Question

如图,已知直线AB、CD相交于点O,∠AOE=∠COF=90°。∠AOF与∠DOE、∠BOF、∠COE有怎样的大小关系，为什么

图不方便画，就是初一数学书P162页的第6题

Answer 1

∠AOF=∠AOE-∠EOF=90°-∠EOF=∠DOE

∠AOF<∠BOF 且∠AOF+∠BOF=180°

∠COE=∠COF+∠EOF=90°+（∠AOE-∠AOF)
=90°+（90°-∠AOF)
=180-∠AOF
即∠AOF+∠COE=180° 且∠AOF<∠COE