# 卡诺定理

{\displaystyle {\begin{aligned}&{}\qquad DG+DH+DF\\&{}=|DG|+|DH|-|DF|\\&{}=R+r\end{aligned}}}

OOA + OOB + OOC = R + r

## 引理

${\displaystyle \triangle ABC}$中，${\displaystyle R}$${\displaystyle \triangle ABC}$之外接圓半徑，且${\displaystyle r}$${\displaystyle \triangle ABC}$之內切圓半徑，則

${\displaystyle r=4R\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})}$

## 證明

${\displaystyle {\overline {DB}}=R}$
${\displaystyle \angle {HDB}=\angle {A}}$

${\displaystyle {\overline {DH}}=R\cos(A)}$

${\displaystyle {\overline {DG}}=R\cos(C)}$
${\displaystyle {\overline {DF}}=R\cos(B)}$

${\displaystyle {\overline {DG}}+{\overline {DH}}+{\overline {DF}}\,}$
${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$
${\displaystyle =R(2\cos({\frac {A+B}{2}})\cos({\frac {A-B}{2}})+1-2\sin ^{2}({\frac {C}{2}}))\,}$
${\displaystyle =R(2\cos({\frac {\pi -C}{2}})\cos({\frac {A-B}{2}})+1-2\sin({\frac {\pi -(A+B)}{2}})\sin({\frac {C}{2}}))\,}$
${\displaystyle =R(2\sin({\frac {C}{2}})\cos({\frac {A-B}{2}})+1-2\cos({\frac {(A+B)}{2}})\sin({\frac {C}{2}}))\,}$
${\displaystyle =R(2\sin({\frac {C}{2}})(\cos({\frac {A-B}{2}})-\cos({\frac {(A+B)}{2}}))+1)\,}$
${\displaystyle =R(4\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})+1)\,}$
${\displaystyle =4R\sin({\frac {A}{2}})\sin({\frac {B}{2}})\sin({\frac {C}{2}})+R\,}$

${\displaystyle {\overline {DG}}+{\overline {DH}}+{\overline {DF}}=R+r}$

${\displaystyle {\overline {DH}}=R\cos(A)\,}$
${\displaystyle {\overline {DF}}=R\cos(\pi -B)=-R\cos(B)\,}$
${\displaystyle {\overline {DG}}=R\cos(C)\,}$

${\displaystyle {\overline {DG}}+{\overline {DH}}-{\overline {DF}}\,}$
${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$
${\displaystyle =R+r\,}$