ヘロンの公式

\(\triangle ABC\)で頂点\(A,B,C\)の対辺の長さを\(a,b,c\)とする。
\begin{align*} S & =\frac{1}{2}\left|\overrightarrow{CA}\times\overrightarrow{CB}\right|\\ & =\frac{1}{2}\left|CA\right|\left|CB\right|\left|\sin\left(\angle BCA\right)\right|\\ & =\frac{1}{2}\left|CA\right|\left|CB\right|\sqrt{1-\cos^{2}\left(\angle BCA\right)}\\ & =\frac{1}{2}\left|CA\right|\left|CB\right|\sqrt{1-\left(\frac{\left|AC\right|^{2}+\left|CB\right|^{2}-\left|AB\right|^{2}}{2\left|AC\right|\left|CB\right|}\right)^{2}}\\ & =\frac{1}{2}ab\sqrt{1-\left(\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)^{2}}\\ & =\frac{1}{2}ab\sqrt{\left(1+\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)\left(1-\frac{a^{2}+b^{2}-c^{2}}{2ab}\right)}\\ & =\frac{1}{4}\sqrt{-\left(\left(a+b\right)^{2}-c^{2}\right)\left(\left(a-b\right)^{2}-c^{2}\right)}\\ & =\frac{1}{4}\sqrt{-\left(a+b+c\right)\left(a+b-c\right)\left(a-b+c\right)\left(a-b-c\right)}\\ & =\frac{1}{4}\sqrt{\left(a+b+c\right)\left(b+c-a\right)\left(a+c-b\right)\left(a+b-c\right)}\\ & =\frac{1}{4}\sqrt{\left(a+b+c\right)\left(a+b+c-2a\right)\left(a+b+c-2b\right)\left(a+b+c-2c\right)}\\ & =\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \end{align*}