# 海伦公式

（重定向自海倫公式

${\displaystyle A={\sqrt {s(s-a)(s-b)(s-c)}}}$，其中${\displaystyle s={\frac {a+b+c}{2}}}$

${\displaystyle A={\sqrt {{\frac {1}{4}}\left[a^{2}c^{2}-\left({\frac {a^{2}+c^{2}-b^{2}}{2}}\right)^{2}\right]}}}$，其中${\displaystyle a\geq b\geq c}$

## 证明

### 利用三角公式和代数式变形来证明

${\displaystyle \cos C={\frac {a^{2}+b^{2}-c^{2}}{2ab}}}$

{\displaystyle {\begin{aligned}\sin C&={\sqrt {1-\cos ^{2}C}}\\&={\sqrt {(1+\cos C)(1-\cos C)}}\\&={\sqrt {\left(1+{\frac {a^{2}+b^{2}-c^{2}}{2ab}}\right)\left(1-{\frac {a^{2}+b^{2}-c^{2}}{2ab}}\right)}}\\&={\sqrt {\left[{\frac {(a+b)^{2}-c^{2}}{2ab}}\right]\left[{\frac {c^{2}-(a-b)^{2}}{2ab}}\right]}}\\&={\frac {\sqrt {(a+b+c)(a+b-c)(c+a-b)(c-a+b)}}{2ab}}\\&={\frac {\sqrt {(2s)(2s-2c)(2s-2b)(2s-2a)}}{2ab}}\\&={\frac {2}{ab}}{\sqrt {s(s-c)(s-b)(s-a)}}\end{aligned}}}
{\displaystyle {\begin{aligned}A&={\frac {1}{2}}ab\sin C\\&={\frac {ab}{2}}\cdot {\frac {2}{ab}}{\sqrt {s(s-a)(s-b)(s-c)}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}\end{aligned}}}

### 利用勾股定理和代数式变形来证明

${\displaystyle b^{2}=h^{2}+d^{2}}$
${\displaystyle a^{2}=h^{2}+(c-d)^{2}}$
${\displaystyle a^{2}-b^{2}=c^{2}-2cd}$
${\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}}$
{\displaystyle {\begin{aligned}h^{2}&=b^{2}-\left({\frac {-a^{2}+b^{2}+c^{2}}{2c}}\right)^{2}\\&={\frac {(2bc-a^{2}+b^{2}+c^{2})(2bc+a^{2}-b^{2}-c^{2})}{4c^{2}}}\\&={\frac {((b+c)^{2}-a^{2})(a^{2}-(b-c)^{2})}{4c^{2}}}\\&={\frac {(b+c-a)(b+c+a)(a+b-c)(a-b+c)}{4c^{2}}}\\&={\frac {2(s-a)\cdot 2s\cdot 2(s-c)\cdot 2(s-b)}{4c^{2}}}\\&={\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}\end{aligned}}}
{\displaystyle {\begin{aligned}A&={\frac {ch}{2}}\\&={\sqrt {{\frac {c^{2}}{4}}\cdot {\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}\end{aligned}}}

### 用旁心來證明

${\displaystyle \bigtriangleup ABC}$ 中，${\displaystyle {\overline {AB}}=c,{\overline {BC}}=a,{\overline {CA}}=b}$

${\displaystyle I}$ 為內心，${\displaystyle I_{a},I_{b},I_{c}}$ 為三旁切圓。

${\displaystyle \because \angle I_{a}BI=\angle I_{a}CI=90^{\mathsf {o}}}$

${\displaystyle \therefore I_{a}CIB}$ 四點共圓，並設此圓為圓${\displaystyle O}$

1. ${\displaystyle I}$ 做鉛直線交${\displaystyle {\overline {BC}}}$ ${\displaystyle P}$ ，再延長${\displaystyle {\overleftrightarrow {IP}}}$ ，使之與圓${\displaystyle O}$ 交於${\displaystyle Q}$ 點。再過${\displaystyle I_{a}}$ 做鉛直線交${\displaystyle {\overline {BC}}}$ ${\displaystyle R}$ 點。
2. 先證明${\displaystyle \Box I_{a}QPR}$ 為矩形：${\displaystyle \because \angle QPR=90^{\mathsf {o}},\angle I_{a}RP=90^{\mathsf {o}}}$ ，又${\displaystyle \angle I_{a}QI=\angle I_{a}BI=90^{\mathsf {o}}}$ (圓周角相等)。${\displaystyle \therefore \Box I_{a}QPR}$ 為矩形。因此，${\displaystyle {\overline {I_{a}R}}={\overline {QP}}}$
3. ${\displaystyle {\overline {PI}}=}$ 內切圓半徑${\displaystyle ={\frac {\bigtriangleup }{\frac {a+b+c}{2}}}}$ ${\displaystyle {\overline {I_{a}R}}=}$ 旁切圓半徑${\displaystyle ={\frac {\bigtriangleup }{\frac {b+c-a}{2}}}}$ 。且易知${\displaystyle {\overline {BP}}={\frac {c+a-b}{2}},{\overline {PC}}={\frac {a+b-c}{2}}}$ 。由圓冪性質得到：${\displaystyle {\overline {PC}}\times {\overline {PB}}={\overline {PQ}}\times {\overline {PI}}={\overline {I_{a}R}}\times {\overline {PI}}}$ 。故${\displaystyle {\frac {a+b-c}{2}}\times {\frac {c+a-b}{2}}={\frac {\bigtriangleup }{\frac {a+b+c}{2}}}\times {\frac {\bigtriangleup }{\frac {b+c-a}{2}}}}$ ${\displaystyle \Rightarrow \bigtriangleup ={\sqrt {{\frac {a+b+c}{2}}\times {\frac {b+c-a}{2}}\times {\frac {a+c-b}{2}}\times {\frac {a+b-c}{2}}}}}$