Heron's Formula - Proof

Proof

A modern proof, which uses algebra and is quite unlike the one provided by Heron (in his book Metrica), follows. Let a, b, c be the sides of the triangle and A, B, C the angles opposite those sides. We have

by the law of cosines. From this proof get the algebraic statement:

The altitude of the triangle on base a has length b·sin(C), and it follows

\begin{align} T & = \frac{1}{2} (\mbox{base}) (\mbox{altitude}) \\ & = \frac{1}{2} ab\sin \widehat C \\ & = \frac{1}{4}\sqrt{4a^2 b^2 -(a^2 +b^2 -c^2)^2} \\ & = \frac{1}{4}\sqrt{(2a b -(a^2 +b^2 -c^2))(2a b +(a^2 +b^2 -c^2))} \\ & = \frac{1}{4}\sqrt{(c^2 -(a -b)^2)((a +b)^2 -c^2)} \\ & = \sqrt{\frac{(c -(a -b))(c +(a -b))((a +b) -c)((a +b) +c)}{16}} \\ & = \sqrt{\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}\frac{(a + b + c)}{2}} \\ & = \sqrt{\frac{(a + b + c)}{2}\frac{(b + c - a)}{2}\frac{(a + c - b)}{2}\frac{(a + b - c)}{2}} \\ & = \sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}. \end{align}

The difference of two squares factorization was used in two different steps.

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