Question

这道数学题怎么解决啊

如图：在△ABC中，∠A=96°，延长BC到点D，∠ABC与∠ACD的平分线相交于点A1，求∠A1的度数。

Answer 1

解：∵A1B、A1C分别平分∠ABC和∠ACD，
∴∠ACD=2∠A1CD，∠ABC=2∠A1BC，
而∠A1CD=∠A1+∠A1BC，∠ACD=∠ABC+∠A，
∴∠A=2∠A1=96°，
∴∠A1=48°

Answer 2

∠A1=∠A的一半=48°，用到的原理是 三角形的外角等于不相邻的两个内角和，角ACD=角A+角ABC，角A1CD=角A1+角A1BC

Answer 3

∠ABC+∠ACB=180-96=84
∠ACD=180-∠ACB
∠A1BC=∠ABC/2, ∠ACA1=∠ACD/2
∠BCA1 = ∠ACB+∠ACA1

∠BA1C=180-∠A1BC-∠BCA1=180-∠ABC/2-(∠ACB+∠ACA1)
= 180-∠ABC/2-(∠ACB+∠ACD/2)
=180-∠ABC/2-(∠ACB+(180-∠ACB)/2)
=180-∠ABC/2-(∠ACB+90-∠ACB/2)
=90- ∠ABC/2-∠ACB/2
=90-(∠ABC+∠ACB)/2
=90-84/2
=90-42
=48

Answer 4

∠A1BC=∠ABC/2
∠A1CD=∠ACD/2 ∠ACD=∠ABC+∠A so∠ACD-∠ABC=∠A
∠ACD=∠A+∠ABC
∠A1CD=∠A1+∠A1BC
∠A1=∠A1CD-∠A1BC=∠ACD/2-∠ABC/2=(∠ACD-∠ABC)/2=∠A/2=48°

Answer 5

没有图啊。