答案
解:∠DOE=∠EOC+∠DOC=
∠AOC+
∠BOC=
∠AOB=90°,
所以∠EOC与∠COD互余,∠AOE与∠BOD互余,∠AOE与∠BOD互余,∠AOE与∠COD互余;
∠AOE与∠COD互补,∠EOC与∠EOB互补,∠AOC与∠COB互补,∠AOD与∠DOB互补,∠AOD与∠COD互补.
解:∠DOE=∠EOC+∠DOC=
∠AOC+
∠BOC=
∠AOB=90°,
所以∠EOC与∠COD互余,∠AOE与∠BOD互余,∠AOE与∠BOD互余,∠AOE与∠COD互余;
∠AOE与∠COD互补,∠EOC与∠EOB互补,∠AOC与∠COB互补,∠AOD与∠DOB互补,∠AOD与∠COD互补.