3. ★★★【问题背景】(1)如图①，$∠ ACB=∠ ADE=90°,AC=BC,AD=DE$.求证:$BE=\sqrt{2}CD$;
【变式迁移】(2)如图②，E为正方形ABCD外一点，$∠ E=45°$,过点D作$DF⊥ BE$,垂足为F,连接CF.求$\frac{BE}{CF}$的值;
【拓展创新】(3)如图③，A是$△ BEF$内一点，$BE=BF,AF=\sqrt{6},∠ EAB=90°,∠ FEA=∠ BFA,AE=2AB$,直接写出AB的长.

(1)【证明】$\because ∠ ACB=∠ ADE=90°,AC=BC,AD=DE$,
$\therefore ∠ DAE=∠ CAB=45°,AB=\sqrt{2}AC,AE=\sqrt{2}AD$.
$\therefore ∠ DAC=∠ EAB,\frac{AB}{AC}=\frac{AE}{AD}=\sqrt{2}$.
$\therefore △ ABE∽ △ ACD.\therefore \frac{BE}{CD}=\frac{AB}{AC}=\sqrt{2}.\therefore BE=\sqrt{2}CD$.
(2)【解】如图①,连接 $BD$,易得$∠ BDC=45°$.
$\because ∠ E=45°,DF⊥ BE,\therefore ∠ E=∠ EDF=45°$.
$\therefore ∠ EDB=45°+∠ FDB=∠ FDC$. 易知 $△ EDF$ 和$△ DBC$ 均为等腰直角三角形,$\therefore \frac{ED}{FD}=\frac{BD}{DC}=\sqrt{2}$.
$\therefore △ EDB∽ △ FDC.\therefore \frac{BE}{CF}=\frac{BD}{DC}=\sqrt{2}$.

(3)【解】$AB=\sqrt{3}$.【点拨】如图②,过点 $A$ 作 $AH⊥ AF$,交 $EF$ 于点 $H$,连接 $BH$,则$∠ FAH=∠ EAB=90°$.
$\because BE=BF,\therefore ∠ BEF=∠ BFE$.
又$\because ∠ FEA=∠ BFA,\therefore ∠ AFE=∠ BEA$.
$\therefore △ AFH∽ △ AEB.\therefore \frac{AH}{AB}=\frac{AF}{AE}$,即$\frac{AH}{AF}=\frac{AB}{AE}$.
$\because AE=2AB,\therefore \frac{AH}{AF}=\frac{AB}{AE}=\frac{1}{2}$.
$\therefore AH=\frac{1}{2}AF=\frac{\sqrt{6}}{2}$.
$\therefore FH=\sqrt{AF^2+AH^2}=\frac{\sqrt{30}}{2}$.
$\because ∠ FAE=∠ HAE+90°=∠ HAB,\frac{AH}{AF}=\frac{AB}{AE}$,
$\therefore △ FAE∽ △ HAB.\therefore ∠ AFE=∠ AHB$.
$\therefore ∠ AHB+∠ AHF=∠ AFE+∠ AHF=180°-90°=90°$,即$∠ BHF=90°$.
又$\because BE=BF,\therefore HE=HF=\frac{\sqrt{30}}{2}$.
$\because △ FAE∽ △ HAB,\therefore \frac{BH}{EF}=\frac{AB}{AE}=\frac{1}{2}$.
$\therefore BH=\frac{1}{2}EF=EH=\frac{\sqrt{30}}{2}$.
$\therefore$ 易得 $BE=\sqrt{2}BH=\sqrt{15}$.
$\because AE=2AB,AB^2+AE^2=BE^2=15,\therefore AB=\sqrt{3}$.

4. ★★ [2025 上海改编] 在$□ ABCD$中，E,F分别是BC，CD上的点.
(1)若E是BC的中点.
①如图①，连接EF，如果$AE=EF$，求证：$∠ BAE=∠ CFE$；
②如图②，如果$CF=DF$，连接BF交AE于点G，求$\frac{S_{△ BEG}}{S_{△ AEF}}$的值；
(2)如图③，若$AD=5$，$AB=3$，$CF=1$，$∠ AEB=∠ AFE=∠ EFC$，请直接写出AF的长.

(1)①【证明】如图①,分别延长 $FE,AB$ 交于点 $H$.
$\because$ 四边形 $ABCD$ 是平行四边形,$\therefore AB// CD$.
$\therefore ∠ EBH=∠ ECF$.
$\because E$ 是 $BC$ 的中点,$\therefore BE=CE$.
又$\because ∠ BEH=∠ CEF,\therefore △ BEH≌ △ CEF$.
$\therefore EH=EF,∠ H=∠ CFE$.
$\because AE=EF,\therefore AE=EH$.
$\therefore ∠ H=∠ BAE.\therefore ∠ BAE=∠ CFE$.
②【解】如图②,分别延长 $BF,AD$ 交于点 $M$.
$\because$ 四边形 $ABCD$ 是平行四边形，
$\therefore AD// BC,AD=BC$.
$\therefore$ 易得$△ BEG∽ △ MAG,△ BCF∽ △ MDF$.
$\therefore \frac{BE}{AM}=\frac{GE}{AG}=\frac{BG}{GM},\frac{BC}{DM}=\frac{BF}{MF}=\frac{CF}{DF}=1$.
$\therefore BF=MF,BC=DM$.
$\because E$ 是 $BC$ 的中点,$\therefore BC=2CE=2BE$.
设 $CE=BE=m$,则 $DM=BC=2m$,
$\therefore AM=AD+DM=4m$.
$\therefore \frac{GE}{AG}=\frac{BG}{MG}=\frac{BE}{AM}=\frac{m}{4m}=\frac{1}{4}.\therefore$ 易知$\frac{BG}{GF}=\frac{2}{3}$.
$\therefore \frac{S_{△ BGE}}{S_{△ BGA}}=\frac{S_{△ GFE}}{S_{△ GFA}}=\frac{GE}{AG}=\frac{1}{4},\frac{S_{△ ABG}}{S_{△ AFG}}=\frac{BG}{FG}=\frac{2}{3}$.
设 $S_{△ ABG}=4n$,则 $S_{△ BGE}=n,S_{△ AFG}=6n$,
$\therefore S_{△ EGF}=\frac{3}{2}n$.
$\therefore \frac{S_{△ BEG}}{S_{△ AEF}}=\frac{S_{△ BEG}}{S_{△ AGF}+S_{△ EGF}}=\frac{n}{6n+\frac{3}{2}n}=\frac{2}{15}$.

(2)【解】$AF=\frac{5\sqrt{3}+6}{3}$.【点拨】如图③,分别延长 $AD$,$EF$ 交于点 $M$. $\because$ 四边形 $ABCD$ 是平行四边形,$\therefore AD// BC$,$CD=AB=3$. $\therefore ∠ AEB=∠ EAD.\because ∠ AEB=∠ AFE=∠ EFC,\therefore ∠ EFA=∠ EAD$. 又 $\because ∠ AEF=∠ MEA$,$\therefore △ AEF∽ △ MEA$. $\because ∠ AEB+∠ AEF+∠ FEC=∠ EFC+∠ FCE+∠ FEC=180°,∠ AEB=∠ EFC$,$\therefore ∠ AEF=∠ FCE. \therefore △ AEF∽ △ ECF. \because AD// BC$,$\therefore$ 易得$△ ECF∽ △ MDF.\therefore \frac{EC}{DM}=\frac{EF}{FM}=\frac{CF}{DF}.\because CF=1$,$\therefore DF=CD-CF=2$. 设 $CE=s,FE=t$,$\because △ ECF∽ △ AEF$,$\therefore \frac{CF}{EF}=\frac{CE}{AE}=\frac{EF}{AF}$,即$\frac{1}{t}=\frac{s}{AE}=\frac{t}{AF}.\therefore AE=st,AF=t^2.\because \frac{EC}{DM}=\frac{EF}{FM}=\frac{CF}{DF}$,即$\frac{s}{DM}=\frac{t}{FM}=\frac{1}{2}$,$\therefore DM=2s$,$FM=2t.\therefore AM=AD+DM=5+2s. \because △ AEF∽ △ MEA$,$\therefore \frac{EF}{AE}=\frac{AE}{EM}=\frac{AF}{AM}$,即$\frac{t}{st}=\frac{st}{t+2t}=\frac{t^2}{5+2s}$.
$\therefore \begin{cases}\frac{t}{st}=\frac{st}{t+2t},\\\frac{st}{t+2t}=\frac{t^2}{5+2s},\end{cases}$ 解得 $\begin{cases}s=\sqrt{3},\\t^2=\frac{5\sqrt{3}+6}{3}\end{cases}$ 或 $\begin{cases}s=-\sqrt{3},\\t^2=\frac{6-5\sqrt{3}}{3}\end{cases}$(舍去). $\therefore AF=\frac{5\sqrt{3}+6}{3}$.