三角形中重心、内心、外心、垂心向量计算公式

三角形中重心、内心、外心、垂心向量计算公式一、对ΔABC重心O来讲有OA⇀+OB⇀+OC⇀=0\mathop{OA}\limits^{\rightharpoonup}+\mathop{OB}\limits^{\rightharpoonup}+\mathop{OC}\limits^{\rightharpoonup}=0OA⇀+OB⇀+OC⇀=0证明:延长CO与线段AB‾\overline{AB}AB交于点D,根据A、D、B三点共线公式OD⇀=mOA⇀+nOB⇀\mathop{OD}\limits^{\rightharpoonup}=m

一、对ΔABC重心O来讲有
O A ⇀ + O B ⇀ + O C ⇀ = 0 ⇀ \mathop{OA}\limits ^{\rightharpoonup}+\mathop{OB}\limits ^{\rightharpoonup}+\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} OA+OB+OC=0
证明:延长CO与线段 A B ‾ \overline{AB} AB交于点D
根据ADB三点共线公式
O D ⇀ = m O A ⇀ + n O B ⇀ \mathop{OD}\limits ^{\rightharpoonup}=m\mathop{OA}\limits ^{\rightharpoonup}+n\mathop{OB}\limits ^{\rightharpoonup} OD=mOA+nOB(其中m+n=1),因为D是线段 A B ‾ \overline{AB} AB的中点,所以有
O A ⇀ + O B ⇀ = 2 O D ⇀ \mathop{OA}\limits ^{\rightharpoonup}+\mathop{OB}\limits ^{\rightharpoonup}=2\mathop{OD}\limits ^{\rightharpoonup} OA+OB=2OD
又因 O C ⇀ = 2 D O ⇀ \mathop{OC}\limits ^{\rightharpoonup}=2\mathop{DO}\limits ^{\rightharpoonup} OC=2DO
所以 O A ⇀ + O B ⇀ + O C ⇀ = 0 ⇀ \mathop{OA}\limits ^{\rightharpoonup}+\mathop{OB}\limits ^{\rightharpoonup}+\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} OA+OB+OC=0,得证。
反过来,如果
O A ⇀ + O B ⇀ + O C ⇀ = 0 ⇀ \mathop{OA}\limits ^{\rightharpoonup}+\mathop{OB}\limits ^{\rightharpoonup}+\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} OA+OB+OC=0
O A ⇀ + O B ⇀ + O C ⇀ = ( O D ⇀ + D A ⇀ ) + ( O D ⇀ + D B ⇀ ) + O C ⇀ \mathop{OA}\limits ^{\rightharpoonup}+\mathop{OB}\limits ^{\rightharpoonup}+\mathop{OC}\limits ^{\rightharpoonup}=(\mathop{OD}\limits ^{\rightharpoonup}+\mathop{DA}\limits ^{\rightharpoonup})+(\mathop{OD}\limits ^{\rightharpoonup}+\mathop{DB}\limits ^{\rightharpoonup})+\mathop{OC}\limits ^{\rightharpoonup} OA+OB+OC=(OD+DA)+(OD+DB)+OC
= ( O C ⇀ + 2 O D ⇀ ) + ( D A ⇀ + D B ⇀ ) =(\mathop{OC}\limits ^{\rightharpoonup}+2\mathop{OD}\limits ^{\rightharpoonup})+(\mathop{DA}\limits ^{\rightharpoonup}+\mathop{DB}\limits ^{\rightharpoonup}) =(OC+2OD)+(DA+DB)
= m O D ⇀ + n D A ⇀ = 0 ⇀ =m\mathop{OD}\limits ^{\rightharpoonup}+n\mathop{DA}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} =mOD+nDA=0
O D ⇀ \mathop{OD}\limits ^{\rightharpoonup} OD D A ⇀ \mathop{DA}\limits ^{\rightharpoonup} DA线性无关,所以上式要取得 0 ⇀ \mathop{0}\limits ^{\rightharpoonup} 0只有

O C ⇀ + 2 O D ⇀ = 0 ⇀ \mathop{OC}\limits ^{\rightharpoonup}+2\mathop{OD}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} OC+2OD=0并且 D A ⇀ + D B ⇀ = 0 ⇀ \mathop{DA}\limits ^{\rightharpoonup}+\mathop{DB}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} DA+DB=0
可得 D A ‾ \overline{DA} DA= B D ‾ \overline{BD} BD,以及 C O ‾ \overline{CO} CO=2 O D ‾ \overline{OD} OD,即D是线段 A B ‾ \overline{AB} AB的中点,O为ΔABC的重心。
三角形中重心、内心、外心、垂心向量计算公式
二、对ΔABC内心O来讲有
a O A ⇀ + b O B ⇀ + c O C ⇀ = 0 ⇀ a\mathop{OA}\limits ^{\rightharpoonup}+b\mathop{OB}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} aOA+bOB+cOC=0
证明:延长CO与线段 A B ‾ \overline{AB} AB交于点D
三角形中重心、内心、外心、垂心向量计算公式
因为 C D ‾ \overline{CD} CD是∠ACB的角平分线,
根据角平分线性质,线段
O A ‾ / O B ‾ = C A ‾ / C B ‾ = D A ‾ / D B ‾ = b / a \overline{OA}/\overline{OB}=\overline{CA}/\overline{CB}=\overline{DA}/\overline{DB}=b/a OA/OB=CA/CB=DA/DB=b/a
并且
C O ‾ / O D ‾ = C A ‾ / A D ‾ = C B ‾ / B D ‾ = ( C A ‾ + C B ‾ ) / ( A D ‾ + B D ‾ ) = ( a + b ) / c \overline{CO}/\overline{OD}=\overline{CA}/\overline{AD}=\overline{CB}/\overline{BD}=(\overline{CA}+\overline{CB})/(\overline{AD}+\overline{BD})=(a+b)/c CO/OD=CA/AD=CB/BD=(CA+CB)/(AD+BD)=(a+b)/c
C O ‾ \overline{CO} CO O D ‾ \overline{OD} OD共线,长度比为 ( a + b ) / c (a+b)/c (a+b)/c,故
( a + b ) O D ⇀ + c O C ⇀ = 0 ⇀ (a+b)\mathop{OD}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} (a+b)OD+cOC=0
再根据A、D、B三点共线性质有
a O A ⇀ + b O B ⇀ = ( a + b ) O D ⇀ a\mathop{OA}\limits ^{\rightharpoonup}+b\mathop{OB}\limits ^{\rightharpoonup}=(a+b)\mathop{OD}\limits ^{\rightharpoonup} aOA+bOB=(a+b)OD,所以

a O A ⇀ + b O B ⇀ + c O C ⇀ = ( a + b ) O D ⇀ + c O C ⇀ = 0 ⇀ a\mathop{OA}\limits ^{\rightharpoonup}+b\mathop{OB}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}=(a+b)\mathop{OD}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} aOA+bOB+cOC=(a+b)OD+cOC=0,得证。
反之,若已知 a O A ⇀ + b O B ⇀ + c O C ⇀ = 0 ⇀ a\mathop{OA}\limits ^{\rightharpoonup}+b\mathop{OB}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}=\mathop{0}\limits ^{\rightharpoonup} aOA+bOB+cOC=0,则
a O A ⇀ + b O B ⇀ + c O C ⇀ = a\mathop{OA}\limits ^{\rightharpoonup}+b\mathop{OB}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}= aOA+bOB+cOC=
a ( O D ⇀ + D A ⇀ ) + b ( O D ⇀ + D B ⇀ ) + c ( O D ⇀ + D C ⇀ ) = a(\mathop{OD}\limits ^{\rightharpoonup}+\mathop{DA}\limits ^{\rightharpoonup})+b(\mathop{OD}\limits ^{\rightharpoonup}+\mathop{DB}\limits ^{\rightharpoonup})+c(\mathop{OD}\limits ^{\rightharpoonup}+\mathop{DC}\limits ^{\rightharpoonup})= a(OD+DA)+b(OD+DB)+c(OD+DC)=
( a + b + c ) O D ⇀ + c D C ⇀ + ( a D A ⇀ + b D B ⇀ ) = (a+b+c)\mathop{OD}\limits ^{\rightharpoonup}+c\mathop{DC}\limits ^{\rightharpoonup}+(a\mathop{DA}\limits ^{\rightharpoonup}+b\mathop{DB}\limits ^{\rightharpoonup})= (a+b+c)OD+cDC+(aDA+bDB)=
( a + b ) O D ⇀ + c O C ⇀ + ( a D A ⇀ + b D B ⇀ ) = 0 ⇀ (a+b)\mathop{OD}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}+(a\mathop{DA}\limits ^{\rightharpoonup}+b\mathop{DB}\limits ^{\rightharpoonup})=\mathop{0}\limits ^{\rightharpoonup} (a+b)OD+cOC+(aDA+bDB)=0
因向量 ( ( a + b ) O D ⇀ + c O C ⇀ ) ((a+b)\mathop{OD}\limits ^{\rightharpoonup}+c\mathop{OC}\limits ^{\rightharpoonup}) ((a+b)OD+cOC) ( a D A ⇀ + b D B ⇀ ) (a\mathop{DA}\limits ^{\rightharpoonup}+b\mathop{DB}\limits ^{\rightharpoonup}) (aDA+bDB)线性无关,所以上式要取得 0 ⇀ \mathop{0}\limits ^{\rightharpoonup} 0,只有
( a D A ⇀ + b D B ⇀ ) = 0 ⇀ (a\mathop{DA}\limits ^{\rightharpoonup}+b\mathop{DB}\limits ^{\rightharpoonup})=\mathop{0}\limits ^{\rightharpoonup} (aDA+bDB)=0,再由 D A ‾ \overline{DA} DA B D ‾ \overline{BD} BD共线,
可得 A D ‾ / D B ‾ = A C ‾ / C B ‾ = b / a \overline{AD}/\overline{DB}=\overline{AC}/\overline{CB}=b/a AD/DB=AC/CB=b/a,即线段 C D ‾ \overline{CD} CD是∠ACB的角平分线,同理可证另两条角平分线 A O ‾ \overline{AO} AO B O ‾ \overline{BO} BOO为ΔABC的内心。另外,
O C ‾ / O D ‾ = ( a + b ) / c \overline{OC}/\overline{OD}=(a+b)/c OC/OD=(a+b)/c
三、对ΔABC外心O来讲有
O A ⇀ 2 = O B ⇀ 2 = O C ⇀ 2 {\mathop{OA}\limits ^{\rightharpoonup}}^2={\mathop{OB}\limits ^{\rightharpoonup}}^2={\mathop{OC}\limits ^{\rightharpoonup}}^2 OA2=OB2=OC2
证明:线段 O A ‾ \overline{OA} OA O B ‾ \overline{OB} OB O C ‾ \overline{OC} OC为外接圆的半径,所以等长,向量 O A ⇀ 2 {\mathop{OA}\limits ^{\rightharpoonup}}^2 OA2内积为长度的平方。
四、对ΔABC垂心O来讲有
O A ⇀ ⋅ O B ⇀ = O B ⇀ ⋅ O C ⇀ = O C ⇀ ⋅ O A ⇀ \mathop{OA}\limits ^{\rightharpoonup}·\mathop{OB}\limits ^{\rightharpoonup}=\mathop{OB}\limits ^{\rightharpoonup}·\mathop{OC}\limits ^{\rightharpoonup}=\mathop{OC}\limits ^{\rightharpoonup}·\mathop{OA}\limits ^{\rightharpoonup} OAOB=OBOC=OCOA
证明:因为线段 A B ‾ ⊥ C O ‾ \overline{AB}⊥\overline{CO} ABCO,所以
O C ⇀ ⋅ A B ⇀ = 0 \mathop{OC}\limits ^{\rightharpoonup}·\mathop{AB}\limits ^{\rightharpoonup}=0 OCAB=0,因
A B ⇀ = A O ⇀ − B O ⇀ \mathop{AB}\limits ^{\rightharpoonup}=\mathop{AO}\limits ^{\rightharpoonup}-\mathop{BO}\limits ^{\rightharpoonup} AB=AOBO,所以
O C ⇀ ⋅ ( A O ⇀ − B O ⇀ ) = 0 \mathop{OC}\limits ^{\rightharpoonup}·(\mathop{AO}\limits ^{\rightharpoonup}-\mathop{BO}\limits ^{\rightharpoonup})=0 OC(AOBO)=0,化简得
O C ⇀ ⋅ A O ⇀ = O C ⇀ ⋅ B O ⇀ \mathop{OC}\limits ^{\rightharpoonup}·\mathop{AO}\limits ^{\rightharpoonup}=\mathop{OC}\limits ^{\rightharpoonup}·\mathop{BO}\limits ^{\rightharpoonup} OCAO=OCBO,即
O C ⇀ ⋅ O A ⇀ = O B ⇀ ⋅ O C ⇀ \mathop{OC}\limits ^{\rightharpoonup}·\mathop{OA}\limits ^{\rightharpoonup}=\mathop{OB}\limits ^{\rightharpoonup}·\mathop{OC}\limits ^{\rightharpoonup} OCOA=OBOC,同理可证
O A ⇀ ⋅ O B ⇀ = O B ⇀ ⋅ O C ⇀ \mathop{OA}\limits ^{\rightharpoonup}·\mathop{OB}\limits ^{\rightharpoonup}=\mathop{OB}\limits ^{\rightharpoonup}·\mathop{OC}\limits ^{\rightharpoonup} OAOB=OBOC,即
O A ⇀ ⋅ O B ⇀ = O B ⇀ ⋅ O C ⇀ = O C ⇀ ⋅ O A ⇀ \mathop{OA}\limits ^{\rightharpoonup}·\mathop{OB}\limits ^{\rightharpoonup}=\mathop{OB}\limits ^{\rightharpoonup}·\mathop{OC}\limits ^{\rightharpoonup}=\mathop{OC}\limits ^{\rightharpoonup}·\mathop{OA}\limits ^{\rightharpoonup} OAOB=OBOC=OCOA
反之也可证,当
O A ⇀ ⋅ O B ⇀ = O B ⇀ ⋅ O C ⇀ = O C ⇀ ⋅ O A ⇀ \mathop{OA}\limits ^{\rightharpoonup}·\mathop{OB}\limits ^{\rightharpoonup}=\mathop{OB}\limits ^{\rightharpoonup}·\mathop{OC}\limits ^{\rightharpoonup}=\mathop{OC}\limits ^{\rightharpoonup}·\mathop{OA}\limits ^{\rightharpoonup} OAOB=OBOC=OCOA时,O为ΔABC垂心。

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