答案：10. 解：$AF = \sqrt{2}DE$
(1) 如题图③，作 $DG// AF$，交 $BC$ 的延长线于点 $G$，连接 $EG$.
$\because$ 四边形 $ABCD$ 是正方形，
$\therefore ∠ ABC = ∠ BCD = 90^{\circ},AB = BC = CD = AD$，$AD// BC$.
$\because DG// AF,AD// BC,\therefore$ 四边形 $AFGD$ 为平行四边形，
$\therefore AF = DG,AD = FG,\therefore FG = CD$.
$\because ∠ ABC = 90^{\circ},AB = BC$，
$\therefore ∠ ACB = 45^{\circ},∠ ACD = 45^{\circ}$.
$\because EF⊥ AC,\therefore ∠ FEC = 90^{\circ},\therefore ∠ EFC = ∠ ECF = 45^{\circ}$，
$\therefore EF = EC,\therefore ∠ EFC = ∠ ECD$，
$\therefore △ CDE≌ △ FGE (SAS)$，
$\therefore ED = EG,∠ FEG = ∠ CED$，
$\therefore ∠ DEG = ∠ FEC = 90^{\circ},\therefore △ DEG$ 是等腰直角三角形，
$\therefore DG^{2} = DE^{2} + EG^{2} = 2DE^{2},\therefore DG = \sqrt{2}DE$，
$\therefore AF = \sqrt{2}DE$.
(2) 如答图，作 $DG⊥ DE$，并截取 $DG = DE$，连接 $AG$，$GE$，则 $△ DEG$ 是等腰直角三角形，
$\therefore EG^{2} = DE^{2} + DG^{2} = 2DE^{2},\therefore EG = \sqrt{2}DE$.
$\because$ 四边形 $ABCD$ 是正方形，$\therefore ∠ ADC = 90^{\circ},CD = AD$，
$\therefore ∠ DAC = ∠ DCA = 45^{\circ}$，同理，$∠ ACB = 45^{\circ}$.
$\because ∠ ADC = ∠ GDE = 90^{\circ},\therefore ∠ GDA = ∠ EDC$，
$\therefore △ GDA≌ △ EDC (SAS)$，
$\therefore ∠ GAD = ∠ ECD = 45^{\circ},AG = EC,\therefore ∠ GAE = 90^{\circ}$.
$\because EF⊥ AC,\therefore ∠ FEC = ∠ FEA = 90^{\circ}$，
$\therefore ∠ EFC = ∠ ECF = 45^{\circ},\therefore EF = EC,\therefore EF = AG$.
$\because ∠ GAE = ∠ FEA = 90^{\circ},\therefore AG// EF$，
$\therefore$ 四边形 $AGEF$ 为平行四边形，
$\therefore AF = EG,\therefore AF = \sqrt{2}DE$.
