证明：连接AH，∵AD⊥BC，BH⊥AC，∴∠FDB=∠AEF=90°，∵∠AFE=∠BFD，∴∠EAF=∠FBD，∵∠HAC=∠HBC，∴∠HAE=∠EAF，∵BH⊥AC，∴∠AEF=∠AEH=90°，在△AEF和△AEH中，∠DAE＝∠HAEAE＝AE∠AEF＝∠AEH∴△AEF≌...