中線定理

用萊布尼茨標量函數約簡，可以容易導出這性質：只需要在兩個平方中引入${\displaystyle I}$：

${\displaystyle |{\vec {AB}}|^{2}+|{\vec {AC}}|^{2}=\left|{\vec {AI}}+{\vec {IB}}\right|^{2}+\left|{\vec {AI}}+{\vec {IC}}\right|^{2}}$

得出

${\displaystyle |{\vec {AB}}|^{2}+|{\vec {AC}}|^{2}=|{\vec {AI}}|^{2}+|{\vec {IB}}|^{2}+2{\vec {AI}}\cdot {\vec {IB}}+|{\vec {AI}}|^{2}+|{\vec {IC}}|^{2}+2{\overrightarrow {AI}}\cdot {\overrightarrow {IC}}}$

${\displaystyle I}$是${\displaystyle BC}$的中點，因此${\displaystyle {\overrightarrow {IB}}}$和${\displaystyle {\overrightarrow {IC}}}$相反，可知式中兩個標積抵消。又因${\displaystyle {\overline {IC}}={\overline {IB}}}$，得出

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}=2{\overline {AI}}^{2}+2{\overline {IB}}^{2}\,}$

這可能是阿波罗尼奥斯的證明方法，因為他不知道萊布尼茨函數。證明如下： 設${\displaystyle H}$是從${\displaystyle A}$到${\displaystyle BC}$的垂足，則${\displaystyle \triangle BHA}$和${\displaystyle \triangle AHC}$是直角三角形。用勾股定理可得

${\displaystyle {\overline {AB}}^{2}={\overline {BH}}^{2}+{\overline {AH}}^{2}\,}$

${\displaystyle {\overline {AC}}^{2}={\overline {AH}}^{2}+{\overline {HC}}^{2}\,}$

${\displaystyle {\overline {AI}}^{2}={\overline {IH}}^{2}+{\overline {AH}}^{2}\,}$

所以

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}={\overline {BH}}^{2}+2{\overline {AH}}^{2}+{\overline {HC}}^{2}\,}$

把${\displaystyle BH}$和${\displaystyle HC}$用${\displaystyle BI}$和${\displaystyle IH}$表達出來（記得${\displaystyle I}$是${\displaystyle BC}$的中點，因此${\displaystyle BI=IC}$）。注意到雖然現在的情形假設${\displaystyle H}$在線段${\displaystyle BI}$上，但其 他情形也可以用這個方法。

${\displaystyle {\overline {BH}}={\overline {BI}}-{\overline {IH}}\,}$

${\displaystyle {\overline {HC}}={\overline {IC}}+{\overline {IH}}={\overline {BI}}+{\overline {IH}}\,}$

代入前式：

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}=({\overline {BI}}-{\overline {IH}})^{2}+2{\overline {AH}}^{2}+({\overline {BI}}+{\overline {IH}})^{2}\,}$

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}={\overline {BI}}^{2}-2{\overline {BI}}\cdot {\overline {IH}}+{\overline {IH}}^{2}+2{\overline {AH}}^{2}+{\overline {BI}}^{2}+2{\overline {BI}}\cdot {\overline {IH}}+{\overline {IH}}^{2}\,}$

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}=2{\overline {BI}}^{2}+2{\overline {IH}}^{2}+2{\overline {AH}}^{2}=2{\overline {BI}}^{2}+2({\overline {IH}}^{2}+{\overline {AH}}^{2})\,}$

${\displaystyle \triangle IHA}$是直角三角形(H為${\displaystyle \triangle ABC}$於${\displaystyle {\overline {BC}}}$之垂足) ，因此

${\displaystyle {\overline {IH}}^{2}+{\overline {AH}}^{2}={\overline {AI}}^{2}\,}$

代入前式得出

${\displaystyle {\overline {AB}}^{2}+{\overline {AC}}^{2}=2{\overline {BI}}^{2}+2{\overline {AI}}^{2}\,}$