I'm having some difficulty wrapping my head around rigid motions in a plane. In particular, I'm trying to solve this following problem:
In a Euclidean plane, show that the product of two rotations around different points is equal to either a rotation around a third point or a translation. Hint: Show that it has at most one fixed point.
I working in a plane $\Pi$ over an Euclidean ordered field, so rotations are transformations defined by \begin{cases} x'=cx-sy \\ y'=sx+cy \end{cases} where $c^2+s^2=1$. I simply take two rotations $\psi$ and $\phi$ such that \phi=\begin{cases} x'=cx-dy\\ y'=dx+cy \end{cases} and \psi=\begin{cases} x'=ex-fy\\ y'=fx+ey \end{cases} where $c^2+d^2=1$ and $e^2+f^2=1$. Composing them, I see $ \begin{align*} \psi\phi(x,y) &= (ecx-edy-fdx-fcy,fcx-fdy+edx+ecy) \\ &= ((ec-fd)x-(ed+fc)y,(fc+ed)x+(ec-fd)y) \end{align*} $ but $ \begin{align*} (ec-fd)^2+(ed+fc)^2 &= e^2c^2-2ecfd+f^2d^2+e^2d^2+2edfc+f^2c^2 \\ &= (e^2+f^2)c^2+(e^2+f^2)d^2 \\ &= c^2+d^2=1 \end{align*} $ so $\psi\phi$ is a rotation. I feel I have done something wrong, since the problem seems to be prodding me in a different direction. What is the correct way to show this? Thanks.