一道向量几何题

刚才刚做过这个题：
1
过Q点做QD平行于AB,交CP于D点,则：|QD|/|AP|=1/2,故：|QD|/|PB|=1/4
而△BPM∽△QMD,故：|MQ|/|BM|=|QD|/|PB|=1/4,即：QM=QB/5
而：QB=QA+AB=AB-AC/2,则：AM=AQ+QM=AC/2+(AB-AC/2)/5
=AB/5+AC/2-AC/10=AB/5+2AC/5=(AB+2AC)/5
2
再做条辅助线吧,过Q点做QE平行MN,交BC于E,|MN|/|QE|=4/5
而：|CE|/|AC|=|QC|/|BC|,故：|CE|=(2/2sqrt(5))*4=4/sqrt(5)=(2/5)|BC|
即：CE=2CB/5,故：QE=CE-CQ=2CB/5+AC/2=2(AB-AC)/5+AC/2=2AB/5+AC/10
故：MN=4QE/5=4(2AB/5+AC/10)/5=(8AB+2AC)/25,故：AN=AM+MN
=(AB+2AC)/5+(8AB+2AC)/25=(13AB+12AC)/25