△ABC≌△ADE,∠DAC=60°,∠BAE=100°,BC,DE相交于点F,则∠DFB的度数

设AC与AD交于M∵△ABC≌△ADE∴∠B=∠D,∠BAC=∠DAE∴∠BAC-∠DAC=∠DAE-∠DAC即∠BAM=∠CAE∵∠BAM+∠DAC+∠CAE=∠BAE=100°    ∠DAC=60°∴∠BAM=∠CAE=20°在△ABM和△FDM中∠B=∠D,∠AMB=∠FMD∴∠DFM=∠BAM=20°即∠DFB=20°