Question

如图所示,在平行四边形ABCD中,E为DC上一点,且DE:EC=3:2,AE交BD于F,AE的延长线交BC的延长线于G

Answer 1

1.易证△AED∽△GEC

又DE:CE=3:2

∴AD:GC=3:2

∴AD:GB=3:5

∴S△AFD:S△GFB=9:25(△AFD∽△GFB)

2.作EH∥DB交BC于H

∵DE:EC=3:2

∴BH:HC=3:2=9:6

又BC:CG=3:2=15:10

∴GE:EF=GH:HB=16:9