I want to prove the statement : A rigid motion of the plane that does not have a fixed point is either translation or glide reflection (i.e., a translation followed by a reflection).
I want a proof which doesn't use linear algebra.
Example: (One can skip reading this part. It is just to illustrate the kind of proof I am trying to find.)
Suppose a rigid motion $f$ fixes two points and does not fix every point in $\mathbb{C}$. $F$ has to fix every point on the line $l$ passing through the two fixed points, since $F$ is a rigid motion. Let $p \in \mathbb{C}$ be a point not fixed by $f$ and $q$ is the closest point to $p$ lying on $l$. Now, as $$|p - q| = |f(p) - f(q)| = |f(p) - q|$$ $f(p)$ should be the reflection of $p$ along $l$. Now, as the two fixed points and $p$ are non-collinear and on them $f$ acts as a reflection along $l$, $f$ has to be the reflection along $l$. The last statement follows because, a rigid motion is completely determined by its action on 3 non-collinear points.
Attempt: Let $f$ be such that it has no fixed point. It is easy to see that if $f$ takes every line to its parallel line, then $f$ is a translation.
Suppose there exists line $l, l^\prime$ such that $f(l) = l^\prime$ and $l \cap l^{\prime} \neq \emptyset$. Let $l \cap l^{\prime}= \{p\}$. There exists a point $q(\neq p) \in l$ such that $f(q) = p$.
We construct a line $m$ and a point $a$ as shown in the GIF. My claim is that $f = R_m \circ T_a$, where $R_m$ is the reflection about $m$ and $T_a$ is the translation by $a$. Is this correct? If yes, my idea is to prove that $T_a^{-1} \circ R_m^{-1} \circ f$ fixes the points $q, q+a, p$. I am unable to prove that.
Note: $q+a$ is reflection of $p$ about the line $m$.
