还是我来解吧~~作DF垂直于BC于F，再延长AD，作EG垂直于AD交AD延长线于G，-----------------------------------------------------DF=AB=1，AD=BF=3，FC=BC-BF=2------------------------------==>∠GDC+∠FDC=90°,∠GDC+∠GDF=90°-----------------------------------==>∠FDC=∠GDF,∠EGD=∠DFC,---------------------------==>ΔEGD∽ΔCFD==>∵ED=DC==>ΔEGD≌ΔCFD----------------------------------------==>DG=DF=1,EG=CF=2==>AG=AD+DG=4----------------------------------==>AE=√（AG^2+EG^2）=2√5明白了吧~~