## Thursday, March 28, 2013

### 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 sthen $$s=x+y+z$$, $$x=s-a$$, $$y=s-b$$, 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 $$\sqrt{s(s-a)(s-b)(s-c)}$$.