This is not the usual request for an intuitive proof, which has been asked already.
Having looked at various sources, I've basically concluded that Heron's formula relies on proving
$$xyz = x+y+z$$
where $x$, $y$ and $z$ are the lengths between the meeting point of the incircle and the sides, and the vertices. If I take the diagram below, $x$ is $CX$, and $z$ is $XA$. $y$ would be the segment from $B$ to the incircle meeting the side, marked by a black dot, which is also $YC$. Hence $x+y+z=s$ where $s$ is the semiperimeter.
From here it is easy to show that if the radius ($r$) of the incircle is $1$, then the area of the triangle is $x+y+z$. The radius does not really matter, because when it comes to proving the formula, you can just proportionally reduce the size of the triangle by a factor of $1/r^2$. So we can work with the $r=1$ case.
Proving $xyz = x+y+z$ visually is not difficult by looking at this diagram:
http://jwilson.coe.uga.edu/emt725/Heron/Heron2/Heron2.html
When $r = 1$, then $EY = xy$ ($x = (s-c)$ and $y = (s-b)$ since $s = x+y+z$ as defined above) and $EY$ is also equal to $s/z$, so it is conceptually not difficult to show that $x+y+z = xyz$ and from this that $A^2 = (x+y+z)xyz = s(s-a)(s-b)(s-c)$.
But to my mind, it would be even better if we could show that $xyz$ corresponds to the area of the triangle directly, rather than messing about with equivalences. If we take the $\Delta AYE$ triangle in the above diagram and form a rectangle, whose new vertex we call $P$, and we extend a line from point $X$ to the side $EP$, calling this new vertex $Q$, then $PQXA$ is meant to be of the same area as the triangle, since $XA=z$ and $EY=xy$. Any ways to prove that rectangle $PQXA$ is equivalent to triangle $\Delta ABC$? Or perhaps there's a better way to prove $xyz$ corresponds to the area of the triangle $\Delta ABC$?
Cheers
Best Answer
First, let's rectify the dimensionality issues by refraining from setting $r=1$.
With liberal re-labeling of the figure ...
... we have
$$\left.\begin{array}{cr} \triangle A'E'C\sim\triangle CEI \to & \dfrac{h}{s-b} =\dfrac{s-c}{r} \\[6pt] \triangle A'E'A\sim\triangle IEA \to & \dfrac{h}{s} = \dfrac{r}{s-a} \end{array}\right\}\to \frac{(s-b)(s-c)}{r}=h=\frac{rs}{s-a} \tag{1}$$ so that $$(s-a)(s-b)(s-c) = s r^2 \tag{2}$$
Now, one may read the right-hand side of $(2)$ as $|\triangle ABC|^2/s$, so that $|\triangle ABC|^2=s(s-a)(s-b)(s-c)$; that's Heron. OP prefers to interpret the right-hand side as $|\triangle ABC|\,r$, and thus seeks to establish directly that $$|\triangle ABC| = \frac{(s-a)(s-b)(s-c)}{r} \tag{3}$$
In particular, since $h=(s-b)(s-c)/r$ (via $(1)$), OP suggests showing $(3)$ via a demonstration that $|\triangle ABC|=h(s-a)$, perhaps by treating the product as twice the area of a triangle with base $s-a$ and height $h$. While OP considers introducing a rectangle, there's a more-natural option:
A little angle-chasing shows that the marked angles at $B$ and $F$ are congruent, as are those at $C$ and $E$. Thus,
$$\left.\begin{array}{r} \overline{A'B}\parallel\overline{DF}\;\to\;|\triangle A'DF|=|\triangle BDF| \\ \overline{A'C}\parallel\overline{DE}\;\to\;|\triangle A'DE|=|\triangle CDE| \end{array}\right\}\;\to\; \begin{align} \\ \\ |\triangle ABC| &= \phantom{2}\;|\square AFA'E| \\ &=2\;|\triangle AA'E| \\ &= \phantom{2}\;|AE|\;|A'E'|\end{align} \tag{$\star$}$$ as desired. $\square$
It's also worth noting that $|\triangle ABC|=h(s-a)=\frac12h(-a+b+c)$ follows immediately from $A'$'s role as an excenter (being equidistant from the side-lines of $\triangle ABC$): $$|\triangle ABC|+|\triangle A'BC| = |\square ABA'C| = |\triangle AA'C|+|\triangle AA'B|$$ $$\begin{align}\to\quad |\triangle ABC| &= -|\triangle A'BC|+|\triangle AA'C|+|\triangle AA'B| \\[4pt] &=-\tfrac12ha+\tfrac12hb+\tfrac12hc \\[4pt] &=\phantom{-}\tfrac12h(-a+b+c) \end{align}$$