证明：连接AD，∵边AB的垂直平分线交BC于D，∴BD=AD，∴∠B=∠BAD=22.5°，∴∠ADE=22.5°+22.5°=45°，∵AE⊥BC，∴∠AEC=∠AED=90°，∴∠DAE=45°=∠ADE，∴DE=AE，∵DF⊥AC，∴∠DFC=90°=∠AEC，∴∠ACE+∠FDC...