∵∠EOC= 2 5 ∠AOC，∴设∠EOC=2x，∠AOC=5x，则∠BOC=180°-5x．∵∠BOE为直角，∴2x+180°-5x=90°，解得x=30°．∴∠BOC=30°，∠AOC=150°．∴∠AOF= 1 2 ∠AOC=75°，∠AOD=∠BOC=30°．∴∠DOF=∠AOF+∠AOD=75°+30°=105°．
