Proving that Angles OBC and CDO are Equal in a Parallelogram

[dirk_mec1]

Homework Statement

A parallellogram ABCD has an interior point O sucht that <br /> \alpha + \beta = 180^o <br />

http://img413.imageshack.us/img413/5636/post102741235319763.png

Prove that:

<br /> \angle{OBC}=\angle{CDO}<br />

Homework Equations

Definitions of a parallellogram.

The Attempt at a Solution

I don't know how to start can some give me a hint?

 

[tiny-tim]

Hi dirk_mec1!

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

 

[dirk_mec1]

Hi dirk_mec1!

Hi!

Think outside the box …

Do you mean box or parallellogram?

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 180^o.

move one of the triangles around

I've thought about this but I don't understand what you mean.

 

[tiny-tim]

Do you mean box or parallellogram?

Both!

Sum of the angles in a triangle is 180^o.

That's sum of three angles … how about sum of two angles?

Hint: circles are very un-boxlike … ​

 

[dirk_mec1]

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

 

[tiny-tim]

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

That's the one!

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

 

[dirk_mec1]

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

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]

I'm sorry Tim I've looked at it and I can't find opposite angles for which the sum is 180 deg.

move OAB up to the top

 

[dirk_mec1]

move OAB up to the top

Actually what do you mean by "shifting a triangle"? You can't switch angles so you probably mean something else.

 

[tiny-tim]

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

 

[dirk_mec1]

Ok, I did that and thus:

(\angle{OAB} + \angle{CDO}) +(\angle{OBA} + \angle{DCO}) =180^o

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