# Plane geometry

1. Feb 27, 2009

### 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: May 4, 2017
2. Feb 27, 2009

### 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

3. Feb 27, 2009

### dirk_mec1

Hi!

Do you mean box or parallellogram? :tongue:

Sum of the angles in a triangle is 180o.

4. Feb 27, 2009

### tiny-tim

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

Hint: circles are very un-boxlike …

5. Feb 28, 2009

### dirk_mec1

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

6. Feb 28, 2009

### tiny-tim

That's the one!

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

7. Mar 1, 2009

### dirk_mec1

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.

8. Mar 1, 2009

### tiny-tim

move OAB up to the top

9. Mar 1, 2009

### dirk_mec1

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

10. Mar 1, 2009

### tiny-tim

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

11. Mar 2, 2009

### 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...