Every finite group of isometries of the plane is isomorphic to $Z_n$ or $D_n$ 
Fraleigh(7th) Theorem 12.5: Every finite group $G$ of isometries of the plane is isomorphic to either $Z_n$ or to a dihedral group $D_n$ for some positive integer $n$. (Note: An isometry of $\mathbb{R}^2$ is a permutation of $\mathbb{R}^2$ that preserves distances.)
Proof: First we show that there is a fixed point by all of $G$. Let $G=\{f_1,\cdots,f_m\}$ and let $(x_i,y_i):=f_i(0,0)$. Then the point $P=(\bar{x},\bar{y})=\left( \dfrac{x_1+\cdots+x_m}{m},\dfrac{y_1+\cdots+y_m}{m}\right)$ is the centroid of the set $S=\{(x_i,y_i)\ |\ i=1,\cdots,m\}$. The isometries in $G$ permute the points in $S$ among themselves, since if $f_i f_j=f_k$ then $f_i(x_j,y_j)=f_if_j(0,0)=f_k(0,0)=(x_k,y_k)$. It can be shown that the centroid of a set of points is uniquely determined by its distances from the points, and since each isometry in $G$ just permutes the set $S$, it must leave the centroid $(\bar{x},\bar{y})$ fixed. Thus $G$ consists of the identity, rotations about $P$, and reflections across a line through $P$. ......

First, I can't understand the bold text. How can I change the phrase "the centroid of a set of points is uniquely determined by its distances from the points" into mathmatical words? And thus why the centroid is left fixed by all of $G$?
Second, What is the name of the above theorem? I saw some sites calling it Leonardo's theorem, but there is no search result of this name in wiki. Is there any other name?
 A: Let $u_1$ be a point in the plane. Think of $u_1$ as a vector, so we won't have to use two coordinates all the time. In the quoted passage, $u_1=(0,0)$.  Then the passage looked at the orbit of $u_1$ under $G$, and considered  the centroid of this orbit.  But we can work with any starting point, and the idea works in the same way for $3$-dimensional space.
So let the orbit of $u_1$ consist of the points $u_1$, $u_2$, $\dots$, $u_n$. (The orbit of $u_1$ is finite, since the group is finite.) An isometry in $G$ permutes the $u_i$.
The centroid $\overline{u}$ of the orbit is defined by
$$\overline{u}=\frac{u_1+u_2+\cdots +u_n}{n}.$$
We want to show that $\overline{u}$ is a fixed point of every isometry in $G$.
Here is a proof, probably a lot longer than necessary. First we need a lemma.
Lemma Let $\phi$ be an isometry, and let $x$ and $y$ be points. Let $0 \le t \le 1$, and suppose that $z=tx+(1-t)y$. Then $\phi(z)=t\phi(x)+(1-t)\phi(y)$.
Proof: The isometry $\phi$ preserves distances.  Let $d$ be the distance between $x$ and $y$.  Then $z$ is at distance $(1-t)d$ from $x$ and distance $td$ from $y$.  It follows that $\phi(z)$ is at distance $(1-t)d$ from $\phi(x)$ and distance $td$ from $\phi(y)$.  However, since the distance between $\phi(x)$ and $\phi(y)$ is $d$, this means that $\phi(z)=t\phi(x)+(1-t)\phi(y)$, since there is exactly one point that is simultaneously at distance $(1-t)d$ from $\phi(x)$ and at distance $td$ from $\phi(y)$.
Theorem: Let $\phi$ be an isometry,  let $u_1$, $u_2$, $\dots$, $u_n$ be points, and let $\overline{u}$ be their centroid. Then
$$\phi(\overline{u})=\frac{\phi(u_1)+\phi(u_2)+\cdots \phi(u_n)}{n}$$
(or, more informally, $\phi$ preserves centroids).
Proof: By induction on $n$.  The case $n=1$ gives a nice easy start, but the lemma also takes care of $n=2$.  We show that if the result holds at $n$, it holds at $n+1$.
Let the centroid of the first $n$ points be $\overline{c}_n$, and let the centroid of the $n+1$ points be $\overline{c}_{n+1}$.
Note that 
$$\overline{c}_{n+1} =\frac{n}{n+1}\overline{c}_n +\frac{1}{n+1}u_{n+1}.$$
(This is easily verified by substituting the definition of centroid, but it is also physically obvious.)
Now we can apply the lemma, since $\overline{c}_{n+1}$ has been expressed in the form $t\overline{c}_n +(1-t)u_{n+1}$.
We find that 
$$\phi(\overline{c}_{n+1})=\frac{n}{n+1}\phi(\overline{c}_n) +\frac{1}{n+1}\phi(u_{n+1}).$$
Now by the induction hypothesis
$$\phi(\overline{c}_n)=\frac{\phi(u_1)+\phi(u_2)+\cdots \phi(u_n)}{n}.$$
Simplify the right-hand side. We obtain the centroid of the $\phi(u_i)$.
Comment: Took longer than I thought it would! Probably wouldn't have done it if I had known. But it is done in gruesome detail, and I have a wordiness problem.
There is undoubtedly a way to short-circuit all this. The above argument used only the definition of isometry.
A: In isolation it does sound like the bold text needs a justification. But do you know at this point that any isometry of the plane is affine? In other words that the elements of the group are of the form $f_j:\vec{x}\mapsto L_j(\vec{x})+\vec{y}_j$ for some linear mapping $L_j$ (that actually also needs to be an isometry) and a constant vector $\vec{y}_j$? If you do, then the claim does follow, because the centroid is an affine linear combination of the points in the orbit with coefficients all equal to $1/m$, and that particular combination is stable. Write $\vec{x}_i=f_i(0,0)$. Then
for all $j$
$$
f_j(P)=L_j(P)+\vec{y}_j=L_j\left(\frac{\sum_{i=1}^m\vec{x}_i}m\right)+\vec{y}_j
=\frac1m\sum_{i=1}^m (L(\vec{x}_i)+\vec{y_j})=\frac1m\sum_{i=1}^mf_j(\vec{x}_i)=P,
$$
because we can split the vector $\vec{y}_j$ into $m$ equal parts, and in the last step the mapping $f_j$ just permutes the vectors $\vec{x}_i$.
I'm not happy with this argument, because it sounds like there is something more elegant out there.
