Suppose that $f$ and $g$ are isometries and $A, B, C$ are non-collinear. Prove that if $f(A) = g(A)$, $f(B) = g(B)$, and $f(C) = g(C)$ then $f = g$.
I was thinking that because every isometry has an inverse which is also an isometry, then $f^{-1}(f(A))=A$, and we know that $g(f^{-1}(f(A)))=f(A)=g(A)$, so then the function $g(f^{-1})$ has three fixed points, namely $f(A)$, $f(B)$, and $f(C)$, then $g(f^{-1}) = i$ (identity transformation) So then the inverse of $g$ is $f^{-1}$, so $g=f$.
Is this a solution to the problem or did I do something incorrect?