Question

如图△ABC的内角平分线或外角平分线交于点P，且∠P=β，∠A=∠α求β和α的关系

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Answer 1

连接AP延长至BC于D
∵∠BPD=1/2∠B+∠BPD
∠CPD=1/2∠ACB+∠CAP
∴∠β=1/2∠B+∠BPD+1/2∠ACB+∠CAP=∠α+1/2∠B+1/2∠ACB=∠α+1/2(180-∠α)=90+1/2α

Answer 2

∠P=1/2∠A
因为BP PC为角平分线
所以∠ABP+∠ACP=∠PBC+∠PCB
因为∠ABC+∠ACB=180-∠A
所以∠PBC+∠PCB=90-1/2∠A
所以∠P=180-90+∠A=90+1/2∠A

追问

为什么∠PBC+∠PCB=90-1/2∠A
所以∠P=180-90+∠A=90+1/2∠A