I am asked to finish the following sentence:
Let $\sigma$ be an isometry on $\mathbb{R}^2$, suppose it fixes the points $A$ and $B$
Suppose $\sigma$ also fixes a third point $C$ which is not on the line $AB$.
Complete the following sentence: If $P$ is a fourth point in the plane which is not fixed by $\sigma$, so $\sigma(P) \neq P$, then $A$, $B$ and $C$ are all on the $\dots$ BLANK
This is a contradiction and so $\sigma(P) = P$. Since P is arbitrary, we conclude that $\sigma$ is the identity.
I am not entirely sure what the author is going for. I think what they want is to make the case that $A,B$ and $C$ are on same reflection line, but I do not see how. This would contradict $C$ not being on the line through $AB$. It just seems like the argument skips a step
You are correct, the term for the blank space would be bisection of the segment $P, \sigma(P)$, since $d(P,X)=d(\sigma(P), X)$ for each $X\in\{A, B, C\}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.