Proof of Heron's Formula for the Area of a Triangle
\[a, \: b, \: c\]
has has area given by Heron's Formula \[A= \sqrt{s(s-a)(s-b)(s-c)}\]
where \[s=\frac{a+b+c}{2}\]
.To prove it start with the simple formula
\[A=\frac{1}{2}ab sin C\]
and using the Cosine Rule \[cosC=\frac{a^2+b^2-c^2}{2ab}\]
. Then\[\begin{equation} \begin{aligned} A &= \frac{1}{2}ab sinC \\ &= \frac{1}{2}ab \sqrt{1-cos^2C} \\ &= \frac{1}{2}ab \sqrt{1-(\frac{a^2+b^2-c^2}{2ab})^2} \\ &= \sqrt{\frac{a^2b^2}{4}-\frac{a^2+b^2-c^2}{16}} \\ &= \sqrt{(\frac{ab}{2} - \frac{a^2+b^2-c^2}{4})(\frac{ab}{2} + \frac{a^2+b^2-c^2}{4})} \\ &= \sqrt{(\frac{2ab -a^2-b^2+c^2}{4})(\frac{2ab +a^2+b^2-c^2}{4})} \\ &= \sqrt{(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2})(\frac{a+b+c}{2}-c)} \\ &= \sqrt{(s-a)(s-b)s(s-c)} \\ &= \sqrt{s(s-a)(s-b)(s-c)} \end{aligned} \end{equation}\]
