What does it mean for one geometrical axiom to be considered equivalent to another geometrical axiom?
For example consider Playfair`s axiom:
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.
This is always described as being equivalent to Euclid's parallel postulate (the 5th postulate) which states:
If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.
However in Euclidean geometry two points are required to define a line. Therefore I would argue Playfair's axiom is not logically equivalent to Euclid's fifth postulate since it posits the existence of lines through a single point. Playfair's axiom is often characterized as a more streamlined version of Euclid's postulate but I suspect this is because it tacitly uses ideas that are not present in Euclidean geometry.
First, there is an implicit presumption that one is working with a background set of axioms that I'll denote $\mathcal S$. In your question, $\mathcal S$ is the set of axioms of what I believe is called "neutral geometry", which means Euclid's axioms (suitably formalized) without the fifth postulate. In particular, $\mathcal S$ includes the axiom that two points define a line, and a bunch more axioms.
Consider now two further axioms under consideration, which I'll denote $A_1$ and $A_2$. In your question $A_1$ is Euclid's fifth postulate and $A_2$ is Playfair's axiom.
To say that $A_1$ and $A_2$ are logically equivalent (under the implicit presumption that $\mathcal S$ holds), means: $$\mathcal S \implies (A_1 \iff A_2) $$ This is tautologically equivalent to the following, which is how I would expect the equivalence of the $5^{\text{th}}$ postulate and Playfair's axiom to be actually proved: $$\bigl((\mathcal S \quad\text{and}\quad A_1) \implies A_2 \bigr) \quad\text{and}\quad \bigl( (\mathcal S \quad\text{and}\quad A_2) \implies A_1 \bigr) $$