(1)连接CB
∴∠OCB+∠OBC=90°
∵AC∥BD
∴∠A=∠ABD
∴∠PCO+∠PBO=1/2(∠ACO+∠A)=45°
∴∠P=180°-∠PCB-∠PBC=180°-90°-45°=45°
(2)∵AC∥BD
∴∠A=∠OBD
∠ACO=∠BDO
∴∠PBD=1/2(180°-∠A)=90°-1/2∠A
∠PCD=1/2∠C
∠PDC=180°-∠C
∴∠P=360°-∠PBD-∠PCD-∠PDC=360°-90°+1/2∠A-1/2∠C-180°+∠C
=360°-270°+1/2(∠A+∠C)=90°+45°=135°
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