“奔驰定理”:对于$\triangle{ABC}$内一点$P$,记$S_A=\triangle{PBC},S_B=\triangle{PAC},S_C=\triangle{PAB}$,则有$S_A\cdot\vec{PA} + S_B\cdot\vec{PB} + S_C\cdot\vec{PC} = \vec{0}$

证明:延长AP交BC于点Q

$则有 \frac{BQ}{CQ} = \frac{\triangle{ABQ}}{\triangle{ACQ}} = \frac{\triangle{BPQ}}{\triangle{CPQ}} = \frac{\triangle{ABQ} - \triangle{BPD}}{\triangle{ACQ} - \triangle{CPQ}} = \frac{S_C}{S_B}$
$\therefore \vec{PQ} = \frac{CQ}{BC} \vec{PB} + \frac{BQ}{BC} \vec{PC} = \frac{S_B}{S_B + S_C} \vec{PB} + \frac{S_C}{S_B + S_C} \vec{PC}$
$\because \frac{PQ}{PA} = \frac{\triangle{BPQ}}{\triangle{BPA}} = \frac{\triangle{CPQ}}{\triangle{CPA}} = \frac{\triangle{BPQ} + \triangle{CPQ}}{\triangle{BPA} + \triangle{CPA}} = \frac{S_A}{S_B + S_C}$
$\therefore \vec{PQ} = -\frac{S_A}{S_B + S_C}\vec{PA}$
$\therefore -\frac{S_A}{S_B + S_C}\vec{PA} = \frac{CQ}{BC}\vec{PB} + \frac{BQ}{BC}\vec{PC} = \frac{S_B}{S_B + S_C} \vec{PB} + \frac{S_C}{S_B + S_C} \vec{PC}$
$\therefore S_A\cdot\vec{PA} + S_B\cdot\vec{PB} + S_C\cdot\vec{PC} = \vec{0}$

最后修改:2021 年 08 月 23 日 01 : 58 AM
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