$(1)由旋转得△ADE≌△ABC,则AC=AE$
$∴∠AEC=∠C=∠AED=65°$
$∴∠DEB=180°-∠AED-∠AEC=50°$
$(2)由旋转得,∠BAC=∠DAE,AB=AD,AC=AE$
$∴∠DAB=∠CAE$
$∴∠ABD=∠ADB=\frac {1}{2}(180°-∠BAD)$
$∠C=∠AEC=\frac {1}{2}(180°-∠CAE)$
$∴∠ABD=∠C$
$∵∠BAC=90°$
$∴∠C+∠ABC=90°$
$∴∠DBC=∠ABD+∠ABC=∠C+∠ABC=90°$
$∴BC⊥BD$