Can someone check if my indirect proof that the diagonals of a parallelogram bisect one another is correct?

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- anonymous

Let's assume that the diagonals of a parallelogram do not bisect one another. A parallelogram has four sides and it's opposite angles and opposite sides are congruent and parallel. If the diagonals of a parallelogram intersect each other, and a parallelogram's opposite angles and opposite sides are congruent and parallel, then the intersecting diagonals must also bisect one another.

- anonymous

- anonymous

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- anonymous

- anonymous

- anonymous

I've been sitting herre for 30 minutes waiting for an answer so ...

- anonymous

You're proof is correct. They have to bisect

- anonymous

I rather not use someone else's work and claim it as my own. I just want to know if this is correct and if not, then if I could get a few tips and hints that can help me.

- Loser66

Hey girl!! although you said that, I still post my opinion. Read or not, It's up to you. :)

- Loser66

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- Loser66

Let ABCD is a parallelogram. By definition, ABCD is a non-self intersect quadrilateral, hence its diagonals intersect.
Let O is the intersection point of the diagonals. We need prove OA = OC and OB= OD to show that the diagonals are bisect.
Prove OA =OC
consider \(\triangle AOD ~~and ~~\triangle BOC\)
we have \(\angle O_1 =\angle O_2\) (vertical opposite angles)
\(\angle A_1=\angle C_1\) (alternate exterior angles)
\(\angle B_1=\angle D_1\) (alternate exterior angles)
Hence \(\triangle AOD = \triangle BOC\) (AAA case)
\(\implies OA = OC \)

- Loser66

Do the same with OB and OD
then conclude that the diagonals are bisect.

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