如图：已知P为∠AOB的平分线OP上一点,PC⊥OA于C,∠OPA+∠OBP=180°.求证：AO+BO=2CO

如图：已知P为∠AOB的平分线OP上一点,PC⊥OA于C,∠OPA+∠OBP=180°.求证：AO+BO=2CO
如图：已知P为∠AOB的平分线OP上一点,PC⊥OA于C,∠OPA+∠OBP=180°.求证：AO+BO=2CO

如图：已知P为∠AOB的平分线OP上一点,PC⊥OA于C,∠OPA+∠OBP=180°.求证：AO+BO=2CO
证明：
作PD⊥OB于D点
∵OP为∠AOB的平分线
∴ PC=PD OC=CD
∵∠OPA+∠OBP=180°
又∵∠DBP+∠OBP=180°
∴∠DBP=∠OPA
又∵∠ACP=∠BDP=90°
∴△PCA≌△PDB (AAS)
∴AC=BD
∴AO+BO=AC+CO+DO-BD
=CO+DO=2CO □

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