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anonymous

  • one year ago

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

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  1. anonymous
    • one year ago
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    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.

  2. anonymous
    • one year ago
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    @ganeshie8 @Abhisar

  3. anonymous
    • one year ago
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    @SolomonZelman

  4. anonymous
    • one year ago
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    @dan815

  5. anonymous
    • one year ago
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    @sammixboo

  6. anonymous
    • one year ago
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    I've been sitting herre for 30 minutes waiting for an answer so ...

  7. anonymous
    • one year ago
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    You're proof is correct. They have to bisect

  8. anonymous
    • one year ago
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    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.

  9. Loser66
    • one year ago
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    Hey girl!! although you said that, I still post my opinion. Read or not, It's up to you. :)

  10. Loser66
    • one year ago
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    |dw:1434656647662:dw|

  11. Loser66
    • one year ago
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    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) \(\implies OA = OC \)

  12. Loser66
    • one year ago
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    Do the same with OB and OD then conclude that the diagonals are bisect.

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