Heron's Formula; an Intuitive or Visual Proof I've found several proofs for Heron's formula for the area of a triangle in term of its sides, but none of them is simple and intuitive enough to show WHY the formula works. 
Do you know an intuitive or visual proof for it?
Thanks.

I noticed that many proofs for Herons formula use the properties of angles, which is why I wanted to show a proof that doesn't involve any,  and seems to be simpler than the usual methods that are applied.
A triangle with side lengths $a, b, c,$ where $c$ is the greatest side length and an altitude($h$) that intercepts $c$ such that $c$ is the  sum of two side lengths ,
$c = m + n$, using Pythagorean theorem and the area formula of a triangle,  we can prove Herons formula.
First we find $a$ and $b$ in terms of $m$ and $n$ using Pythagorean theorem:
$a^2 - m^2 = h^2$
$b^2 - n^2 = h^2$
$a^2 - m^2 = b^2 - n^2$
$a^2 - b^2 = m^2 - n^2$
Then using $c = m + n$, eliminate $m$:
$a^2 - b^2 = (c-n)^2 - n^2$
$a^2 - b^2 = c^2 - 2cn$
$n = \frac{1}{2c}(c^2 + b^2 - a^2)$
We can Eliminate $n$ from $b^2 - n^2 = h^2$ to find $h$:
$h = \sqrt{b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2}$
Plugging this into the area formula ($A = \frac{1}{2}ch$) gives:
$A = \frac{1}{2}c\sqrt{ b^2 - (\frac{1}{2c}(c^2 + b^2 - a^2))^2} $
$A = \frac{1}{2}c\sqrt{ \frac{1}{4c^2}(2a^2b^2+2a^2c^2-a^4-b^4+2b^2c^2-c^4)} $
$A = \sqrt{\frac{1}{16}(c^2 - (a - b)^2)(( a + b)^2 - c^2)} $
$A = \sqrt{\frac{1}{16}(a + b - c)( a + b + c)( b + c - a)(a + c - b)} $
$A = \sqrt{s(s - a)(s- b)(s- c)}$
Q.E.D.
