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Plane geometry

  1. Feb 27, 2009 #1
    1. The problem statement, all variables and given/known data

    A parallellogram ABCD has an interior point O sucht that [tex]
    \alpha + \beta = 180^o
    [/tex]

    http://img413.imageshack.us/img413/5636/post102741235319763.png [Broken]

    Prove that:

    [tex]
    \angle{OBC}=\angle{CDO}
    [/tex]

    2. Relevant equations

    Definitions of a parallellogram.

    3. The attempt at a solution
    I don't know how to start can some give me a hint?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 27, 2009 #2

    tiny-tim

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    Hi dirk_mec1! :smile:

    Think outside the box …

    Hint: what theorem do you know (nothing to do with parallelograms) about two triangles with angles adding to 180º?

    move one of the triangles around :wink:
     
  4. Feb 27, 2009 #3
    Hi!

    Do you mean box or parallellogram? :tongue:

    Sum of the angles in a triangle is 180o.

    I've thought about this but I don't understand what you mean.
     
  5. Feb 27, 2009 #4

    tiny-tim

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    Both! :biggrin:
    That's sum of three angles … how about sum of two angles?

    Hint: circles are very un-boxlike … :wink:
     
  6. Feb 28, 2009 #5
    Do you mean: ''in a cyclic quadrilateral, opposite angles are supplementary (their sum is π radians)''?
     
  7. Feb 28, 2009 #6

    tiny-tim

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    That's the one! :biggrin:

    Now shift one of the triangles around so as to make that cyclic quadrilateral, and then draw a … ? :wink:
     
  8. Mar 1, 2009 #7
    I'm sorry Tim I've looked at it and I can't find opposite angles for which the sum is 180 deg. I do know that the opposite angles in the parallelogram are equal.
     
  9. Mar 1, 2009 #8

    tiny-tim

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    move OAB up to the top :wink:
     
  10. Mar 1, 2009 #9
    Actually what do you mean by "shifting a triangle"? You can't switch angles so you probably mean something else.
     
  11. Mar 1, 2009 #10

    tiny-tim

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    make a copy of OAB and and put it at the top
     
  12. Mar 2, 2009 #11
    Ok, I did that and thus:

    [tex](\angle{OAB} + \angle{CDO}) +(\angle{OBA} + \angle{DCO}) =180^o[/tex]

    but you mention drawing something I guess it's a circle but I'm not sure...
     
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