# 中線

## 性質1

### 證明

${\displaystyle [ABO]=[ABE]-[BEO]}$
${\displaystyle [ACO]=[ACE]-[CEO]}$

## 性質2

${\displaystyle \triangle }$  ABC中，連接角A的中線記為${\displaystyle m_{a}}$ ，連接角B的中線記為${\displaystyle m_{b}}$ ，連接角C的中線記為${\displaystyle m_{c}}$ ，它們長度的公式為：

${\displaystyle m_{a}={\frac {1}{2}}{\sqrt {2(b^{2}+c^{2})-a^{2}}}}$
${\displaystyle m_{b}={\frac {1}{2}}{\sqrt {2(c^{2}+a^{2})-b^{2}}}}$
${\displaystyle m_{c}={\frac {1}{2}}{\sqrt {2(a^{2}+b^{2})-c^{2}}}}$

### 證明

${\displaystyle \triangle }$ ABD中，${\displaystyle AD=m_{a}}$
${\displaystyle (m_{a})^{2}=(AB)^{2}+(BD)^{2}-2(AB)(BD)\cos \angle ABD}$ 餘弦定理

${\displaystyle i.e.\ \cos \angle ABD={\frac {c^{2}+a^{2}-b^{2}}{2ca}}}$  & ${\displaystyle BD={\frac {a}{2}}}$

${\displaystyle i.e.\ (m_{a})^{2}=(c)^{2}+({\frac {a}{2}})^{2}-2(c)({\frac {a}{2}}){\frac {c^{2}+a^{2}-b^{2}}{2ca}}}$
${\displaystyle =(c)^{2}+({\frac {a^{2}}{4}})-({\frac {c^{2}+a^{2}-b^{2}}{2}})}$
${\displaystyle ={\frac {4c^{2}+a^{2}-2c^{2}-2a^{2}+2b^{2}}{4}}}$
${\displaystyle ={\frac {2b^{2}+2c^{2}-a^{2}}{4}}}$
${\displaystyle m_{a}={\frac {1}{2}}{\sqrt {2(b^{2}+c^{2})-a^{2}}}}$

Q.E.D.