Rigid motion vs Isometry Does anyone know the difference between rigid motion and isometry?
In the real plane these two definitions coincide, but do always coincide?
 A: Yes, if the space you're working in has enough structure to talk about lengths and angles, they coincide. Here's more information, and a proof!

A rigid motion (in the plane) is considered "a motion which is a combination of reflections, translations, and rotations". This can be generalised for other inner product spaces as "a transformation that preserves distances between points and angles between points".
On the other hand, an isometry is just something that preserves distances.
The type of structure we want to work in is called an "inner product space". This is the abstract structure that has a notion of angles and lengths. The "distance" between two points $a, b$ is given by the norm, $\Vert a - b \Vert $. The "angle" between two points $a$ and $b$ are given by the inner product (a.k.a dot product), $\langle a,b\rangle$. They are related by $\langle a,a\rangle = \Vert a\Vert^2$.
This means a function $f$ is an isometry if $\Vert f(a)-f(b)\Vert = \Vert a - b \Vert$ for all $a, b$ in your space, and it is a rigid motion if it has the additional property that $\langle f(a), f(b) \rangle = \langle a, a \rangle$ for all $a, b$. (Take a look at the wikipedia page for more info.)

Let $A$ be an inner product space.
Let $f:A \rightarrow A$ be an isometry.
Without loss of generality we can suppose $f(0) = 0$, otherwise we can just compose $f$ with a translation which we already know preserves distances and angles.
We will now prove that an isometry of an inner product space must also preserve angles. (That is, isometry = rigid motion.)

Let $a\in A$. Then, as $f$ is an isometry, $\Vert f(a)-f(0)\Vert = \Vert a - 0 \Vert$. Because $f(0) = 0$, this tells us that $\Vert f(a)\Vert = \Vert a \Vert$. From the above relation between inner products and norms, we now obtain $\langle f(a), f(a) \rangle = \langle a,a\rangle$ .
Therefore, for $x, y \in A$, we have \begin{align}
\Vert f(x)-f(y)\Vert^2 &= \langle f(x)-f(y), f(x) - f(y) \rangle\\
&= \langle f(x), f(x) \rangle - 2\langle f(x), f(y) \rangle + \langle f(y), f(y) \rangle\\
&= \langle x, x \rangle - 2\langle f(x), f(y) \rangle + \langle y, y \rangle\\
&= \langle x, x \rangle - 2\langle x, y \rangle+ 2\langle x, y \rangle- 2\langle f(x), f(y) \rangle + \langle y, y \rangle\\
&= \langle x-y, x - y \rangle + 2\langle x, y \rangle- 2\langle f(x), f(y) \rangle\\
&= \Vert x-y\Vert^2 + 2\langle x, y \rangle- 2\langle f(x), f(y) \rangle\\
\end{align}
Because we know that $\Vert f(x)-f(y)\Vert^2 = \Vert x-y\Vert^2$, this tells us that $\langle x, y \rangle = \langle f(x), f(y) \rangle$, as required.

Long story short, in a mathematical space that is rich enough to talk about distances and angles, isometries preserve angles, so all isometries are in fact rigid motions.
