# Could an alternative proof of Heron's formula follow from writing $s^2\frac{\sqrt{(b+c)^2-a^2}}s\frac{\sqrt{(a+c)^2-b^2}}s\frac{\sqrt{(a+b)^2-c^2}}s$?

I've looked at a number of proofs of Heron's formula and other related questions on this forum, but I was just wondering if the following half-baked idea could relate to yet another proof. Rearranging the formula:

\begin{align} K &= \sqrt{s(s-a)(s-b)(s-c)}\\[0.5em] &= s^2\frac{\sqrt{s(s-a)}}s\frac{\sqrt{s(s-b)}}s\frac{\sqrt{s(s-c)}}s \\[0.5em] &= s^2\frac{\sqrt{(b+c)^2-a^2}}{a+b+c}\frac{\sqrt{(a+c)^2-b^2}}{a+b+c}\frac{\sqrt{(a+b)^2-c^2}}{a+b+c} \end{align}

Now the first ratio, for example, is the ratio of final to initial height if you take a line segment of length $$(a+b+c)$$ and collapse it into a right triangle with base $$a$$ and hypotenuse $$b+c$$. So I'm thinking, maybe if you start with a triangle of base $$2s$$ and height $$s$$ or vice versa (hence area $$s^2$$), and perform three such sequential "collapse" operations, you could get the triangle $$ABC$$, while reducing the area by the corresponding ratio each time? Or something roughly in that vein...

I thought that could be nice because it would be a pretty short proof that more closely hews to the symmetry between $$a$$, $$b$$, and $$c$$.