證法一：∵ABCD是平行四邊形,∴AB=CD,∠B=∠D,∠BAC=∠BCD,AD=BC,
∵AE、CF分別平分∠BAD、∠BCD,
∴∠BAE=1/2∠BAD,∠DCF=1/2∠BCD,∴∠BAE=∠DCF,
∴ΔABE≌ΔCDF(ASA),∴AE=CF,BE=DF,
∴BC-BE=AD-DF,即CE=AF,
∴四邊形AECF是平行四邊形(兩組對邊分別相等的四邊形是平行四邊形).
證法二：∵ABCD是平行四邊形,∴AD∥BC,∠BAD=∠BCD,∴∠DFC=∠FCE,
∵AE、CF分別平分∠BAD、∠BCD,
∴∠DAE=1/2∠BAD,∠DFC=1/2∠BCD,∴∠DAE=∠FCE,
∴∠DFC=∠DAE,∴AE∥CF,∴四邊形AECF是平行四邊形(定義).
證法三：∵ABCD是平行四邊形,∴AD=BC,AB=CD,AD∥BC,
∴∠DAE=∠AEB,∵AE平分∠BAD,∴∠DAE=∠BAE,
∴∠BAE=∠AEB,∴BA=BE,同理：DF=DC,
∴BE=DF,∴AD-DF=BC-BE,即AF=CE,
∴四邊形AECF是平行四邊形(一組對邊平行且相等).