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Math Help - Transitivity of parallel lines

  1. #1
    Senior Member OReilly's Avatar
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    Transitivity of parallel lines

    I need to prove that a\parallel b \wedge b\parallel c \Rightarrow a\parallel c if a,b,c are not in the same plane.

    See attachment.

    If we put point B in plane \alpha then we get line d that is parallel to line b. Also if we put point C then e is parallel to c.

    Now we must prove that a\parallel d \wedge d\parallel e \Rightarrow a\parallel e.

    If e is not parallel to a then a and e must intersect but then we would have two lines that are parallel to d which is contradictory so its must be a\parallel c.

    Is that proof ok?
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  2. #2
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by OReilly
    Is that proof ok?
    To be frank, no. let me rephrase your proof and you might see why not:
    a\parallel e \wedge e\parallel c \Rightarrow a\parallel c

    heres an idea, draw a line from a to b so that all angles formed are ninety degrees, do the same for b and c, but starting the line at the point where the other line touches b. now connect the points that form on a and c and you'll find that all angles created are ninety degrees, proving a and c are parrallel.
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  3. #3
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Quick
    To be frank, no. let me rephrase your proof and you might see why not:
    a\parallel e \wedge e\parallel c \Rightarrow a\parallel c

    heres an idea, draw a line from a to b so that all angles formed are ninety degrees, do the same for b and c, but starting the line at the point where the other line touches b. now connect the points that form on a and c and you'll find that all angles created are ninety degrees, proving a and c are parrallel.

    No, I need to prove without using angles, just using this axioms and theorems (last is theorem):
    1) If two different planes have one common point, then they have one common line.
    2) Every line a and every point B which is out of line a has only one line that goes through point B and is parallel to line a (axiom of parallelness)
    3) Line and point out of line forms one plane.


    Now let me explain a\parallel e \wedge e\parallel c \Rightarrow a\parallel c.

    Since a and e are in same plane and c is not, lets put point A into line a. Now line c and line a forms one plane. If c isn't parallel to line a then they intersect so line e would have two parallel lines a and c through point A which is not allowed by axiom of parallelness so it must be a\parallel c
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  4. #4
    MHF Contributor Quick's Avatar
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    Quote Originally Posted by OReilly
    Since a and e are in same plane and c is not, lets put point A into line a. Now line c and line a forms one plane. If c isn't parallel to line a then they intersect so line e would have two parallel lines a and c through point A which is not allowed by axiom of parallelness so it must be a\parallel c
    alright, this looks fine, I would recommend using line b like you used line e, in other words, rephrase your post like this:

    Since a and b are in same plane and c is not, lets put point A into line a. Now line c and point A forms one plane. If c isn't parallel to line a then they intersect so line b would have two parallel lines, a and c, intersect which is not allowed by axiom of parallelness so it must be a\parallel c
    Last edited by Quick; August 6th 2006 at 01:13 PM.
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  5. #5
    Senior Member OReilly's Avatar
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    Quote Originally Posted by Quick
    alright, this looks fine, I would recommend using line b like you used line e, in other words, rephrase your post.
    Yes, I could't do that. Actually, I don't need line d and e at all. That was my first conception. I can form plane with line b and point A on line a and then form another plane with line c and point A and prove it as I did.
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