如图,△ABC中,角ACE=90度,sinA=五分之四,BC=8
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∠AFD=∠AFE.理由:过A作AM⊥DC于M,AN⊥BE于N.∵∠BAD=∠CAE=90°,∴∠BAD+∠BAC=∠CAE+∠BAC,即∠DAC=∠BAE;在△ABE和△ADC中,AB=AD(已知)
作∠B的角平分线,交AD于F不难证明∠FBD=1/2∠B因为∠ACE=∠ABC+∠BAC(外角)∠BFD=1/2(∠ABC+∠BAC)(外角)所以∠BFD=1/2∠ACE所以∠ADC=∠BFD+∠FB
很简单啊,由已知条件可得:∠ACE=∠A+∠ABC,∠A=90°∠ACD=∠DCE=(1/2)∠ACE,∠ABD=∠DBC=(1/2)∠ABC,所以在△BCD中,∠DBC+∠D+∠BCD=180°,∠
我网上搜了下,找到图了,顺便答案也发给你把.连结DE,如下图红线所示由于△BCD为等边三角形,BC=BD这样BC和BE都已经变换到△BDE中,因此我们现在只要想办法证明出AB=DE且∠BDE=90°即
证明:∵∠BAC=∠DAE,…(3分)∴∠BAC+∠CAD=∠DAE+∠CAD,即∠EAC=∠DAB,…(4分)在△AEC和△ADB中AD=AE∠DAB=∠EACAB=AC,∴△AEC≌△ADB(SA
同学,你好以下是解答的过程∵BD平分∠ABC∴∠CBD=1/2∠ABC∵CD平分∠ACE∴∠ECD=1/2∠ACE∵∠ACE=∠A+∠ABC∴∠ECD=1/2∠ACE=1/2(∠A+∠ABC)=1/2
∠DCE=∠BCD-∠BCE=∠BCD-(∠AEC-∠B)=∠BDC-∠AEC+∠B=∠BDC-∠ACE+∠B=∠BDC-(∠ACD+∠DCE)+∠B=∠BDC-∠ACD-∠DCE+∠B=∠A-∠DC
证明:∵AB=AC,∴∠B=∠C∵AD=AE,∴∠ADE=∠AED,又∵点D、E在BC边上,∴∠ADB=∠AEC,在△ABD和△ACE中,∠ADB=∠AEC∠B=∠CAB=AC∴△ABD≌△ACE
证明:∵AB=AC∴ΔABC是等腰Δ∵等腰Δ三线合一∴AD平分∠BAC∴∠BAD=∠CAD又∵CA=BAAE=AE∴ΔABE≌ΔACE(SAS)如仍有疑惑,欢迎追问.祝:学习进步!
因为角ACE=角A+角ABC(1)角DCE=角D+角DBC(2)角DCE=角ACE/2角DBC=角ABC/2所以(2)式可表示成:角ACE/2=角D+角ABC/2(3)由(1)(3)式可得角A=2*角
8cmCDE和BDE全等CE=BEAC+CE+AE=AC+BE+AE=AB+AC=8cm
∵如图,△ABC中,AB=3,AC=4,BC=5,∴BC2=AB2+AC2,∴∠BAC=90°,∵△ABD,△ACE都是等边三角形,∴∠DAB=∠EAC=60°,∴∠DAE=150°.∵△ABD和△F
∠ACE=∠A+∠ABC,∠BCD=180°-∠DCE=180°-∠ACE/2=180°-(∠A+∠ABC)/2,∠D+∠DBC+∠BCD=180°20°+∠ABC/2+180°-∠A/2-∠ABC/
(1)∵CD平分∠ACE∴∠ACD=∠ECD∵∠ECD=∠D+∠CBD∴2∠ECD=2∠D+2∠CBD∴∠ACE=2∠D+2∠CBD∵BD平分∠ABC,∠ACE=∠A+∠ABC∴2∠D=∠A(1)当∠
证明:在△ABD与△ACE中,∵AB=ACBD=CEAD=AE∴△ABD≌△ACE(SSS)∴∠ABD=∠ACE
证明:作AD⊥BC于D.∵AB=AC∴∠B=∠ACDCD=1/2BC∵CE=1/2BC∴CD=CE∵∠ADC=∠E=90°CA=CA∴⊿ACE≌⊿ACD∴∠ACE=∠ACD∴∠ACE=∠B手机提问的朋
解:∵∠DCE=∠DBC+∠BDC.∴2∠DCE=2∠DBC+2∠BDC.即∠ACE=∠ABC+2∠BDC;又∠ACE=∠ABC+∠A.∴2∠BDC=∠A=70度,∠BDC=35度.
这样做:∵△ABC全等于△ACE∴∠B=∠C∵中间那组对顶角相等然后∠B=∠C中间那组对顶角相等两个三角形中两组角相等剩下的∠1=∠2
设∠BAD=∠DAC=x则∠ADC+x=∠ACE∠ADC=∠B+x两者相加2∠ADC+x=∠B+∠ACE+x∠ADC=1/2(∠B+∠ACE)
∵∠DCE是⊿BCD的外角∴∠DCE=∠CBD+∠D即2∠DCE=2∠CBD+2∠D∵BD,CD分别平分∠ABC,ACE∴∠ABE=2∠CBD,∠ACE=2∠DCE∴∠ACE=∠ABE+2∠D∵∠AC