△ABC中∠ABC的平分线与∠ACB的外角平分线交于点D,∠A等于50°
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![△ABC中∠ABC的平分线与∠ACB的外角平分线交于点D,∠A等于50°](/uploads/image/f/935496-0-6.jpg?t=%E2%96%B3ABC%E4%B8%AD%E2%88%A0ABC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%B8%8E%E2%88%A0ACB%E7%9A%84%E5%A4%96%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9D%2C%E2%88%A0A%E7%AD%89%E4%BA%8E50%C2%B0)
(1)证明:∵AD平分∠BAC∴∠BAD=∠CAD∵BE平分∠ABC∴∠ABE=∠CBE∵∠BED=∠BAD+∠ABE∴∠BED=∠CAD+∠CBE∵弧CD=弧CD∴∠CAD=∠CBD(同弧的圆周角相
∠D的度数为:70/2=35°.设,∠CAD=∠DAB=∠1,∠CBD=∠DBE=∠2.∠ABC=180-(∠C+2∠1),而,∠ABC=180-2∠2,则有∠C+2∠1=2∠2,∠2-∠1=∠C/2
解题思路:根据内角和定理解答解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/rea
/>∵∠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
∠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
假设AC与BE焦点为F问题1:∠ABC=40°,那么角∠EBC=20°,∠ACB=80°,那么角∠ACD=100°,∠ACE=50°,∠EFC=∠EBC+∠BCF=80°+20°=100°,那么∠E=
∠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?
证明:∵AD平分∠BAC∴∠BAD=∠CAD∵BE平分∠ABC∴∠ABE=∠CBE∵∠BED=∠BAD+∠ABE∴∠BED=∠CAD+∠CBE∵弧CD=弧CD∴∠CAD=∠CBD(同弧的圆周角相等)∴
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:
∠AN=∠A/2=α/2^N,∠A=32,∠A4=32/2⁴=2
用塞瓦定理来证:三角形ABC内先引两条角分线设为AOBO交于O点然后连接CO并由塞瓦三角形式sin∠OAB/sin∠OAC*sin∠OCA/sin∠OCB*sin∠OBC/sin∠OBA=1因为AOB
(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°.
证明:∵∠ABC与∠ACB的平分线相交于点O,∴∠OBC=12∠ABC,∠OCB=12∠ACB,∴∠OBC+∠OCB=12(∠ABC+∠ACB),在△OBC中,∠BOC=180°-(∠OBC+∠OCB
解题思路:利用三角形内角和定理求解。解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include
①∵∠ABC=40°,∠ACB=60°,∠ABC与∠ACB的平分线交于点I,∴∠IBC=20°∠ICB=30°,∴∠BIC=180°-∠IBC-∠ICB=130°;②∵∠ABC+∠ACB=80°,∠A
1、角D=110度,角P=70度角A=40度,角B+角C=180-40=140度,1/2∠B+1/2∠C=70°,在△BDC中,∠D=180-70=110°∠B的外角+∠C的外角=360°-140°=
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
∵∠ACA1=∠A1CD=12∠ACD=12(∠A+∠ABC),又∵∠ABA1=∠A1BD=12∠ABD,∠A1CD=∠A1BD+∠A1,∴∠A1=12∠A=12α.同理∠A2=12∠A1,…即每次作
∠BEC=180°-∠EBC-∠ECB=180°-1/2∠B-(∠BCA+1/2∠ACD)=180°-1/2∠B-{(180°-∠A-∠B)+1/2(∠A+∠B)}=180°-1/2∠B-{180°-
设AC与BD的交点为O则在△DOC中,∠D+∠DCO+∠DOC=180度在△DOC中,∠A+∠ABO+∠AOB=180度因为∠DOC与∠AOB是对顶角,所以∠DOC=∠AOB所以,∠D+∠DCO=∠A