证明:∵AE平分∠BAC∴∠BAE=∠CAE∴∠BAC=∠BAE+∠CAE=2∠CAE∵CE平分∠ACD∴∠ACE=∠DCE∴∠ACD=∠ACE+∠DCE=2∠ACE∵AB‖CD∴∠BAC+∠ACD=180 (两直线平行,同旁内角互补)∴2∠CAE+2∠ACE=180∴∠CAE+∠ACE=90∵∠AEC+∠CAE+∠ACE=180 (三角形内角和特性)∴∠AEC=180-(∠CAE+∠ACE)=180-90=90∴AE⊥CF