I have seen an interesting proof of Heron's formula here. It is very simple, but I do not understand one point. The author demands, that the formula should contain factor $(a+b+c)$, because when we take $a=b=c=0$, the area of the triangle should be zero. And because of other reasons the formula should be like this: $$S=k\cdot\sqrt{ (a+b+c)(b+c-a)(a+c-b)(a+b-c) }$$ where $k$ is a certain konstant. I understand, that this formula is homogeneous of degree 2, which it is supposed to be.
But my question is - why it could not be for example like this: $$S=k\cdot \sqrt[4]{a^2+b^2+c^2} \cdot \sqrt{ (b+c-a)(a+c-b)(a+b-c) }$$ because it's also homogeneous of degree 2 and colapses in case, when $a=b=c=0$?
Or is this not a proof, but just an analysis of the Heron's formula?
Here's one way to explain it. Let $s$ denote the triangle's semiperimeter. The $a+b-c$ factor is $2s-2c$, but let's talk about $s-c$ instead because we have a proportionality constant anyway. Now, for a cyclic quadrilateral we get an area of $\sqrt{(s-a)(s-b)(s-c)(s-d)}$ (with $s:=(a+b+c+d)/2$ for sides $a,\,b,\,c,\,d$), of which Heron's formula is the $d=0$ special case obtained by sliding one vertex onto another. So instead of trying to motivate Heron, let's try to motivate Brahmagupta's formula. The linked "proof" has a geometry justification for $s-c$; all the other factors come the same way. What's more, at that point the square-rooted product already has the units of area, so we need no more factors with units of length.