海伦公式

与希羅在他的著作《Metrica》中的原始证明不同，在此我们用三角公式和公式变形来证明。设三角形的三边${\displaystyle a,b,c}$的对角分别为${\displaystyle A,B,C}$，则余弦定理为

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

利用和平方、差平方、等公式，从而有

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

${\displaystyle {\begin{aligned}A&={\frac {1}{2}}ab\sin C\\&={\frac {ab}{2}}\cdot {\frac {2}{ab}}{\sqrt {s(s-a)(s-b)(s-c)}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}\end{aligned}}}$

${\displaystyle b^{2}=h^{2}+d^{2}}$

${\displaystyle a^{2}=h^{2}+(c-d)^{2}}$

${\displaystyle a^{2}-b^{2}=c^{2}-2cd}$

${\displaystyle d={\frac {-a^{2}+b^{2}+c^{2}}{2c}}}$

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

${\displaystyle {\begin{aligned}A&={\frac {ch}{2}}\\&={\sqrt {{\frac {c^{2}}{4}}\cdot {\frac {4s(s-a)(s-b)(s-c)}{c^{2}}}}}\\&={\sqrt {s(s-a)(s-b)(s-c)}}\end{aligned}}}$

設${\displaystyle \bigtriangleup ABC}$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. 過${\displaystyle I}$做鉛直線交${\displaystyle {\overline {BC}}}$於${\displaystyle P}$，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${\displaystyle {\overline {BC}}}$於${\displaystyle R}$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}}}}}$