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I cam across a question (in my textbook) about proofs with parallel lines. The question is: Prove that the property that || is transitive implies that for any point P and line l, there is at the most 1 line through P that is parallel to the line l.

In other words the question is asking me to prove that (P is on q & P is on s & l||q & l||s) --> q = s, using the transitive property.

I know that the transitivity property tells me that l||m & m||n --> n||l, but I am not sure how to do this proof.

Any hints or solutions would be much appreciated!! :)

Thank you in advance

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  • $\begingroup$ The notation in your second paragraph is messed up. On the left side, you have q,s,n but on the right side you have m and n. Once you correct this, you should be able to see the solution. $\endgroup$ – Ted Sep 27 '13 at 6:49
  • $\begingroup$ @Ted After correcting it, I can see that q || s, but I can't think of the next step $\endgroup$ – Michael Ferashire Silva Sep 27 '13 at 7:01
  • $\begingroup$ q||s, but they both go through P, so ... $\endgroup$ – Ted Sep 27 '13 at 16:38
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Let's say it is not true.

take 2 lines a and b with p on them, both parallel to l. than a||l and b||l, because of symmetry of ||, l||b. because of transitivity a||b. a||b and p on both a and b, a=b.

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You want to show that the transitivity of parallelism is equivalent to the Playfair's Axiom. It's explained perfectly here:

http://www.mathteacherctk.com/blog/2011/10/is-parallelism-an-equivalence-relation/

Michael

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