(1)以向量AB和向量BC表示向量AQ和向量CP;(2)求AX:AQ和CX:CP.及(3)證明B、X、R三點共線

Hello

2008-04-21 5:03 am

ABC為一三角形，P、Q、R分別為AB、BC和AC的中點.

(A)以向量AB和向量BC表示向量AQ和向量CP

(B)若AQ和CP相交於X,求AX:AQ和CX:CP.
(提示:向量AC=向量AB+向量BC=向量AX+向量XC)

(C)證明B、X、R三點共線

更新1:

請問知識長(2/3)(AX+XQ+CX+XP)=AX+XC怎樣得來?實在不明白.

更新2:

請問知識長為何CP=(1/2)AB+BC??? 我的算式: CP=CB+BP CP=-BC+(1/2)BA CP=-BC-(1/2)AB

回答 (1)

myisland8132

2008-04-21 9:27 pm

✔ 最佳答案

(A)
AQ=AB+(1/2)BC
CP=(1/2)AB+BC

(B)
AC=AB+BC=AX+XC

So
(2/3)(AX+XQ+CX+XP)=AX+XC
(2/3)(XQ+XP)=(1/3)(AX+XC)
(2/3)(XQ)-(1/3)AX=(1/3)(XC)-(2/3)XP

Since these two vectors are not parallel
(2/3)(XQ)-(1/3)AX=(1/3)(XC)-(2/3)XP=0
AX:XQ=CX:XP=2:1

(C)
BR=BC+(1/2)CA

Also
BX=(1/2)(BC)+(1/3)QA
XR=(2/3)(QA)+(1/2)AC

BX+XR
=(1/2)(BC)+(1/3)QA+(2/3)(QA)+(1/2)AC
=(1/2)(BC)+(1/2)AC+QA
=(1/2)AC+(1/2)(BC)-AB-(1/2)BC
=(1/2)AC-AB
=(-1/2)CA-(AC+CB)
=(-1/2)AC+CA+BC
=BC+(1/2)CA
=BR

So B、X、R三點共線

2008-04-23 16:44:53 補充：
(2/3)(AX+XQ+CX+XP)=AX+XC怎樣得來

用

AQ=AB+(1/2)BC
CP=(1/2)AB+BC

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