数学
在△ABC中,点M、N分别在边AB、AC上,且AM=2MB,AN=35AC,线段CM与BN相交于点P,且AB=a,AC=b,则AP用a和b表示为(  )A.AP=49a+13bB.AP=49a+23bC.AP=29a+43bD.AP=47a+37b

2020-02-07

在△ABC中,点M、N分别在边AB、AC上,且
AM
=2
MB
AN
=
3
5
AC
,线段CM与BN相交于点P,且
AB
=
a
AC
=
b
,则
AP
a
b
表示为(  )

A.
AP
=
4
9
a
+
1
3
b

B.
AP
=
4
9
a
+
2
3
b

C.
AP
=
2
9
a
+
4
3
b

D.
AP
=
4
7
a
+
3
7
b
优质解答
∵AM=2MB,AN=35AC,∴AM=23AB;又∵B,P,N三点共线,∴存在实数m,使得AP=mAB+(1-m)AN,即AP=mAB+35(1-m)AC;同理M,P,C三点共线,∴存在实数n,使得AP=nAC+(1-n)AM,即AP=nAC+23(1-n)AB;由向量相等得,... ∵AM=2MB,AN=35AC,∴AM=23AB;又∵B,P,N三点共线,∴存在实数m,使得AP=mAB+(1-m)AN,即AP=mAB+35(1-m)AC;同理M,P,C三点共线,∴存在实数n,使得AP=nAC+(1-n)AM,即AP=nAC+23(1-n)AB;由向量相等得,...
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