三角形中重心、内心、外心、垂心向量计算公式 | 您所在的位置:网站首页 › 重心的向量定理是什么 › 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 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A、D、B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⇀ 则 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⇀, 因 O D ⇀ \mathop{OD}\limits ^{\rightharpoonup} OD⇀与 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。 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⇀)与 ( 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与 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和 B O ‾ \overline{BO} BO,O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 OA⇀2=OB⇀2=OC⇀2, 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 OA⇀2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} OA⇀⋅OB⇀=OB⇀⋅OC⇀=OC⇀⋅OA⇀ 证明:因为线段 A B ‾ ⊥ C O ‾ \overline{AB}⊥\overline{CO} AB⊥CO,所以 O C ⇀ ⋅ A B ⇀ = 0 \mathop{OC}\limits ^{\rightharpoonup}·\mathop{AB}\limits ^{\rightharpoonup}=0 OC⇀⋅AB⇀=0,因 A B ⇀ = A O ⇀ − B O ⇀ \mathop{AB}\limits ^{\rightharpoonup}=\mathop{AO}\limits ^{\rightharpoonup}-\mathop{BO}\limits ^{\rightharpoonup} AB⇀=AO⇀−BO⇀,所以 O C ⇀ ⋅ ( A O ⇀ − B O ⇀ ) = 0 \mathop{OC}\limits ^{\rightharpoonup}·(\mathop{AO}\limits ^{\rightharpoonup}-\mathop{BO}\limits ^{\rightharpoonup})=0 OC⇀⋅(AO⇀−BO⇀)=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} OC⇀⋅AO⇀=OC⇀⋅BO⇀,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} OC⇀⋅OA⇀=OB⇀⋅OC⇀,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} OA⇀⋅OB⇀=OB⇀⋅OC⇀,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} OA⇀⋅OB⇀=OB⇀⋅OC⇀=OC⇀⋅OA⇀。 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} OA⇀⋅OB⇀=OB⇀⋅OC⇀=OC⇀⋅OA⇀时,O为ΔABC垂心。 |
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