According to Heath, this proof is really due to Archimedes.
It should be mentioned that it is of course a lot easier to prove the result
using trigonometry! The area of a triangle is 1/2ab.sinC, and using c^{2}=a^{2}+b^{2}-2ab.cosC,
and sin^{2}+cos^{2}=1, the result follows in a few lines.

Here they are:

_{}

Note that the *ab* outside cancels that inside, so all
we have to do is rearrange the numerator inside. For convenience, I change the
overall sign of the expression to get:

(a^{2} + b^{2} – c^{2})^{2}
- 4a^{2}b^{2} = (a^{2}-b^{2})^{2} –
2c^{2}(a^{2} +b^{2}) + c^{4}

=
((a + b)^{2} – c^{2})((a – b)^{2} – c^{2})

= (a + b + c)(a + b – c)(a – b + c)(a – b – c)

QED