Question

有关三角形的外角的题证明，怎么证

如图所示，在△ABC中，∠A＝α，△ABC的内角平分线或外角平分线交于点P，且∠P＝β，试探求下列各图中α与β的关系，并全部说明理由

Answer 1

解：
(1)
β=∠P
=180°-∠PBC-∠PCB
=180°-∠ABC/2-∠ACB/2
=180°-(∠ABC+∠ACB)/2
=180°-(180°-∠BAC)/2
=180°-(180°-α)/2
=90°+α/2

(2)
β=∠P
=∠PCD-∠PBC (假设D在BC延长线上)
=∠ACD/2-∠ABC/2
=(∠ACD-∠ABC)/2
=∠A/2
=α/2

(3)
β=∠P
=180°-∠PBC-∠PCB
=180°-∠DBC/2-∠ECB/2 (假设D，E分别在AB，AC延长线上)
=180°-(∠DBC+∠ECB)/2
=180°-[(180°-∠ABC)+(180°-∠ACB)]/2
=180°-[360°-(∠ABC+∠ACB)]/2
=180°-[360°-(180°-∠BAC)]/2
=180°-(180°+∠BAC)/2
=90°-α/2