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9.如图,在菱形ABCD中,点E在BC边上,连接AE并延长交DC的延长线于点F.若CE:AD=1:4,CF=2,则菱形ABCD的周长为（
A.30
B.24
C.18
D.12
B
）A.30
B.24
C.18
D.12
答案
9.B
10.如图，$AB// CD$，$AC$与$BD$相交于点$E$，作$EF// AB$，交$BC$于点$F$，$AB=1$，$CD=2$，则$EF=\_\_\_\_\_\_$.
答案
10.$\frac{2}{3}$
11.如图,AC是$□ ABCD$的对角线,在AD边上取一点F,连接BF交AC于点E,延长BF交CD的延长线于点G.已知$DG=DC$,$BE=7$,求EF的长.
答案
11. 解:在$□ ABCD$ 中,$AB=CD$,$AB// CD$,
$\because DG=DC,\therefore AB=CD=DG$,
易证$△ ABF≌△ DGF$,
$\therefore AF=DF,\therefore AF=\frac{1}{2}AD=\frac{1}{2}BC$.
$\because AF// BC,\therefore △ AFE∽△ CBE$,
$\therefore \frac{EF}{BE}=\frac{AF}{BC}=\frac{1}{2}$,
$\therefore EF=\frac{1}{2}BE=\frac{7}{2}$.
$\because DG=DC,\therefore AB=CD=DG$,
易证$△ ABF≌△ DGF$,
$\therefore AF=DF,\therefore AF=\frac{1}{2}AD=\frac{1}{2}BC$.
$\because AF// BC,\therefore △ AFE∽△ CBE$,
$\therefore \frac{EF}{BE}=\frac{AF}{BC}=\frac{1}{2}$,
$\therefore EF=\frac{1}{2}BE=\frac{7}{2}$.
12. 如图,在边长为1的正方形ABCD中,E为AD的中点,连接BE,将△ABE沿BE折叠得到△FBE,且BF交AC于点G.求CG的长.
答案
12. 解:延长 $BF$ 交 $CD$ 于点 $H$,连接 $EH$.
易证 $\mathrm{Rt}△ DEH≌\mathrm{Rt}△ FEH$,$\therefore DH=FH$.
设 $DH=FH=x(x＞0)$,则 $CH=1-x$,$BH=1+x$.
在 $\mathrm{Rt}△ CHB$ 中,$BC^2+CH^2=BH^2$,
即 $1^2+(1-x)^2=(1+x)^2$,解得 $x=\frac{1}{4}$,
$\therefore CH=DC-DH=1-\frac{1}{4}=\frac{3}{4}$.
$\because DC// AB,\therefore △ HGC∽△ BGA$,
$\therefore \frac{CG}{AG}=\frac{CH}{AB}=\frac{3}{4}$,即$\frac{CG}{AC-CG}=\frac{3}{4}$,
$\because AC=\sqrt{2},\therefore \frac{CG}{\sqrt{2}-CG}=\frac{3}{4}$,解得 $CG=\frac{3\sqrt{2}}{7}$.
易证 $\mathrm{Rt}△ DEH≌\mathrm{Rt}△ FEH$,$\therefore DH=FH$.
设 $DH=FH=x(x＞0)$,则 $CH=1-x$,$BH=1+x$.
在 $\mathrm{Rt}△ CHB$ 中,$BC^2+CH^2=BH^2$,
即 $1^2+(1-x)^2=(1+x)^2$,解得 $x=\frac{1}{4}$,
$\therefore CH=DC-DH=1-\frac{1}{4}=\frac{3}{4}$.
$\because DC// AB,\therefore △ HGC∽△ BGA$,
$\therefore \frac{CG}{AG}=\frac{CH}{AB}=\frac{3}{4}$,即$\frac{CG}{AC-CG}=\frac{3}{4}$,
$\because AC=\sqrt{2},\therefore \frac{CG}{\sqrt{2}-CG}=\frac{3}{4}$,解得 $CG=\frac{3\sqrt{2}}{7}$.
