# 海伦公式

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

### 用旁心来证明

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

${\displaystyle I}$为内心，${\displaystyle I_{a},I_{b},I_{c}}$为三旁切圆。

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

${\displaystyle \therefore I_{a}CIB}$四点共圆，并设此圆为圆${\displaystyle O}$

1. ${\displaystyle I}$做铅直线交${\displaystyle {\overline {BC}}}$${\displaystyle P}$，再延长${\displaystyle {\overleftrightarrow {IP}}}$，使之与圆${\displaystyle O}$交于${\displaystyle Q}$点。再过${\displaystyle I_{a}}$做铅直线交${\displaystyle {\overline {BC}}}$${\displaystyle R}$点。
2. 先证明${\displaystyle \Box I_{a}QPR}$为矩形：${\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}}}$(圆周角相等)。${\displaystyle \therefore \Box I_{a}QPR}$为矩形。因此，${\displaystyle {\overline {I_{a}R}}={\overline {QP}}}$
3. ${\displaystyle {\overline {PI}}=}$内切圆半径${\displaystyle ={\frac {\bigtriangleup }{\frac {a+b+c}{2}}}}$${\displaystyle {\overline {I_{a}R}}=}$旁切圆半径${\displaystyle ={\frac {\bigtriangleup }{\frac {b+c-a}{2}}}}$。且易知${\displaystyle {\overline {BP}}={\frac {c+a-b}{2}},{\overline {PC}}={\frac {a+b-c}{2}}}$。由圆幂性质得到：${\displaystyle {\overline {PC}}\times {\overline {PB}}={\overline {PQ}}\times {\overline {PI}}={\overline {I_{a}R}}\times {\overline {PI}}}$。故${\displaystyle {\frac {a+b-c}{2}}\times {\frac {c+a-b}{2}}={\frac {\bigtriangleup }{\frac {a+b+c}{2}}}\times {\frac {\bigtriangleup }{\frac {b+c-a}{2}}}}$${\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}}}}}$

## 资料来源

1. ^ 香港大學教育學院母語教學教師支援中心：數學科詞彙表. [2009-07-06]. （原始内容存档于2009-06-16）.
2. ^ Weisstein, Eric W. (编). Heron's Formula. at MathWorld--A Wolfram Web Resource. Wolfram Research, Inc. [2009-07-06]. （原始内容存档于2015-09-05） （英语）.