# 欧拉定理 (几何)

${\displaystyle d=|IO|={\sqrt {R(R-2r)}}}$
${\displaystyle d^{2}=R(R-2r)\,}$

${\displaystyle R}$ ≥ ${\displaystyle 2r.}$

## 证明

(1)當${\displaystyle d=0}$ 時，表示外心${\displaystyle O}$ 與內心${\displaystyle I}$ 重合，此時易證三角形${\displaystyle \displaystyle ABC}$ 為正三角形，且${\displaystyle R=2r}$ ，因此${\displaystyle \displaystyle d^{2}=R(R-2r)}$

(2)當${\displaystyle d}$ 大於${\displaystyle 0}$ 時，請參考右下圖：

(a)设三角形${\displaystyle ABC}$ 的外心为${\displaystyle O}$ ，内心为${\displaystyle I}$ ，延长${\displaystyle AI}$ 交外接圆于${\displaystyle L}$ ，则${\displaystyle L}$ 为弧${\displaystyle BC}$ 的中点。连${\displaystyle LO}$ 延长交外接圆于${\displaystyle M}$ ，过${\displaystyle I}$ ${\displaystyle ID}$ 垂直于${\displaystyle AB}$ ${\displaystyle D}$ 为垂足，则${\displaystyle ID=r}$ 。易证三角形${\displaystyle \displaystyle ADI}$ 与三角形${\displaystyle \displaystyle MBL}$ 相似，故${\displaystyle {\frac {ID}{BL}}={\frac {AI}{ML}}}$ ，即${\displaystyle ID\times ML=AI\times BL}$ 。所以${\displaystyle 2Rr=AI\times BL}$

(b)连接${\displaystyle \displaystyle BI}$ ，因

${\displaystyle \angle BIL=\angle BAI+\angle ABI={\frac {\angle BAC}{2}}+{\frac {\angle ABC}{2}}}$
${\displaystyle \angle IBL=\angle IBC+\angle CBL={\frac {\angle ABC}{2}}+{\frac {\angle BAC}{2}}}$

(c)設${\displaystyle \displaystyle OI}$ 延长线交外接圆于${\displaystyle \displaystyle P,\;\displaystyle Q}$  两点，则${\displaystyle PI\cdot QI=AI\cdot IL=2Rr}$ ，所以${\displaystyle \displaystyle (R+d)(R-d)=2Rr}$ ，即${\displaystyle \displaystyle d^{2}=R(R-2r)}$