BE=CE,∠AEB=∠MECР∴△BEA≌△CEMР∴CM=AB,∠1=∠BР∴AB∥CMР∴∠M=∠MAD,∠MCF=∠ADFР∴△MCF∽△ADFР∴Р∵CM=AB,AD=ACР∴Р(方法二)Р过D作DG∥BC交AE于GР则△ABE∽△ADG,△CEF∽△DGFР∴,Р∵AD=AC,BE=CEР∴Р12.答案:证明:Р过点D作DF∥AB交AC的延长线于点F,则∠2=∠3Р∵AC平分∠DABР∴∠1=∠2Р∴∠1=∠3Р∴AD=DFР∵∠DEF=∠BEA,∠2=∠3Р∴△BEA∽△DEFР∴Р∵AD=DFР∴Р∵AC为AB、AD的比例中项Р∴Р即Р又∵∠1=∠2Р∴△ACD∽△ABCР∴Р∴Р∴Р13.答案:解:Р证明:Р过点E作PQ∥BC分别交BA延长线和DC于点P和点QР∵AB∥CD,PQ∥BCР∴四边形PQCB和四边形EQCF是平行四边形Р∴PB=EF=CQ,Р又∵AB=b,CD=aР∴AP=PB-AB=EF-b,DQ=DC-QC=a-EFР∴Р∴Р14.答案:解:Р连接MFР∵M是AC的中点,EF=FCР∴MF∥AE且MF=AE??∴△BEN∽△BFM??∴BN:BM=BE:BF=NE:MF??∵BE=EF??∴BN:BM=NE:MF=1:2??∴BN:NM=1:1??设NE=x,则MF=2x,AE=4x??∴AN=3x??∵MF∥AE??∴△NAQ∽△MFQ??∴NQ:QM=AN:MF=3:2??∵BN:NM=1:1,NQ:QM=3:2??∴BN:NQ:QM=5:3:2Р15.答案:证明:(1)Р如图1,AD、BE为△ABC的中线,且AD、BE交于点OР过点C作CF∥BE,交AD的延长线于点FР∵CF∥BE且E为AC中点Р∴∠AEO=∠ACF,∠OBD=∠FCD,AC=2AEР∵∠EAO=∠CAFР∴△AEO∽△ACFР∴Р∵D为BC的中点,∠ODB=∠FDC
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