Showing that an Isometry on the Euclidean Plane fixing the origin is Linear Suppose $f$ is an isometric (i.e., distance preserving) function on $\mathbb{E}^2$ such that $f(0,0) = (0,0)$.  Then I want to show that $f$ is necessarily linear.  Now $f$ is linear iff $f$ is both additive and homogenous.  The following is an attempted proof for the homogeneity of f (missing the last step); still more, I have no idea how to argue for the additivity of $f$.  Any ideas?
Let $x \in \mathbb{E}^2$ and $\alpha \in \mathbb{R}$ 
We know that $\forall x \in \mathbb{E}^2$, $\Vert x - 0 \Vert = \Vert f(x) - f(0)\Vert = \Vert f(x) - 0 \Vert$ so that $\Vert x\Vert = \Vert f(x)\Vert$.
From this we immediately have the following facts:
$\Vert x \Vert = \Vert f(x) \Vert$
$\Vert \alpha x \Vert = \Vert f(\alpha x) \Vert$
We can then argue that since $\Vert \alpha x\Vert = |\alpha| \Vert x \Vert = |\alpha| \Vert f(x) \Vert = \Vert \alpha f(x) \Vert$, we also have that $\Vert f(\alpha x) \Vert = \Vert \alpha f(x)\Vert$.
Finally, we have that
$\Vert \alpha x - x \Vert = \Vert \alpha f(x) - f(x) \Vert$ 
iff $ \Vert (\alpha - 1)x \Vert = \Vert (\alpha - 1) f(x) \Vert$ 
iff $|\alpha - 1| \Vert x \Vert = |\alpha - 1| \Vert f(x) \Vert$
Since the last of these statements is in fact true, we now have $\Vert \alpha x - x \Vert = \Vert \alpha f(x) - f(x) \Vert$ as desired.
Now at this point it seems like I have all of the facts required to assert that $f(\alpha x) = \alpha f(x)$, but I can't figure out how to formally state why without illegally appealing to visual intuition.  Any ideas?
 A: It suffices to prove that $||f(\alpha x) - \alpha f(x) || = 0$. Now
$$\begin{eqnarray*}  ||f(\alpha x) - \alpha f(x) || &=& ||f(\alpha x)||^2-2\alpha\langle f(\alpha x),f(x) \rangle + \alpha^2 ||f(x)||^2 \\
&=& 2\alpha^2 ||f(x)||^2 - 2\alpha\langle f(\alpha x),f(x) \rangle \end{eqnarray*}$$
where $\langle \cdot,\cdot \rangle$ denotes the usual Euclidean inner product. From tbe last line in your question and using the fact that $f$ is an isometry, we get that
$$||f(\alpha x) - f(x) || = ||\alpha f(x) - f(x)||. \tag{*}$$
Squaring both sides of the equation $(*)$ we obtain
$$\|f(\alpha x)\|^2+\|f(x)\|^2-2\langle f(\alpha x),f(x)\rangle=\alpha^2\|f(x)\|^2+\|f(x)\|^2-2\langle \alpha f(x),f(x)\rangle.$$
So by cancellation we obtain $\alpha||f(x)||^2 = \langle f(\alpha x),f(x) \rangle$. It now follows that
$$\begin{eqnarray*} ||f(\alpha x) - \alpha f(x)|| &=& 2\alpha^2 ||f(x)||^2 - 2\alpha\langle f(\alpha x),f(x) \rangle \\
&=& 2\alpha^2 ||f(x)||^2 - 2\alpha ( \alpha||f(x)||^2 )\\
&=& 2\alpha^2 ||f(x)||^2 - 2\alpha^2 ||f(x)||^2 \\
&=& 0\end{eqnarray*}$$
from which we conclude that $f(\alpha x) - \alpha f(x) = 0$, i.e.
$$f(\alpha x) = \alpha f(x).$$
A: Any line $g\subset{\mathbb E}^2$ can be viewed as being the median of two suitably chosen points. It follows that an isometry $f:\ {\mathbb E}^2\to{\mathbb E}^2$ maps lines onto lines.
Given an $x\ne0$ and an $\alpha\ne1$ the three different points $0$, $x$, and $y:=\alpha x$ are collinear; therefore their images $0$, $x'$, and $y'$ are collinear as well. As $x'\ne0$ we necessarily have $y'=\beta x'$ for some $\beta\in{\mathbb R}$.
Since $f$ is an isometry it follows that $$|\beta|\ \|x'\|=\|\beta x'\|=\|y'\|=\|y\|=\|\alpha x\|=|\alpha|\ \|x\|\ ,$$
whence $|\beta|=|\alpha|$, or $\beta=\pm\alpha$. Assume that $\beta=-\alpha$. Then
$$|\alpha+1|\ \|x'\|=|1-\beta|\ \|x'\|=d(x',y')=d(y,x)=|\alpha-1|\ \|x\|\ .$$
This says that $\alpha$ is equidistant from $-1$ and $1$; whence $\alpha=0$. It follows that $\beta=\alpha$ in all cases.
A: I will give a more geometric point of view (for $\mathbb{R}^n$):
1) Since lines are geodesics, defined by minimization of distances, isometries map lines to lines.
2) In addition, isometries respect the distance between any two lines, so isometries map parallel lines to parallel lines. Hence isometries map parallelograms to parallelograms.
3) An isometry that maps $0$ to $0$ then maps parallelograms with $0$ as a vertex to parallelograms with $0$ as a vertex. But these are the parallelograms used to define addition in $\mathbb{R}^n$; therefore, the two ways of constructing the diagonal of the image parallelogram show that $f(a+b)=f(a)+f(b)$ for any isometry $f$ such that $f(0)=0$.
4) Now, by 3), $f(na)=nf(a)$ for every natural number $n$, which implies that $f(qa)=qf(a)$ for every rational number $q$.
5) Since $f$ is isometry, it is continuous, hence by 4) we get $f(\lambda a)=\lambda f(a)$ for every $\lambda\in\mathbb{R}$.
QED
By the way, if $f$ is an isometry such that $f(0)=a$, then $g:=T_{-a}\circ f$ (where $T_x$ is the translation by $x$) is an isometry such that $g(0)=0$. Then, as $f=T_a\circ g$, every isometry is the composition of a translation and an orthogonal linear map (i.e., a linear isometry).
