3. ★★【问题背景】(1)如图①,在$△ ABC$和$△ ADE$中,$∠ ACB=∠ ADE=90°,AC=BC,AD=DE$,连结$BE,CD$.求证:$BE=\sqrt{2}CD$;

(1) 证明：$\because ∠ ACB=∠ ADE=90°,AC=BC,AD=DE$，
$\therefore$ 易得 $∠ DAE=∠ CAB,AB=\sqrt{2}AC,AE=\sqrt{2}AD$，
$\therefore ∠ EAB=∠ DAC,\frac{AB}{AC}=\frac{AE}{AD}=\sqrt{2}$，
$\therefore △ ABE∽△ ACD,\therefore \frac{BE}{CD}=\frac{AE}{AD}=\sqrt{2},\therefore BE=\sqrt{2}CD$.

【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，连结CF.求$\frac{BE}{CF}$的值；

(2) 解：如图①,连结 BD.
$\because ∠ E=45°,DF⊥ BE,\therefore ∠ EDF=∠ E=45°$.
易知 $∠ BDC=45°$，
$\therefore ∠ EDB=∠ FDC=45°+∠ FDB$.
易知 $\frac{ED}{FD}=\frac{BD}{DC}=\sqrt{2}$，
$\therefore △ EDB∽△ FDC,\therefore \frac{BE}{CF}=\frac{BD}{DC}=\sqrt{2}$.

【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.

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

4. [上海普陀区期末] 在梯形 $ABCD$ 中，$AD// BC$，点 $E$ 在边 $AB$ 上，且 $AE=\dfrac{1}{3}AB$。
(1) 如图①，若点 $F$ 在边 $CD$ 上，且 $DF=\dfrac{1}{3}CD$，连结 $EF$，求证: $EF// BC$；
(2) 如图②，已知 $AD=AE=1$，点 $M$ 在边 $BC$ 上，连结 $EM,DM,EC$，$DM$ 与 $EC$ 交于点 $N$。若 $BC=4$，$CD^2=DM· DN$，$∠ DMC=∠ CEM$，求边 $CD$ 的长。

(1) 证明 连结DE并延长交CB的延长线于点G,如图①.
$\because AD// BC,\therefore \frac{AE}{BE}=\frac{DE}{EG}$.
$\because AE=\frac{1}{3}AB,DF=\frac{1}{3}CD,\therefore \frac{AE}{BE}=\frac{1}{2},\frac{DF}{FC}=\frac{1}{2}$，
$\therefore \frac{DE}{EG}=\frac{DF}{FC},\therefore \frac{DE}{DG}=\frac{DF}{DC}$，
$\because ∠ EDF=∠ GDC,\therefore △ DEF∽△ DGC$，
$\therefore ∠ DEF=∠ G,\therefore EF// BC$.
(2) 解 如图②,延长 BA,CD交于点P,过点E作$EQ⊥ BC$于点Q.
$\because AD// BC,\therefore △ PAD∽△ PBC,\therefore \frac{PD}{PC}=\frac{PA}{PB}=\frac{AD}{BC}=\frac{1}{4}$.
$\because AE=\frac{1}{3}AB,AE=1,\therefore AB=3,\therefore \frac{PA}{PA+3}=\frac{1}{4},\therefore PA=1$，
$\therefore EP=2$.
$\because CD^2=DM· DN,\therefore \frac{CD}{DM}=\frac{DN}{CD}$.
又 $\because ∠ CDN=∠ MDC,\therefore △ DCN∽△ DMC$，
$\therefore ∠ DCN=∠ DMC$.
$\because ∠ DMC=∠ CEM,\therefore ∠ CEM=∠ DCN,\therefore EM// CD$，
$\therefore \frac{BE}{EP}=\frac{BM}{MC}$.
$\because AB=3,AE=1,\therefore BE=2,\therefore \frac{BE}{EP}=1=\frac{BM}{MC}$，
$\therefore$ 易得 $BM=MC=2$，
易知 $△ BEM∽△ BPC,\therefore \frac{BM}{BC}=\frac{ME}{PC}=\frac{1}{2}$.
$\therefore$ 设 $ME=2a$,则 $PC=4a$.
$\because \frac{PD}{PC}=\frac{1}{4},\therefore PD=a,\therefore DC=3a$.
$\because EM// CD,\therefore △ ENM∽△ CND,\therefore \frac{EN}{CN}=\frac{EM}{DC}=\frac{2}{3}$，
$\therefore$ 设 $EN=2b$,则 $CN=3b,\therefore EC=5b$.
$\because ∠ NMC=∠ CEM,∠ MCN=∠ ECM$，
$\therefore △ CNM∽△ CME,\therefore \frac{CN}{CM}=\frac{CM}{CE},\therefore CM^2=CN· CE$，
$\therefore 4=3b· 5b$,解得 $b=\frac{2\sqrt{15}}{15}$(负值已舍去),$\therefore CE=\frac{2\sqrt{15}}{3}$.
由勾股定理可得 $BE^2-BQ^2=CE^2-CQ^2$，
即 $4-BQ^2=(\frac{2\sqrt{15}}{3})^2-(4-BQ)^2$,解得 $BQ=\frac{5}{3}$，
$\therefore EQ^2=BE^2-BQ^2=\frac{11}{9}. \because QM=BM-BQ=2-\frac{5}{3}=\frac{1}{3}$，
$\therefore$ 在 $\mathrm{Rt}△ EQM$ 中,$EM=\sqrt{EQ^2+QM^2}=\frac{2\sqrt{3}}{3}$.
$\because \frac{EM}{DC}=\frac{2}{3},\therefore DC=\sqrt{3}$.