余弦定理
$$\begin{eqnarray}
|BH|^2&=&c^2-|AH|^2
\\&=&c^2-c^2 \cos^2(A)
\\&=&a^2-|CH|^2
\\&=&a^2-\{b-c \cos(A)\}^2
\\&=&a^2-b^2+2bc\cos(A)-c^2\cos^2(A)
\end{eqnarray}$$
$$\begin{eqnarray}
c^2\cancel{-c^2 \cos^2(A)}&=&a^2-b^2+2bc\cos(A)\cancel{-c^2\cos^2(A)}
\\c^2&=&a^2-b^2+2bc\cos(A)
\\b^2+c^2-a^2&=&2bc\cos(A)
\\\cos(A)&=&\frac{b^2+c^2-a^2}{2bc}
\end{eqnarray}$$
ヘロンの公式
$$\begin{eqnarray}
S^2&=&\left(\frac{1}{2}|AC||BH|\right)^2
\\&=&\left\{\frac{1}{2}bc\sin{(A)}\right\}^2
\\&=&\frac{1}{4}b^2c^2\sin^2{(A)}
\\&=&\frac{1}{4}b^2c^2\left\{1-\cos^2{(A)}\right\}\;\cdots\;\sin^2{(x )}+\cos^2{(x)}=1
\\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\cos^2{(A)}
\\&=&\frac{1}{4}b^2c^2-\frac{1}{4}b^2c^2\left(\frac{b^2+c^2-a^2}{2bc}\right)^2\;\cdots\;\cos{(A)}=\frac{b^2+c^2-a^2}{2bc}
\\&=&\frac{1}{4}b^2c^2-\frac{1}{4}\cancel{b^2c^2}\frac{\left(b^2+c^2-a^2\right)^2}{4\cancel{b^2c^2}}
\\&=&\frac{1}{4}b^2c^2-\frac{1}{16}\left(b^2+c^2-a^2\right)^2
\\&=&\frac{1}{16}\left\{4b^2c^2-\left(b^2+c^2-a^2\right)^2\right\}
\\&=&\frac{1}{16}\left\{\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\right\}
\\&=&\frac{1}{16}\left[\left\{2bc+\left(b^2+c^2-a^2\right)\right\}\left\{2bc-\left(b^2+c^2-a^2\right)\right\}\right]
\\&=&\frac{1}{16}\left\{\left(2bc+b^2+c^2-a^2\right)\left(2bc-b^2-c^2+a^2\right)\right\}
\\&=&\frac{1}{16}\left[\left\{\left(b^2+2bc+c^2\right)-a^2\right\}\left\{a^2-\left(b^2-2bc+c^2\right)\right\}\right]
\\&=&\frac{1}{16}\left[\left\{\left(b+c\right)^2-a^2\right\}\left\{a^2-\left(b-c\right)^2\right\}\right]\;\cdots\;(a \pm b)^2=a^2\pm2ab+b^2
\\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-\left(b-c\right)\right)\right\}\right]
\\&=&\frac{1}{16}\left[\left\{\left(b+c+a\right)\left(b+c-a\right)\right\}\left\{\left(a+b-c\right)\left(a-b+c\right)\right\}\right]
\\&=&\frac{1}{16}\left(a+b+c\right)\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)
\\&=&\frac{a+b+c}{2}\frac{-a+b+c}{2}\frac{a-b+c}{2}\frac{a+b-c}{2}
\\&=&\frac{a+b+c}{2}\frac{-a+b+c\color{red}{+a-a}}{2}\frac{a-b+c\color{red}{+b-b}}{2}\frac{a+b-c\color{red}{+c-c}}{2}
\\&=&\frac{a+b+c}{2}\frac{a+b+c-2a}{2}\frac{a+b+c-2b}{2}\frac{a+b+c-2c}{2}
\\&=&\frac{a+b+c}{2}\left(\frac{a+b+c}{2}-a\right)\left(\frac{a+b+c}{2}-b\right)\left(\frac{a+b+c}{2}-c\right)
\\&=&s\left(s-a\right)\left(s-b\right)\left(s-c\right)\;\cdots\;s=\frac{a+b+c}{2}
\end{eqnarray}$$
$$\begin{eqnarray}
S&=&\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}\;\cdots\;s=\frac{a+b+c}{2}
\end{eqnarray}$$
0 件のコメント:
コメントを投稿