9. [2026 六安九中期中] 如图，AB, CD 相交于点 E, 且 AC//EF//DB, 点 C,F,B 在同一条直线上. 已知 AC = p, EF = r, DB = q, 则 p,q,r 之间满足的数量关系式是 ()


A.$\frac{1}{r} + \frac{1}{q} = \frac{1}{p}$
B.$\frac{1}{p} + \frac{1}{r} = \frac{2}{q}$
C.$\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$
D.$\frac{1}{q} + \frac{1}{r} = \frac{2}{p}$

C 见平行线,得三角形相似,从而得出对应边成比例.
$ \begin{array}{l}AC// EF\rightarrow△ BEF∽△ BAC\rightarrow\dfrac{EF}{AC}=\dfrac{BF}{BC}\\EF// BD\rightarrow△ CFE∽△ CBD\rightarrow\dfrac{EF}{BD}=\dfrac{CF}{BC}\end{array}\}\xrightarrow{\mathrm{两式相加}}\dfrac{EF}{AC}+\dfrac{EF}{BD}=1\rightarrow\dfrac{1}{p}+\dfrac{1}{q}=\dfrac{1}{r}$.

10. 教材P122T1变式 如图,已知在梯形ABCD中, $AD// BC,\frac{S_{△ ABD}}{S_{△ BCD}}=\frac{1}{2}$,则$\frac{S_{△ BOC}}{S_{△ BCD}}=$.

$\dfrac{2}{3}$ $\because \dfrac{S_{△ ABD}}{S_{△ BCD}}=\dfrac{1}{2},AD// BC,\therefore \dfrac{AD}{BC}=\dfrac{1}{2},△ OAD∽△ OCB,\therefore \dfrac{OD}{OB}=\dfrac{AD}{BC}=\dfrac{1}{2},\therefore \dfrac{OB}{BD}=\dfrac{2}{3},\therefore \dfrac{S_{△ BOC}}{S_{△ BCD}}=\dfrac{OB}{BD}=\dfrac{2}{3}$.

11. [2026安庆期中]如图,在正方形网格中,每个小正方形的边长均为1,A,B,C,D都是格点,AB与CD相交于点O,则$\frac{AO}{OB}=$.

$\dfrac{3}{10}$ 如图,$\because EH// CF,\therefore △ DEH∽△ DCF,\therefore \dfrac{EH}{CF}=\dfrac{DH}{DF}$,即$\dfrac{EH}{3}=\dfrac{1}{2},\therefore EH=\dfrac{3}{2},\therefore AE=AH-EH=3-\dfrac{3}{2}=\dfrac{3}{2}.\because AE// BC,\therefore △ AOE∽△ BOC,\therefore \dfrac{AO}{OB}=\dfrac{AE}{BC}=\dfrac{\dfrac{3}{2}}{5}=\dfrac{3}{10}$.

12. [2026亳州段考]如图,直线$l_1$,$l_2$,$l_3$分别交直线$l_4$于点$A$,$B$,$C$,交直线$l_5$于点$D$,$E$,$F$,且$l_1// l_2// l_3$.已知$DE:DF=3:8$,$AC=24$.
(1)求$BC$的长.
(2)当$AD=4$,$CF=20$时,求$BE$的长.

解:(1)$\because l_1// l_2// l_3,\therefore \dfrac{AB}{AC}=\dfrac{DE}{DF}$,即$\dfrac{AB}{24}=\dfrac{3}{8}$,解得$AB=9$,
$\therefore BC=AC-AB=24-9=15$.
(2)如图,过点$A$作$AN// DF$交$CF$于点$N$,交$EB$于点$M$.
易得四边形ADEM和四边形ADFN为平行四边形,
$\therefore EM=FN=AD=4,\therefore CN=CF-FN=20-4=16$.
$\because BM// CN,\therefore △ ABM∽△ ACN$,
$\therefore \dfrac{BM}{CN}=\dfrac{AB}{AC}$,即$\dfrac{BM}{16}=\dfrac{9}{24}$,解得$BM=6$,
$\therefore BE=EM+BM=4+6=10$.

13. 推理能力 [2024 无锡中考]
如图, 在 $△ ABC$ 中, $AC = 2$, $AB = 3$, 直线 $CM // AB$, $E$ 是 $BC$ 上的动点(端点除外), 射线 $AE$ 交 $CM$ 于点 $D$. 在射线 $AE$ 上取一点 $P$, 使得 $AP = 2ED$, 作 $PQ // AB$, 交射线 $AC$ 于点 $Q$. 设 $AQ = x$, $PQ = y$. 当 $x = y$ 时, $CD =$ ; 在点 $E$ 运动的过程中, $y$ 关于 $x$ 的函数表达式为 .

2 $y=\dfrac{3x^2}{8-2x}$ 解题思路:第1问,易得$CD// PQ$,则$△ APQ∽△ ADC$,得出$\dfrac{AQ}{AC}=\dfrac{PQ}{CD}$,代入数据即可求出$CD=2$.第2问,由$△ APQ∽△ ADC$,得出$CD=\dfrac{2y}{x}$.设$DE=t$,则$AP=2t$,再通过证得$△ CDE∽△ BAE$,得出$\dfrac{CD}{AB}=\dfrac{DE}{AE}$,则$AE=\dfrac{3xt}{2y}$,进而得出$AD=AE+DE=\dfrac{t(3x+2y)}{2y}$,最后由$△ APQ∽△ ADC$得出$\dfrac{AQ}{AC}=\dfrac{AP}{AD}$,代入各个数据,即可得出$y$关于$x$的函数表达式.
$\because CM// AB,PQ// AB,\therefore CD// PQ,\therefore △ APQ∽△ ADC,\therefore \dfrac{AQ}{AC}=\dfrac{PQ}{CD}$,即$\dfrac{x}{2}=\dfrac{y}{CD}$.又$\because x=y,\therefore CD=2.\because △ APQ∽△ ADC,\therefore \dfrac{AQ}{AC}=\dfrac{PQ}{CD}$,即$\dfrac{x}{2}=\dfrac{y}{CD}$,得$CD=\dfrac{2y}{x}$.设$DE=t$.$\because AP=2ED,\therefore AP=2t$.
$\because CM// AB,\therefore △ CDE∽△ BAE,\therefore \dfrac{CD}{AB}=\dfrac{DE}{AE}$,即$\dfrac{\dfrac{2y}{x}}{3}=\dfrac{t}{AE}$,得$AE=\dfrac{3xt}{2y}$,$\therefore AD=AE+DE=\dfrac{3xt}{2y}+t=\dfrac{t(3x+2y)}{2y}$.$\because △ APQ∽△ ADC,\therefore \dfrac{AQ}{AC}=\dfrac{AP}{AD}$,即$\dfrac{x}{2}=\dfrac{2t}{\dfrac{t(3x+2y)}{2y}}$,得$y=\dfrac{3x^2}{8-2x}$.

1. 直接利用8字型 如图，，$△ ABC$中，$∠ ACB = 90°, AB = 12$，$D,E$分别是边$AB,BC$的中点，$CD$与$AE$交于点$O$，则$OD$的长是（）

A.$1.5$
B.$1.8$
C.$2$
D.$2.4$

1 C 在Rt△ABC中,D是边AB的中点,$\therefore CD = \frac{1}{2}AB = 6$.
又$\because E$是边$BC$的中点,$\therefore DE$是$△ ABC$的中位线,$\therefore DE // AC, DE = \frac{1}{2}AC$(中位线定理),$\therefore △ EOD ∽ △ AOC, \therefore \frac{OD}{OC} = \frac{DE}{AC} = \frac{1}{2}, \therefore OD = \frac{1}{3}CD = 2$.

2. 间接构造8字型 [2026滁州期中]
如图,在$△ ABC$中,点$D,F$在边$AB$上,点$E$在边$BC$上,连接$CF,DE$,交于点$G$.
若$DE// AC$,$BD=2AD$,$BF=2DF$,则$\frac{DG}{EG}$的值为 ()

A.$1$
B.$\frac{3}{2}$
C.$\frac{5}{3}$
D.$\frac{7}{4}$

2 B 如图,过点$B$作$BP// AC$,交$CF$的延长线于点$P$(构造8字型).$\because DE// AC, \therefore BP// DE// AC$(依据:如果两条直线都和第三条直线平行,那么这两条直线也互相平行),$\therefore △ DBE ∽ △ ABC, △ GCE ∽ △ PCB, \therefore \frac{BD}{AB} = \frac{BE}{BC}, \frac{CE}{BC} = \frac{EG}{BP}. \because BD = 2AD$,
$\therefore AB = BD + AD = 3AD, \therefore \frac{BD}{AB} = \frac{BE}{BC} = \frac{2}{3}, \therefore \frac{CE}{BC} = \frac{EG}{BP} = \frac{1}{3}$,
$\therefore BP = 3EG. \because BP// DG, \therefore △ BFP ∽ △ DFG, \therefore \frac{BF}{DF} = \frac{BP}{DG}$.
$\because BF = 2DF, \therefore \frac{BF}{DF} = \frac{BP}{DG} = 2, \therefore BP = 2DG.$ 又$\because BP = 3EG$,
$\therefore 3EG = 2DG, \therefore \frac{DG}{EG} = \frac{3}{2}$.