Proof of Heron's Formula for the area of a triangle 
Let $a,b,c$ be the lengths of the sides of a triangle. The area is given by Heron's formula:
$$A = \sqrt{p(p-a)(p-b)(p-c)},$$
where $p$ is half the perimeter, or $p=\frac{a+b+c}{2}$.

Could you please provide the proof of this formula?
Thank you in advance.

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.
