P在三角形ABC內部，xPA+yPB+zPC=0，求x, y, z

若\(P\)為\(\triangle ABC\)內部一點且\(x\overrightarrow{PA}+ y\overrightarrow{PB} +z\overrightarrow{PC}=0\)，求\(x:y:z \)。
類似的題型有許多種，用漸近的方式來求取不同的P點所帶來的不同x:y:z .........

範例1：\(P在\triangle ABC\)的底邊\(\overline{BC}\)上，且\(\overline{BP}:\overline{PC}=m:n\)，則\( \overrightarrow {AP}=\frac{m}{m+n}\overrightarrow{AC}+\frac{n}{m+n}\overrightarrow{AB}\)。

證明：$$\overrightarrow{AP}=\overrightarrow{AB}+\overrightarrow{BP}=\overrightarrow{AB}+\frac{m}{m+n}\overrightarrow{BC}= \overrightarrow{AB}+\frac{m}{m+n}\left(\overrightarrow{BA}+\overrightarrow{AC}\right)\\= \overrightarrow{AB}+\frac{m}{m+n}\left(-\overrightarrow{AB}+\overrightarrow{AC}\right)
=\left(1-\frac{m}{m+n}\right) \overrightarrow{AB}+ \frac{m}{m+n}\overrightarrow{AC}=\frac{n}{m+n}\overrightarrow{AB} +\frac{m}{m+n}\overrightarrow{AC}$$

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範例2：若\(P\)為\(\triangle ABC\)的重心，則\(\overrightarrow{PA}+ \overrightarrow{PB} +\overrightarrow{PC}=0\)。

證明：$$\begin{cases} \overrightarrow { PA } =\overrightarrow { PD } +\overrightarrow { DA } =-\frac { 1 }{ 2 } \overrightarrow { PC } -\frac { 1 }{ 2 } \overrightarrow { AB }  \\ \overrightarrow { PB } =\overrightarrow { PE } +\overrightarrow { EB } =-\frac { 1 }{ 2 } \overrightarrow { PA } -\frac { 1 }{ 2 } \overrightarrow { BC }  \\ \overrightarrow { PC } =\overrightarrow { PF } +\overrightarrow { FC } =-\frac { 1 }{ 2 } \overrightarrow { PB } -\frac { 1 }{ 2 } \overrightarrow { CA }  \end{cases}\\ \Rightarrow \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC } =-\frac { 1 }{ 2 } \left( \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC }  \right) -\frac { 1 }{ 2 } \left( \overrightarrow { AB } +\overrightarrow { BC } +\overrightarrow { CA }  \right) \\ \Rightarrow \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC } =-\frac { 1 }{ 2 } \left( \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC }  \right) -0\\ \Rightarrow \frac { 3 }{ 2 } \left( \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC }  \right) =0\Rightarrow \overrightarrow { PA } +\overrightarrow { PB } +\overrightarrow { PC } $$這特性反推也成立，即\(\overrightarrow{PA}+ \overrightarrow{PB} +\overrightarrow{PC}=0\Rightarrow P\)為重心。

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範例3：若\(P\)為\(\triangle ABC\)內部一點且\(x\overrightarrow{PA}+ y\overrightarrow{PB} +z\overrightarrow{PC}=0\)，則\(x:y:z = \triangle PBC:\triangle PAC:\triangle PAB\)。

證明：

$$令\cases{\overrightarrow{PA'} =x\overrightarrow{PA} \\\overrightarrow{PB'} =y\overrightarrow{PB} \\ \overrightarrow{PC'} =z\overrightarrow{PC} } \Rightarrow \overrightarrow{PA'}+ \overrightarrow{PB'} +\overrightarrow{PC'} = x\overrightarrow{PA} +y\overrightarrow{PB} +z\overrightarrow{PC}=0 \Rightarrow P為\triangle A'B'C'之重心\\ 又\cfrac{\triangle PAB}{ \triangle PA'B'} =\cfrac{{1\over 2} \cdot\overline{PA} \cdot \overline{PB} \cdot\sin \angle APB}{{1\over 2} \cdot\overline{PA'} \cdot \overline{PB'} \cdot\sin \angle APB} = \cfrac{   \overline{PA} \cdot \overline{PB} }{ \overline{PA'} \cdot \overline{PB'}  }= \cfrac{   \overline{PA} \cdot \overline{PB} }{ \overline{xPA} \cdot y\overline{PB}  } =\cfrac{1}{xy};\\ 同理，\cfrac{\triangle PBC}{ \triangle PB'C'} =\cfrac{1}{yz}及\cfrac{\triangle PCA}{ \triangle PC'A'} =\cfrac{1}{xz};\\ 因此\triangle PAB: \triangle PBC: \triangle PCA =\cfrac{1}{xy} : \cfrac{1}{yz}: \cfrac{1}{zx} =z:x:y \\ \Rightarrow x:y:z = \triangle PBC: \triangle PCA: \triangle PAB$$

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範例4：若\(P\)為\(\triangle ABC\)的內心且\(x\overrightarrow{PA}+ y\overrightarrow{PB} +z\overrightarrow{PC}=0\)，則\(x:y:z = \overline{BC}:  \overline{AC}: \overline{AB}\)。

證明：
\(P\)是內心\(\Rightarrow \triangle PAB:\triangle PBC:\triangle PCA= \overline{AB}: \overline{BC}: \overline{CA}\)
\(\Rightarrow x:y:z = \triangle PBC:\triangle PAC:\triangle PAB =\overline{BC}: \overline{AC}: \overline{AB}\)
\(\Rightarrow x:y:z = \overline{BC}: \overline{AC}: \overline{AB}\)

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範例5：若\(P\)為\(\triangle ABC\)的外心且\(x\overrightarrow{PA}+ y\overrightarrow{PB} +z\overrightarrow{PC}=0\)，則\(x:y:z = \overline{BC}\cos{\angle A}:  \overline{AC}\cos{\angle B}: \overline{AB}\cos{\angle C}\)。

證明：

$$\triangle PBC=\frac { 1 }{ 2 } \overline { BC } \times \overline { PD } =\frac { 1 }{ 2 } \overline { BC } \times \left( R\cos { \angle BPD }  \right) =\frac { 1 }{ 2 } \overline { BC } \times \left( R\cos { \left( \frac { 1 }{ 2 } \angle BPC \right)  }  \right) \\ =\frac { 1 }{ 2 } \overline { BC } \times \left( R\cos { \left( \frac { 1 }{ 2 } \times 2\angle A \right)  }  \right) =\frac { 1 }{ 2 } \overline { BC } \times \left( R\cos { \angle A }  \right) =\frac { R }{ 2 } \overline { BC } \times \cos { \angle A } \\ 同理\triangle PCA=\frac { R }{ 2 } \overline { AC } \times \cos { \angle B } ,\triangle PAB=\frac { R }{ 2 } \overline { AB } \times \cos { \angle C } \\ \Rightarrow x:y:z=\triangle PBC:\triangle PAC:\triangle PAB=\frac { R }{ 2 } \overline { BC } \cos { \angle A } :\frac { R }{ 2 } \overline { AC } \cos { \angle B } :\frac { R }{ 2 } \overline { AB } \cos { \angle C } \\ =\overline { BC } \cos { \angle A } :\overline { AC } \cos { \angle B } :\overline { AB } \cos { \angle C } $$