奔驰定理与三角形四心

一、奔驰定理

设点\(O\)是\(\Delta ABC\)所在平面上且与\(A,B,C\)不重合的一点，若则

​\( {S_{\Delta OBC}} \cdot \overrightarrow {OA} + {S_{\Delta OAC}} \cdot \overrightarrow {OB} + {S_{\Delta OAB}} \cdot \overrightarrow {OC} = \overrightarrow 0 \)​

【推导】​\( \dfrac{{AO}}{{OD}} = \dfrac{{{S_{\Delta OAB}}}}{{{S_{\Delta OBD}}}} = \dfrac{{{S_{\Delta OAC}}}}{{{S_{\Delta OCD}}}} = \dfrac{{{S_{\Delta OAB}} + {S_{\Delta OAC}}}}{{{S_{\Delta OBC}}}} \)​

\( \dfrac{{BD}}{{DC}} = \dfrac{{{S_{\Delta ABD}}}}{{{S_{\Delta ACD}}}} = \dfrac{{{S_{\Delta OBD}}}}{{{S_{\Delta OCD}}}} = \dfrac{{{S_{\Delta ABD}} – {S_{\Delta OBD}}}}{{{S_{\Delta ACD}} – {S_{\Delta OCD}}}} = \dfrac{{{S_{\Delta OAB}}}}{{{S_{\Delta OAC}}}} \)​

\( \therefore \overrightarrow {AO} = \dfrac{{{S_{\Delta OAB}} + {S_{\Delta OAC}}}}{{{S_{\Delta OBC}}}}\overrightarrow {OD} ， \)

​\( \overrightarrow {OD} = \dfrac{{{S_{\Delta OAC}}}}{{{S_{\Delta OAB}} + {S_{\Delta OAC}}}}\overrightarrow {OB} + \dfrac{{{S_{\Delta OAB}}}}{{{S_{\Delta OAB}} + {S_{\Delta OAC}}}}\overrightarrow {OC} \)​​．

\( \therefore \overrightarrow {AO} = \dfrac{{{S_{\Delta OAC}}\overrightarrow {OB} + {S_{\Delta OAB}}\overrightarrow {OC} }}{{{S_{\Delta OBC}}}} \)​，

即​\( {S_{\Delta OBC}} \cdot \overrightarrow {OA} + {S_{\Delta OAC}} \cdot \overrightarrow {OB} + {S_{\Delta OAB}} \cdot \overrightarrow {OC} = \overrightarrow 0 \)​

二、三角形四心

奔驰定理\({S_{\Delta OBC}} \cdot \overrightarrow {OA} + {S_{\Delta OAC}} \cdot \overrightarrow {OB} + {S_{\Delta OAB}} \cdot \overrightarrow {OC} = \overrightarrow 0 \)的四种应用：

应用一：\(O\)为重心时，\({S_{\Delta OBC}}:{S_{\Delta OAC}}:{S_{\Delta OAB}} = 1:1:1\)
故\(\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O\)是\(\Delta ABC\)的重心

应用二：\(O\)为内心时，\({S_{\Delta OBC}}:{S_{\Delta OAC}}:{S_{\Delta OAB}} = (\dfrac{1}{2}ar):(\dfrac{1}{2}br):(\dfrac{1}{2}cr) = a:b:c\)
故\(a \cdot \overrightarrow {OA} + b \cdot \overrightarrow {OB} + c \cdot \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O\)是\(\Delta ABC\)的内心
\(\sin A \cdot \overrightarrow {OA} + \sin B \cdot \overrightarrow {OB} + \sin C \cdot \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O\)是\(\Delta ABC\)的内心

应用三：\(O\)为外心时，
\[\begin{array}{l}
{S_{\Delta OBC}}:{S_{\Delta OAC}}:{S_{\Delta OAB}}\\
= (\dfrac{1}{2}{R^2}\sin \angle BOC):(\dfrac{1}{2}{R^2}\sin \angle AOC):(\dfrac{1}{2}{R^2}\sin \angle AOB)\\
= \sin 2A:\sin 2B:\sin 2C
\end{array}\]

故\(\sin 2A \cdot \overrightarrow {OA} + \sin 2B \cdot \overrightarrow {OB} + \sin 2C \cdot \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O\)是\(\Delta ABC\)的外心

应用四：\(O\)为垂心时，
\[\begin{array}{l}
{S_{\Delta OBC}}:{S_{\Delta OAC}} = (\dfrac{1}{2}OC \cdot BF):(\dfrac{1}{2}OC \cdot AF)\\
= BF:AF = (\dfrac{{BF}}{{CF}}):(\dfrac{{AF}}{{CF}}) = (\dfrac{1}{{\tan B}}):(\dfrac{1}{{\tan A}}) = \tan A:\tan B
\end{array}\]

故\({S_{\Delta OBC}}:{S_{\Delta OAC}}:{S_{\Delta OAB}} = \tan A:\tan B:\tan C\)

故\(\tan A \cdot \overrightarrow {OA} + \tan B \cdot \overrightarrow {OB} + \tan C \cdot \overrightarrow {OC} = \overrightarrow 0 \Leftrightarrow O\)是\(\Delta ABC\)的垂心

三、与四心有关的轨迹

【应用一】​\( AP\)​为中线，则​\(BP:PC = 1:1\)

​\( （1）\overrightarrow {AP}  = \dfrac{1}{2}\overrightarrow {AB}  + \dfrac{1}{2}\overrightarrow {AC}  = \dfrac{1}{2}(\overrightarrow {AB}  + \overrightarrow {AC} ) = \lambda (\overrightarrow {AB}  + \overrightarrow {AC} )\)

​\( \begin{array}{c} （2）\overrightarrow {AP}  = \dfrac{{AH}}{2}(\dfrac{{\overrightarrow {AB} }}{{AH}} + \dfrac{{\overrightarrow {AC} }}{{AH}}) = \dfrac{{AH}}{2}(\dfrac{{\overrightarrow {AB} }}{{c\sin B}} + \dfrac{{\overrightarrow {AC} }}{{b\sin C}})\\ = \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\sin B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\sin C}}) \end{array} \)​

【例1】（1）若\(\overrightarrow {OP} = \overrightarrow {OA} + \lambda (\overrightarrow {AB} + \overrightarrow {AC} )\)（\(\lambda >0\)），则点\(P\)的轨迹一定通过\(\Delta ABC\)的______心．
（2）若​\( \overrightarrow {OP} = \overrightarrow {OA} + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\sin B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\sin C}}) \)​（\(\lambda >0\)），则点\(P\)的轨迹一定通过\(\Delta ABC\)的______心．

【解析】（1）\(\overrightarrow {OP} = \overrightarrow {OA} + \lambda (\overrightarrow {AB} + \overrightarrow {AC} ) \Leftrightarrow \overrightarrow {AP} = \lambda (\overrightarrow {AB} + \overrightarrow {AC} )\)

因此\(P\)在线段\(BC\)中线所在直线上，即点\(P\)的轨迹一定通过\(\Delta ABC\)的重心．

（2）\(\overrightarrow {AP} = \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\sin B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\sin C}})\)

由正弦定理可知\(|\overrightarrow {AB} |\sin B = |\overrightarrow {AC} |\sin C\)，从而\[\overrightarrow {AP} = \dfrac{\lambda }{{\left| {\overrightarrow {AB} } \right|\sin B}}(\overrightarrow {AB} + \overrightarrow {AC} )\]

因此\(P\)在线段\(BC\)中线所在直线上，即点\(P\)的轨迹一定通过\(\Delta ABC\)的重心．

【应用二】\(AP\)为角平分线：\(BP:PC = c:b\)

\(\begin{array}{c}
\overrightarrow {AP} = \dfrac{b}{{b + c}}\overrightarrow {AB} + \dfrac{c}{{b + c}}\overrightarrow {AC} = \dfrac{{bc}}{{b + c}}(\dfrac{{\overrightarrow {AB} }}{c} + \dfrac{{\overrightarrow {AC} }}{b})\\
= \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}})
\end{array}\)

【例2】若\(\overrightarrow {OP} = \overrightarrow {OA} + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}})\)（\(\lambda >0\)），则点\(P\)的轨迹一定通过\(\Delta ABC\)的______心．

【解析】\(\overrightarrow {AP} = \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}})\)，\(\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}},\dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}}\)都是单位向量，

因此\(\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|}}\)表示菱形的对角线，从而点\(P\)在\(\angle BAC\)的平分线上

即点\(P\)的轨迹一定通过\(\Delta ABC\)的_内心．

【应用三】\(AP\)为高：\(BP:PC = \tan C:\tan B\)
\(\begin{array}{c}
\overrightarrow {AP} = \dfrac{{\tan B}}{{\tan B + \tan C}}\overrightarrow {AB} + \dfrac{{\tan C}}{{\tan B + \tan C}}\overrightarrow {AC} \\
= \dfrac{1}{{\tan B + \tan C}}(\dfrac{{\sin B}}{{\cos B}}\overrightarrow {AB} + \dfrac{{\sin C}}{{\cos C}}\overrightarrow {AC} )
\end{array}\)
\( = \dfrac{{bc}}{{2R(\tan B + \tan C)}}(\dfrac{{\overrightarrow {AB} }}{{c\cos B}} + \dfrac{{\overrightarrow {AC} }}{{b\cos C}}) \)

\( = \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)

【例3】（1）若\(\overrightarrow {OP} = \overrightarrow {OA} + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)（\(\lambda >0\)），则点\(P\)的轨迹一定通过\(\Delta ABC\)的______心．
（2）若\(\overrightarrow {OP} = \dfrac{{\overrightarrow {OB} + \overrightarrow {OC} }}{2} + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)（\(\lambda >0\)），则点\(P\)一定通过\(\Delta ABC\)的______心．

【解析】（1）注意到\(\left| {\overrightarrow {AB} } \right|\cos B,\left| {\overrightarrow {AC} } \right|\cos C\)都是向量的投影，如果分子为数量积，则可约分

于是，两边同“乘”一个向量，根据\(B,C\)为向量夹角可知，应同“乘”\(\overrightarrow {BC} \)

\(\overrightarrow {OP} = \overrightarrow {OA} + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)

\(\Leftrightarrow \overrightarrow {AP} = \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)

\(\begin{array}{c}
\overrightarrow {AP} \bullet \overrightarrow {BC} = \lambda (\dfrac{{\overrightarrow {AB} \bullet  \overrightarrow {BC} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} \bullet \overrightarrow {BC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\\
= \lambda (\dfrac{{|\overrightarrow {AB} | \cdot |\overrightarrow {BC} |\cos (\pi – B)}}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{|\overrightarrow {AC} | \cdot |\overrightarrow {BC} |\cos C}}{{\left| {\overrightarrow {AC} } \right|\cos C}})\\
= \lambda ( – |\overrightarrow {BC} | + |\overrightarrow {BC} ) = 0
\end{array}\)

故\(\overrightarrow {OP} \bot \overrightarrow {BC} \)，即\(P\)在\(BC\)边高所在直线上

从而点\(P\)的轨迹一定通过\(\Delta ABC\)的垂心

（2）\(\overrightarrow {BP} + \overrightarrow {CP} = + \lambda (\dfrac{{\overrightarrow {AB} }}{{\left| {\overrightarrow {AB} } \right|\cos B}} + \dfrac{{\overrightarrow {AC} }}{{\left| {\overrightarrow {AC} } \right|\cos C}})\)

同（1）可得\((\overrightarrow {BP} + \overrightarrow {CP} ) \bullet \overrightarrow {BC} = 0\)，设\(\overrightarrow {BP} + \overrightarrow {CP} = \mu \overrightarrow {PD} \)，则\(\overrightarrow {PD} \bot \overrightarrow {BC} \)

即\(PD\)为线段\(BC\)的垂直平分线

从而点\(P\)一定通过\(\Delta ABC\)的外心