题目

如图,在三角形abc中,外角角acd的平分线

提问时间：2020-10-16

答案

(1)
∠ACD = ∠A + ∠ABC
∠BCA1 = ∠ACD / 2 + ∠BCA = ∠A/2 + ∠ABC/2 + ∠BCA
∠A1 = 180° - ∠ABC/2 - ∠BCA1 = ∠A + ∠ABC + ∠BCA - ∠ABC/2 - (∠A/2 + ∠ABC/2 + ∠BCA)
= ∠A/2
(2)
∠An = ∠A x (1/2)^n
(3) 看不清是4 还是6
∠A4 = 64° x (1/2)^4 = 64° x (1/16) = 4°
∠A6 = 64° x (1/2)^6 = 64° x (1/64) = 1°