# Plane geometry

#### dirk_mec1

1. The problem statement, all variables and given/known data

A parallellogram ABCD has an interior point O sucht that $$\alpha + \beta = 180^o$$

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

Prove that:

$$\angle{OBC}=\angle{CDO}$$

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

Homework Helper
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? :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

#### tiny-tim

Homework Helper
Do you mean box or parallellogram? :tongue:
Both!
Sum of the angles in a triangle is 180o.
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

Homework Helper
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

Homework Helper
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

Homework Helper
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...

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