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Math Help - analytic proof(s)

  1. #1
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    Post analytic proof(s)

    We are asked to show the analytic proof of the following:
    "The diagonals of a parallelogram are equal in length. The figure is a rectangle."

    >What I did was that I plotted a rectangle on a coordinate plane, and I used the distance formula to show that the diagonals are equal. The result was that the diagonals were indeed equal. But my question is this: "Can I already assume that what I've plotted was a rectangle (because it's obvious)? Or Should I still have to show that the figure is indeed a rectangle (through analytic proof) to be more precise?

    thanks in advance..
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  2. #2
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    Hello, mamen!

    You did it backwards.
    You should have stated the theorem clearly


    If the diagonals of a parallelogram are equal in length, then the figure is a rectangle.

    First, graph a parallelogram.

    Code:
          |   (r,h)           (b+r,h)
          |   D o - - - - - - - - o C
          |    /: *           *  /
          |   / :   *     *     /
          |  /  :h    *        /
          | /   : *     *     /
          |/  * :         *  /
        A o - - + - - - - - o B - -
       (0,0) r            (b,0)

    We have parallelogram ABCD.

    The base is: .b = AB, .b ≠ 0

    Vertex D is (r, h), where r is the horizontal displacement of D and h is the height.
    . . Then vertex C is (b+r, h).

    Then: .AC^2 .= .(b + r)^2 + h^2

    And: . BD^2 .= .(r - b)^2 + h^2


    Equate: .(b + r)^2 + h^2 .= .(r - b)^2 + h^2

    . b^2 + 2br + r^2 + h^2 .= .r^2 - 2br + b^2 + h^2

    And we have: .4br = 0 . . b = 0, r = 0


    Since b ≠ 0, we have: r = 0.
    That is, D is directly above A: ./ DAB = 90 degrees.

    Therefore, ABCD is a rectangle.
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  3. #3
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    Post analytic proof

    Dear teacher,
    This is the actual sentence that was given to us:
    "The diagonals of a parallelogram are equal length, the figure is a rectangle."

    We are asked to prove that statement.

    thanks in advance..
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