# Surjectivity of isometry

I am looking for the proof Prove of "any isometry S is a surjective mapping".

My attempt:

pick any two points $A, B$, consider their images $S(A) = A'$ , $S(B) = B'$ . To prove surjectivity, I need to find, for any point $X'$ in the plane, a point $X$ such that $S(X) = X'$ . I know from the definition of isometry that in this case we would have $XA = X'A'$ , $XB = X'B'$ .

Now my question is how to use this to construct possible candidates for X such that $S(X) = X'$.

• Isometries are onto by definition. Otherwise are called isometric embeddings. Ex. A line in a plane is an isometrica embedding. thus you cannot prove surjectivity by using just the isometry preperty. Apr 7 '15 at 7:35