Does rotation in plane involve Euclid's parallel postulate?

Question

It is given a line $l$, a point $P$ not on $l$, and $PQ$ perpendicular to $l$, i.e., $Q$ is on $l$. Let $R$ be a point on $l$ with $QR = PQ$. Therefore, $<PQR = 90^{\rm o}$.

I can't figure out if Euclid's parallel postulate is involved when I rotate the object $PQR$ clockwise by $90^{\rm o}$. Any hint or reference is highly appreciated.

Answer 1

It seems that rotation indeed involves Euclids fifth postulate. Section 8 of this paper is all about that.

Answer 2

I'm quite positive rotation does not involve Euclid's fifth postulate. I'll try to present strict definition of rotation in synthetic setting.

Definition 1. An ordered pair $(A,B)$ of halflines having the same origin will be called a directed angle. Also denote $AB:=(A,B)$

Definition 2. We introduce a relation $\simeq$ among directed angles: We say $AB\simeq CD$ iff these angles are congruent (or equivalently have the same measure) and have the same orientation (in case $AB$ and $CD$ are not $0$ or $180$ degrees). Having the same orientation roughly speaking means that going from $A$ to $B$ and from $C$ to $D$ are both clockwise or both counterclockwise.

It can be proven that

Proposition 1. $\simeq$ is an equivalence relation.

Proposition 2. Given a halfline $A$ of origin $o$ and a directed angle $PQ$, there is a unique halfline $B$ of origin $o$ such that $AB\simeq PQ$.

Now if we fix a directed angle $PQ$ and a point $o$ on a plane, we can define a rotation around $o$ by a directed angle $PQ$ the following way: Obviously we map $o$ to $o$. Take $a\neq o$. By proposition 2 we can find a halfline $B$ of origin $o$ such that $AB\simeq PQ$, where $A$ is a halfline $\overrightarrow{oa}$. Next we find a unique point $b$ on $B$ such that $oa\equiv ob$. Finally we map $a$ to $b$.

What is the most important thing about rotations is that it can be proven in neutral geometry that they are isometries. It follows quite easily from lemma

Lemma. Let halflines $P,Q,P_1,Q_1$ all have the same origin. If $PP_1\simeq QQ_1$, then $PQ\simeq P_1Q_1$.

To prove that rotations are isometries you have to consider few cases but basically you apply the lemma and SAS congruence rule.