利用「向量比」或「分點比」,可以產生許多簡潔有力的推論。這個章節,我們收集了許多可以利用這種方法來證明的美麗定理。
圖示
摘要
PD→DA→+PE→EB→+PF→FC→=−1\dfrac{\overrightarrow{P{\color{red}D}}}{\overrightarrow{{\color{red}D}A}}+ \dfrac{\overrightarrow{P{\color{red}E}}}{\overrightarrow{{\color{red}E}B}}+ \dfrac{\overrightarrow{P{\color{red}F}}}{\overrightarrow{{\color{red}F}C}}=-1DAPD+EBPE+FCPF=−1 或
[APD]+[BPE]+[CPF]=2[A{\color{red}P}D]+[B{\color{red}P}E]+[C{\color{red}P}F] = {\color{blue}2}[APD]+[BPE]+[CPF]=2
AP→PD→=AE→EC→+AF→FB→\dfrac{\overrightarrow{A{\color{red}P}}}{\overrightarrow{{\color{red}P}D}}= \dfrac{\overrightarrow{A{\color{red}E}}}{\overrightarrow{{\color{red}E}C}}+ \dfrac{\overrightarrow{A{\color{red}F}}}{\overrightarrow{{\color{red}F}B}}PDAP=ECAE+FBAF
或
(A,P,D)=(A,F,B)+(A,E,C)(A,{\color{red}P},D)=(A,{\color{red}F},B)+(A,{\color{red}E},C)(A,P,D)=(A,F,B)+(A,E,C)
P=11+1β+γA+11+1γ+αB+11+1α+βCP= \frac{1}{1+\frac{1}{\beta}+\gamma}A + \frac{1}{1+\frac{1}{\gamma}+\alpha}B + \frac{1}{1+\frac{1}{\alpha}+\beta}CP=1+β1+γ1A+1+γ1+α1B+1+α1+β1C 或
11+(A,P,D)A+11+(B,P,E)B+11+(C,P,F)C \frac{1}{1+(A,{\color{red}P},D)}A + \frac{1}{1+(B,{\color{red}P},E)}B + \frac{1}{1+(C,{\color{red}P},F)}C1+(A,P,D)1A+1+(B,P,E)1B+1+(C,P,F)1C
DM→MA→=11+α(BP→PA→+α⋅CQ→QA→)\dfrac{\overrightarrow{D\color{red}{M}}}{\overrightarrow{{\color{red}M}A}}= \dfrac{ 1 }{ 1+{\color{blue}\alpha} } \left( \dfrac{\overrightarrow{B\color{red}{P}} }{\overrightarrow{{\color{red}P}A}} + {\color{blue}\alpha} \cdot\dfrac{\overrightarrow{C\color{red}{Q}}}{\overrightarrow{{\color{red}Q}A}} \right) MADM=1+α1(PABP+α⋅QACQ) 或
1[AMD]=11+α(1[APB]+α⋅1[AQC])\dfrac{1}{[A{\color{red}M}D]}=\dfrac{ 1 }{ 1+{\color{blue}\alpha} } \left( \dfrac{1}{[A{\color{red}P}B]} + {\color{blue}\alpha} \cdot \dfrac{1}{[A{\color{red}Q}C]} \right)[AMD]1=1+α1([APB]1+α⋅[AQC]1)
X,Y,ZX,Y,ZX,Y,Z 三點共線
Last updated 4 years ago
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