11. 如图,平行四边形$ABCD$的对角线$AC$与$BD$相交于点$O$,点$E$为$AO$的中点,过点$A$作$AF// BD$交$BE$的延长线于点$F$,连接$DF$.求证:四边形$AODF$是平行四边形.

11. 证明:$\because AF// BD$,
$\therefore ∠ EAF=∠ EOB,∠ AFE=∠ OBE$.
又$\because$点$E$为$AO$的中点,$\therefore AE=OE$.
在$△ AEF$和$△ OEB$中,
$\begin{cases}∠ AFE=∠ OBE, \\∠ EAF=∠ EOB, \\AE=OE,\end{cases}$
$\therefore △ AEF≌△ OEB(\mathrm{AAS})$.$\therefore AF=OB$.
$\because$四边形$ABCD$是平行四边形,
$\therefore OB=OD$.$\therefore AF=OD$.
又$\because AF// OD$,$\therefore$四边形$AODF$是平行四边形.

12. $△ ABC$是等边三角形,$D$是$BC$边上的一点,以$AD$为边作等边$△ ADE$,过点$C$作$CF// DE$交$AB$于点$F$.
(1)当点$D$是$BC$边的中点时,如图①,求证:$EF=CD$;
(2)如图②,当点$D$是$BC$边上的任意一点时(除点$B,C$外),(1)中的结论是否仍然成立? 若成立,请给出证明.若不成立,请说明理由.

12.(1)证明:$\because △ ABC$是等边三角形,
$\therefore ∠ B=∠ ACB=60°,BC=AC$.
$\because D$是$BC$的中点,
$\therefore AD⊥ BC,∠ CAD=\dfrac{1}{2}∠ BAC=30°$.
$\because △ AED$是等边三角形,
$\therefore AD=DE,∠ ADE=60°$.
$\therefore ∠ BDE=90°-∠ ADE=90°-60°=30°$.
$\because CF// DE$,
$\therefore ∠ BCF=∠ BDE =30°$.
在$△ BCF$和$△ CAD$中,
$\because ∠ BCF=∠ CAD =30°,∠ B=∠ ACB,BC=AC$,
$\therefore △ BCF≌△ CAD$.
$\therefore CF=AD$.
$\because AD=DE$,
$\therefore CF = DE$.
又$\because CF// DE$,
$\therefore$四边形$EDCF$是平行四边形,
$\therefore EF=CD$.
(2)解:$EF=CD$成立.
理由:$\because CF// DE$,
$\therefore ∠ BDE=∠ BCF$.
$\because ∠ BDE +∠ ADE =∠ CAD+∠ ACB$,
由(1)知$∠ ADE=∠ ACB=60°$,
$\therefore ∠ BDE =∠ CAD$,
$\therefore ∠ BCF =∠ CAD$.
在$△ BCF$和$△ CAD$中,
$\because ∠ BCF=∠ CAD ,∠ B=∠ ACB,BC=AC$,
$\therefore △ BCF≌△ CAD$.
$\therefore CF=AD$.
$\because AD=DE$,
$\therefore CF = DE$.
又$\because CF// DE$,
$\therefore$四边形$EDCF$是平行四边形,
$\therefore EF=CD$.