# 卡诺定理

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

OOA + OOB + OOC = R + r

## 引理

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

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

## 證明

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

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

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

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

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

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

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