∠BOE=75°证明：在矩形ABCD中AO=BO=CO=DO ∠ABC=90°∵∠CAE=15° AE平分∠BAD∴∠BAE=∠BEA=45° AB=BE∴∠BAC=60° 等边△AOB∴∠BCA=30° AB=1/2AC=BO∴BE=BO又∵∠DBC=∠ACB=30°在△BOE中∠BOE=（180°-∠DBC）÷2=75°
