I understand that Bohr postulated that electrons can only occupy orbits at certain radii, and that in order to move from one orbit (or stationary state) to another, it would have to absorb or emit a quantum of energy from a photon. This would lead to the fact that the change in energy between stationary states is $$\Delta E = E_2 - E_1 = hf,$$ where $h$ is Planck's constant and $f$ is the frequency.
And though I conceptually understand how this would also imply that the angular momentum of and electron is also quantized, I can't quite derive how the angular momentum is an integer multiple of $h/2\pi$.
I read an answer on another Phys.SE post:
It made sense to me, but I did not have the reputation points in order to comment and ask a follow up question. That is why I am posting here. In the top ranked answer by Kenshin, it reads:
$$\Delta E = h\nu = \frac{hv}{2\pi r} \tag 1$$
Also, we know the kinetic energy at a particular energy level is given by
\begin{align} \text{K.E.} & = \frac{mv^2}{2} = \frac{Lv}{2r}, \quad \text{so therefore} \\ -U & = 2KE = \frac{Lv}{r} \end{align}
Again, taking $r$ and $v$ to be the average radius and velocity during the transition, we get
$$\Delta E = \frac{(L_2 - L_1)v}{r}. \tag 2$$
Equating $(1)$ and $(2)$ gives
$$\frac{(L_2 - L_1)v}{r} = \frac{hv}{2\pi r}.$$
My question is why is potential energy, $U$, used to find the $ΔE$? Why not kinetic energy, or why not total energy?