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?