余弦定理を組み合わせて対角線の長さを辺の長さだけで表す
$$\begin{eqnarray}
|BD|^2&=&a^2+d^2-2ad\cos{(A)}\;\cdots\;余弦定理
\\&=&b^2+c^2-2bc\cos{(C)}\;\cdots\;余弦定理
\\&=&b^2+c^2-2bc\cos{(\pi-A)}\;\cdots\;A+C=\pi,\;C=\pi-A
\\&&\cdots\;円に内接する四角形の向かい合う角の和は180度(=\pi)
\\&=&b^2+c^2+2bc\cos{(A)}\;\cdots\;\cos{(\pi-x)}=-\cos{(x)}
\end{eqnarray}$$
$$\begin{eqnarray}
a^2+d^2-2ad\cos{(A)}&=&b^2+c^2+2bc\cos{(A)}
\\a^2+d^2-b^2-c^2&=&2\left(ad+bc\right)\cos{(A)}
\\\cos{(A)}&=&\frac{a^2+d^2-b^2-c^2}{2\left(ad+bc\right)}
\end{eqnarray}$$
ブラーマグプタの公式
$$\begin{eqnarray}
S=S_1+S_2&=&\frac{1}{2}|AD||BE|+\frac{1}{2}|BC||DF|
\\&=&\frac{1}{2}|AD||AB|\sin{(A)}+\frac{1}{2}|BC||CD|\sin{(C)}
\\&=&\frac{1}{2}da\sin{(A)}+\frac{1}{2}bc\sin{(A)}\;\cdots\;C=\pi-A,\;\sin{(\pi-x)}=\sin{(x)}
\\&=&\frac{1}{2}\left(ad+bc\right)\sin{(A)}
\end{eqnarray}$$
$$\begin{eqnarray}
S^2&=&\left\{\frac{1}{2}\left(ad+bc\right)\sin{(A)}\right\}^2
\\&=&\left\{\frac{1}{2}\left(ad+bc\right)\sin{(A)}\right\}^2
\\&=&\frac{1}{4}\left(ad+bc\right)^2\sin^2{(A)}
\\&=&\frac{1}{4}\left(ad+bc\right)^2\left\{1-\cos^2{(A)}\right\}\;\cdots\;\sin^2{(x )}+\cos^2{(x)}=1
\\&=&\frac{1}{4}\left(ad+bc\right)^2\left\{1-\left(\frac{a^2+d^2-b^2-c^2}{2\left(ad+bc\right)}\right)^2\right\}
\;\cdots\;\cos{(A)}=\frac{a^2+d^2-b^2-c^2}{2\left(ad+bc\right)}
\\&=&\frac{1}{4}\cancel{\left(ad+bc\right)^2}
\frac{4\left(ad+bc\right)^2-\left(a^2+d^2-b^2-c^2\right)^2}{4\cancel{\left(ad+bc\right)^2}}
\\&=&\frac{1}{16}\left\{4\left(ad+bc\right)^2-\left(a^2+d^2-b^2-c^2\right)^2\right\}
\\&=&\frac{1}{16}\left\{2\left(ad+bc\right)+\left(a^2+d^2-b^2-c^2\right)\right\}\left\{2\left(ad+bc\right)-\left(a^2+d^2-b^2-c^2\right)\right\}
\\&=&\frac{1}{16}\left\{\left(a^2+2ad+d^2\right)-\left(b^2-2bc+c^2\right)\right\}\left\{\left(b^2+2bc+c^2\right)-\left(a^2-2ad+d^2\right)\right\}
\\&=&\frac{1}{16}\left\{\left(a+d\right)^2-\left(b-c\right)^2\right\}\left\{\left(b+c\right)^2-\left(a-d\right)^2\right\}
\\&=&\frac{1}{16}
\left\{
\left(a+d\right)+\left(b-c\right)
\right\}
\left\{
\left(a+d\right)-\left(b-c\right)
\right\}
\left\{
\left(b+c\right)+\left(a-d\right)
\right\}
\left\{
\left(b+c\right)-\left(a-d\right)
\right\}
\\&=&\frac{1}{16}
\left(a+b-c+d\right)
\left(a-b+c+d\right)
\left(a+b+c-d\right)
\left(-a+b+c+d\right)
\\&=&\frac{1}{16}
\left(-a+b+c+d\color{red}{+a-a}\color{black}{}\right)
\left(a-b+c+d\color{red}{+b-b}\color{black}{}\right)
\left(a+b-c+d\color{red}{+c-c}\color{black}{}\right)
\left(a+b+c-d\color{red}{+d-d}\color{black}{}\right)
\\&=&
\frac{a+b+c+d-2a}{2}
\frac{a+b+c+d-2b}{2}
\frac{a+b+c+d-2c}{2}
\frac{a+b+c+d-2d}{2}
\\&=&
\left(\frac{a+b+c+d}{2}-a\right)
\left(\frac{a+b+c+d}{2}-b\right)
\left(\frac{a+b+c+d}{2}-c\right)
\left(\frac{a+b+c+d}{2}-d\right)
\\&=&
\left(s-a\right)
\left(s-b\right)
\left(s-c\right)
\left(s-d\right)\;\cdots\;s=\frac{a+b+c+d}{2}
\end{eqnarray}$$
$$\begin{eqnarray}
S&=&\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-d\right)}\;\cdots\;s=\frac{a+b+c+d}{2}
\\\left.S\right|_{d=0}&=&\sqrt{\left(s-a\right)\left(s-b\right)\left(s-c\right)\left(s-0\right)}\;\cdots\;s=\frac{a+b+c+0}{2}
\\&=&\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\;\cdots\;s=\frac{a+b+c}{2}\;\cdots\;ヘロンの公式
\end{eqnarray}$$
