证明：过E作GH//AB,交AD于G,交BC于H,则∠BAE＝∠AEG
∵正方形ABCD,∴EH＝BH＝AG,HC＝GD＝GE
又∠EHC＝∠AGE＝90°,∴△EHC≌△AGE
∴∠ECH＝∠AEG,∴∠BAE＝∠ECH
又∵正方形ABCD,∴FD//BC,∴∠F＝∠ECH
∴∠F＝∠BAE