卡諾定理 (內切圓、外接圓)

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设ABC为三角形，O为其外心。则O到ABC各边的距离之和为

${\displaystyle OO_{A}+OO_{B}+OO_{C}=R+r}$，

其中r为内切圆半径，R为外接圆半径。这个定理叫做卡诺定理（法語：Théorème de Carnot），以拉扎尔·卡诺為名。

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

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

假設${\displaystyle \triangle ABC}$為銳角三角形，${\displaystyle D}$為${\displaystyle \triangle ABC}$之外接圓圓心，${\displaystyle D}$至${\displaystyle \triangle ABC}$三邊之距離分別為${\displaystyle {\overline {DG}$、${\displaystyle {\overline {DH}$、${\displaystyle {\overline {DF}$，其中${\displaystyle {\overline {DG}$為${\displaystyle D}$至${\displaystyle {\overline {AB}$之距離，${\displaystyle {\overline {DH}$為${\displaystyle D}$至${\displaystyle {\overline {BC}$之距離，${\displaystyle {\overline {DF}$為${\displaystyle D}$至${\displaystyle {\overline {AC}$之距離。連接${\displaystyle D}$與${\displaystyle B}$，在${\displaystyle \triangle HDB}$中，根據三角形外心性質，可以得到

${\displaystyle {\overline {DB}=R}$

${\displaystyle \angle {HDB}=\angle {A}$

所以，可以得到${\displaystyle {\overline {DH}$的表示式，

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

同理，亦可得到${\displaystyle {\overline {DG}$和${\displaystyle {\overline {DF}$的表示式，

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

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

因此，

${\displaystyle {\overline {DG}+{\overline {DH}+{\overline {DF}\,}$

${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$

${\displaystyle =R(2\cos({\frac {A+B}{2})\cos({\frac {A-B}{2})+1-2\sin ^{2}({\frac {C}{2}))\,}$

${\displaystyle =R(2\cos({\frac {\pi -C}{2})\cos({\frac {A-B}{2})+1-2\sin({\frac {\pi -(A+B)}{2})\sin({\frac {C}{2}))\,}$

${\displaystyle =R(2\sin({\frac {C}{2})\cos({\frac {A-B}{2})+1-2\cos({\frac {(A+B)}{2})\sin({\frac {C}{2}))\,}$

${\displaystyle =R(2\sin({\frac {C}{2})(\cos({\frac {A-B}{2})-\cos({\frac {(A+B)}{2}))+1)\,}$

${\displaystyle =R(4\sin({\frac {A}{2})\sin({\frac {B}{2})\sin({\frac {C}{2})+1)\,}$

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

根據引理，即可得證，

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

此外，若${\displaystyle \triangle ABC}$為鈍角三角形，且${\displaystyle \angle {B}$大於${\displaystyle 90}$度，其餘符號假設均與上面相同，則可以得到，

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

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

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

所以，

${\displaystyle {\overline {DG}+{\overline {DH}-{\overline {DF}\,}$

${\displaystyle =R(\cos(A)+\cos(B)+\cos(C))\,}$

${\displaystyle =R+r\,}$

故得證卡諾定理。

• Perrier, Frédéric. Carnot's Theorem in Trigonometric Disguise. The Mathematical Gazette. 2007-03, 91 (520): 115-117 [2023-05-15]. （原始内容存档于2020-10-20）.