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Geometry - Help with theorem proof please

  1. Feb 11, 2014 #1
    Geometry -- Help with theorem proof please

    1. The problem statement, all variables and given/known data

    Let ##A,B,C,D## be points. If ##\vec{AB} = \vec{CD}## then ##A=C##.

    2. Relevant equations

    None

    3. The attempt at a solution

    This question was a theorem in my book that wasn't proved. I am wondering how to prove it?

    It is saying that the vertex ##A## must equal ##C## if the ray ##\vec{AB} = \vec{CD}##.

    The definition I have for ray is:

    ##\vec{AB} = \vec{AB} \cup \{ C \in P \ | \ A-B-C\}.## Where ##A-B-C## means ##B## is between ##A## and ##C##. And ##P## is the set of points.
     
    Last edited: Feb 12, 2014
  2. jcsd
  3. Feb 12, 2014 #2

    tiny-tim

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    Hi Lee33! :smile:

    (your definition doesn't look quite correct)

    Suppose A ≠ C

    A is in ##\vec{CD}##, so … ? :wink:
     
  4. Feb 13, 2014 #3
    tiny-tim - Can you elaborate a bit more please?
     
  5. Feb 14, 2014 #4

    tiny-tim

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    Hi Lee33! :smile:

    Apply the definition you were given …
    Suppose A ≠ C

    C is in ##\vec{AB}##, so what can you say about A B and C ? :wink:
     
  6. Feb 14, 2014 #5
    If C is in ##\vec{AB}## and ##C\ne A## then B is between A and C?
     
  7. Feb 15, 2014 #6

    Mark44

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    This definition makes no sense to me. First off, why would ##\vec{AB}## be equal to itself union some other thing (unless the other thing happened to be the empty set).

    Second, how do you interpret ##\{ C \in P \ | \ A-B-C\}##? Does | have its usual meaning of "such that" or am I missing something? An explanation, in words, would be helpful.

    Third, where are these points? Are they on a line or are they in the plane?

    Fourth, how do you get that A - B - C means that B is between A and C?
     
  8. Feb 15, 2014 #7
    Sorry, I will elaborate.

    First question: If A and B are distinct points in a metric geometry then the line segment from A to B is the set ##\vec{AB}=\{C \in P \ | \ A-C-B \ or \ C = A \ or \ C = B\}##.

    If A and B are distinct points in a metric geometry then the ray from A toward B is the set ##\vec{AB}=\vec{AB}\cup \{C\in P \ | \ A-B-C\}.##

    Second question: Yes, it means such that. Let P be the set of points in a metric geometry, and let C be a point in P such that B is between A and C.

    Third question: They are on a line.

    Fourth: That is just a notation for convenience. ##A-B-C## just means B is between A and C.

    I will add the definition of between-ness: B is between A and C if the distance ##d(A,B)+d(B,C) = d(A,C)##.
     
  9. Feb 15, 2014 #8

    tiny-tim

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    Hi Lee33! :smile:

    (just got up :zzz:)
    nooo, C is (strictly) between A and B :wink:

    ok, and if A is in ##\vec{CD}##, then … ? :smile:
     
  10. Feb 15, 2014 #9
    If A is in ##\vec{CD}## then A is between C and D.
     
  11. Feb 15, 2014 #10

    tiny-tim

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    yes (strictly between) :smile:

    ok, now you have two statements, and you should be able to prove a contradiction (thereby showing that "A ≠ C" was false) :wink:

    (drawing yourself a diagram might help)
     
  12. Feb 15, 2014 #11
    Alright thanks for the help! I will use your hints.

    Question. Do I use both statements in my proof? That is, suppose ##A\ne C## and A is in ##\vec{CD}## then A is bewteen C and D. Also, I will use if ##A\ne C## and C is in ##\vec{AB}## then C is between A and B?
     
  13. Feb 15, 2014 #12

    tiny-tim

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    yes :smile:
     
  14. Feb 15, 2014 #13
    You are using the same notation for line segment and ray, and it's confusing the bejeesus out of the people who are trying to help you.

    Might I suggest ##\overline{AB}## for the segment and ##\overrightarrow{AB}## for the ray so that ##\overrightarrow{AB}=\overline{AB}\cup \{C\in P \ | \ A-B-C\}.##
     
  15. Feb 15, 2014 #14

    tiny-tim

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    it's not confusing me :smile:
     
  16. Feb 15, 2014 #15
    Are you sure? :confused: Like, really sure? :cool: Because when Lee asked

    you replied

    which is generally false :mad: regardless of which of Lee's two definitions of ##\vec{AB}## you're using. :tongue:
     
  17. Feb 15, 2014 #16

    tiny-tim

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    well, Lee33 :smile: didn't contradict me, sooo i assume i got it right! o:)
     
  18. Feb 17, 2014 #17
    gopher_p - Sorry about that, you're right!

    tiny-tim - If ##A\ne C## then ##C\in \vec{AB}## thus ##A-C-B## but where will the point ##D## be?
     
  19. Feb 18, 2014 #18

    tiny-tim

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    but you haven't used …
     
  20. Feb 18, 2014 #19
    So my proof should go like this:

    Suppose ##A\ne C##, now since ##\vec{AB}=\vec{CD}## then ##A\in \vec{CD}## and ##C\in \vec{AB}##. Thus ##C-A-D## and ##A-C-B## which is a contradiction?
     
  21. Feb 18, 2014 #20

    tiny-tim

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    yes!!

    if i'm understanding the terminology correctly, you can't have both ##A-C## and ##C-A## unless C = A :smile:
     
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