Question

如图所示,在△ABC中,△ABC的内角平分线或外角平分线交于点P,试探求下列各图中∠A与∠P的关系并加以说明

Answer 1

图（1）∠P=180°-1/2∠ABC-1/2∠ACB

=180°-1/2(∠ABC+1/2∠ACB)

=180°-1/2(180°-∠A)

=180°-90°+1/2∠A

=90°+1/2∠A

图（2）BC延长至点E,假设AC线和BP线相交,交点为D.

∠A=180°-∠ADB-1/2∠ABC

=180°-∠ADB-1/2(180°-∠A-∠ACB)

=180°-∠ADB-90°+1/2∠A+1/2∠ACB

1/2∠A =90°-∠ADB+1/2∠ACB

∠A =180°-2∠ADB+∠ACB

同样道理

∠P =180°-∠PDC-1/2∠ACE

=180°-∠PDC-1/2(180°-∠ACB)

=180°-∠PDC-90°+1/2∠ACB

=90°-∠PDC+1/2∠ACB

由于∠ADB=∠PDC

所以∠A =180°-2∠PDC+∠ACB

而∠P =90°-∠PDC+1/2∠ACB

所以∠A =2∠P

图（3）

∠P=180°-∠CBP-∠BCP

=180°-1/2(180°-∠ABC)-1/2(180°-∠ACB)

=180°-90°+1/2∠ABC-90°+1/2∠ACB

=1/2∠ABC+1/2∠ACB

=1/2(∠ABC+∠ACB)

=1/2(180°-∠A)

=90°-1/2∠A