如图,点C在直线BE上,∠ABC与∠ACE的角平分线交于点A1.(1)若∠A=50°,求∠A1的度数;(2)若∠A=α,求∠A1的度数;(3)在(2)的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2;再作∠A2BE、∠A2
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![如图,点C在直线BE上,∠ABC与∠ACE的角平分线交于点A1.(1)若∠A=50°,求∠A1的度数;(2)若∠A=α,求∠A1的度数;(3)在(2)的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2;再作∠A2BE、∠A2](/uploads/image/z/2574273-57-3.jpg?t=%E5%A6%82%E5%9B%BE%2C%E7%82%B9C%E5%9C%A8%E7%9B%B4%E7%BA%BFBE%E4%B8%8A%2C%E2%88%A0ABC%E4%B8%8E%E2%88%A0ACE%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9A1.%EF%BC%881%EF%BC%89%E8%8B%A5%E2%88%A0A%3D50%C2%B0%2C%E6%B1%82%E2%88%A0A1%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E2%88%A0A%3D%CE%B1%2C%E6%B1%82%E2%88%A0A1%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B%EF%BC%883%EF%BC%89%E5%9C%A8%EF%BC%882%EF%BC%89%E7%9A%84%E6%9D%A1%E4%BB%B6%E4%B8%8B%2C%E8%8B%A5%E5%86%8D%E4%BD%9C%E2%88%A0A1BE%E3%80%81%E2%88%A0A1CE%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2C%E4%BA%A4%E4%BA%8E%E7%82%B9A2%EF%BC%9B%E5%86%8D%E4%BD%9C%E2%88%A0A2BE%E3%80%81%E2%88%A0A2)
如图,点C在直线BE上,∠ABC与∠ACE的角平分线交于点A1.(1)若∠A=50°,求∠A1的度数;(2)若∠A=α,求∠A1的度数;(3)在(2)的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2;再作∠A2BE、∠A2
如图,点C在直线BE上,∠ABC与∠ACE的角平分线交于点A1.
(1)若∠A=50°,求∠A1的度数;
(2)若∠A=α,求∠A1的度数;
(3)在(2)的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2;再作∠A2BE、∠A2CE的平分线,交于点A3;…;依次类推,则∠A2,∠A3,…,∠An分别为多少度?
如图,点C在直线BE上,∠ABC与∠ACE的角平分线交于点A1.(1)若∠A=50°,求∠A1的度数;(2)若∠A=α,求∠A1的度数;(3)在(2)的条件下,若再作∠A1BE、∠A1CE的平分线,交于点A2;再作∠A2BE、∠A2
∠A+∠B+∠BCA=180°
∠A1+∠A1BE+∠A1CB=180°
∠A1BE=1/2∠B
∠A1CB=∠BCA+∠A1CA
∠A1CA=1/2∠ACE=∠A1CE=∠A1+∠A1BE
所以∠A1+∠A1BE+∠A1CB=∠A1+1/2∠B+∠BCA+∠A1+1/2∠B
=2∠A1+∠B+∠BCA=180°
所以∠A+∠B+∠BCA=2∠A1+∠B+∠BCA
所以∠A=2∠A1
所以第一题 ∠A=60° ∠A1=30°
第二题 ∠A=m,若∠A1=m/2
缺少会发现的眼睛
∠A+∠B+∠BCA=180° ∠A1+∠A1BE+∠A1CB=180° ∠A1BE=1/2∠B ∠1、如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1,(1)若∠A
∵∠A1=∠A1CE-∠A1BC
=∠ACE-∠ABC
=(∠ACE-∠ABC)
=∠A.
∴(1)当∠A=60°时,∠A1=30°;
(2)当∠A=m时,∠A1=m;
(3)依次类推∠A2=m,∠A3=m,∠An=m.