8.（重庆中考）如图，在$△ ABC$中，延长$AC$至点$D$，使$CD=CA$，过点$D$作$DE// CB$，且$DE=DC$，连接$AE$交$BC$于点$F$。若$∠ CAB=∠ CFA$，$CF=1$，则$BF=$。

8.3 解析:$\because CD=CA,DE// CB,DE=DC,\therefore \frac{FA}{FE}=\frac{CA}{CD}=1,CD=CA=DE,\therefore AF=EF,\therefore DE=CD=AC=2CF=2,\therefore AD=AC+CD=4.\because DE// CB,\therefore ∠ CFA=∠ E,∠ ACB=∠ D.\because ∠ CAB=∠ CFA,\therefore ∠ CAB=∠ E,\therefore △ CAB≌△ DEA,\therefore BC=AD=4,$
$\therefore BF=BC-CF=3$.故答案为3.

9. (2026·青岛期中)如图,在$△ ABC$中,$AB=AC=5$,$BC=6$,$D$为$AB$上一点($D$不与$A$,$B$重合),过点$D$分别作$BC$和$AC$的平行线,交$AC$于点$E$,交$BC$于点$F$,过点$E$作$AB$的平行线,交$BC$于点$G$,交$DF$于点$O$,连接$DG$,$EF$,$BO$.下列结论:①$BG=CF$;②四边形$ADOE$是菱形;③当$\frac{AD}{BD}=\frac{2}{5}$时,四边形$DEFG$是矩形;④当$\frac{AD}{BD}=\frac{5}{11}$时,$BO$平分$∠ ABC$.正确的是(填写序号).

9.①②④ 解析:$\because BC// DE,AC// DF,\therefore$四边形$BGED,DECF$都是平行四边形,$\therefore DE=BG,DE=CF,\therefore BG=CF$,故①正确;$\because EG// AB,AC// DF,\therefore$四边形$ADOE$是平行四边形.$\because BC// DE,\therefore ∠ ADE=∠ ABC,∠ AED=∠ C.\because AB=AC$,$\therefore ∠ ABC=∠ C,\therefore ∠ ADE=∠ AED,\therefore AD=AE,\therefore$四边形$ADOE$是菱形,故②正确;$\because \frac{AD}{BD}=\frac{2}{5},BC// DE,AC// DF,AB=AC=5,BC=6,\therefore \frac{DE}{BC}=\frac{AD}{AB}=\frac{2}{5+2}=\frac{2}{7},\therefore DE=\frac{2}{7}×6=\frac{12}{7},\therefore BG=CF=DE=\frac{12}{7},\therefore GF=BC-BG-CF=\frac{18}{7},\therefore DE≠ GF$,$\therefore$四边形$DEFG$不是矩形,故③错误;当$\frac{AD}{BD}=\frac{5}{11}$时,$BD=\frac{11}{5+11}AB=\frac{55}{16}$,$\because$四边形$BGED,DECF$都是平行四边形,$\therefore EG=BD=\frac{55}{16}$,同理可得$\frac{DE}{BC}=\frac{AD}{AB}=\frac{5}{5+11}=\frac{5}{16},\therefore DE=\frac{5}{16}×6=\frac{15}{8},\therefore BG=CF=DE=\frac{15}{8},\therefore GF=BC-BG-CF=\frac{9}{4}.\because DE// BC,AB// GE,\therefore \frac{EO}{OG}=\frac{DE}{GF}=\frac{5}{6},∠ DBO=∠ GOB$,即$OG=\frac{6}{5+6}EG=\frac{15}{8},\therefore BG=OG,\therefore ∠ GBO=∠ GOB,\therefore ∠ DBO=∠ GBO$,故④正确.综上所述,①②④正确.故答案为①②④.

10. 如图,在$△ ABC$中,点$D$为$BC$上一点,点$P$在$AD$上,过点$P$作$PM// AC$交$AB$于点$M$,作$PN// AB$交$AC$于点$N$.
(1)若点$D$是$BC$的中点,且$AP:PD=2:1$,求$AM:AB$的值;
(2)若点$D$是$BC$的中点,试证明:$\frac{AM}{AB}=\frac{AN}{AC}$.

10.(1)如图①,过点$D$作$DE// PM$交$AB$于点$E$.$\because PM// AC$,
$\therefore DE// AC$.$\because$点$D$为$BC$中点,$\therefore \frac{BD}{DC}=\frac{BE}{EA}=1$,$\therefore$点$E$是$AB$中点,且$\frac{AM}{AE}=\frac{AP}{AD}=\frac{2}{3}$,$\therefore \frac{AM}{AB}=\frac{AM}{2AE}=\frac{1}{3}$,即$AM:AB=1:3$.

(2)如图②,延长$AD$至点$Q$,使$DQ=AD$,连接$BQ,CQ$,则四边形$ABQC$是平行四边形,$\therefore PM// BQ,PN// CQ$,
$\therefore \frac{AM}{AB}=\frac{AP}{AQ},\frac{AN}{AC}=\frac{AP}{AQ},\therefore \frac{AM}{AB}=\frac{AN}{AC}$.

易错提醒 已知一边中点和平行线可以推得另一边中点,但不能直接使用,尤其在解答题中.

11. (2026·成都月考)△ABC中,点D是BC边上的一点,点F在AD上,连接BF并延长交AC于点E.
(1)如图①,点D是BC中点,点F是AD中点,DG//BE交AC于点G,求证:$\frac{AE}{EC}=\frac{1}{2}$;
(2)如图②,若BD:DC=1:4,AF:FD=3:2,求AE:EC的值;
(3)若F为AD的中点,设$\frac{BD}{BC}=m$,$\frac{AE}{AC}=n$,请求出m,n之间的等量关系.

11.(1)$\because$点$D$是$BC$中点,$\therefore BD=CD$.$\because DG// BE$交$AC$于点$G$,$\therefore EG=CG=\frac{1}{2}EC$.又$\because$点$F$是$AD$的中点,$\therefore AF=FD$.
$\because DG// BE,\therefore AE=EG=\frac{1}{2}EC.\therefore \frac{AE}{EC}=\frac{1}{2}.$
(2)如图①,过点$D$作$DG// BE$交$AC$于点$G$,$\because \frac{BD}{DC}=\frac{1}{4}$,
$\therefore \frac{EG}{GC}=\frac{BD}{DC}=\frac{1}{4},\therefore \frac{EG}{EC}=\frac{1}{5}.\because \frac{AF}{FD}=\frac{3}{2},DG// EF,\therefore \frac{AE}{EG}=\frac{AF}{FD}=\frac{3}{2}$,即$EG=\frac{2}{3}AE,\therefore \frac{EG}{EC}=\frac{\frac{2}{3}AE}{EC}=\frac{1}{5},\therefore \frac{AE}{EC}=\frac{3}{10}$,即$AE:EC=3:10.$
(3)如图②,过点$D$作$DH// BE$交$AC$于点$H$,$\because \frac{BD}{BC}=m$,
$\therefore \frac{EH}{EC}=\frac{BD}{BC}=m.\because$点$F$是$AD$的中点,$\therefore AF=FD.\because DH// BE,\therefore AE=EH,\therefore \frac{AE}{EC}=\frac{EH}{EC}=m,\therefore \frac{AE}{AC}=\frac{m}{m+1}.\because \frac{AE}{AC}=n$,
$\therefore \frac{m}{m+1}=n.$

技法点拨 构造辅助线是解题的关键,过点$D$作$DG// BE$交$AC$于点$G$,根据平行线分线段成比例定理证得$\frac{EG}{EC}=\frac{1}{5}$,$\frac{AE}{EG}=\frac{3}{2}$,进而可求得$\frac{AE}{EC}$的值.