如图，直线AB，CD交于O，OE平分∠AOD，OF⊥OE于O，若∠BOC=80°，求∠DOF的度数．

解：如图，
  ∵∠BOC=80°
  ∴∠AOC=180°-80°
      =100°
  ∴∠AOD=180°-100°
      =80°
  ∵OE平分∠AOD
  ∴∠AOE=∠DOE=40°
  ∵OF⊥OE
  ∴∠EOF=90°
  ∴∠DOF=∠DOE+∠EOF
      =40°+90°
      =130°

略