答案：(1)证明:$\because △ ABD$和$△ ACE$都为等边三角形,
$\therefore AD=AB,AE=AC,∠ DAB=∠ EAC=60°$,
$\therefore ∠ DAB+∠ BAC=∠ EAC+∠ BAC$,
即$∠ DAC=∠ BAE$.
在$△ DAC$和$△ BAE$中,$\begin{cases} AD=AB,\\ ∠ DAC=∠ BAE,\\ AC=AE, \end{cases}$
$\therefore △ DAC≌△ BAE(\mathrm{SAS})$.
(2)证明:由(1)知$△ DAC≌△ BAE$,$\therefore BE=DC$.
(3)解:由(1)知$△ DAC≌△ BAE$,$\therefore ∠ ACD=∠ AEB$,
则$∠ DFE=∠ FEC+∠ FCE=∠ FEC+∠ ACD+∠ ACE=∠ FEC+∠ AEB+∠ ACE=∠ AEC+∠ ACE=120°$.

答案：(1)证明:$\because △ ABC$和$△ ADE$是等边三角形,
$\therefore AB=AC,AD=AE,∠ BAC=∠ DAE=60°$,
$\therefore ∠ BAC-∠ DAC=∠ DAE-∠ DAC$,即$∠ BAD=∠ CAE$.
在$△ ABD$和$△ ACE$中,$\begin{cases} AB=AC,\\ ∠ BAD=∠ CAE,\\ AD=AE, \end{cases}$
$\therefore △ ABD≌△ ACE(\mathrm{SAS})$.
(2)解:$AB// CE$,$EC=AC+CD$.证明如下:
由(1)得$△ ABD≌△ ACE$,
$\therefore ∠ B=∠ ACE=60°$,$CE=BD$.
$\because ∠ B=∠ BAC$,$\therefore ∠ BAC=∠ ACE$,$\therefore AB// CE$.
$\because CE=BD$,$AC=BC$,
$\therefore CE=BD=BC+CD=AC+CD$.

答案：
(1)证明:$\because △ ABD$和$△ ACE$都是等腰直角三角形,
$\therefore AB=AD,AC=AE,∠ BAD=∠ CAE=90°$,
$\therefore ∠ BAD+∠ BAC=∠ CAE+∠ BAC$,即$∠ CAD=∠ EAB$.
在$△ CAD$和$△ EAB$中,$\begin{cases} AC=AE,\\ ∠ CAD=∠ EAB,\\ AD=AB, \end{cases}$
$\therefore △ CAD≌△ EAB(\mathrm{SAS})$,$\therefore BE=CD$.
(2)如答图,作$DG⊥ EA$交$EA$的延长线于点$G$,$BH⊥ AC$于点$H$,则$∠ AGD=∠ AHB=90°$.
$\because ∠ CAE=90°$,$\therefore ∠ CAG=90°=∠ BAD$,
$\therefore ∠ DAG=∠ BAH$.
在$△ ADG$和$△ ABH$中,$\begin{cases} ∠ AGD=∠ AHB,\\ ∠ DAG=∠ BAH,\\ AD=AB, \end{cases}$
$\therefore △ ADG≌△ ABH(\mathrm{AAS})$,$\therefore DG=BH$.
$\because S_{△ ABC}=\frac{1}{2}AC· BH$,$S_{△ ADE}=\frac{1}{2}AE· DG$,
$\therefore S_{△ ABC}=S_{△ ADE}$.