如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1(1) 若∠A=60°,求∠A1的度数;(2) 若∠A=m,求若∠A1的度数;
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![如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1(1) 若∠A=60°,求∠A1的度数;(2) 若∠A=m,求若∠A1的度数;](/uploads/image/z/2574275-59-5.jpg?t=%E5%A6%82%E5%9B%BE%2CC%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%3D60%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%3Dm%2C%E6%B1%82%E8%8B%A5%E2%88%A0A1%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B)
如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1(1) 若∠A=60°,求∠A1的度数;(2) 若∠A=m,求若∠A1的度数;
如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1
(1) 若∠A=60°,求∠A1的度数;
(2) 若∠A=m,求若∠A1的度数;
如图,C在直线BE上,∠ABC与∠ACE的角平分线交于点A1(1) 若∠A=60°,求∠A1的度数;(2) 若∠A=m,求若∠A1的度数;
1.∠A1的度数为30°
2.∠A1的度数为 m/2度
3.∠A2的度数为 m/4度.∠A3的度数为 m/8度,∠A4的度数为 m/16度,依此类推
∵∠A1=∠A1CE-∠A1BC
=1 2 ∠ACE-1 2 ∠ABC
=1 2 (∠ACE-∠ABC)
=1 2 ∠A.
∴(1)当∠A=60°时,∠A1=30°;
(2)当∠A=m时,∠A1=1 2 m;
(3)依次类推∠A2=1 4 m,∠A3=1 8 m,∠An=(1 2 )nm.
∠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...
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∠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
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