《2025年滚动迁移中考总复习数学山西专版》

5. 如图，在△ABC中，点D，E为边AB的三等分点，过点E作EF//AC交BC于点F，过点D作DG//AC交BC于点G，连接AF交DG于点H.若AC = 12，则DH的长为( )

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

答案： C

6. 教材变式 如图，□ABCD中，点E在AD上，且AE = 2DE，连接CE并延长交BA的延长线于点F.

(1)判断AF与AB的数量关系，并说明理由.


(2)点G是线段AF上一点，满足∠FCG = ∠FCD，CG交AD于点H.
①若AG = 1，FG = 4，求GH的长.
②若$S_{\triangle AHG}=4$，$\frac{AG}{FG}=\frac{1}{4}$，则$S_{四边形ABCH}=$______.

答案： 解:
(1) 一题多解
$AF = 2AB$.理由如下:
方法一,$\because$ 四边形$ABCD$是平行四边形,
$\therefore AB// CD,AB = CD.\therefore \angle DCE = \angle F,\angle D = \angle EAF$.
$\therefore \triangle DCE\sim\triangle AFE.\therefore \frac{DC}{AF}=\frac{DE}{AE}=\frac{1}{2}$.
$\therefore AF = 2DC.\therefore AF = 2AB$.
方法二,先证$\triangle DCE\sim\triangle AFE$,得$\frac{FE}{CE}=\frac{AE}{DE}=2$,则$\frac{FE}{FC}=\frac{2}{3}$.
再证$\triangle AFE\sim\triangle BFC$,得$\frac{FA}{FB}=\frac{FE}{FC}=\frac{2}{3}$,则$AF = 2AB$.
(2)①$\because AG = 1,FG = 4,\therefore AF = AG + FG = 5$.
由
(1)可知$AF = 2AB = 2CD,\therefore AB = CD=\frac{5}{2}$.
$\because AB// CD,\therefore \angle FCD = \angle F$.
$\because \angle FCG = \angle FCD,\therefore \angle F = \angle FCG.\therefore CG = FG = 4$.
$\because AG// CD,\therefore \angle GAH = \angle D,\angle HGA = \angle HCD$.
$\therefore \triangle HGA\sim\triangle HCD.\therefore \frac{GH}{CH}=\frac{AG}{DC}=\frac{1}{\frac{5}{2}}=\frac{2}{5}$.
$\therefore GH=\frac{2}{7}CG=\frac{2}{7}\times4=\frac{8}{7}$.
②45
点拨:由①易得$\frac{AG}{CD}=\frac{2}{5}$,由$AB = CD$可得$\frac{AG}{GB}=\frac{2}{7}$,
可证$\triangle GAH\sim\triangle GBC$,则$\frac{S_{\triangle GAH}}{S_{\triangle GBC}}=\frac{4}{49}$,
因为$S_{\triangle AHG}=4$,所以$S_{\triangle GBC}=49$,
所以$S_{四边形ABCH}=S_{\triangle GBC}-S_{\triangle GAH}=49 - 4 = 45$.