$解：(2)过点C作CG⊥AB交AB$
$的延长线于点G，过点F作\ $
$FH⊥DE交DE的延长线于点H$
$∵∠ABC=∠DEF，$
$且∠ABC、∠DEF 都是钝角$
$∴180°-∠ABC=180°-∠DEF，$
$即∠CBG=∠FEH$
$在△CBG 和△FEH 中$
$\begin{cases}{∠G=∠H}\\{∠CBG=∠FEH}\\{BC=EF}\end{cases}$
$∴△CBG≌△FEH(\mathrm {AAS})$
$∴CG=FH$
$在Rt△ACG 和Rt△DFH中$
$\begin{cases}{AC=DF}\\{CG=FH}\end{cases}$
$∴Rt△ACG≌Rt△DFH(\mathrm {HL})$
$∴∠A=∠D$
$在△ABC和△DEF中$
$\begin{cases}{∠ABC=∠DEF}\\{∠A=∠D}\\{AC=DF}\end{cases}$
$∴△ABC≌△DEF(\mathrm {AAS})$
