正方形ABCD中,E,F是AD,DC上的点,且DE=DF,若∠EBF=45度,且EF=2根号2,求正方形ABCD周长

延长DA至G,使AG＝FC；令BD与EF的交点为H.
∵ABCD是正方形,∴AB＝CB、∠GAB＝∠FCB＝90°,又AG＝FC,∴△ABG≌△CBF,
∴BG＝BF、∠ABG＝∠CBH.······①
∵ABCD是正方形,∴∠CBD＝∠ABD＝45°,又∠EBF＝45°,∴∠ABD＝∠EBF,
∴∠ABE＋∠DBE＝∠DBE＋∠DBF,∴∠ABE＝∠DBF.······②
①＋②,得：∠ABG＋∠ABE＝∠CBH＋∠DBF,∴∠EBG＝∠CBD＝45°,
∴∠EBG＝∠EBF＝45°,而BG＝BF、BE＝BE,∴△EBG≌△EBF,∴EG＝EF＝2√2,
∴AE＋AG＝2√2,∴AE＋CF＝2√2.······③
∵ABCD是正方形,∴AD＝CD、又DE＝DF,∴AD－DE＝CD－DF,∴AE＝CF.······④
由③、④,得：AE＝√2.
∵AE＝CF、AB＝CB、∠EAB＝∠FCB＝90°,∴△ABE≌△CBF,∴BE＝BF,
∴B在EF的垂直平分线上.
∵DE＝DF,∴D在EF的垂直平分线上.
∵B、D都在EF的垂直平分线上,∴BD是EF的垂直平分线,∴DH⊥EH、EH＝EF/2＝√2.
∵ABCD是正方形,∴DE⊥DF,又EH＝FH,∴DH＝EH＝√2.
∵DH⊥EH、EH＝DH＝√2,∴DE＝2,而AE＝√2,∴AD＝DE＋AE＝2＋√2,
∴正方形ABCD的周长＝4AD＝8＋8√2.