ad平分∠bac,de∥ac交ab于e,df∥ab交ac于f
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∵DE∥AC,EF∥BC,所以四边形EFCD是平行四边形.设ED=x,则AC=4+x.∵AD平分∠BAC,由三角形内角平分线定理,得出ABAC=BDDC=154+x又DECA=BDBC.∴xx+4=1
1.连结OD,角EDA=角AFB角AFB+角FAB=角EDA+角ADO=90度,DE垂直于圆ODE是圆O的切线;2.连接BD,角ADB=90度=角E,由相似,由勾股定理求AE=9,再由相似求BF=10
(1)由AD平分∠BAC,得到∠1=∠2,而OD=OA,∠2=∠3,所以∠1=∠3,则有OD∥AE,而DE⊥AC,所以OD⊥DE;(2)过D作DP⊥AB,P为垂足,则DP=DE=3,由⊙O的半径为5,
(1)证明:连接DM.在Rt△ADE中,MD为斜边AE的中线,则DM=MA,∴∠MDA=∠MAD,∵AD平分∠BAC,∴∠MAD=∠DAC,∴∠MDA=∠DAC,∴MD∥AC,∵AC⊥BC,BF⊥BC
1.证明:连结BD,CD.因为DG垂直于BC,且DG平分BC于G,所以BD=CD,因为DE垂直于AB于E,DF垂直于AC于F,且AD平分角BAC,所以DE=DF,角BED=角CFD=90度,所以直角三
证明:延迟CD交AB于点F∵AD平分∠BAC∴∠BAD=∠CAD∵AD⊥CF∴∠ADF=∠ADC∵∠BAD=∠CADAD=AD∠ADF=∠ADC∴△ADF≌△ADC(ASA)∴AF=AC∴BF=AB-
1、△CDF≌△BDE证明:∵AD平分∠BAC∴∠BAD=∠CAD∵DE⊥AB,DF⊥AC∴∠AED=∠AFD∠BED=90∵AD=AD∴△AED≌△AFD(AAS)∴DE=DF∵BD=CD∴△CDF
⑴连接DB,DC证明:∵AD平分∠BAC,DE⊥AB,DF⊥AF,∴DE=DF,∠DAE=∠DAF又∵DG垂直平分BC∴DB=DC在Rt△BDE与Rt△CDF中DE=DFDB=DC∴Rt△BDE≌Rt
AD平分∠BAC,角1=角2DE‖AC,角2=角ADEAE=DEBE/AB=DE/ACBE/AB=AE/AC(AB-AE)/AB=AE/AC1-AE/AB=AE/ACAE/AB+AE/AC=1
连接BC、OD,交BC于点F∵AB是直径则∠ACB=90°∵OA=OD∴∠OAD=∠ODA∵∠OAD=∠DAD∴∠CAD=∠ODA∴OD∥AE∴OF⊥BC∴四边形CEDF是矩形∴DF=CE,OF是△A
连接BD,三角形ADB相似于三角形AEDAE:AD=AD:AB求出AD勾股定理求DB
(1)正确,理由:AD平分∠BAC,所以∠EAD=∠DAC,又∠ADE和∠ACD都是直角,所以∠AED+∠EAD=∠ADC+∠DAC=90º,所以∠AED=∠ADC(2)错误,理由:Rt△A
∵AE平分∠BAC∴∠BAE=∠CAE又∵DE//AC∴∠DEA=∠CAE∴∠BAE=∠DEA∴DE=AD∵DE//ACDF//BC∴四边形DECF为平行四边形∴DE=FC又∵DE=AD∴AD=FC
证明:∵AD平分∠BAC,∴∠BAD=∠CAD,∵DE∥AC,∴∠EDA=∠CAD,∴∠EDA=∠EAD,∴AE=ED,又∵EF⊥AD,∴EF是AD的垂直平分线,∴AF=DF,∴∠FAD=∠FDA,又
证明:∵EF∥AD,∴∠F=∠BAD,∠AEF=∠DAC.∵AD平分∠BAC,∴∠BAD=∠DAC,∴∠F=∠AEF,∴AE=AF,即△AEF为等腰三角形.
DE‖AC有∠EDA=∠DAC=∠EAD所以EA=EDEO垂直于AD,可知∠AEO=∠DEO所以有:△AEF全等于△DEF有AF=DF有∠FAD=∠FDA∠FAD=∠DAC+∠CAF∠FDA=∠B+∠
因为AE=AC,AD平分∠BAC,即∠CAD=∠CAB,所以△ADE与△ADC全等,所以CD=因为EF平行BC,所以∠FEC=∠ECD,所以∠CED=∠FEC所以CE平分∠DEF
证明:∵点D在∠BAC的平分线上,∴∠1=∠2.(1分)又∵DE∥AC,∴∠2=∠3,∴∠1=∠3.(2分)∴AE=DE.(3分)又∵BD⊥AD于点D,∴∠ADB=90°.(4分)∴∠EBD+∠1=∠
证明:设EF分别交AD、AC于G、H,连接DH.DE‖AC,∠DAC=∠ADE【内错角】AD平分∠bac,∠DAC=∠DAB所以∠DAB=∠ADE,则AE=DE又因为EF平分∠AED,则EG⊥AD,且