Question

如图，M、P分别为△ABC的边AB、AC上的点，且AM=BM，AP=2CP，若BP与CM相交于点N.求证：BN=3PN.

Answer 1

证明：过点P作PD∥MC,交AB于D,
∵PD∥MC,
∴AD/DM=AP/PC,
∵AP/PC=2,
∴AD/DM=2,
∴DM/AM=1/3,
∵AM=BM
∴DM/BM=1/3
∵PD∥MC,
∴NP/BN=DM/BM=1/3
即BN=3NP

追问

http://zhidao.baidu.com/question/431371769940712044.html?sort=6#answer-1508830284
能帮看看嘛？

追答

好的，已经回答了

Answer 2

2CP=AP 则S△CPN / S△ANP =1/2
S△ANC = 3 S△CNP

M 为中点则S△BNM =S△ AMN

S△AMN / S△ANC =MN / NC =S△BNM / 3S△CNP
S△BNM / S△CNP = NM *BN SinBNM / CN*NPSinCNP

所以 BN=3NP