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,...