Parallel axiom. If a line $n$ falling on lines $l$ and $m$ makes angles $\alpha$ and $\beta$ with $\alpha+\beta$ less than two right angles, then $l$ and $m$ meet on the side on which $\alpha$ and $\beta$ occur.
It follows $(*)$ that if $\alpha+\beta$ equals two right angles (that is, a straight angle) then $l$ and $m$ do not meet. Because if they meet on one side (forming a triangle), they must meet on the other (forming a congruent triangle, by ASA), since there are angles $\alpha$ and $\beta$ on both sies and one side in common. This contradicts uniqueness of the line through any two points.
I don't follow how the parallel axiom implies $(*)$. In particular, it is not clear to me how/when the argument presented for the statement "$\alpha+\beta=180°\Rightarrow l,m\text{ do not meet}$" makes use of the parallel axiom (relatedly, nor do I see the relevance of the similarity of the formed triangles).
What is wrong with simply stating "suppose $\alpha+\beta=180°$ are the angles on one side of $n$; if $l,m$ meet at the side of $\alpha,\beta$, then $(\dagger)$ $l,m$ must also meet at the opposite side (because here $n$ makes angles of $\beta$ and $\alpha$ with $l$ and $m$ respectively), thus uniqueness of lines through pairs of points is contradicted"?
The only thing I can see myself to be assuming here, is at the point where I make the statement $(\dagger)$, where I assume that any two adjacent angles formed by intersecting lines must be supplementary.
Best Answer
You do not need the parallel axiom to show the statement that if $\alpha+\beta=180^\circ$ then the two lines do not meet. It's a theorem of so-called neutral geometry (geometry with the usual axioms except the parallel axiom), and the proof is as given in your post, and as you say they do not use the parallel axiom in that proof.