$E=mc^2$ resembles non-relativistic kinetic energy formula $E_K = \frac{1}{2}mv^2$? The simplest equation expressing mass–energy equivalence is the famous $E=mc^2$ where $c$ represents the speed of light. Compare this with $E_K = \frac{1}{2}mv^2$. 
Since $E=mc^2$ can be applied to rest mass ($m_0$) and rest energy ($E_0$) to show their proportionality as  $E_0=m_0c^2$, I ask whether this resemblance is just a coincidence created by the need of any equation to be homogeneous for units, or are the two equations fundamentally related?
I know about Mass-Energy Equivalency but I could not find the answer I'm looking for there.
 A: In short, no, it is not a coincidence, they are related. Namely, you may derive the kinetic energy as the first order approximation to the relativistic energy.
We have,
$$ E_0 = mc^2 $$
as you say correctly. Then 
$$ E = \gamma m c^2 = \left( 1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} m c^2 $$
or using a binomial expansion
$$ E \simeq \left( 1 + \frac{1}{2} \frac{v^2}{c^2} + \dots \right) m c^2 \simeq mc^2 + \frac{1}{2} mv^2 $$
So subtacting the rest energy $E_0$ we get
$$ E_k = E - E_0 \simeq \left( mc^2 + \frac{1}{2} mv^2  \right) - \left( mc^2 \right) = \frac{1}{2} m v^2$$
Note that we can of course only use this expansion when $v \ll c$. This makes sense, because that is exactly the case in Newtonian mechanics, which is where we use the more familiar kinetic energy formula.
A: Every relationship between mass and energy will contain two factors of velocity, for dimensional consistency.
In special relativity we have the more exact relationship
$$
E^2 - p^2c^2 = m^2c^4
$$
where the momentum $p$ is
$$
p = \frac{mv}{\sqrt{1-v^2/c^2}}.
$$
You can do a little algebra to show that the total energy is always
$$
E = \frac{ mc^2}{\sqrt{1-v^2/c^2}}.
$$
In mathematics we have a tool called the "binomial theorem," which tells us that 
$$
(1+\epsilon)^n = 1 + n\epsilon + \frac{n(n-1)}{2!}\epsilon^2 + \cdots
$$
This expression turns out to hold even if the power $n$ is not a positive integer!  If $\epsilon\ll1$, we also have the luxury of being able to throw away the higher powers.  For small speeds, then, the Einstein equation becomes
$$
E = \frac{ mc^2}{\sqrt{1-v^2/c^2}} = mc^2 + mc^2 \cdot \frac{-1}{2}\frac{-v^2}{c^2} + mc^2 \cdot\mathcal O\left( (v/c)^4 \right)
$$
which is clearly the rest energy, the classical kinetic energy, and a relativistic correction that becomes large when $v\approx c$.
