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.

Nobody can provide the proof but many can provide a proof or perhaps many proofs.
Notice that the area must be a 2nd-degree homogeneous function of $a$, $b$, and $c$, for example, if you multiply $a$, $b$, and $c$ by $9$ then you multiply the area by $9^2=81$, etc.
Next, notice that the area must be $0$ if $a+b=c$: if the distance along one side plus the distance along another side is equal to the distance along the third side, then the three corners are on a straight line, so the area is $0$.  For that reason $a+b-c$ should appear as a factor, i.e. as something you multiply by.
For the same reason, $b+c-a$ and $c+a-b$ should be factors.
Next, notice that if $a=b=c=0$, then the area must be $0$, so that's why $a+b+c$ should be a factor.
Now we have $(a+b+c)(a+b-c)(b+c-a)(c+a-b)$.  That is homogeneous of degree $4$ rather than $2$, so take its square root and we have $\sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}$, which is homogenous of degree $2$.  That, then, should be proportional to the area.
Now let's see what the constant of proportionality is: The area of a right triangle with legs of length $1$ and hypotenuse of length $\sqrt2$ is $1/2$.  Plugging in those three numbers for $a$, $b$, and $c$ we get
$$
\frac 1 2 =\text{constant}\times\sqrt{\left(1+1+\sqrt 2\right) \left(1+1-\sqrt 2\right)\left(1-1+\sqrt 2\right)\left(-1+1+\sqrt2\right)} = \text{constant}\times 2.
$$
So the "constant" is $\dfrac 1 4$ and finally we have
$$
\text{area} = \frac 1 4 \sqrt{(a+b+c)(a+b-c)(b+c-a)(c+a-b)}.
$$
That's Heron's formula.
