C在直线BF上,∠ABC与∠ACE的角平分线交于A1

题中"C在直线BF上",应为"C在直线BE上"
(1)∠A1=∠A1CE-∠A1BC
=(1/2)∠ACE-(1/2)∠ABC
=(1/2)(∠A+∠ABC)-(1/2)∠ABC
=(1/2)∠A
若∠A＝60°,则:∠A1=30°
(2)由上面,
∠A1=(1/2)∠A
若∠A＝m,则:∠A1=m/2
(3)在（2）的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2
则:∠A2=(1/2)∠A1=m(1/2)^2
∠A3=(1/2)∠A2=m(1/2)^3
...
依此类推,
∠An=m(1/2)^n