What is proof of Heron's Formula?

Answer 1

This is a proof that uses the cosine rule and Pythagoras' theorem.

As on any triangle with c being the opposite side of θ and a and b are the other sides:

c^2=a^2+b^2-2abcosθ

We can rearrange this for θ:

θ=arccos[(a^2+b^2-c^2)/(2ab)]

On a right-angle triangle cosθ=a/h. We can therefore construct a right-angle triangle with θ being one of the angles, the adjacent side being a^2+b^2-c^2 and the hypotenuse being 2ab. As the formula for the area of a triangle is also absinθ/2, when a and b being two sides and θ the angle between them, the opposite side of θ on the right-angle triangle we have constructed is 4A, with A being the area of the original triangle, as it is 2absinθ.

Therefore, according to Pythagoras' theorem:

(2ab)^2=(a^2+b^2-c^2)^2+(4A)^2

4a^2*b^2=(a^2+b^2-c^2)^2+16A^2

16A^2=4a^2*b^2-(a^2+b^2-c^2)^2

This is where it will start to get messy:

16A^2=4a^2*b^2-(a^2+b^2-c^2)(a^2+b^2-c^2)

=4a^2*b^2-(a^4+a^2*b^2-a^2*c^2+a^2*b^2+b^4-b^2*c^2- a^2*c^2-b^2*c^2+c^4)

=4a^2*b^2-(a^4+2a^2*b^2-2a^2*c^2+b^4-2b^2*c^2+c^4)

=-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.1)

We will now see:

(a+b+c)(-a+b+c)(a-b+c)(a+b-c)

=(-a^2+ab+ac-ab+b^2+bc-ac+bc+c^2)(a^2+ab-ac-ab-b^2+bc+ac+bc-c^2)

=(-a^2+b^2+2bc+c^2)(a^2-b^2+2bc-c^2)

=-a^4+a^2*b^2-2a^2*bc+a^2*c^2+a^2*b^2-b^4+2b^3*c-b^2*c^2+2a^2*bc-2b^3*c+(2bc)^2-2bc^3+a^2*c^2-b^2*c^2+2bc^3-c^4

=-a^4+2a^2*b^2+2a^2*c^2-b^4+(2bc)^2-c^4-2b^2*c^2

=-a^4+2a^2*b^2+2a^2*c^2-b^4+2b^2*c^2-c^4 (Eq.2)

And now that we know that Eq.1=Eq.2, we can make Eq.1=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)

Therefore:

16A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)

A^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)/16

=[(a+b+c)/2][(-a+b+c)/2][(a-b+c)/2][(a+b-c)/2]

And so if we let s=(a+b+c)/2

A^2=s(s-a)(s-b)(s-c)