如图,在△bad
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△BAD全等于△ACEBA=ACAD=CEBD=AE要证明BD=DE+CE只需证AE=AD+DE据题意,A、D、E在一条直线上AE=AD+DE成立BD=DE+CE成立根本不用看图
△BAD全等于△ACE所以AD=CEBD=AEAE=AD+DE所以BD=CE+DE
(1)△ABC∽△ADE,△ABD∽△ACE(2分)(2)①证△ABC∽△ADE,∵∠BAD=∠CAE,∠BAD+∠DAC=∠CAE+∠DAC,即∠BAC=∠DAE.(4分)又∵∠ABC=∠ADE,∴
证明:在Rt△ABC和Rt△BAD中,AB=BAAC=BD,∴Rt△ABC≌Rt△BAD,∴∠BAD=∠ABC,∴AE=BE.
求∠BAD=∠2=?已知条件:∠1=∠2=∠5∠3=∠4∠3+∠4+∠5=180°∵∠3=∠4∴2∠3+∠5=180°∴∠3=90°-1/2∠5∵∠1+(∠2+∠3)+∠5=180°∴∠1+(∠2+9
证明:过点D作DE⊥AB于E,∵DE⊥AB,∴∠AED=90°,∴∠ACB=∠AED=90°,又∵∠CAD=∠BAD,AD=AD,∴△ACD≌△AED,∴CD=ED,AC=AE,∵∠ACB=90°,A
解题思路:构造全等三角形进行证明.解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/
证明:∵四边形ABCD是菱形∴AD//BC(菱形对边平行)∴∠B+∠BAD=180°∵∠BAD=2∠B∴3∠B=180°∠B=60°∵AB=BC(菱形邻边相等)∴△ABC是等边三角形(有一个角是60°
因为∠ADC=∠BAD+∠B(三角形外角性质)∠ADC=70°(已知)所以∠BAD+∠B=70°(等量代换)因为∠BAD=∠B(等边三角形底角相等)所以∠B=1/2*70°=35°
(1)证明:∵∠AEC与∠BED是对顶角,∴∠AEC=∠BED,在△ACE和△BDE中,∠AEC=∠BED∠C=∠D=90°AC=BD∴△ACE≌△BDE(AAS),(3分)∴AE=BE;(4分)(2
在菱形ABCD中AB=BC,AD∥BC∴∠BAD+∠B=180°∵∠BAD=2∠B∴∠B=180°÷(1+2)=60°∴△ABC是等边三角形
BD=1Xsinθ/2Abd的面积:1/2xBDXAB.cotθ/2=0.5cosθ/2Bcd面积:1/2xBDXBD.sin60=√3/4(sinθ/2)2S=0.5cosθ/2+√3/4(sinθ
∵AB=AC,∴∠B=∠C∵∠BAD=∠CAE,∴∠ADE=∠AED,∴AD=AE∴△ADE是等腰三角形.
相似因为∠BAD=∠CAE,所以∠BAC=∠DAE又因为∠ABC=∠ADE所以△ABC∽△ADE所以AD/AE=AB/AC在△ABD和△ACE中AD/AE=AB/AC,∠BAD=∠CAE所以△ABD∽
(1)∵∠BAD=15°,∠ADC=4∠BAD,∴∠ADC=60°,∴∠B=60°-15°=45°;(2)证明:过C作CE⊥AD于E,连接EB.∵∠ECD=90°-60°=30°∴DC=2ED,∵DC
∵AB=AD∠BAD=32°∴∠ADB=∠ABD=(180º-32º)/2=74º∵AD=DC∠ADB=∠DAC+∠DCA∴∠DAC=∠DCA=∠ADB/2=37
∠D+∠BCD=180°60°+∠D+(180°-∠BCD)/2=180°∴∠BCD=100°,∠D=80°∴∠BAD=100°再问:60°+∠D+(180°-∠BCD)/2=180°这是啥意思勒再答
过A做AE⊥BC于E,∵∠ADC=4∠BAD=60º∠ADC=∠BAD+∠B∴∠B=45º∴AE=BE设DE=X∴AD=2X,AE=BE=√3X∵DC=2BD,CD=2BD=1∴1
根据已知条件:角CBE=角ABD,角BCE=角BAD可以判定△ABD∽△CBD,所以AB:BD=CB:BE且∠ABD=∠CBE;而∠ABC=∠ABC+∠DBC;∠DBE=∠CBE+∠DBC,故∠ABC
(1)∵∠BAD=∠CAE,∠DAC=∠DAC.∴∠BAC=∠DAE,又∵∠ABC=∠ADE.∴△ABC∽△ADE,(AA)∴AB:AC=AD:AE°∵∠BAD=∠CAE∴△ABD∽ACE(SAS)(