如图,AB∥CD,DF交AC于点E,交AB于点F,DE=EF
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证明:作DM平行BC,交AE于M.则CF:DF=CE:DM.又BE=CE,则CF:DF=BE:DM=AB:AD;又AD=AC.所以CF:DF=AB:AC.
证明:∵DE∥AC,DF∥AB,∴四边形AEDF是平行四边形,∴DE=AF,又AB=AC,∴∠B=∠C,∵DF∥AB,∴∠CDF=∠B,∴∠CDF=∠C,∴DF=CF,∴AC=AF+FC=DE+DF.
∵AB∥CD,AC∥BD,∴∠ABC=∠DCB,∠ACB=∠DBC.∵BC=CB,∴△CAB≌△CDB,∴AB=CD,AC=BD.∵AB∥CD,AC∥BD,∴∠BAO=∠CDO,∠OBA=∠OCD,∠
∠1=∠2证明:∵AD平分∠BAC∴∠BAD=∠CAD∵DE//AC∴∠1=∠CAD∵DF//AB∴∠2=∠BAD∴∠1=∠2
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
数学天才团为您不难证明△BEC∽△BCA∵∠A=30°∴∠BCE=30°BC=2BE∵DF∥BC∴DF⊥AC∠FDC=30°根据“角边角”△BEC≌△GED∴GE=BE∴BC=BG在Rt△ABC中∵∠
1正相似2通过相似知DF/BG=DF/GC所以相等
1、∵AD=ADAB=ACBD=CD∴△ABD≌△ACD(SSS)∴∠BAD=∠CAD即∠EAD=∠FAD∵DE⊥AB于点E,DF⊥AC于点F∴∠AED=∠AFD=90°∵AD=AD∴△ADE≌△AD
(1)证明:∵DF∥BC,∠ACB=90°,∴∠CFD=90°.∵CD⊥AB,∴∠AEC=90°.在Rt△AEC和Rt△DFC中,∠AEC=∠CFD=90°,∠ACE=∠DCF,DC=AC,∴Rt△A
∵∠ACB=90°,CD是中线,∴AD=BD=CD=6,∵DF⊥AB,∴∠F+∠B=90°,∵∠ACB=90°,∴∠A+∠B=90°,∴∠F=∠A,又∠FDB=∠ADE=90°,∴ΔADE∽ΔFDB,
(1)证明:在△ABC和△ADC中,AB=ADBC=DCAC=AC,∴△ABC≌△ADC(SSS),∴∠BAC=∠DAC,在△ABF和△ADF中,AB=AD∠BAF=∠DAFAF=AF,∴△ABF≌△
证明:∵∠ABD=∠ACD∴∠EBD=∠FCD(等角的补角相等)∵BD=CD(已知),∠E=∠F=90°∴△BDE≌△CDF(AAS)∴DE=DF(全等三角形对应边相等)
∵AB∥CD∴﹤BAC﹦再问:DE﹦EF怎么求的!!
证明:∵CD⊥AE∴∠AGC=∠AGD=90∵AE平分∠BAC∴∠BAE=∠CAE∵AG=AG∴△AGC≌△AGD(ASA)∴AC=AD∴∠ACD=∠ADC∵AE=AE∴△AEC≌△AED(SAS)∴
AE=BF?CE平行DH则∠CEO=∠DFO180-∠CEO=180-∠DFO即是∠BFD=∠AEC又CE=DFAE=BF则三角形AEC与BFD全等则∠ACE=∠BDF得证!
证明:∵AD平分∠BAC∴∠BAD=∠CAD∵DE∥AC∴∠ADE=∠CAD∴∠ADE=∠BAD∴AE=DE∵DF∥AB∴平行四边形AEDF(两组对边平行)∴AF=DE,DF=AE∴AE=DE=AF=
证明:过点D作DG//CB交AB于点G,则有:DF/EF=BG/BE,AB/BG=AC/CD因为AB/BG=AC/CD,所以AB/AC=BG/CD,因为BE=CD,所以DF/EF=AB/AC.
证明:∵AB∥CD.∴∠AFE=∠D;又FE=DE;∠AEF=∠CED.∴⊿AEF≌⊿CED(ASA),AE=EC.
CD平分∠ACB=>∠DCE=∠DCF-----(1)DE//AC=>∠DCF=∠CDE-----(2)(1)(2)=>∠DCE=∠CDE=>EC=ED-----(3)DE//AC、DF//BC=>□
很简单:△bdf全等于△adc从ad=df入手直角三角型的全等最好证了