Prove that if two isometries $f,g$ of the Euclidean 2D space have the same image at 3 non-colinear points $x_i,i=1,2,3$, then they are identical.
Source: Nastasescu C., Nita C., Vraciu C., "Bazele algebrei, vol 1" (Romanian) (translation: Fundamentals of Algebra), page 40.
You want to use the following fact: If $x_1, x_2, x_3 \in \mathbb{R}^2$ are in general position (not colinear) and $a,b,c$ are real numbers, then there is at most one point $p$ in the plane so that $d(p,x_1) = a$, $d(p,x_2)=b$, and $d(p,x_3) = c$. Draw a picture. Circles in the plane with different centers intersect at most twice.
With this in mind, let $p \in \mathbb{R}^2$ be given. Then if $a=d(p,x_1)$, $b=d(p,x_2)$, and $c = (p, x_3)$, then $$d(q,x_1) = a, d(q,x_2)=b, d(q, x_3) = c \Rightarrow q = p$$ Then if $f,g$ are isometries, we have $$d(f(p), f(x_1)) = a, d(f(p), f(x_2)) = b, d(f(p), f(x_3)) = c$$ $$d(g(p), g(x_1)) = a, d(g(p), g(x_2)) = b, d(g(p), g(x_3)) = c$$ if $f,g$ agree at $x_1,x_2,x_3$, then $$d(g(p), f(x_1)) = a, d(g(p), f(x_2)) = b, d(g(p), f(x_3)) = c$$ Which implies that $f(p) = g(p)$.