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Perpendicular bisectors of the sides in a quadrangle

  1. Jun 12, 2005 #1
    1. What reqiurements does a quadrangle need that all the perpendicular bisectors of the sides crosses in one point??

    2. And how can you proof it?

    To 1. - I think each angle have to be 90° but I can't proof it :blushing:
  2. jcsd
  3. Jun 12, 2005 #2


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    Off the top of my head, I couldn't tell you the requirements, but all angles equal to 90 definitely isn't one. The perpendicular bisectors of a trapezoid will meet in one place if both angles on the base are equal and both angles on the top are equal.

    Edit: Once I think about it, the requirements are pretty obvious (in fact, I think I did know this off the top of my head at one point). Start from the intersection, drawing the perpendicular bisectors for two of the sides. However long your sides have to extend in one direction in order to form an intersection, they have to extend in the opposite direction as well. Then start working on the last two sides.

    The relationship between the angles is pretty straight forward - A rectangle with 4 90 degree angles meet them, but it's not the only quadrangle to meet them. (That should limit the possible relationships down to a pretty small number).
    Last edited: Jun 12, 2005
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