Question

如图,已知在三角形ABC中,角C=60度,角BAC与角ABC的平分线AD、BE交于F,若AE=3,BD=4,求AB

Answer 1

在AB上截取AH=AE,连接FH∵j6C=60°∴∠CAB+∠CBA=120°∵AD、BE是角平分线∴∠FAB+∠FBA=60°∴∠AFB=120°∵AE=AH,∠EAF=∠HAF,AF=AF∴△AHF≌△AEF∴∠AFH=∠AFE=60°∴∠AFH=∠BFH=∠BFD=60°∵∠FBH=∠FBD,BF=BF...