∵四边形ABCD是正方形，∴∠ACD=∠ACB=45°．∵∠ACB=∠CAE+∠AEC，∴∠CAE+∠AEC=45°．∵CE=AC，∴∠CAE=∠AEC，∴∠CAE=22.5°．∵∠CAE+∠ACD+∠AFC=180°，∴∠AFC=112.5°．故答案为：112.5°．
