12. 如图,在矩形ABCD中,点E、F分别在边AD、DC上,△ABE∽△DEF,AB=6,AE=9,DE=2.
(1) 求EF的长.
(2) 求证:∠BEF=90°.

12. (1) $\because△ ABE∽△ DEF,\therefore \dfrac{AB}{DE}=\dfrac{AE}{DF}.\because AB=6,AE=9,DE=2,\therefore \dfrac{6}{2}=\dfrac{9}{DF}.\therefore DF=3.$ 在矩形$ABCD$中,$∠ D=90°.$ $\therefore$ 在$\mathrm{Rt}△ DEF$中,$EF=\sqrt{DE^2+DF^2}=\sqrt{2^2+3^2}=\sqrt{13}$
(2) $\because△ ABE∽△ DEF,\therefore∠ ABE=∠ DEF.$ $\because$ 在$\mathrm{Rt}△ ABE$中,$∠ ABE+∠ AEB=90°,\therefore∠ DEF+∠ AEB=90°.\therefore∠ BEF=90°$

13. 如图，$AD // EG // BC$，$EG$分别交$AB$、$DB$、$AC$于点$E$、$F$、$G$。若$AD=6$，$BC=10$，$AE=3$，$AB=5$，求$EG$、$FG$的长。

13. 在$△ ABC$中，$\because EG// BC,\therefore \dfrac{EG}{BC}=\dfrac{AE}{AB}.\because BC=10,AE=3,AB=5,\therefore \dfrac{EG}{10}=\dfrac{3}{5}.\therefore EG=6.$ $\because$ 在$△ BAD$中,$EF// AD$，$\therefore \dfrac{EF}{AD}=\dfrac{BE}{AB}.\because AD=6,AE=3,AB=5,\therefore \dfrac{EF}{6}=\dfrac{5-3}{5}.\therefore EF=\dfrac{12}{5}.\therefore FG=EG-EF=\dfrac{18}{5}$

14. 如图，在$\mathrm{Rt}△ ABC$中，$∠ C=90°$，$AC=6$，$BC=8$，沿折痕$EF$折叠$△ AEF$，使点$A$落在线段$BC$上的点$D$处.
（1）当点$D$是线段$BC$的中点时，求线段$CE$的长；
（2）若$△ DFB$与$△ ACB$相似，求线段$CD$的长.

14. (1) $\because D$是线段$BC$的中点，$BC=8,\therefore CD=\dfrac{1}{2}BC=4.$ 由折叠可知,$AE=ED=AC-CE=6-CE.$ $\because∠ C=90°,AC=6,\therefore ED^2=CE^2+CD^2,$即$(6-CE)^2=CE^2+4^2.$ 解得$CE=\dfrac{5}{3}$
(2) $\because∠ C=90°,AC=6,BC=8,\therefore AB=\sqrt{AC^2+BC^2}=10.$ $△ DFB$与$△ ACB$相似分两种情况:① $\dfrac{FD}{BD}=\dfrac{AC}{BC}=\dfrac{6}{8}=\dfrac{3}{4}$,则$FD// AC,∠ BDF=90°$,如图①所示. 由折叠性质可知,$AF=DF.$ 设$AF=DF=x$,则$BF=AB-AF=10-x.$ $\because FD// AC,\therefore△ DFB∽△ CAB.$ $\therefore \dfrac{DF}{CA}=\dfrac{BF}{BA}=\dfrac{BD}{BC}$,即$\dfrac{x}{6}=\dfrac{10-x}{10}.$ 解得$x=\dfrac{15}{4}.\therefore \dfrac{BD}{8}=\dfrac{\dfrac{15}{4}}{6}.\therefore BD=5.\therefore CD=BC-BD=3.$ ② $\dfrac{FD}{BF}=\dfrac{CA}{BC}=\dfrac{3}{4}$,则$∠ BFD=90°$,如图②所示,则$△ DFB∽△ ACB.$ $\therefore \dfrac{DF}{AC}=\dfrac{BF}{BC}=\dfrac{BD}{BA}$,即$\dfrac{x}{6}=\dfrac{10-x}{8}.$ 解得$x=\dfrac{30}{7}.\therefore \dfrac{BD}{10}=\dfrac{\dfrac{30}{7}}{6}.\therefore BD=\dfrac{50}{7}.\therefore CD=BC-BD=\dfrac{6}{7}.$ 综上,若$△ DFB$与$△ ACB$相似,则线段$CD$的长为$3$或$\dfrac{6}{7}$