1.如图，在$△ ABC$中，D，E分别是BC，AC上的点，连接DE并延长与BA的延长线交于点F，且$BD=DC$.求证：$\frac{AE}{EC}=\frac{FA}{FB}$.

过点A作AM//DF交BD于点M，
则$\frac{AE}{EC}=\frac{MD}{DC},\frac{FA}{FB}=\frac{DM}{BD}$.
$\because BD=DC,\therefore \frac{AE}{EC}=\frac{FA}{FB}$.

2. 如图,在$△ ABC$中,D是AB上一点,E是$△ ABC$内一点,$DE// BC$,过点D作AC的平行线交CE的延长线于点F,CF与AB交于点P.
求证:$\frac{PE}{PF}=\frac{PA}{PB}$.

$\because DE// BC,\therefore \frac{PD}{PB}=\frac{PE}{PC}$.
$\therefore PD· PC=PE· PB$.
$\because DF// AC,\therefore \frac{PF}{PC}=\frac{PD}{PA}$.
$\therefore PD· PC=PF· PA$.
$\therefore PE· PB=PF· PA.\therefore \frac{PE}{PF}=\frac{PA}{PB}$.

3. 如图，已知四边形ABCD是菱形，点E是对角线AC上的一点，连接BE并延长交AD于点F，交CD的延长线于点G，连接DE.
求证:(1)$∠ G=∠ ADE$；
(2)$EB^2=EF· EG$.

(1) 在菱形 ABCD 中，$AB// CD, AB = AD$，$∠ BAE=∠ DAE$.
又$\because AE=AE,\therefore △ ABE≌△ ADE. \therefore ∠ ABE=∠ ADE$.
$\because AB// CG,\therefore ∠ G=∠ ABE. \therefore ∠ G=∠ ADE$.
(2)$\because AF// BC,\therefore \frac{EB}{EF}=\frac{EC}{EA}$.
$\because AB// CG,\therefore \frac{EG}{EB}=\frac{EC}{EA}$.
$\therefore \frac{EB}{EF}=\frac{EG}{EB}$,即 $EB^2=EF· EG$.

4. 如图，$DE // BC$，$EF // CG$，$AD:AB=1:3$，$AE=3$.
(1)求$EC$的值；
(2)求证：$AD · AG=AF · AB$.

(1) $\because DE// BC,\therefore \frac{AD}{AB}=\frac{AE}{AC}$.
又$\because \frac{AD}{AB}=\frac{1}{3},AE=3,\therefore \frac{3}{AC}=\frac{1}{3}.\therefore AC=9$.
$\therefore EC=AC-AE=9-3=6$.
(2)【证明】$\because EF// CG,\therefore \frac{AE}{AC}=\frac{AF}{AG}$.
又$\because \frac{AD}{AB}=\frac{AE}{AC},\therefore \frac{AD}{AB}=\frac{AF}{AG}.\therefore AD· AG=AF· AB$.