Let $A,B,C,D$ be four points in $\Bbb R^n$ with $dist(A,B)=dist(C,D)$. Prove that there exists an isometry $f$ of $\Bbb R^n$ such that $f(A)=C$ and $f(B)=D$.
I want to use the fact that each isometry is of the form $f = Mx + n$ where $M$ is an orthogonal matrix. Intuitively, $f$ rotates the vector defined by $A,B$ and then translates it so it is aligned with the vector defined with $C,D$. Any ideas how can I formalize it into a proof?
You could express your intuitive idea as follows:
Let $t$ be the translation mapping $A$ to $C$. Then $t(B)$ and $D$ are equidistant from $C$. Therefore there is a rotation $r$, centre $C$, mapping $t(B)$ to $D$.
Then $rt(A)=r(C)=C$ and $rt(B)=D$, as required.
ADDED EXPLANATION
If we translate by $a$ (to the origin), multiply by an orthogonal matrix $M$ and then translate by $b$, these transformations do the following to any point v.
$M(v+a)+b=M(v)+M(a)+b$ and $Ma+b$ is your $n$.