Question

已知OD平分∠AOB,DC⊥OA于C,AO+BO=2OC,求证∠OAD+∠OBD=180°

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Answer 1

如图，过D作DE⊥OB交OB的延长线于E

∵OD平分∠AOB,DC⊥OA

∴CD=ED,OC=OE

又∵AO+BO=2OC,即AO-OC=OC-BO

∴AC=AO-OC=OC-BO=OE-BO=BE

∴RtΔACD≌RtΔBED

∴∠DBE=∠OAD

∴∠OAD+∠OBD=180°

Answer 2

作DE⊥OB，如图，∵AO+BO=2OC，AO＞OC∴AO-OC=OC-BO∴AC=BE又∵OD是角平分线，∴DC=DE∴△DEB≌△DCA（直角△2条直角边相等）∴∠DAC=∠DBE∵∠DBE+∠OBD=180°∴∠DAC+∠OBD=180°