AE=1/2√65j解答题……不是填空、、谢谢过点E作EF⊥BC于点F；延长AE交BC延长线于点G。设EF=4K；CF=3K；易证△AED≌△GEC∴AE=EG；GC=AD=2；∴BG=BC+GC=7。易证AB//EF；AE=EG→BF=FG→BF=1/2BG=7/2CF=BC-BF=5-7/2=3/2 →3K=3/2→K=1/2→EF=4K=2→AB=2EF=2X2=4 AE=1/2AG=1/2√(AB²+BG²)=1/2√(4²+7²)=1/2√65