# 中線定理

${\text{AB}}^{\text{2}}+{\text{AC}}^{\text{2}}=2\left({\text{AM}}^{\text{2}}+{\text{BM}}^{\text{2}}\right)$

である．

## ■証明

${\text{AB}}^{\text{2}}+{\text{AC}}^{\text{2}}$

$=\left({\text{BH}}^{\text{2}}+{\text{AH}}^{\text{2}}\right)+\left({\text{CH}}^{\text{2}}+{\text{AH}}^{\text{2}}\right)$

$={\text{BH}}^{2}+{\text{CH}}^{2}+2{\text{AH}}^{\text{2}}$

${\text{BH}}^{\text{2}}+{\text{CH}}^{\text{2}}$

$={\left(\text{BM}+\text{MH}\right)}^{2}+{\left(\text{CM}-\text{MH}\right)}^{2}$

$={\text{BM}}^{2}+2\text{BM}\cdot \text{MH}+{\text{MH}}^{2}$$+{\text{CM}}^{2}-2\text{CM}\cdot \text{MH}+{\text{MH}}^{2}$

$=2{\text{BM}}^{\text{2}}+2{\text{MH}}^{\text{2}}$

$\left(\because \text{BM}=\text{CM}\right)$

したがって

${\text{AB}}^{\text{2}}+{\text{AC}}^{\text{2}}$$=2{\text{BM}}^{2}+2{\text{MH}}^{2}+2{\text{AH}}^{2}$

$=2{\text{BM}}^{\text{2}}+2\left({\text{MH}}^{\text{2}}+{\text{AH}}^{2}\right)$

$=2{\text{BM}}^{\text{2}}+2{\text{AM}}^{\text{2}}$

（∵三平方の定理より， ${\text{MH}}^{\text{2}}+{\text{AH}}^{2}={\text{AM}}^{2}$

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