(1)①∵∠AOB=90°,∠AOC=45°
∴∠COB=∠AOB-∠AOC=45°=∠AOC
∴CO平分∠AOB
②∵∠C=45°=∠AOC
∴OA∥CD(内错角相等,两直线平行)
(2)∠AOC=45°成立.
理由如下:
∵CD∥OB
∴∠BOD=∠D=45°
∵∠AOB=90°
∴∠AOD=∠AOB-∠BOD
=90°-45°
=45°
又∵∠COD=90°
∴∠AOC=∠COD-∠AOD
=90°-45°
=45°
(1)①∵∠AOB=90°,∠AOC=45°
∴∠COB=∠AOB-∠AOC=45°=∠AOC
∴CO平分∠AOB
②∵∠C=45°=∠AOC
∴OA∥CD(内错角相等,两直线平行)
(2)∠AOC=45°成立.
理由如下:
∵CD∥OB
∴∠BOD=∠D=45°
∵∠AOB=90°
∴∠AOD=∠AOB-∠BOD
=90°-45°
=45°
又∵∠COD=90°
∴∠AOC=∠COD-∠AOD
=90°-45°
=45°