1. 新考法分类讨论法 如图，在$△ ABC$中，$AB=6$，$CA=4$，D为AC的中点，点E在AB上，当AE为时，$△ ABC$与以点A，D，E为顶点的三角形相似。

1. 3或$\frac{4}{3}$ 【点拨】$\because D$为$AC$的中点，$CA=4$，$\therefore AD=\frac{1}{2}CA=2$。当$\frac{AE}{AB}=\frac{AD}{AC}$时，$\because ∠ A=∠ A$，$\therefore △ AED ∽ △ ABC$。$\therefore AE=\frac{AB· AD}{AC}=\frac{6×2}{4}=3$；当$\frac{AE}{AC}=\frac{AD}{AB}$时，$\because ∠ A=∠ A$，$\therefore △ ADE ∽ △ ABC$。$\therefore AE=\frac{AC· AD}{AB}=\frac{4×2}{6}=\frac{4}{3}$。综上，当$AE=3$或$\frac{4}{3}$时，$△ ABC$与以点A，D，E为顶点的三角形相似。

2.[邵阳模拟]如图,在$□ ABCD$中,连结对角线AC,延长AB至点E,使$BE=AB$,连结DE,分别交BC,AC于点F,G.
(1)求证:$BF=CF$;
(2)若$BC=6,DG=4$,求FG的长.

2. (1)【证明】$\because$ 四边形$ABCD$是平行四边形，$\therefore AD// BC,AD=BC$。$\therefore ∠ EBF=∠ EAD$。又$\because ∠ BEF=∠ AED$，$\therefore △ EBF ∽ △ EAD$。$\therefore \frac{BF}{AD}=\frac{EB}{EA}$。又$\because BE=AB$，$\therefore$ 易得$BF=\frac{1}{2}AD=\frac{1}{2}BC$。$\therefore BF=CF$。
(2)【解】由(1)知$AD// BC$，$\therefore ∠ FCG=∠ DAG$。又$\because ∠ FGC=∠ DGA$，$\therefore △ FGC ∽ △ DGA$。$\therefore \frac{FG}{DG}=\frac{FC}{AD}$。$\because AD=BC=6$，$FC=\frac{1}{2}BC=3$，$DG=4$，$\therefore \frac{FG}{4}=\frac{3}{6}$。$\therefore FG=2$。

3. 新考法·构造法 山西中考 如图，在四边形ABCD中，AD//BC，∠B=90°，AB=8，BC=4，点E在边AB上，AE=3，连结CE，且∠DCE=∠BCE。点F在BC的延长线上，连结DF。若DF=DC，则线段CF的长为。

3. $\frac{18}{5}$ 【点拨】如图，延长$CE$交$DA$的延长线于点$G$，过点$D$作$DH⊥ BF$于点$H$，则$∠ BHD=90°$。$\because DF=DC$，$\therefore CH=FH=\frac{1}{2}CF$。$\because AD// BC$，$∠ B=90°$，$\therefore ∠ B=∠ GAE=90°$，$∠ B+∠ BAD=180°$。$\therefore ∠ B=∠ BAD=∠ BHD=90°$。$\therefore$ 四边形$ABHD$是矩形，$\therefore AB=DH=8$，$AD=BH$。$\because ∠ AEG=∠ BEC$，$∠ B=∠ GAE$，$\therefore △ AEG ∽ △ BEC$。$\therefore \frac{AG}{BC}=\frac{AE}{BE}$。$\because AB=8$，$AE=3$，$\therefore BE=5$。$\therefore \frac{AG}{4}=\frac{3}{5}$。$\therefore AG=\frac{12}{5}$。$\because AD// BC$，$\therefore ∠ G=∠ BCE$。$\because ∠ DCE=∠ BCE$，$\therefore ∠ DCE=∠ G$。$\therefore CD=GD$。设$CH=FH=x$，则$AD=BH=4+x$，$\therefore CD=GD=4+x+\frac{12}{5}=x+\frac{32}{5}$。在$\mathrm{Rt}△ DCH$中，由勾股定理得$CD^2=CH^2+DH^2$，$\therefore (x+\frac{32}{5})^2=x^2+8^2$，解得$x=\frac{9}{5}$，即$CH=\frac{9}{5}$，$\therefore CF=2CH=\frac{18}{5}$。

4.[上海静安区期末] 如图,在$△ ABC$中,$∠ ACB=90°$,$CA=CB$,$CD$是$AB$边上的高,点$E$为线段$CD$上一点(不与点$C,D$重合),连结$BE$,作$EF⊥ BE$与$AC$的延长线交于点$F$,与$BC$交于点$G$,连结$BF$.求证:
(1)$△ CFG∽△ EBG$;
(2)$∠ CEF=∠ CBF$.

4. 【证明】(1)$\because ∠ ACB=90°$，$EF⊥ BE$，$\therefore ∠ FCG=∠ BEG=90°$。又$\because ∠ CGF=∠ EGB$，$\therefore △ CFG ∽ △ EBG$。
(2)由(1)知$△ CFG ∽ △ EBG$，$\therefore \frac{CG}{EG}=\frac{FG}{BG}$。$\therefore \frac{CG}{FG}=\frac{EG}{BG}$。又$\because ∠ CGE=∠ FGB$，$\therefore △ CGE ∽ △ FGB$。$\therefore ∠ CEG=∠ FBG$，即$∠ CEF=∠ CBF$。

5. 小强在学习《图形的相似》一章中对“直角三角形斜边上作高”这一基本图形(如图①)产生了如下问题,请同学们帮他解决.
在$△ ABC$中,$D$为边$AB$上一点,连结$CD$.
(1)如图②,若$∠ ACD=∠ B$,求证:$AC^2=AD· AB$;
(2)如图③,在(1)的条件下,若$D$为$AB$的中点,$BC=4$,求$CD$的长;
(3)如图④,$E$为$CD$的中点,连结$BE$,若$∠ CDB=∠ CBD=30°,∠ ACD=∠ EBD,AC=2\sqrt{7}$,求$BE$的长.

5. (1)【证明】$\because ∠ ACD=∠ B$，$∠ A=∠ A$，$\therefore △ ACD ∽ △ ABC$。$\therefore \frac{AC}{AB}=\frac{AD}{AC}$，即$AC^2=AD· AB$。
(2)【解】$\because D$为$AB$的中点，$\therefore$ 设$AD=BD=m$。$\because AC^2=AD· AB$，$\therefore AC^2=m· (m+m)=2m^2$。$\therefore AC=\sqrt{2}m$。$\therefore \frac{AD}{AC}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$。由(1)知$△ ACD ∽ △ ABC$，$\therefore \frac{CD}{BC}=\frac{AD}{AC}=\frac{\sqrt{2}}{2}$。$\because BC=4$，$\therefore CD=2\sqrt{2}$。
(3)【解】如图，过点$C$作$CH// EB$交$AB$的延长线于点$H$，过点$C$作$CG⊥ AB$。
$\because E$为$CD$的中点，$\therefore$ 设$CE=DE=a$。$\therefore CD=2a$。$\because ∠ CDB=∠ CBD=30°$，$\therefore CB=CD=2a$，$∠ DCB=120°$，在$\mathrm{Rt}△ BCG$中，$CG=\frac{1}{2}BC=a$，$\therefore BG=\sqrt{3}a$。$\therefore$ 易知$BD=2\sqrt{3}a$。过点$B$作$BF⊥ EC$交$EC$的延长线于点$F$。易知$∠ FCB=60°$，$\therefore ∠ CBF=30°$。$\therefore CF=\frac{1}{2}BC=a$，$\therefore BF=\sqrt{3}a$，$EF=2a$。$\therefore BE=\sqrt{7}a$。$\because CH// BE$，$E$为$CD$的中点，$\therefore CH=2BE=2\sqrt{7}a$，$DH=2DB=4\sqrt{3}a$，$∠ EBD=∠ H$。又$\because ∠ ACD=∠ EBD$，$\therefore ∠ ACD=∠ H$。又$\because ∠ A=∠ A$，$\therefore △ ACD ∽ △ AHC$。$\therefore \frac{AD}{AC}=\frac{AC}{AH}=\frac{CD}{CH}=\frac{2a}{2\sqrt{7}a}=\frac{\sqrt{7}}{7}$。又$\because AC=2\sqrt{7}$，$\therefore AD=2$，$AH=14$。$\therefore DH=12$，即$4\sqrt{3}a=12$，$\therefore a=\sqrt{3}$。$\therefore BE=\sqrt{21}$。