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.

Consider a triangle in the 3D cartesian system whose side lengths are a, b and c. Assume the three vertices of the triangle are on three coordinate axes. The area of this triangle can be represented as the sum of three area vectors where each vector corresponds to the projections of the given triangle in each of the three planes XY, YZ and ZX respectively. Let x,y be the base and altitude respectively for the XY-plane triangle, y,z for the YZ-plane triangle and x, z for the XZ-plane triangle. By Pythagoras theorem,
$$ a^2=x^2+y^2 $$
$$ b^2=y^2+z^2 $$
$$ c^2=z^2+x^2 $$
The area vector of the first triangle would be
$ \frac{1}{2} xy \hat{k} $
Similarly, we can form the area vectors of the other two projected triangles. The resultant vector, i.e., the area vector of the triangle in consideration would be
$ \frac{1}{2} xy \hat{k} + \frac{1}{2} yz \hat{i} + \frac{1}{2} zx \hat{j} $.
The magnitude of this vector(in effect the area of the triangle in consideration) is given by
$$ A=\frac{1}{2} \sqrt{x^2y^2+y^2z^2+z^2x^2} $$
By using the three relations we derived earlier connecting x, y and z with a, b and c as well as some expansions and cancellations, we can rewrite the above equation as
$$ A=\frac{1}{4} \sqrt{2c^2a^2+2c^2b^2+2a^2b^2-b^4-a^4-c^4} $$
This is what we obtain on expanding the original heron's formula in terms of s, a, b and c after substituting
$$ s=\frac{(a+b+c)}{2} $$
