《2025年全程检测单元测试卷九年级数学全一册湘教版》

18. ($12$分)【综合实践】如图$9$①，$l_1// l_2// l_3$，直线$AB$与$HC$相交于点$O$，直线$AB$分别交$l_1$，$l_2$，$l_3$于点$A$，$D$，$B$，直线$HC$分别交$l_1$，$l_2$，$l_3$于点$H$，$E$，$C$. 已知$CE = 6$，$HE = 3$，$AB = 12$.
【尝试探究】(1)($4$分)在图$9$①中，$BD =$8.
【类比延伸】(2)($8$分)平移直线$AB$，使点$A$与点$H$重合，如图$9$②，过点$D$作$DF// AC$，交$BC$于点$F$. 若$DE = 5$，求$BF$的长.

答案： 18.
(1)$8$ 【解析】$\because l_{1} // l_{2} // l_{3}$，$\therefore \frac{AD}{BD} = \frac{HE}{CE}$，即$\frac{AB - BD}{BD} = \frac{HE}{CE}$. $\therefore \frac{12 - BD}{BD} = \frac{3}{6}$. $\therefore BD = 8$.
(2)$\because l_{2} // l_{3}$，$DF // AC$，$\therefore$四边形$CEDF$是平行四边形. $\therefore FC = DE = 5$，$DF = CE = 6$. $\because l_{1} // l_{2} // l_{3}$，$\therefore \frac{AD}{BD} = \frac{HE}{CE}$，即$\frac{AD}{BD} = \frac{3}{6} = \frac{1}{2}$. $\because DF // AC$，$\therefore \frac{FC}{BF} = \frac{AD}{BD} = \frac{1}{2}$. $\therefore BF = 10$.