I am in the final step of a proof on classifying the symmetries of $\mathbb{R}^2$.
Suppose we have some symmetry $\sigma$ that fixes at least two points, say $A$ and $B$. Then consider $C$ which is the point that does not lie on the line $AB$, and we have $\sigma(C)\neq C$, we also have the reflection $\tau$ in the line $AB$. Prove that $\tau \cdot \sigma $ fixes $A,B$ and $C$ and, as it fixes three points, it is the identity.
I have already concluded that $\tau$ fixes $A$ and $B$ as they are points on the reflection line. So $\tau \cdot \sigma (A)=\tau(A) $ and $\tau \cdot \sigma (B)=\tau(B)=B $. I am uncertain about how to prove that $C$ gets fixed. We want to show that $\tau \cdot \sigma(C)=C$
Notice that as $\sigma(C) \neq C$, we know that $d(A, C)=d(A, \sigma (C))$ and also that $d(B, c)=d(B, \sigma (C))$. This also means that both $A$ and $B$ are on the perpendicular bisector of line segment $C \sigma (C)$. We observe that $C$ is simply mirrored in the line through $A$ and $B$, hence, if we apply $\tau$, we mirror it back. We conclude $\tau (\sigma (C))=C$, this means $C$ is also a fixed point of $\tau \cdot \sigma$