Heron's Formula/Proof 3

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Theorem

Let $\triangle ABC$ be a triangle with sides $a$, $b$ and $c$ opposite vertices $A$, $B$ and $C$ respectively.


Then the area $\AA$ of $\triangle ABC$ is given by:

$\AA = \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }$

where $s = \dfrac {a + b + c} 2$ is the semiperimeter of $\triangle ABC$.


Proof

Heron3.png

Let $\AA$ 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:

$(1): \quad \AA = \dfrac {r \paren {c + a + b} } 2 = r s$

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.

By Length of Tangent to Excircle:

$s = CD = CE$

Therefore:

$DA = s - b$
$EB = s - a$

Note that:

$AF + DA = BG + EB$

and:

$AF + BG = C$

We need to show that $\triangle MAF \sim \triangle DNA$.

Note that:

$AM$ bisects $\angle BAC$
$AN$ bisects $\angle BAD$

But $\angle BAC$ and $\angle BAD$ are supplementary angles.

Hence the half-angles $\angle MAF$ and $\angle DAN$ are complementary angles.

It follows that:

$\angle MAF = \angle DNA$

Since $\triangle MAF$ and $\triangle DNA$ are both right triangles:

$\triangle DNA \sim \triangle MAF$

As right triangles sharing an acute angle:

$\triangle NDC \sim \triangle MFC$

from which:

\(\ds \dfrac R r\) \(=\) \(\ds \dfrac s {s - c}\)
\(\ds \dfrac R {s - b}\) \(=\) \(\ds \dfrac {s - a} r\)
\(\ds \leadsto \ \ \) \(\ds R\) \(=\) \(\ds \dfrac {r s} {s - c}\) substituting for $R$
\(\ds \) \(=\) \(\ds \dfrac {\paren {s - a} \paren {s - b} } r\)
\(\ds \leadsto \ \ \) \(\ds r^2\) \(=\) \(\ds \dfrac {\paren {s - a} \paren {s - b} \paren {s - c} } s\) rearranging
\(\ds \leadsto \ \ \) \(\ds \AA\) \(=\) \(\ds s \sqrt {\dfrac {\paren {s - a} \paren {s - b} \paren {s - c} } s}\) from $(1)$
\(\ds \) \(=\) \(\ds \sqrt {s \paren {s - a} \paren {s - b} \paren {s - c} }\) simplifying

$\blacksquare$


Source of Name

This entry was named for Heron of Alexandria.


Sources