如图,若△ABC≌△ADE,C和E,B和D分别是对应顶点
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1、AB=8,∵ΔABC∽ΔADE,∴AD/AB=AE/AC,4/8=3/AC,AC=6,∴CD=AC-AE=3,2、∵D、E分别为AC、BC中点,∴DE∥AB,∴ΔABC∽ΔDEC.3、∵∠A=∠B
证法一:连接CE,∵Rt△ABC≌Rt△ADE,∴AC=AE.∴∠ACE=∠AEC.又∵Rt△ABC≌Rt△ADE,∴∠ACB=∠AED.∴∠ACE=∠ACB=∠AEC-∠AED.即∠BCE=∠DEC
(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,∴
20°因为△ABC≌△ADE,所以∠BAC=∠DAE∠BAD=∠BAC-∠DAC∠CAE=∠DAE-∠DAC所以∠BAD=∠CAE=20°再问:咳咳,求步骤咯~再答:望采纳,O(∩_∩)O谢谢!祝学习
因为BD=CE,BC=BD-CD,DE=CE-CD,所以BC=DE.又因为AB=AE,AC=AD,所以:△ABC≌ADE(边边边)
证明:(1)∵△ABC、△ADE是等边三角形,∴AE=AD,BC=AC=AB,∠BAC=∠DAE=60°,∴∠BAC+∠CAD=∠DAE+∠CAD,即:∠BAD=∠CAE,∴△BAD≌△CAE,∴BD
证明:∵∠1=∠2∴∠1+∠EAC=∠2+∠EAC即∠BAC=∠DAE又∵AB=AD,AC=AE∴△ABC≌△ADE(SAS)
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∽
因为三角形全等,所以角bac等于角dae所以角bad等于角cae
20°因为△ABC≌△ADE,所以∠BAC=∠DAE∠BAD=∠BAC-∠DAC∠CAE=∠DAE-∠DAC=20
∠BAC=∠DAE所以∠CAE=∠BAD再问:等于多少度
直角三角形∠AED=180°-∠A-∠ADE∠C=180°-∠A-∠B∵∠ADE=∠B两个等式相减,得∠AED=∠C=90°∴△ADE是直角三角形
过D作AC的平行线交AB于P∴△BDP为等边三角形,BD=BP,∴AP=CD,∵∠BPD为△ADP的外角,∴∠ADP+∠DAP=∠BPD=60°而∠ADP+∠EDC=180°-∠BDP-∠ADE=60
因为∠BAD=∠CAE,所以∠BAD+∠CAD=∠CAE+∠CAD,即∠BAC=∠DAE.在△ABC和△ADE中,因为AC=AE,∠C=∠E,∠BAC=∠DAE,由角边角定理,△ABC≌△ADE.
在△COD和△BOE中,OC=OE∠COD=∠EOBOD=OB,∴△COD≌△BOE,∴∠D=∠B,∵OC=OE,OD=OB,∴DE=BC在△ADE和△ABC中,∠A=∠A∠B=∠DDE=BD,∴△A
(1)△ABE≌△ACB∵,△ADE、△ABC是等腰直角三角形,∴AB=ACAD=AE角BAC=∠EAD=45°∵AB=ACAD=AE角BAC=∠EAD=45°∴△ABE≌△ACB(SAS)(2)∵△
(1)∵∠BAD=∠CAE,∠DAC=∠DAC.∴∠BAC=∠DAE,又∵∠ABC=∠ADE.∴△ABC∽△ADE,(AA)∴AB:AC=AD:AE°∵∠BAD=∠CAE∴△ABD∽ACE(SAS)(