如图，已知三角形ABC的角ABC点角平分线与角ACB的外角角ACD的平分线交于点P，试探究角P与角

△ABC中，∠A + ∠ABC + ∠ACB = 180°
△PBC中，∠P + ∠PBC + ∠PCB = 180°
∠PBC = ∠ABC/2 (角平分线)
∠PCB = ∠ACB + ∠ACP
= ∠ACB + ∠ACD/2 (角平分线)
= ∠ACB + (∠A + ∠ABC)+/2 (两角和 = 第三角外角)
∠P + ∠PBC + ∠PCB = ∠P + ∠ABC/2 + ∠ACB + (∠A + ∠ABC)/2 = ∠P + ∠ABC + ∠ACB + ∠A/2 = 180° = ∠A + ∠ABC + ∠ACB
∠P = ∠A/2