A simple proof that a plane similarity without fixed points is an isometry Is there a (relatively) simple proof of the fact that a plane similarity without fixed points is an isometry?
In the literature it is usually formulated as "a non-isometric similarity has a fixed point" and the proof is somewhat involved, using the concept of directed distance. I am looking for something simpler, to use for the classification of plane similarities.
A simple proof of the above-mentioned classification will also do.
 A: A similarity can be expressed as:
$$
s(\vec x)=rA\vec x+\vec t,
$$
where $r$ is the scale factor, $A$ is an orthogonal matrix and $\vec t$ is a translation vector. If $s(\vec x)=\vec x$ then:
$$
\vec x=(I-rA)^{-1}\vec t,
$$
where $I$ is the identity matrix. Hence a fixed point doesn't exist only if $I-rA$ is a singular matrix. Plugging in the explicit form of $A$ in ${\mathbb R}^2$ it is easy to see that $\det(I-rA)=0$ only if $r=1$.
EDIT.
In the special case $s=i\circ\sigma$, where $i$ is a translation of vector $\vec t$ and $\sigma$ a stretch of scale factor $r$, it is easy to construct a fixed point $P$ of $s$ if $r\ne1$:
$$
P=O+{1\over1-r}\vec t,
$$
where $O$ is the fixed point of $\sigma$. In fact:
$$
P'=\sigma(P)=O+{r\over1-r}\vec t
\quad\text{and}\quad
s(P)=i(P')=P'+\vec t=O+{1\over1-r}\vec t=P.
$$
If $i$ is a glide reflection, we can set up a coordinate system such that $x$ axis is the reflection line, $\vec t=(t,0)$ is the translation vector, while the fixed point of $\sigma$ is on $y$ axis at $O=(0,d)$.
A simple computation shows then that $P=(x,y)$ is a fixed point if:
$$
x={r\over1-r} t,\quad y={1-r\over1+r}d.
$$
