$解：作?DG⊥DF?交?BC?的延長(zhǎng)線于點(diǎn)?G?$
$則?∠CDG=∠ADF=90°-∠FDC?$
$又?AD=CD，??∠A=∠DCG=90°?$
$∴?△DAF≌△DCG?$
$∴?AF=CG，??DF=DG?$
$∵?∠EDF=45°?$
$∴?∠EDG=90°-45°=45°=∠EDF?$
$又?DE=DE?$
$∴?△DFE≌△DGE?$
$∴?EF=EG?$
$設(shè)?AF?的長(zhǎng)為?x，?則?BF=6-x，??EF=3+x?$
$由?BF^2+BE^2=EF^2，??(6-x)^2+3^2=(3+x)^2?$
$解得?x=2?$
$∴?EF=5?$
