Cfrgtkky

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 $tau cdot sigma$