解:∵∠BDC=∠A+∠ABD, ∠BDC=95, ∠A=60∴∠ABD=∠BDC-∠A=95-60=35°∵BD平分∠ABC∴∠CBD=∠ABD=35°∵DE∥BC∴∠BDE=∠CBD=35°∴∠BED=180-(∠ABD+∠BDE)=180-(35+35)=110°
