卡诺定理

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

设ABC为三角形，O为其外心。则O到ABC各边的距离之和为

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：），以拉扎尔·卡诺為名。

引理

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

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

證明

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。