ブラーマグプタの公式

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

$$\begin{array}{rcl}|BD{|}^2& =& a^2+d^2-2ad\cos (A)\cdots \mathrm{余}\mathrm{弦}\mathrm{定}\mathrm{理}\\ & =& b^2+c^2-2bc\cos (C)\cdots \mathrm{余}\mathrm{弦}\mathrm{定}\mathrm{理}\\ & =& b^2+c^2-2bc\cos (\pi -A)\cdots A+C=\pi ,C=\pi -A\\ & & \cdots \mathrm{円}\mathrm{に}\mathrm{内}\mathrm{接}\mathrm{す}\mathrm{る}\mathrm{四}\mathrm{角}\mathrm{形}\mathrm{の}\mathrm{向}\mathrm{か}\mathrm{い}\mathrm{合}\mathrm{う}\mathrm{角}\mathrm{の}\mathrm{和}\mathrm{は}180\mathrm{度}(=\pi )\\ & =& b^2+c^2+2bc\cos (A)\cdots \cos (\pi -x)=-\cos (x)\end{array}$$ $$\begin{array}{rcl}{a}^2+d^2-2ad\cos (A)& =& b^2+c^2+2bc\cos (A)\\ a^2+d^2-b^2-c^2& =& 2(ad+bc)\cos (A)\\ \cos (A)& =& \frac{a^2+d^2-b^2-c^2}{2(ad+bc)}\end{array}$$

ブラーマグプタの公式

$$\begin{array}{rcl}S=S_1+S_2& =& \frac{1}{2}|AD||BE|+\frac{1}{2}|BC||DF|\\ & =& \frac{1}{2}|AD||AB|\sin (A)+\frac{1}{2}|BC||CD|\sin (C)\\ & =& \frac{1}{2}da\sin (A)+\frac{1}{2}bc\sin (A)\cdots C=\pi -A,\sin (\pi -x)=\sin (x)\\ & =& \frac{1}{2}(ad+bc)\sin (A)\end{array}$$ $$\begin{array}{rcl}{S}^2& =& {\{\frac{1}{2}(ad+bc)\sin (A)\}}^2\\ & =& {\{\frac{1}{2}(ad+bc)\sin (A)\}}^2\\ & =& \frac{1}{4}{(ad+bc)}^2{\sin }^2(A)\\ & =& \frac{1}{4}{(ad+bc)}^2\{1-{\cos }^2(A)\}\cdots {\sin }^2(x)+{\cos }^2(x)=1\\ & =& \frac{1}{4}{(ad+bc)}^2\{1-{\left(\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}\right)}^2\}\cdots \cos (A)=\frac{a^2+d^2-b^2-c^2}{2(ad+bc)}\\ & =& \frac{1}{4}\cancel{{(ad+bc)}^2}\frac{4{(ad+bc)}^2-{(a^2+d^2-b^2-c^2)}^2}{4\cancel{{(ad+bc)}^2}}\\ & =& \frac{1}{16}\{4{(ad+bc)}^2-{(a^2+d^2-b^2-c^2)}^2\}\\ & =& \frac{1}{16}\{2(ad+bc)+(a^2+d^2-b^2-c^2)\}\{2(ad+bc)-(a^2+d^2-b^2-c^2)\}\\ & =& \frac{1}{16}\{(a^2+2ad+d^2)-(b^2-2bc+c^2)\}\{(b^2+2bc+c^2)-(a^2-2ad+d^2)\}\\ & =& \frac{1}{16}\{{(a+d)}^2-{(b-c)}^2\}\{{(b+c)}^2-{(a-d)}^2\}\\ & =& \frac{1}{16}\{(a+d)+(b-c)\}\{(a+d)-(b-c)\}\{(b+c)+(a-d)\}\{(b+c)-(a-d)\}\\ & =& \frac{1}{16}(a+b-c+d)(a-b+c+d)(a+b+c-d)(-a+b+c+d)\\ & =& \frac{1}{16}(-a+b+c+d+a-a)(a-b+c+d+b-b)(a+b-c+d+c-c)(a+b+c-d+d-d)\\ & =& \frac{a+b+c+d-2a}{2}\frac{a+b+c+d-2b}{2}\frac{a+b+c+d-2c}{2}\frac{a+b+c+d-2d}{2}\\ & =& (\frac{a+b+c+d}{2}-a)(\frac{a+b+c+d}{2}-b)(\frac{a+b+c+d}{2}-c)(\frac{a+b+c+d}{2}-d)\\ & =& (s-a)(s-b)(s-c)(s-d)\cdots s=\frac{a+b+c+d}{2}\end{array}$$ $$\begin{array}{rcl}S& =& \sqrt{(s-a)(s-b)(s-c)(s-d)}\cdots s=\frac{a+b+c+d}{2}\\ {S|}_{d=0}& =& \sqrt{(s-a)(s-b)(s-c)(s-0)}\cdots s=\frac{a+b+c+0}{2}\\ & =& \sqrt{s(s-a)(s-b)(s-c)}\cdots s=\frac{a+b+c}{2}\cdots \mathrm{ヘ}\mathrm{ロ}\mathrm{ン}\mathrm{の}\mathrm{公}\mathrm{式}\end{array}$$

ヘロンの公式

余弦定理

$$\begin{array}{rcl}|BH{|}^2& =& c^2-|AH{|}^2\\ & =& c^2-c^2{\cos }^2(A)\\ & =& a^2-|CH{|}^2\\ & =& a^2-\{b-c\cos \left(A\right){\}}^2\\ & =& a^2-b^2+2bc\cos \left(A\right)-c^2{\cos }^2(A)\end{array}$$ $$\begin{array}{rcl}{c}^2\cancel{-c^2{\cos }^2(A)}& =& a^2-b^2+2bc\cos \left(A\right)\cancel{-c^2{\cos }^2(A)}\\ c^2& =& a^2-b^2+2bc\cos \left(A\right)\\ b^2+c^2-a^2& =& 2bc\cos \left(A\right)\\ \cos \left(A\right)& =& \frac{b^2+c^2-a^2}{2bc}\end{array}$$

ヘロンの公式

$$\begin{array}{rcl}{S}^2& =& {(\frac{1}{2}|AC||BH|)}^2\\ & =& {\{\frac{1}{2}bc\sin (A)\}}^2\\ & =& \frac{1}{4}{b}^2{c}^2{\sin }^2(A)\\ & =& \frac{1}{4}{b}^2{c}^2\{1-{\cos }^2(A)\}\cdots {\sin }^2(x)+{\cos }^2(x)=1\\ & =& \frac{1}{4}{b}^2{c}^2-\frac{1}{4}{b}^2{c}^2{\cos }^2(A)\\ & =& \frac{1}{4}{b}^2{c}^2-\frac{1}{4}{b}^2{c}^2{\left(\frac{b^2+c^2-a^2}{2bc}\right)}^2\cdots \cos (A)=\frac{b^2+c^2-a^2}{2bc}\\ & =& \frac{1}{4}{b}^2{c}^2-\frac{1}{4}\cancel{b^2{c}^2}\frac{{(b^2+c^2-a^2)}^2}{4\cancel{b^2{c}^2}}\\ & =& \frac{1}{4}{b}^2{c}^2-\frac{1}{16}{(b^2+c^2-a^2)}^2\\ & =& \frac{1}{16}\{4b^2{c}^2-{(b^2+c^2-a^2)}^2\}\\ & =& \frac{1}{16}\{{\left(2bc\right)}^2-{(b^2+c^2-a^2)}^2\}\\ & =& \frac{1}{16}\left[\{2bc+(b^2+c^2-a^2)\}\{2bc-(b^2+c^2-a^2)\}\right]\\ & =& \frac{1}{16}\left\{(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)\right\}\\ & =& \frac{1}{16}\left[\{(b^2+2bc+c^2)-a^2\}\{a^2-(b^2-2bc+c^2)\}\right]\\ & =& \frac{1}{16}\left[\{{(b+c)}^2-a^2\}\{a^2-{(b-c)}^2\}\right]\cdots (a\pm b{)}^2=a^2\pm 2ab+b^2\\ & =& \frac{1}{16}\left[\left\{(b+c+a)(b+c-a)\right\}\left\{(a+b-c)(a-(b-c))\right\}\right]\\ & =& \frac{1}{16}\left[\left\{(b+c+a)(b+c-a)\right\}\left\{(a+b-c)(a-b+c)\right\}\right]\\ & =& \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\ & =& \frac{a+b+c}{2}\frac{-a+b+c}{2}\frac{a-b+c}{2}\frac{a+b-c}{2}\\ & =& \frac{a+b+c}{2}\frac{-a+b+c+a-a}{2}\frac{a-b+c+b-b}{2}\frac{a+b-c+c-c}{2}\\ & =& \frac{a+b+c}{2}\frac{a+b+c-2a}{2}\frac{a+b+c-2b}{2}\frac{a+b+c-2c}{2}\\ & =& \frac{a+b+c}{2}(\frac{a+b+c}{2}-a)(\frac{a+b+c}{2}-b)(\frac{a+b+c}{2}-c)\\ & =& s(s-a)(s-b)(s-c)\cdots s=\frac{a+b+c}{2}\end{array}$$ $$\begin{array}{rcl}S& =& \sqrt{s(s-a)(s-b)(s-c)}\cdots s=\frac{a+b+c}{2}\end{array}$$