9.[易错题]如图，$AC // BD$，AD与BC交于点E，过点E作$EF // BD$，交线段AB于点F，则下列各式中错误的是（）


A.$\frac{AF}{BF}=\frac{AE}{DE}$
B.$\frac{BF}{AF}=\frac{BE}{CE}$
C.$\frac{AE}{AD}+\frac{BE}{BC}=1$
D.$\frac{AF}{BF}=\frac{CE}{DE}$

9.D ［解析］$\because AC// BD,EF// BD,\therefore BD// EF// AC,\therefore \frac {AF}{BF}=\frac {AE}{ED},\frac {BF}{AF}=\frac {BE}{EC}$,故 A,B 正确.$\because \frac {AE}{AD}=\frac {AF}{AB},\frac {BE}{BC}=\frac {BF}{AB},\therefore \frac {AE}{AD}+\frac {BE}{BC}=\frac {AF}{AB}+\frac {BF}{AB}=\frac {AF+BF}{AB}=\frac {AB}{AB}=1$,故 C 正确.$\because \frac {AF}{BF}=\frac {CE}{EB}$,且$DE≠EB,\therefore \frac {AF}{BF}≠\frac {CE}{DE}$,故 D 错误.

10.[跨学科·物理]物理课上我们学习过凸透镜成像规律.如图,蜡烛AB的高为15 cm,蜡烛AB与凸透镜的距离BE为32 cm,蜡烛的像CD与凸透镜的距离DE为8 cm,则像CD的高为cm.

10.$\frac{15}{4}$

11.如图，菱形ABCD的边长为1，直线l过点C，交AB的延长线于点M，交AD的延长线于N，则$\frac{1}{AM}+\frac{1}{AN}=$。

11.1 ［解析］$\because$ 四边形 ABCD 是菱形,$\therefore CD// AM,BC// AN$,
$\therefore △ NDC∽ △ NAM,△ MCB∽ △ MNA,\therefore \frac {DC}{AM}=\frac {NC}{NM},\frac {BC}{AN}=\frac {MC}{MN}$,即$\frac {1}{AM}=\frac {NC}{MN},\frac {1}{AN}=\frac {MC}{MN},\therefore \frac {1}{AM}+\frac {1}{AN}=\frac {NC}{NM}+\frac {MC}{MN}=1.$

12.如图,在$△ ABC$中,点M为AC边的中点,点E为AB上一点,且$AB=4AE$,连接EM并延长,交BC的延长线于点D.求证:$BC=2CD$.

12.证明:如图,作 $CF// DE$ 交 AB 于 F.
$\because ME// CF,\therefore \frac {AE}{EF}=\frac {AM}{MC}.\because M$ 为 AC 边的中点,$\therefore AM=MC,\therefore AE=EF.\because AB=4AE,\therefore EF=\frac {1}{4}AB,\therefore BF=\frac {1}{2}AB,$
$\therefore BF=2EF.\because CF// DE,\therefore \frac {BC}{CD}=\frac {BF}{EF}=2,\therefore BC=2CD.$

13.已知:如图,在平行四边形ABCD中,E,F分别是边BC,CD上的点,且EF//BD,AE,AF分别交BD于点G和点H,BD=12,EF=8.求:
(1)$\frac{DF}{AB}$的值;
(2)线段GH的长.

13.解:(1)$\because EF// BD,\therefore \frac {CF}{CD}=\frac {EF}{BD}.\because BD=12,EF=8,\therefore \frac {CF}{CD}=\frac {2}{3},\therefore \frac {DF}{CD}=\frac {1}{3}.\because$ 四边形 ABCD 是平行四边形,$\therefore AB=CD$,
$\therefore \frac {DF}{AB}=\frac {1}{3}.$
(2)$\because DF// AB,\therefore \frac {FH}{AH}=\frac {DF}{AB}=\frac {1}{3},\therefore \frac {AH}{AF}=\frac {3}{4}.\because EF// BD,$
$\therefore \frac {GH}{EF}=\frac {AH}{AF}=\frac {3}{4},\therefore \frac {GH}{8}=\frac {3}{4},\therefore GH=6.$

14.[情境题·数学文化]“今有邑方二百步，各中开门，出东门一十五步有木，问出南门几何步而见木？”这段话摘自古代数学著作《九章算术》，译文：如图，正方形城池ABCD，城墙AB长200步，东门点E、南门点F分别是AB，AD的中点，EG⊥AB，FH⊥AD，EG=15步，HG经过A点，则FH=步。

14.$\frac{2000}{3}$