Non-equivalent phrasings of Playfair's Axiom which are in use

Question

For example on ProofWiki Playfair's Axiom is given as

Exactly one straight line can be drawn through any point not on a given line parallel to the given straight line in a plane.

but for example Wikipedia give it as

In a plane, given a line and a point not on it, at most one line parallel to the given line can be drawn through the point.

but both are not equivalent, which an be seen for example the incidence geometry consisting of three points $P, Q, R$ and the lines $\{ P, Q \}, \{ Q, R \}$ and $\{ P, R \}$. It fullfils the last, but not the first. So what is the real parallel postulate, and as they are both not equivalent they can not be equivalent to all the other forms, for example that being parallel is transitive and so on?

Answer 1

Both axioms you wrote are equivalent if you use the whole set of Hilbert's absolute geometry axioms.

It can be proved that at least one parallel line can be drawn through a point not lying on the line from absolute geometry axioms without any equivalent of Euclid's fifth postulate.

Lemma. An exterior angle of a triangle is greater than interior angle not adjacent to it.

With this lemma you can prove that if alternate angles are equal then the lines are parallel.

To prove that you can draw a parallel line through a point you lay off an alternate angle (you can do this by one of the axioms)

Answer 2

The notion of equivalence depends on the theory and the logic.

The two versions of Playfair's postulate you are giving are not equivalent in absolute geometry (a geometry without any assumption about existence and uniqueness of parallels) but are equivalent in neutral geometry (a geometry where parallel always exist but are not necessarily unique).

In the context of Tarski's neutral geometry, and without any continuity axiom, the uniqueness of parallel can be shown to be equivalent to the transitivity of parallelism.