∠bad与∠acd的平分线交于点o且oe=5
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/12 04:15:07
![∠bad与∠acd的平分线交于点o且oe=5](/uploads/image/f/932509-37-9.jpg?t=%E2%88%A0bad%E4%B8%8E%E2%88%A0acd%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9o%E4%B8%94oe%3D5)
/>∵∠A+∠ABC+∠ACB=180∴∠ABC+∠ACB=180-∠A∵∠ACD=180-∠ACB,CA1平分∠ACD∴∠A1CD=∠ACD/2=(180-∠ACB)/2=90-∠ACB/2∵BA1
根据角平分线的定义,三角形的外角性质及三角形的内角和定理可知∠A1=12∠A=a21,∠A2=12∠A1=a22,…,依此类推可知∠A2010的度数.∵∠ABC与∠ACD的平分线交于点A1,∴∠A1=
在△ABC中,∠A=96°,延长BC到D,∠ABC与∠ACD平分线交于A1则ABCA1为平行四边形,所以∠A1BC=∠abc/2=48以此类推∠A2BC=24,∠A3BC=12,∠A4BC=6,∠A5
因为∠ACD=∠A+∠ABC∠A1CD=∠A1+∠A1BC∠A1CD=(1/2)(∠A+∠ABC)=1/2∠A+∠A1BC所以∠A=2∠A1依此类推:∠A=2∠A1=4∠A2=8∠A4=16∠A5所以
∠P=30°∵∠ACD为△ABC的外角∴∠ACD=∠ABC+∠A又BP平分∠ABC.CP平分∠ADC∴∠PBD=1/2∠ABC,∠PCD=1/2∠ADC又∠PCD为△PBC的外角∴∠PCD=∠P+∠P
∠A1=∠A1CD-∠A1BC/2=(∠A+∠ABC)/2-∠A1BC=∠A/2.同样可得∠A2=∠A/4;∠A3=∠A/8;∠A4=∠A/16;.∠An=∠A/2^n;再问:最后一个问题嘞?
解题思路:先得出∠A1=1/2∠A,∠A2=1/4∠A,可得∠An=(1/2)^n∠A,再将∠A=32°,n=4,代入即可得∠A4的度数。解题过程:
△ABC中,∵∠A=∠ACD-∠ABC,A1是∠ABC角平分与∠ACD的平分线的交点,∠A=α,∴∠A1=∠A1CD-∠A1BC=1/2×(∠ACD-∠ABC)=1/2×∠A;同理可得,∠A2=1/2
∠AN=∠A/2=α/2^N,∠A=32,∠A4=32/2⁴=2
还是不明白题目,你究竟是把∠A设成什么?设∠A=0°吗?还是什么再问:当它是未知数就好了再答:http://zhidao.baidu.com/question/1957325072129443500.
(1)∠A+1/2∠ABC=1/2∠ACD+∠Aι∠ACD=∠A+∠ABC∴∠Aι=1/2∠A(2)同理:∠An=(1/2)^n∠A(3)A4=(1/2)^4*∠A=64°/16=4°.
(1)∵A1B是∠ABC的平分线,A1C是∠ACD的平分线,∴∠A1BC=12∠ABC,∠A1CD=12∠ACD,又∵∠ACD=∠A+∠ABC,∠A1CD=∠A1BC+∠A1,∴12(∠A+∠ABC)
(1)∵∠A+∠ABC+∠ACB=180∴∠ABC+∠ACB=180-∠A∵∠ACD=180-∠ACB,CA1平分∠ACD∴∠A1CD=∠ACD/2=(180-∠ACB)/2=90-∠ACB/2∵BA
∠PCD=∠PBC+∠BPC=∠PBC+40°;(1)PB平分∠ABC,得∠PBC=∠ABC/2;PC平分∠ACD,得∠PCD=∠ACD/2;代入(1)得∠ACD-∠ABC=80°;在△ABC中,∠B
△ABC中,∵∠A=∠ACD-∠ABC,A1是∠ABC角平分与∠ACD的平分线的交点,∠A=α,∴∠A1=∠A1CD-∠A1BC=1/2×(∠ACD-∠ABC)=1/2×∠A;同理可得,∠A2=1/2
(1)分别过P点别作BC延长线、BE、AC的的垂线,垂足分别为F,H、G因为CP为角ACF的平分线,所以PF=PG因为BP为角EBF的角平分线,所以PF=PH所以PH=PG,AP平分角CAE(2)因为
2∠BPC=∠BAC证:∠ACD=∠BAC+ABC=∠BAC+2∠PBC ∠PCD=∠PBC+∠BPC∵∠acd的平分线cp与内角∠abc的平分线bp交于点p∴∠PCD=∠ACP
以A和A1两个角为例,∠ACD=∠A+∠ABC,∠A1CD=1/2*∠ACD=1/2*∠A+1/2*∠ABC=1/2*∠A+∠A1BC,∠A1CD为外角=∠A1+∠A1BC所以,∠A1=1/2∠A,因
(1)1.30度(1)2.35度(1)3.0.5n度(2)α/2^2010(度)再问:过程。。。再答:(1)1.∠BA1C=180度-0.5∠ABC-∠ACB-0.5(180度-∠ACB)=90度-0
百度知道羽灵飞雪很高兴为您解答.以A和A1两个角为例,∠ACD=∠A+∠ABC,∠A1CD=1/2*∠ACD=1/2*∠A+1/2*∠ABC=1/2*∠A+∠A1BC,∠A1CD为外角=∠A1+∠A1