解：如图，作FG∥EA交DA的延长线于点G，∵AD∥BC，∴四边形AEFG是平行四边形，∴AE=GF，∠EAD=∠FGD．∵AE=DF，∴FG=FD，∴∠FGD=∠FDA，∴∠EAD=∠FDA，∵AE平分∠BAD，DF平分∠ADC，∠BAD+∠CDA=180°，∴∠BAD=∠CDA=90°，∴四边形ABCD是矩形．
