Plane geometry

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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:

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:
 
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Hi dirk_mec1! :smile:
Hi!

Think outside the box …
Do you mean box or parallellogram? :tongue:

Hint: what theorem do you know (nothing to do with parallelograms) about two triangles with angles adding to 180º?
Sum of the angles in a triangle is 180o.

move one of the triangles around :wink:
I've thought about this but I don't understand what you mean.
 

tiny-tim

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Do you mean: ''in a cyclic quadrilateral, opposite angles are supplementary (their sum is π radians)''?
 

tiny-tim

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Do you mean: ''in a cyclic quadrilateral, opposite angles are supplementary (their sum is π radians)''?
That's the one! :biggrin:

Now shift one of the triangles around so as to make that cyclic quadrilateral, and then draw a … ? :wink:
 
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Now shift one of the triangles around so as to make that cyclic quadrilateral, and then draw a … ? :wink:
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.
 

tiny-tim

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move OAB up to the top :wink:
Actually what do you mean by "shifting a triangle"? You can't switch angles so you probably mean something else.
 

tiny-tim

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make a copy of OAB and and put it at the top
 
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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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