∵AH⊥BC,DF⊥BC,∴AH∥DF,易证：ΔBEA≌ΔBEF,得∠BAE=∠BFE,∵∠BAE+∠CAH=∠CAH+∠C=90°,∴∠BAE=∠C,∴∠BFE=∠C,∴EF∥AC,∴四边形AEFD是平行四边形.∵BD平分∠ABC,∠BAC＝∠DFB＝90°,∴AD＝AF,∴平行四边形A...