《2025年通城学典活页检测九年级数学下册人教版》

9. (12分)如图，四边形ABCD是平行四边形，E是BA延长线上一点，CE分别交AD，BD于点G，F.求证：CF² = GF·EF.
　　

答案： $\because$ 四边形$ABCD$是平行四边形，$\therefore AD// BC$，$AB// CD$. $\therefore \frac{GF}{CF}=\frac{DF}{BF}$，$\frac{CF}{EF}=\frac{DF}{BF}$. $\therefore \frac{GF}{CF}=\frac{CF}{EF}$，即$CF^{2}=GF\cdot EF$

10. (14分)如图，AD//EG//BC，EG分别交AB，DB，AC于点E，F，G.已知AD = 6，BC = 10，AE = 3，AB = 5，求EF，EG的长.
　　　　

答案： $\because AB = 5$，$AE = 3$，$\therefore BE = 5 - 3 = 2$. $\because AD// EG$，$\therefore \triangle BEF\sim\triangle BAD$. $\therefore \frac{EF}{AD}=\frac{BE}{BA}$，即$\frac{EF}{6}=\frac{2}{5}$. $\therefore EF=\frac{12}{5}$. $\because EG// BC$，$\therefore \triangle AEG\sim\triangle ABC$. $\therefore \frac{EG}{BC}=\frac{AE}{AB}$，即$\frac{EG}{10}=\frac{3}{5}$. $\therefore EG = 6$

11. ★(14分)如图，AD，BC相交于点E，AB//CD//EF，点B，F，D在一条直线上，AB = 10，CD = 15.求：
　　(1) $\frac{BF}{DF}$的值；
　　(2) EF的长.
　　　　　　

答案：
(1) $\because AB// CD$，$\therefore \frac{AE}{DE}=\frac{AB}{DC}=\frac{10}{15}=\frac{2}{3}$. $\because AB// EF$，$\therefore \frac{BF}{DF}=\frac{AE}{DE}=\frac{2}{3}$
(2) $\because CD// EF$，$\therefore \triangle BEF\sim\triangle BCD$. $\therefore \frac{EF}{CD}=\frac{BF}{BD}$. $\because CD = 15$，$\frac{BF}{DF}=\frac{2}{3}$，$\therefore EF = CD\cdot\frac{BF}{BD}=15\times\frac{2}{2 + 3}=6$

12. ★(16分)如图，在△ABC中，M为边AC的中点，E为边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.
　　　　

答案：
如图，过点$C$作$CF// DE$，交$AB$于点$F$. $\because ME// CF$，$\therefore \frac{AE}{EF}=\frac{AM}{MC}$. $\because M$为边$AC$的中点，$\therefore AM = MC$. $\therefore AE = EF$. $\because AB = 4AE$，$\therefore EF = AE=\frac{1}{4}AB$. $\therefore BF = AB - AE - EF=\frac{1}{2}AB$. $\therefore BF = 2EF$. $\because CF// DE$，$\therefore \frac{BC}{CD}=\frac{BF}{EF}=2$. $\therefore BC = 2CD$