《2026年启东中学作业本九年级数学下册苏科版》

1. (2024·通州区月考)如图,在$△ ABC$中,$AB = 24$,$AC = 18$,$D$是$AC$上一点,$AD = 12$.在$AB$上取一点$E$,使$△ ADE$与$△ ABC$相似,则$AE$的长为(D)

A.16
B.14
C.16 或 14
D.16 或 9

答案： 1. D

2. (2024·常熟期中)如图,在$△ ABC$中,点$D$在$AC$边上,$AD:DC = 1:3$,$BO:DO = 1:2$,连接$AO$并延长交$BC$于点$E$,则$BE:EC =$$ 1:8 $.

答案： 2. $ 1:8 $

3. 如图,在$△ ABC$中,点$D$,$E$分别在边$AB$,$AC$上,$∠ AED = ∠ B$,射线$AG$分别交线段$DE$,$BC$于点$F$,$G$,且$\frac{AD}{AC} = \frac{DF}{CG}$.
(1)求证:$△ ADF ∽ △ ACG$;
(2)若$AC = 3AD$,求$\frac{AF}{FG}$的值.

答案： 3.
(1) 证明: $ \because ∠ AED = ∠ B, ∠ BAC = ∠ EAD $,
$ \therefore ∠ ADE = ∠ C $.
$ \because \frac{AD}{AC} = \frac{DF}{CG}, \therefore △ ADF ∼ ACG $.
(2) 解: 由
(1)可知 $ △ ADF ∼ △ ACG, \therefore \frac{AD}{AC} = \frac{DF}{CG} = \frac{AF}{AG} $.
$ \because AC = 3AD, \therefore \frac{AD}{AC} = \frac{AF}{AG} = \frac{1}{3}, \therefore AG = 3AF $,
$ \therefore FG = AG - AF = 2AF, \therefore \frac{AF}{FG} = \frac{1}{2} $.

4. (2023·雅安)如图,在$□ ABCD$中,$F$是$AD$上一点,$CF$交$BD$于点$E$,$CF$的延长线交$BA$的延长线于点$G$,$EF = 1$,$EC = 3$,则$GF$的长为(C)

A.4
B.6
C.8
D.10

答案： 4. C

5. (2024·宜兴一模)如图,在矩形$ABCD$中,$AB = 4$,$BC = 5$,$E$为$BC$边延长线上一点,且$CE = 3$.连接$AE$交边$CD$于点$F$,过点$D$作$DH ⊥ AE$于点$H$,则$DH =$$ \sqrt{5} $.

答案： 5. $ \sqrt{5} $