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)
