State and prove law of conservation of energy in case of freely falling body?

Statement : Energy can neither be created nor be destroyed. It can be converted from one form of energy to another but total energy remains constant.
Proof : Let us consider a body mass m is falling freely through a height h above the ground
At point A :

$\mathrm{u}=0;\mathrm{a}=\mathrm{g}$
Height = h
Potential energy = PE = mgh
$\text{ Kinetic energy }=\mathrm{KE}=\frac{1}{2}\mathrm{m}(\mathrm{u}{)}^2$
$\mathrm{KE}=\frac{1}{2}\mathrm{m}(0{)}^2=0$
Total energy at A = E_A = PE + KE
$\therefore {\mathrm{E}}_{\mathrm{A}}=\mathrm{mgh}+0=\mathrm{mgh} ...\left(1\right)$
At B :
Travelled distance = S = X
u =0 and velocity = V = V_B
Height = (h – x)
$\therefore$PE = mg (h-x) = mgh –mgx
$\mathrm{KE}=\frac{1}{2}{\mathrm{mV}}_{\mathrm{B}}^2$
From ${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as};{\mathrm{V}}_{\mathrm{B}}^2-{\mathrm{o}}^2=2\mathrm{gx}$
$\begin{array}{l}\therefore \mathrm{KE}=\frac{1}{2}{\mathrm{mV}}_{\mathrm{B}}^2=\frac{1}{2}\mathrm{m}\left(2\mathrm{gx}\right)\\ \therefore \mathrm{KE}=\mathrm{mgx}\end{array}$
Total energy at B = E_B = PE + KE
${\mathrm{E}}_{\mathrm{B}}=\mathrm{mgh}-\mathrm{mgx}+\mathrm{mgx}\Rightarrow {\mathrm{E}}_{\mathrm{B}}=\mathrm{mgh}-\left(2\right)$
At C :
Let the body hits the ground t C.
$\mathrm{u}=0\text{ and final velocity }\mathrm{V}={\mathrm{V}}_{\mathrm{c}}$
Height = h = 0 and distance = s = h
PE = mgh = mg (0) = 0
PE = 0
From eqn ${\mathrm{v}}^2-{\mathrm{u}}^2=2\mathrm{as}$
${\mathrm{v}}_{\mathrm{c}}^2-{\mathrm{o}}^2=2\mathrm{gh};{\mathrm{v}}_{\mathrm{c}}^2=2\mathrm{gh}$
$\begin{array}{l}\mathrm{KE}=\frac{1}{2}{\mathrm{mV}}_{\mathrm{c}}^2;\mathrm{KE}=\frac{1}{2}\mathrm{m}\times 2\mathrm{gh}\\ \mathrm{KE}=\mathrm{mgh}\end{array}$
Total Energy at C = E_C = PE + KE
${\mathrm{E}}_{\mathrm{c}}=0+\mathrm{mgh}\therefore {\mathrm{E}}_{\mathrm{C}}=\mathrm{mgh} ...\left(3\right)$
From equation (1), (2), and (3)
The total energy at A, B and C are equal
$\therefore {\mathrm{E}}_{\mathrm{A}}={\mathrm{E}}_{\mathrm{B}}={\mathrm{E}}_{\mathrm{C}}=\mathrm{mgh}$
Hence proved law of conservation of energy incase freely falling body.