Is this another equivalent version of the parallel postulate?

Question

Define, for every point $x$ in Euclidean space:

• $Iso(x)$ to be the set of all isometries fixing $x$ (congruence transformations fixing $x$).
• $Sim(x)$ to be the set of all similitudes fixing $x$ (similarity transformations fixing $x$).

Wikipedia says that the following condition is equivalent to the parallel postulate:

There exists a pair of similar, but not congruent, triangles.

Question: Does this equivalent formulation of the parallel postulate imply the following?

For every point $x$, the set $Iso(x)$ is a strict subset of $Sim(x)$.

The converse appears to be true trivially, so this would be an equivalent reformulation of the parallel postulate, provided that the forward implication is actually correct.

Attempt: My confusion is about how much I can assume in trying to prove this -- ostensibly nothing more than the postulates of absolute geometry and their consequences. So in other words, I am unsure if my argument follows from the axioms of absolute geometry. If the facts I use depend on the parallel postulate, that would mean I'm wading into circular reasoning.

The argument: one could compose the similarity transformation whose existence is guaranteed by the postulate with at least one translation (and maybe also a rotation), using the fact that translations (as well as rotations) are isometries in Euclidean space and that the composition of a similarity transformation with an isometry is again a similarity transformation.

I.e. my argument would be that the sets $Iso(x_1)$ and $Iso(x_2)$ for $x_1 \not=x_2$ should be related to each other by conjugation via translations. Same thing for $Sim(x_1) \not= Sim(x_2)$.

The converse should follow from the fact (which I would probably also have to show) that both isometries and similarities map triangles to triangles, and since isometries/similarities which fix the point $x$ are still isometries/similarities, they should still map triangles to triangles. So just take any triangle containing $x$, apply any of the elements of $Sim(x) \setminus Iso(x)$ to it, and then you'll get a new triangle which is similar but not congruent to the first one.

Without using the fact that translations conjugate similarities and isometries to each other, I think that it would only be possible to show that this is equivalent to

There exists a point $x$ such that the set $Iso(x)$ is a strict subset of $Sim(x)$.

Answer 1

Your argument works fine in any reasonable version of absolute geometry once you define what "translations" are, prove that translations are isometries, and prove that for any two points $x$ and $y$ there is a translation sending $x$ to $y$. In fact, there is no need to use translations as opposed to some different kind of isometry, and the easiest kind of isometry to use is a reflection. Given a line $\ell$, the reflection across $\ell$ is defined as the map that sends a point $x$ to the other point on the perpendicular line from $x$ to $\ell$ which is the same distance from $\ell$ as $x$. You can then show in absolute geometry that reflections are isometries. Given two points $x$ and $y$, the reflection across the perpendicular bisector of the segment from $x$ to $y$ is then an isometry that interchanges $x$ and $y$.

If you want a more detailed argument, you'll need to specify exactly what your "axioms of absolute geometry" are.