AB//CD,OE平分∠BOC,OF⊥OE,OP⊥CD,∠ABO=40°,则下列结论：①∠BOE=70°②OF平分∠BO

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AB//CD,OE平分∠BOC,OF⊥OE,OP⊥CD,∠ABO=40°,则下列结论：①∠BOE=70°②OF平分∠BOD ③∠POE=∠BOE ④∠POB=2∠DOF,其中正确结论有（）理由

正确结论有（① ② ③）理由
①
∵ AB∥CD
∴ ∠ABO+∠BOC = 180
∠BOC = 180 -∠ABO = 180-40 = 140
∵ OE平分∠BOC
∴ ∠BOE = 1/2∠BOC = 70°
②
∵ AB∥CD
∴ ∠BOD = ∠ABO=40°
∴ ∠BOF=1/2∠BOD = 20
∵ OF⊥OE ∠BOE+∠BOF=90
∴ ∠BOF = 90 -∠BOE = 90-70 = 20 = 1/2∠BOD
OF平分∠BOD
③
∵ OP⊥CD
∴ ∠COE+POE = 90
∵ ∠COE = 1/2∠BOC = 70
∴ ∠POE = 90 -∠COE = 20
∵ ∠BOF = 1/2∠BOD = 20
∴ ∠POE=∠BOE
④ (不成立)
∵ ∠POB = ∠BOE -∠POE = 70 - 20 = 50
∴ ∠DOF = 1/2∠BOD = 20
∵ ∠POB ≠ 2∠DOF

AB//CD,OE平分∠BOC,OF⊥OE,OP⊥CD,∠ABO=40°,则下列结论：①∠BOE=70°②OF平分∠BO 直线AB、CD相交于点O,OE平分∠BOC,OF⊥CD于O,∠BOD:∠BOE=2:3,求∠AOF的度数直线AB、CD相交于点O,OE平分∠BOC,OF⊥CD,若∠BOD:∠BOE=2:3,求∠AOF 如图,直线AB,CD相交于点O,OE平分∠BOC,OF⊥OE,且∠AOD=66°,求∠BOF的度数如图,直线AB、CD相交于点O,OE平分∠BOC,射线OF⊥OE,诺∠COF=36°,求∠AOD 直线ab,cd相交于点O,OE平分∠BOC,OF⊥OE于点O,若∠COF=36°求∠DOE的度数直线AB,CD交于点O,OE平分∠AOD,OF⊥OE于O,若∠BOC=80°,求∠DOF的度数. 已知直线AB,CD相交于点O,OE平分∠BOD,OF平分∠BOC,∠AOD：∠BOE=4:1,求∠AOF的度数. 已知直线ab、cd相交于点o,且∠boc=80°,oe平分∠boc,of为oe的反向延长线. 12.如图所示,直线AB,CD相交于点O,OE⊥OF,OD平分∠AOE.（1）若∠BOE=60° 如图,AB,CD相交于点O,OE⊥OF,∠BOF=2∠BOE,OC平分∠AOE.求∠BOE的度数. 如图所示,直线AB,CD相交于点O,OE平分∠BOC,OF⊥CD,若OB将∠DOE分成两部分,且∠BOD:∠BOE=2: