QED)
     

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    Last edited: Apr 2, 2005
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