Incidence Axiom 1: for every point P and for every point Q not equal to P there exists a unique line that passes through P and Q.
Incidence Axiom 2: for every line there are at least two distinct points on the line.
Incidence Axiom 3: there exist three distinct points not all on the same line.
Well, they obvious do have at least one point in common. If they had no points in common, they'd be parallel. So it should suffice to show that if the two lines share two distinct points, they are not distinct lines. And obviously if they share more than 2 points in common, they still share two in common and that means they cannot be distinct lines. So by contradiction they cannot have 2 or more in common and by hypothesis they do not have none in common. Pretty sure that leaves only one.
I disagree. Why is it obvious? Do you know the exact axioms the post is using?
That is why I asked in my first. In axiomatic geometry one cannot assume common understandings.
Now I grant you it is likely the case the parallel lines are probability defined as not having any point in common.
Now if that is indeed the case, a proper proof of this statement cannot leave that bit out of any proof.
Really I think that unless the post includes all relevant definitions and axioms, it is hard to give adequate answer.
That is exactly how parallel is defined in incidence geometry; that the lines share no points in common. I wasn't giving a full proof either, that is the key essence to the proof though, since the parallel part is just by definition; there's nothing to actually prove there.