State and prove parallel axes theorem.

Statement :The moment of inertia of a rigid body about any axis is equal to the moment of inertia about a parallel axis through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes.
$\therefore \mathrm{I}={\mathrm{I}}_0+{\mathrm{Mr}}^2$
Proof:
1) consider a rigid body of mass M rotating about an axis AB.
2) Let C be the centre of mass of the body and CD be the parallel axis through C.
3) Let I and I0 be the moments of inertia of the body about AB and CD respectively. r is the perpendicular distance between the parallel axes.
4) Let a particle of mass m is situated at P.
M.I. of the rigid body about AB is $\mathrm{I}=\mathrm{\Sigma m}{\mathrm{AP}}^2$
M.I. of the rigid body about CD is
${\mathrm{I}}_0=\mathrm{\Sigma }{\mathrm{mCP}}^2$
From diagram,

In the right angled
$\begin{array}{l}\mathrm{△}\mathrm{APQ}\Rightarrow {\mathrm{AP}}^2={\mathrm{AQ}}^2+{\mathrm{PQ}}^2\\ \Rightarrow {\mathrm{AP}}^2=(\mathrm{AC}+\mathrm{CQ}{)}^2+{\mathrm{PQ}}^2\\ \Rightarrow {\mathrm{AP}}^2={\mathrm{AC}}^2+{\mathrm{CQ}}^2+2\cdot \mathrm{AC}\cdot \mathrm{CQ}+{\mathrm{PQ}}^2\\ \Rightarrow {\mathrm{AP}}^2={\mathrm{AC}}^2+{\mathrm{PC}}^2+2\cdot \mathrm{AC}\cdot \mathrm{CQ}\\ \left[\begin{array}{l}\because ∆\mathrm{CPQ};\\ {\mathrm{PC}}^2={\mathrm{CQ}}^2+{\mathrm{PQ}}^2\end{array}\right]\end{array}$
Multiplied with
$$\begin{gathered} \begin{array}{l}{\mathrm{\Sigma mAP}}^2={\mathrm{\Sigma mAC}}^2+{\mathrm{\Sigma mPC}}^2+\mathrm{\Sigma m}\cdot 2\mathrm{AC}\cdot \mathrm{CQ}\\ \Rightarrow \mathrm{I}={\mathrm{\Sigma mr}}^2+{\mathrm{I}}_0+0\\ [\because \mathrm{\Sigma m}2\mathrm{r}\cdot \mathrm{CQ}=0 \mathrm{themomentofallparticles} \mathrm{is} \mathrm{zero}] \\ \Rightarrow \mathrm{I}={\mathrm{I}}_0+{\mathrm{Mr}}^2\end{array} \end{gathered}$$
Hence, theorem is proved.