$(1)证明：∵CD//AB，∴∠BAE=∠ACD$
$在△ABE和△CAD中，\begin{cases}{∠BAE=∠ACD}\\{AB=AC}\\{∠ABE=∠CAD}\end{cases}$
$∴△ABE≌△CAD(\mathrm {ASA})$
$(2)解：∵∠ACB=∠ABC=70°，∴∠BAC=180°- ∠ABC-∠ACB=180°-70°-70°=40°$
$∵∠ABE=25°，∴∠CAD=∠ABE=25°$
$∴∠BAD=∠BAC+∠CAD=40°+25°=65°$
$∵CD//AB，∴∠D=180°-∠BAD=180°-65°=115°$
