Transitivity of parallel lines

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August 6th 2006, 09:21 AM
OReilly

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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?
August 6th 2006, 12:35 PM
Quick

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.
August 6th 2006, 12:50 PM
OReilly

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 $a\parallel c$
August 6th 2006, 01:01 PM
Quick

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:

Quote:

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 $a\parallel c$
August 6th 2006, 01:21 PM
OReilly

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.