Question

已知三角形ABC中,BD/DC=2/3,AE/EC=3/4,AD,BE交于F

Answer 1

过D作DG∥AC交BE于G,

则DG/CE=BD/BC=2/5,

∵AE/CE=3/4,∴AE=3/4CE,

∴DG/AE=4/3(DG/CE)=4/3*2/5=8/15,

∴AF/FD=AE/DG=15/8,

∵BG/GE=BD/CD=2/3,

FG/EF=DG/AE=8/15,

设GF=8X,则EF=15X,∴GE=23X,∴BG=46/3X,

∴BF=46/3X+8X=70/3X,

∴BF/EF=14/9,

∴AF/FD*BF/FE=15/8*14/9=35/12.