Every reflection is an isometry proof The theorem is that every reflection $R_{S}$ in an affine subspace $S$ of $\mathbb{E}^{n}$ is an isometry:
$R_S:\ \mathbb{E}^{n} \rightarrow \mathbb{E}^{n}:\ x \mapsto R_{S}(x) = x + 2 \overrightarrow{x\pi_{S}(x)}$
I'm horrendously stuck with the proof.
I get that I'm trying to prove that $R_{S}$ preserves the distance between the two points, but I'm lost as to how.
EDIT:
$\pi_{S}(x)$ is defined as the intersection of $S$ and the euclidean subspace $T_x$ through $x$, perpendicular to $S$.
 A: Pick an origin for $\Bbb E^n$, and call the resulting vector space $\Bbb R^n$. To save notation, let's put the origin in $S$, so that $S$ now becomes a (vector) subspace. Write $\Bbb R^n=S\oplus S^\perp$. Now, identifying points $x\in\Bbb E^n$ with the corresponding vectors in $\Bbb R^n$, write $x=x_1+x_2$, where $x_1\in S$ and $x_2\in S^\perp$. Then $\pi_S(x)=x_1$ and $x-\pi_S(x) = x_2\in S^\perp$. 
Here's a lemma you need to prove, using dot products. (It's just a higher-dimensional Pythagorean Theorem.)
Lemma: Writing $x=x_1+x_2$, $x_1\in S$, $x_2\in S^\perp$, we have $\|x\|^2 = \|x_1\|^2 + \|x_2\|^2$.
The vector $\overrightarrow{x\pi_S(x)}$ can now be written as $\pi_S(x)-x=-x_2$, so our formula for $R_S$ becomes $R_S(x) = (x_1+x_2)+2(-x_2)= x_1-x_2$. Now it's clear that for any $x,y\in\Bbb R^n$, we have
$$R_S(x)-R_S(y) = (x_1-x_2)-(y_1-y_2) = (x_1-y_1) + (y_2-x_2).$$
Because $x_1-y_1\in S$ and $y_2-x_2\in S^\perp$, we have 
\begin{align*}
\|R_S(x)-R_S(y)\|^2 &= \|x_1-y_1\|^2 + \|y_2-x_2\|^2 = \|x_1-y_1\|^2 + \|x_2-y_2\|^2 \\&= \|(x_1+x_2)-(y_1+y_2)\|^2 = \|x-y\|^2,
\end{align*}
as desired.
A: Let $\{u_1,\dotsc,u_k\}$ be an orthonormal basis for $T_x$.
Now pick a point $p$ on $S$.
$\pi_S(x)$ can then be written as follows:
$$ \pi_S(x)$ = x + \sum_{i=1}^k ((p-x) \cdot u_i)u_i$$
For all points $x$ and $y$ of $\mathbb{E}^n$, watch what happens to the distance:
First (for simplicity later), I will work out the vector between the reflection of $x$ and the reflection of $y$.
$$      \begin{array}{rll}
        R_{S}(x) - R_{S}(y) &= (x + 2\overrightarrow{x\pi_{S}(x)}) - (y + 2\overrightarrow{y\pi_{S}(y)}) &\\
                           &= (x + 2(\pi_{S}(x) - x)) - (y + 2(\pi_{S}(y) - y)) &\\
                           &= (-x + 2\pi_{S}(x)) - (-y + 2\pi_{S}(y)) &\\
                           &= (-x + 2(x + \sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i})) - (-y + 2(y + \sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i})) &\\
                           &= (x + 2\sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i}) - (y + 2\sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i}) &\\
                           &= (x-y) + 2\left(\sum_{i=1}^{k}((p-x)\cdot u_{i})u_{i} -  \sum_{i=1}^{k}((p-y)\cdot u_{i})u_{i}\right) &\\
                           &= (x-y) + 2\left(\sum_{i=1}^{k}\left(((p-x)\cdot u_{i})u_{i} - ((p-y)\cdot u_{i})u_{i}\right)\right) &\\
                           &= (x-y) + 2\left(\sum_{i=1}^{k}\left((((p-x)\cdot u_{i})- ((p-y)\cdot u_{i}))u_{i}\right)\right) &\\
                           &= (x-y) + 2\left(\sum_{i=1}^{k}(((p-x)-(p-y))\cdot u_{i} )u_{i}\right) &\\
                           &= (x-y) + 2\left(\sum_{i=1}^{k}((y-x)\cdot u_{i} )u_{i}\right) &\\
                           &= (x-y) - 2\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right) &\\
      \end{array}$$
Now we will look at the quadrate of the distance betwee the reflection of the points:
$$
\begin{array}{rll}
        d(R_{S}(x),R_{S}(y))^2 &= \left\| R_{S}(x) - R_{S}(y)\right\|^{2} & \\
                            &= (R_{S}(x) - R_{S}(y))^{2} &\\
                            &= \left((x-y) - 2\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)\right)^{2} &\\
                            &= (x-y)^{2} -4(x-y)\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\
                            &= (x-y)^{2} -4(x-y)\left(\sum_{i=1}^{k}u_{i}((x-y)\cdot u_{i} )\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\
                            &= (x-y)^{2} -4\left(\sum_{i=1}^{k}((x-y) \cdot u_{i})\cdot((x-y)\cdot u_{i} )\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )u_{i}\right)^{2} &\\
                            &= (x-y)^{2} -4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )^{2}\right) + 4\left(\sum_{i=1}^{k}((x-y)\cdot u_{i} )^{2}\right) &\\
                            &= (x-y)^{2} &= d(x,y)^2
      \end{array}$$
Notice that the third to last equality only holds because the $u_i$ form an orthonormal basis.
A: Here's a partial answer which is suitable for Euclidean space (not quite as general as an affine space).  Given a vector $a$ in Euclidean space $\mathbb{R}^n$, the formula for the reflection in the hyperplane through the origin, orthogonal to $a$, is given by
$$\text{Ref}_a(v) = v - 2\frac{v\cdot a}{a\cdot a}a$$ 
Recall that the distance between two vectors $v$ and $w$ is $d(v,w)=|v-w|$.  So we can show that 
$$d(\text{Ref}_a(v),\text{Ref}_a(w))=| v - 2\frac{v\cdot a}{a\cdot a}a - (w - 2\frac{w\cdot a}{a\cdot a}a)| = |v-w+2(wa-va)\frac{a}{a\cdot a}| = $$
$$=|v-w+2(w-v)\frac{a\cdot a}{a\cdot a}|=d(v,w)$$
