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

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anonymous
  • anonymous
anonymous
  • anonymous
anonymous
  • anonymous
I've been sitting herre for 30 minutes waiting for an answer so ...
anonymous
  • anonymous
You're proof is correct. They have to bisect
anonymous
  • 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
  • Loser66
Hey girl!! although you said that, I still post my opinion. Read or not, It's up to you. :)
Loser66
  • Loser66
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Loser66
  • 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
  • Loser66
Do the same with OB and OD then conclude that the diagonals are bisect.

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