$证明：(1)∵∠BAC= ∠EAD$
$∴∠BAC+∠DAC= ∠EAD+∠DAC，$
$即 ∠DAB=∠EAC\ $
$在△ABD和△ACE中$
$\begin{cases}{AD=AE}\\{∠DAB=∠EAC}\\{AB=AC}\end{cases}$
$∴△ABD≌△ACE(\mathrm {SAS})$
$(2)解：∵∠BAC=∠EAD，∠CAD=120°$
$∴∠BAC=∠EAD=\frac{180°-∠CAD}{2}=\frac{180°-120°}{2}=30°$
$∵∠BAC是△EAC的外角$
$∴∠BAC=∠AEC+∠ACE=30°$
$∵△ABD≌△ACE$
$∴∠ECA=∠DBA$
$∵∠DME是△BME的外角$
$∴∠DME=∠AEC+∠ABD=∠AEC+∠ACE=30°$
