(1) 证明:∵在△ABC和△ADC中,∴△ABC≌△ADC(SSS),
∴∠BAC=∠DAC,∵在△ABF和△ADF中,∴△ABF≌△ADF,∴∠AFD=∠AFB,∵∠AFB=∠CFE,∴∠AFD=∠CFE
(2)证明:∵AB∥CD,∴∠BAC=∠ACD,又∵∠BAC=∠DAC,∴∠CAD=∠ACD,∴AD=CD,∵AB=AD,CB=CD,∴AB=CB=CD=AD,∴四边形ABCD是菱形.
(3)解:当EB⊥CD时,∠EFD=∠BCD,理由:∵四边形ABCD为菱形,
∴BC=CD,∠BCF=∠DCF,在△BCF和△DCF中,
∴△BCF≌△DCF(SAS),∴∠CBF=∠CDF,∵BE⊥CD,∴∠BEC=∠DEF=90°,∴∠EFD=∠BCD.