设正方形边长=4,则：AD=AB=4,DE=CE=2,BF=3,CF=1,由勾股定理得：AE=2√5,EF=√5,AF=5,在△AEF中,由AE²＋EF²=AF²,∴△AEF是直角△,且∠AEF=90°,考察直角△ADE与直角△AEF,AD∶DE=AE∶EF=2,∴△ADE∽△AEF,∴∠DAE=∠EAF,∴AE平分∠DAF