<s>[INST] Prove that $\sqrt{s(s-a)(s-b)(s-c)}=A$ In 50 AD, the Heron of Alexandria came up with the well-known formula, that, given the three side lengths of a triangle (or even two and an angle, thanks to trigonometry) you can get the area of said triangle by using this formula:
$$
\text{if } s=\frac{a+b+c}{2},\\
\text{then} A=\sqrt{s(s-a)(s-b)(s-c)}
$$
But the book it came with (mathematics 1001) did not give a mathematical proof to this. So I tried and failed to get a proof to this. This is how my train wreck went…
$$\begin{align}
\text{equation }&\sqrt{s(s-a)(s-b)(s-c)}\\
\text{substitute }&\sqrt{\frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)}\\
\text{eliminate fractions}&\sqrt{a+b+c(a+b+c-2a)(a+b+c-2b)(a+b+c-2c)}\\
\text{comb. Like terms }&\sqrt{a+b+c(b+c-a)(a-b+c)(a+b-c)}\\
\text{associative property }&\sqrt{(a+(b+c)((b+c)-a))((a-(b+c))(a+(b-c)))}\\
\text{difference of 2 squares }&\sqrt{((b+c)^2-a^2)(a^2-b^2-c^2)}\\
\text{square of two numbers }& \sqrt{((b^2+bc+c^2)-a^2)(a^2-b^2-c^2)}\\
\text{polynomial multiplication}& \sqrt{2(a^2b^2-b^2c^2+a^2c^2)+a^2(b-c+a^2)-(b+c)(b^3+c^3)}\\
\end{align}$$
The issue is, I don't know how to get to $A$, the area. Does anyone know how to continue the proof train?? Thanks in advance.
 [/INST] Consider a triangle $ABC$.
$[ABC]=\frac{ab}{2}\sin C$
$=\frac{ab}{2}\sqrt{1-\cos^2 C}$
$=\frac{ab}{2}\sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}$
$=\sqrt{\frac{a^2b^2}{4}\left[1-\frac{(a^2+b^2-c^2)^2}{4a^2b^2}\right]}$
$=\sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{16}}$
$=\sqrt{\frac{(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)}{16}}$
$=\sqrt{\frac{((a+b)^2-c^2)(c^2-(a-b)^2)}{16}}$
$=\sqrt{\frac{(a+b+c)(a+b-c)(b+c-a)(a+c-b)}{16}}$
$=\sqrt{s(s-a)(s-b)(s-c)}$
Hence proved. Hope it helps
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