Heron's Formula/Proof 3

Proof

 * Heron3.png

Let $T$ be the area of $\triangle ABC$.

Construct the incircle of $\triangle ABC$.

Let the incenter of $\triangle ABC$ be $M$.

Let the inradius of $\triangle ABC$ be $r$.

$\triangle ABC$ is made up of three triangles: $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$.

From Area of Triangle in Terms of Side and Altitude, the areas of $\triangle AMB$, $\triangle BMC$ and $\triangle CMA$ are given by:
 * $\Area \paren {\triangle AMB} = \dfrac {r c} 2$


 * $\Area \paren {\triangle BMC} = \dfrac {r a} 2$


 * $\Area \paren {\triangle CMA} = \dfrac {r b} 2$

Thus:
 * $\paren 1: \quad T = \dfrac {r \paren {c + a + b} } 2 = r s$

where $s$ is the semiperimeter of $\triangle ABC$.

Construct the excircle of $\triangle ABC$ with excenter $N$ tangent to $AB$, and to $AC$ and $BC$ produced at $D$ and $E$ respectively.

We note that $s = CD = CE$.

Therefore:
 * $DA = s - b$


 * $EB = s - a$

Note that:
 * $AF + DA = BG + EB$

and:
 * $AF + BG = C$

Note also that:
 * $\triangle NDC$ is similar to $\triangle MFC$


 * $\triangle NDA$ is similar to $\triangle AFM$

from which:
 * $\dfrac R r = \dfrac s {s - c}$


 * $\dfrac R {s - b} = \dfrac {s - a} r$

Substituting for $R$:

$R = \dfrac {r s} {s - c} = \dfrac {\paren {s - a} \paren {s - b} } r$

and so:
 * $r^2 = \dfrac {\paren {s - a} \paren {s - b} \paren {s - c}} s$

Thus $\paren 1$ becomes:
 * $T = s \sqrt {\dfrac {\paren {s - a} \paren {s - b} \paren {s - c}} s} = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c}}$