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例1 如图，直线$l_1 // l_2 // l_3$，$AB = 3$，$BC = 5$，$DF = 12$，求$DE$和$EF$的长.
答案
例1 解：$\because l_1 // l_2 // l_3$，$\therefore \frac{AB}{BC} = \frac{DE}{EF}$.设$DE = x$，则$EF = 12 - x$.
$\because AB = 3$，$BC = 5$，$\therefore \frac{3}{5} = \frac{x}{12 - x}$.解得$x = \frac{9}{2}$.$\therefore DE = \frac{9}{2}$，$EF = \frac{15}{2}$.
$\because AB = 3$，$BC = 5$，$\therefore \frac{3}{5} = \frac{x}{12 - x}$.解得$x = \frac{9}{2}$.$\therefore DE = \frac{9}{2}$，$EF = \frac{15}{2}$.
例2 如图，在$\bigtriangleup ABC$中，$D$为边$BC$的中点，$E$为$AD$的中点，延长$BE$交$AC$于点$F$，求$\frac{AF}{AC}$的值.
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.答案
例2 解：过点$D$作$DG // BF$交$AC$于点$G$，$\therefore \frac{AE}{ED} = \frac{AF}{FG}$，$\frac{BD}{DC} = \frac{FG}{GC}$.
又$\because D$为边$BC$的中点，$E$为$AD$的中点，$\therefore AF = FG$，$FG = GC$，$\frac{AF}{AC} = \frac{1}{3}$.
又$\because D$为边$BC$的中点，$E$为$AD$的中点，$\therefore AF = FG$，$FG = GC$，$\frac{AF}{AC} = \frac{1}{3}$.
1.(临沂中考)如图，已知$AB// CD$，$AD$与$BC$相交于点$O$，若$\frac{BO}{OC} = \frac{2}{3}$，$AD = 10$，则$AO =$
4
.答案
1.4 解析：由$AB // CD$可得$\bigtriangleup DOC \backsim \bigtriangleup AOB$，
$\therefore \frac{AO}{DO} = \frac{BO}{CO}$，$\frac{AO}{10 - AO} = \frac{2}{3}$，解得$AO = 4$.
$\therefore \frac{AO}{DO} = \frac{BO}{CO}$，$\frac{AO}{10 - AO} = \frac{2}{3}$，解得$AO = 4$.
2.(长沙中考)如图，在矩形$ABCD$中，点$E$为$AB$的中点，点$F$在边$BC$上，且$BF = 2FC$，$AF$分别与$DE$，$DB$相交于点$M$，$N$，求$\frac{EM}{DM}$的值.
答案
2.解析：延长$AF$，$DC$交于点$G$.$\because AB // DG$，$\therefore \frac{CG}{AB} = \frac{CF}{BF} = \frac{1}{2}$.
$\because AE = \frac{1}{2}AB$，$\therefore AE = CG$.$\therefore \frac{EM}{DM} = \frac{AE}{DG} = \frac{1}{3}$.
$\because AE = \frac{1}{2}AB$，$\therefore AE = CG$.$\therefore \frac{EM}{DM} = \frac{AE}{DG} = \frac{1}{3}$.
