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anonymous

  • one year ago

Proofs..

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  1. anonymous
    • one year ago
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    @jim_thompson5910 mind helping with one more? :/

  2. jim_thompson5910
    • one year ago
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    what's your question?

  3. anonymous
    • one year ago
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    oops wrong one

  4. anonymous
    • one year ago
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    1 Attachment
  5. jim_thompson5910
    • one year ago
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    what's the reasoning for line 2? any idea?

  6. anonymous
    • one year ago
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    No. I mean they are talking about one angle being equal to itself..

  7. jim_thompson5910
    • one year ago
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    which property makes that true?

  8. anonymous
    • one year ago
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    Angle of a triangle? :l I'm clueless

  9. jim_thompson5910
    • one year ago
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    when you look into a mirror, you see your ______

  10. anonymous
    • one year ago
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    REFLECTION

  11. anonymous
    • one year ago
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    This is why I like your help. you make in understandable.

  12. jim_thompson5910
    • one year ago
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    yep REFLEction so the REFLExive property is why A = A is true. it's trivial and seems kinda stupid (of course something is equal to itself, how could it not?) but at the same time it's good to have a rigorous set of rules

  13. jim_thompson5910
    • one year ago
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    see the first line http://www.regentsprep.org/regents/math/geometry/gpb/theorems.htm

  14. anonymous
    • one year ago
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    I feel like you have a folder of math websites. You have so many helpful ones o.o

  15. jim_thompson5910
    • one year ago
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    sometimes, but others I google

  16. jim_thompson5910
    • one year ago
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    and regents prep tends to pop up a lot (esp with geometry)

  17. anonymous
    • one year ago
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    ahh.

  18. anonymous
    • one year ago
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    So what about the last line?

  19. anonymous
    • one year ago
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    Corresponding sides?

  20. jim_thompson5910
    • one year ago
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    I'm checking your line 3 and line 4

  21. anonymous
    • one year ago
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    alright

  22. jim_thompson5910
    • one year ago
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    from this link we visited earlier http://www.regentsprep.org/regents/math/geometry/gp11/LsimilarProof.htm I'm going to focus on the SAS similarity theorem. See attached

    1 Attachment
  23. jim_thompson5910
    • one year ago
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    That theorem says if you have a pair of corresponding congruent angles, and you have that proportion mentioned in the attachment, then the triangles are similar

  24. anonymous
    • one year ago
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    okay

  25. anonymous
    • one year ago
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    was one of the lines wrong or are we on the last line?

  26. jim_thompson5910
    • one year ago
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    so that's the reason for line 3.

  27. jim_thompson5910
    • one year ago
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    we only use the AA theorem IF we had 2 congruent corresponding angles. We had that last time, but we don't have that this time

  28. anonymous
    • one year ago
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    oh..

  29. anonymous
    • one year ago
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    so would it be sas?

  30. jim_thompson5910
    • one year ago
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    SAS similarity theorem, yes

  31. anonymous
    • one year ago
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    is the fourth line okay?

  32. jim_thompson5910
    • one year ago
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    no, but luckily you might know the theorem

  33. jim_thompson5910
    • one year ago
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    you mentioned the AA theorem. What exactly does the AA theorem say?

  34. anonymous
    • one year ago
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    AA? xD

  35. anonymous
    • one year ago
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    Angle Angle

  36. jim_thompson5910
    • one year ago
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    specifically what does the entire theorem say? (other than just Angle Angle)

  37. anonymous
    • one year ago
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    To show two triangles are similar, it is sufficient to show that two angles of one triangle are congruent (equal) to two angles of the other triangle. Theorem: If two angles of one triangle are congruent to two angles of another triangle, the triangles are similar.

  38. jim_thompson5910
    • one year ago
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    So IF the angles are congruent THEN the triangles are similar flip that around to say... IF the triangles are similar THEN the angles are congruent

  39. jim_thompson5910
    • one year ago
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    the theorem you wrote out is the original AA similarity theorem the flipped version is the converse of that said theorem

  40. anonymous
    • one year ago
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    Alright.. so would it be a converse AA theorem?

  41. jim_thompson5910
    • one year ago
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    converse of the AA similarity theorem, yep

  42. anonymous
    • one year ago
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    And then the last line

  43. jim_thompson5910
    • one year ago
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    any ideas?

  44. anonymous
    • one year ago
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    Parallel lines?

  45. jim_thompson5910
    • one year ago
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    well that's what you want to prove

  46. jim_thompson5910
    • one year ago
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    how can you use the previous line?

  47. jim_thompson5910
    • one year ago
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    https://www.mathsisfun.com/geometry/parallel-lines.html

  48. anonymous
    • one year ago
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    I dont understand how to use the line before

  49. anonymous
    • one year ago
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    Traversal something?

  50. jim_thompson5910
    • one year ago
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    this might be of better help http://www.nhvweb.net/nhhs/math/mschuetz/files/2012/11/Section-3-3-2012-2013.pdf

  51. anonymous
    • one year ago
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    Which one :/

  52. jim_thompson5910
    • one year ago
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    first page

  53. anonymous
    • one year ago
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    Corresponding angles?

  54. jim_thompson5910
    • one year ago
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    the converse of the corresponding angles theorem

  55. anonymous
    • one year ago
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    Once again thank you!

  56. jim_thompson5910
    • one year ago
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    np

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