在△ABC中,∠ABC的外角平分线与∠ACB的平分线相交于点P,已知∠A=80°求∠BPC的度

2025-05-23 14:34:04
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回答1:

∠ ABP=1/2(∠ A+∠ ACB)∠ BPC=180-(∠PBC+1/2∠PCB)=180-(∠ABC+1/2(∠ A+∠ ACB)+1/2∠ACB)=180-(∠ABC+∠ABC+1/2∠A)=180-(180-∠A+1/2A)=1/2∠A=40°