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.

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Fig. 1

By Pythagorean theorem,



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

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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