Chasles' theorem, Euler's theorem, Mozzi's theorem I am confused about the important concept expressed by these three, I think equivalent, theorems. Let me use an example: a rigid body is initially in a certain position and orientation in space (configuration A). 
After 2 seconds, the same body occupies a different position and has a different orientation (let's call it configuration B). The body arrived to the new configuration B through different intermediate configurations that we don't know unless we took a video of the entire motion and have all the instantaneous configurations between A and B. However, the mentioned theorem state that it is possible, mathematically, to bring the rigid body from configuration A to configuration B with just a specific translation followed by a specific rotation (or vice versa). Is that correct?  That may be mathematically/geometrically true but it does not consider the actual, physical motion and sequence of actual configurations that the rigid body passed through as it went from configuration A to configuration B, right? Why are these theorems so useful?
thank you!
 A: 
"Why are these theorems so useful?"

I find the use in the kinematics of a rigid body when dealing with infinitesimal motion and/or velocities. Fundamentally, at any instant, the motion of a rigid body can be decomposed into a rotation about an arbitrary axis, coupled with a parallel translation along that axis.
First of you recognize that given the rotational velocity $\boldsymbol{\omega}$ of a rigid body and the velocity $\boldsymbol{v}_A$ of a specific point A you can find the velocity of any other point B by $$\boldsymbol{v}_B = \boldsymbol{v}_A + \boldsymbol{\omega} \times ( \boldsymbol{r}_B - \boldsymbol{r}_A )$$ where $\boldsymbol{r}_A$ and $\boldsymbol{r}_B$ are the location vectors. And the same for any other point riding on the rigid body.
So the entire motion of a rigid body is specified by the three rotation components $\boldsymbol{\omega}$ and the three velocity components $\boldsymbol{v}_A$ of any arbitrary point A. Hence it is common to say that a rigid body has 6 degrees of freedom.
Now consider the case where the body is rotating about the origin. Additionally, it moves parallel to the rotation axis by $\boldsymbol{v}_\parallel = h\, \boldsymbol{\omega}$. The scalar quantity $h$ (called the screw pitch) is the ratio of the translational velocity to the rotational velocity. Again, with that we can specify the velocity of any other point
$$ \boldsymbol{v}_A = h\,\boldsymbol{\omega} + \boldsymbol{\omega} \times \boldsymbol{r}_A  \tag{1}$$ 
The above states that the three components of $\boldsymbol{v}_A$ can be composed by the scalar pitch $h$ and the two components of $\boldsymbol{r}_A$ that are not parallel to the rotation axis. So knowing where the rotation axis is and what the pitch is we know the motion of the body.
The interesting part is working the problem in reverse.  Given the velocity of the body at a point, where is the axis of rotation? In the system above, given $\boldsymbol{v}_A$ we need to recover $\boldsymbol{r}_A$ as well as $h$.
This is done as follows:
$$\begin{aligned}
  h & = \frac{ \boldsymbol{v}_A \cdot \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \|^2}  \\
 \boldsymbol{r}_A & = \frac{ \boldsymbol{v}_A \times \boldsymbol{\omega} }{ \| \boldsymbol{\omega} \|^2}
\end{aligned} \
\tag{2}$$
In summary, Chasle's Theorem allows us to extract the geometry of motion at any instance but extracting the location of the rotation axis, as well as the screw pitch. The 6 degrees of freedom of a rigid body can be specified by either $(\boldsymbol{\omega},\,\boldsymbol{v}_A)$ or by


*

*Rotation axis direction $\boldsymbol{\hat{z}}$ (two components)

*Rotation speed $\omega$ (one component), such that $\boldsymbol{\omega} = \omega \boldsymbol{\hat{z}}$

*Location of screw axis $\boldsymbol{r}_A$ (two components)

*Screw pitch $h$ (one component)


For a total of 6 degrees of freedom. The decomposition of the motion is summarized as
$$ \begin{aligned}
  \boldsymbol{\omega} & = \omega \, (\boldsymbol{\hat{z}}) \\
  \boldsymbol{v}_A & = \omega \left( h \boldsymbol{\hat{z}} + \boldsymbol{\hat{z}} \times \boldsymbol{r}_A \right)
\end{aligned} \tag{3}$$
The above pair $(\boldsymbol{\hat{z}}, \, h\,\boldsymbol{\hat{z}} + \boldsymbol{\hat{z}}\times \boldsymbol{r}_A)$ designate the screw axis of the motion, in what is called Plucker line coordinates.

Proof
Use (2) in (1) to get
$$ \begin{aligned} \boldsymbol{v}_A & = \frac{ (\boldsymbol{v}_A \cdot \boldsymbol{\omega}) }{ \| \boldsymbol{\omega} \|^2} \boldsymbol{\omega} + \boldsymbol{\omega} \times \frac{ (\boldsymbol{v}_A \times \boldsymbol{\omega}) }{ \| \boldsymbol{\omega} \|^2}  \\
 & = \frac{ (\boldsymbol{v}_A \cdot \boldsymbol{\omega}) \boldsymbol{\omega}}{ \| \boldsymbol{\omega} \|^2} + \frac{ \boldsymbol{v}_A ( \boldsymbol{\omega} \cdot \boldsymbol{\omega}) - \boldsymbol{\omega} ( \boldsymbol{\omega} \cdot \boldsymbol{v}_A)  }{ \| \boldsymbol{\omega} \|^2} \\
 & = \frac{ \boldsymbol{v}_A \| \boldsymbol{\omega} \|^2}{\| \boldsymbol{\omega} \|^2} = \boldsymbol{v}_A \; \checkmark
\end{aligned} $$
