# An Algebraic Proof of Heron’s Formula

Heron’s formula concerning $A$, the area of any triangle states:

$A = \sqrt{s(s-a)(s-b)(s-c)}\quad\quad\quad\quad\quad\quad\quad(1)$

where a,b,c are the three sides of the triangle and, $s=\frac{a+b+c}{2}$.

We are going to prove it with the aid of a CAS:

Substituting $s$ into (1), the formula becomes

$A = \frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}\quad\quad\quad\quad\quad(2)$

A triangle with three known sides is shown in Fig.1 where $x$ is part of the base of the triangle.

Fig. 1

By Pythagorean theorem,

$h^2+x^2=b^2$

$h^2+(a-x)^2=c^2$

To obtain $h^2$, we will use Omega CAS Explorer (see Fig. 2)

The function ‘eliminate’ eliminates variable $x$, returns the value of $h^2$

Fig. 2

The result is $h^2 = \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{4a^2}$, i.e.,

$h = \frac{\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{2a}$.

Therefore, using the standard formula for triangle,

$A = \frac{1}{2}a h =\frac{ \sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}}{4}$

which is (2)

This is the 1st example in my presentation at ACA 2013 titled “An Algebraic Approach to Geometric Proof Using a Computer Algebra System”.

In Fig. 3, $BA \perp CA, BD \perp CD$. Can you find the area of $\Delta ACD$?

Fig. 3