# Heron's Formula

## Theorem

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

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

$A = \sqrt{s \left({s - a}\right) \left({s - b}\right) \left({s - c}\right)}$

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

## Proof 1

Construct the altitude from $A$.

Let the length of the altitude be $h$ and the foot of the altitude be $D$.

Let the distance from $D$ to $B$ be $z$.

From Pythagoras's Theorem:

$(1): \quad h^2 + \left({a - z}\right)^2 = b^2$

and:

$(2): \quad h^2 + z^2 = c^2$

By subtracting $(1)$ from $(2)$:

$2az - a^2 = c^2 - b^2$

which can be expressed in terms of $z$ as:

$z = \dfrac{a^2 + c^2 - b^2}{2a}$

Substituting for $z$ in $(2)$ and simplifying yields:

$h = \sqrt{c^2 - \left({\dfrac{a^2 + c^2 - b^2}{2a}}\right)^2}$

and so:

 $\displaystyle Q$ $=$ $\displaystyle \frac 1 2 a \sqrt{c^2 - \left({\frac{a^2 + c^2 - b^2}{2a} }\right)^2}$ $\quad$ Area of Triangle in Terms of Side and Altitude $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\frac{4 c^2 a^2 - \left({a^2 + c^2 - b^2}\right)^2} {16} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\frac{\left({2ac - a^2 - c^2 + b^2}\right) \left({2ac + a^2 + c^2 - b^2}\right)} {16} }$ $\quad$ Difference of Two Squares $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\frac{\left({b^2 - \left({a - c}\right)^2}\right)\left({\left({a + c}\right)^2 - b^2}\right)} {16} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\frac{\left({b - a + c}\right) \left({b + a - c}\right) \left({a + c - b}\right) \left({a + b + c}\right)} {16} }$ $\quad$ Difference of Two Squares $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\frac{\left({a + b + c}\right) \left({a + b - c}\right) \left({a - b + c}\right) \left({-a + b + c}\right)} {16} }$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{\left({\frac{a + b + c} 2}\right) \left({\frac{a + b + c} 2 - c} \right) \left({\frac{a + b + c} 2 - b}\right) \left({\frac{a + b + c} 2 - a}\right)}$ $\quad$ $\quad$ $\displaystyle$ $=$ $\displaystyle \sqrt{s \left({s - c}\right) \left({s - b}\right) \left({s - a}\right)}$ $\quad$ Definition of semiperimeter $\quad$

$\blacksquare$

## Proof 2

A triangle can be considered as a cyclic quadrilateral one of whose sides has degenerated to zero.

From Brahmagupta's Formula, the perimeter of a cyclic quadrilateral is given by:

$\sqrt{\left({s - a}\right) \left({s - b}\right) \left({s - c}\right) \left({s - d}\right)}$

where $s$ is the semiperimeter:

$s = \dfrac{a + b + c + d} 2$

The result follows by letting $d$ tend to zero.

$\blacksquare$

## Proof 3

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:

$\mathcal A \left({\triangle AMB}\right) = \dfrac {rc} 2$
$\mathcal A \left({\triangle BMC}\right) = \dfrac {ra} 2$
$\mathcal A \left({\triangle CMA}\right) = \dfrac {rb} 2$

Thus:

$(1): \quad T = \dfrac {r \left({c + a + b}\right)} 2 = r s$

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 $R$:

$R = \dfrac {r s} {s - c} = \dfrac {\left({s - a}\right) \left({s - b}\right)} r$

and so:

$r^2 = \dfrac {\left({s - a}\right) \left({s - b}\right)\left({s - c}\right)} s$

Thus $(1)$ becomes:

$T = s \sqrt {\dfrac {\left({s - a}\right) \left({s - b}\right)\left({s - c}\right)} s} = \sqrt {s \left({s - a}\right) \left({s - b}\right)\left({s - c}\right)}$

$\blacksquare$

## Source of Name

This entry was named for Heron of Alexandria.

## Historical Note

Arabic sources from the Middle Ages inform us that Heron's Formula was actually due to Archimedes of Syracuse.

However, Heron's is the earliest proof that survives.