Heron's formula for the area of a triangle

The angle bisectors of the triangle meet at the center of the inscribed circle of radius r.  If we let $2\alpha=A$, $2\beta = B$, and $2\gamma=C$, we have $\alpha+\beta+\gamma=\frac{\pi}{2}$. 

Let x be the distance from the vertex at A to points of tangency, y the distance from B, and z the distance from C.  Then then lengths of the triangle sides opposite A, B, and C are respectively $a=y+z$, $b=x+z$, and $c=x+y$.

Thus if we name the semiperimeter s, then $s=x+y+z$, $x=s-a$, $y=s-b$, and $z=s-c$.

$\tan \alpha =\frac{r}{x}$, $\tan \beta =\frac{r}{y}$, and $\tan \gamma =\frac{r}{z}$.  Because $\gamma$ and $(\alpha+\beta )$ are complementary angles, we obtain

$$\tan\left( \frac{\pi}{2}  - (\alpha+\beta) \right)   = \frac{r}{z}$$
$$\tan\left( \alpha+\beta \right)   = \frac{z}{r}$$

$$\frac{ \tan \alpha+\tan\beta}{1-\tan\alpha \tan\beta}   = \frac{z}{r}$$

$$r \left(\tan \alpha+\tan\beta  \right)= z (1-\tan\alpha\tan\beta)$$

$$r \left( \frac{r}{x}+\frac{r}{y} \right) = z \left( 1 - \frac{r}{x}\frac{r}{y} \right)$$

$$r^2 y + r^2 z = xyz - r^2 z$$

$$r^2 ( x+y+z) = xyz$$

$$r^2 s = xyz$$

The radii at the points of tangency and the angle bisectors form 3 pairs of congruent triangles.  The area of $\Delta ABC$ is $xr+yr+zr= r(x+y+z)$, so area $=rs$, and $(\text{area})^2=r^2s^2$.  Using results we have above, we obtain
$$(\text{area})^2 = s\cdot xyz = s(s-a)(s-b)(s-c)$$
so the area is $\sqrt{s(s-a)(s-b)(s-c)}$.