Every isometry that fixes the origin is either a rotation about the origin or a reflection in a line through the origin.
How can I prove this theorem? Proving the converse is easy and I know how to do that.
What I know: The rotation about the origin through an angle $\theta$ in the positive direction is described by the matrix $$\begin{pmatrix}\cos\theta &-\sin\theta\\ \sin\theta &\cos\theta\end{pmatrix}$$ and the reflection in the line through the origin that makes an angle $\theta/2$ with the $x$-axis is described by $$\begin{pmatrix}\cos\theta &\sin\theta\\ \sin\theta &-\cos\theta\end{pmatrix}.$$
If $A$ is an isometry of $\mathbb{R}^2$ with the standard inner product, then it must take the standard basis vector $e_1$ to a unit vector, ie of the form $(\cos \theta, \sin \theta)$ for some $\theta$. Similarly $Ae_2 = (\cos \phi, \sin \phi)$ for some $\phi$. Since $\langle e_1, e_2 \rangle = 0$, we must also have $\langle Ae_1, Ae_2\rangle = 0$, ie $$0 = \cos\theta \cos \phi + \sin \theta \sin \phi = \cos(\theta - \phi).$$ So without loss of generality, $\phi = \pi/2 + \theta$ or $\phi = 3\pi/2 + \theta$. Plugging these in and writing $Ae_1$ and $Ae_2$ as the columns gives you those two matrices.