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Math Help - Proof: Confusing geometry problem

  1. #1
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    Proof: Confusing geometry problem

    I really have no idea how to prove this

    In quadrialteral ABCD the point of intersection of the perpendicular bisectors of AD and BC lies on AB. Prove that AC = BD if and only if <A = <B
    if you didn't realize I meant to have angle symbols but i didn't know how to make them
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Your jsut tryign to

    prove that the quadrilateral is a parallelogram...so that the diagonals which are already perpendicular will make a square proving that the diagonals are congruent...I think thats what it wants?
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  3. #3
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    Hello,

    If ABCD is a parallelogram, how is it possible to have the intersection on AB ? Since AD and BC are opposit, they may be parallel if it's a parallelogram. If they're parallel, their perpendicular bisectors will be parallel, and the point of intersection won't exist...
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Yeah

    once again...for the like fourth time...I didnt read the question...I'm sorry I will be more studious in the future...I can do these things...I am just a little...uh...uncareful =D
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  5. #5
    Moo
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    I can understand that, i'm too often like this too

    Now, for the problem, i admit it seems tricky oO
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    I think

    this might be a tric question...because if you draw it out it seems like an impossible scenario
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  7. #7
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    Its a trapezoid (Im pretty sure)

    Yea well the idea is to prove that its an isosceles trapezoid which is the only way AC = BD and <A has to equal <B

    but I have no idea how to go about that
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  8. #8
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    Well, i've got something approaching : AC is not exactly equal to BD, but it's possible
    Attached Files Attached Files
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  9. #9
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    Yea even when I assume AC = BD or <A = <B I just don't know what I can prove.
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