5. 如图,已知$AC// FE// BD$.求证:$\frac{BE}{BC}+\frac{AE}{AD}=1$.

$\because AC// FE,\therefore \frac{BE}{BC}=\frac{BF}{BA}$. ①
$\because FE// BD,\therefore \frac{AE}{AD}=\frac{AF}{AB}$. ②
①+②,得$\frac{BE}{BC}+\frac{AE}{AD}=\frac{BF}{BA}+\frac{AF}{AB}=\frac{AB}{AB}=1$.

6. 如图,已知B,C,E三点在同一条直线上,△ABC与△DCE都是等边三角形,其中线段BD交AC于点G,线段AE交CD于点F,连接GF.求证:(1)$△ ACE ≌ △ BCD$;(2)$\frac{AG}{GC}=\frac{AF}{FE}$.

(1)$\because △ ABC$与$△ DCE$都是等边三角形，
$\therefore AC=BC,CE=CD,∠ ACB=∠ DCE=60°$.
$\therefore ∠ DCE+∠ ACD=∠ ACB+∠ ACD$,
即$∠ ACE=∠ BCD. \therefore △ ACE≌△ BCD$.
(2)由(1)知$△ ACE≌△ BCD,\therefore ∠ BDC=∠ AEC$.
又$\because ∠ GCD=180°-∠ ACB-∠ DCE=60°=∠ FCE$,
$CD=CE,\therefore △ GCD≌△ FCE$.
$\therefore CG=CF. \therefore △ CFG$为等边三角形.
$\therefore ∠ CFG=∠ DCE=60°. \therefore GF// CE. \therefore \frac{AG}{GC}=\frac{AF}{FE}$.

7.如图,在△ABC中,点D为边AB的中点,DE//BC,DE交AC于点E,CF//BA,CF交DE的延长线于点F.求证:DE=EF.

$\because DE// BC,\therefore \frac{AD}{DB}=\frac{AE}{EC}$.
$\because$ 点 D 为 AB 的中点,$\therefore AD=DB$,即$\frac{AD}{DB}=1$.
$\because CF// BA,\therefore \frac{DE}{EF}=\frac{AE}{EC}=\frac{AD}{DB}=1. \therefore DE=EF$.

8. 如图,在$△ ABC$中,$M$为$BC$的中点,$O$为$AM$上一点,$BO$的延长线交$AC$于点$D$,$CO$的延长线交$AB$于点$E$,$PQ// BC$,且$PQ$过点$O$与$AB$,$AC$分别交于点$P$和点$Q$,连接$ED$.
求证:(1)$PO=OQ$;
(2)$ED// BC$.

(1)$\because PQ// BC,\therefore PO// BM,OQ// CM$.
$\therefore \frac{PO}{BM}=\frac{AO}{AM},\frac{OQ}{CM}=\frac{AO}{AM}. \therefore \frac{PO}{BM}=\frac{OQ}{CM}$.
$\because M$ 为 BC 的中点,$\therefore BM=CM. \therefore PO=OQ$.
(2)$\because PQ// BC,\therefore \frac{EO}{EC}=\frac{PO}{BC},\frac{DO}{DB}=\frac{OQ}{BC}$.
由(1)知 $PO=OQ,\therefore \frac{EO}{EC}=\frac{DO}{DB}$.
$\therefore \frac{EO}{OC}=\frac{DO}{OB}. \therefore ED// BC$.