∴∠AED=∠BAC=85°
6解:1过点D作DF∥AE交BC的延长线于点F,∵AD∥BC,∴四边形AEFD为平行四边形,∴EF=AD=10,DF=AE=9,∵E是BC的中点,∴BF=12AD+AD=15,∴BD2+DF2=122+92=225=BF2,∴∠BDF=90°,即BD⊥DF,∵AE∥DF,∴AE⊥BD2过点D作DM⊥BF于点M,∵BDDF=BFDM,∴DM=9×1512=356,∴SABCD=BCDM=727分析:1证AB=BE,AB=CD,即可得到结论;2将ABCD的面积转化为△ABE的面积求解即可.
解:1∵四边形ABCD是平行四边形,∴AB=CD,AD∥BE,∴∠DAE=∠E,∵∠BAE=∠DAE,∴∠BAE=∠E,∴AB=BE,∴BE=CD2∵AB=BE,BF⊥AE,∴AF=FE,又∵∠DAF=∠CEF,∠AFD=∠EFC,∴△AFD≌△EFCASA,∴SABCD=S△ABE,∵AB=BE,∠BEA=60°,∴△ABE是等边三角形,由勾股定理得BF=23,∴S△ABE=12AEBF=
43,∴SABCD=438分析:可通过证BECF来得到结论.解:∵BE⊥AD,CF⊥AD,∴∠AEB=∠DFC=90°,∴BE∥CF,∵AB∥CD,∴∠A=∠D,又∵AE=DF,∴△AEB≌△DFCASA,∴BE=CF,∴四边形BECF是平行四边形9平行四边形10解:1∵AB∥DE,∴∠B=∠DEF,∵BE=EC=CF,∴BC=EF,又∵∠ACB=∠F,∴△ABC≌△DEFASA2四边形AECD是平行四边形.证明:∵△ABC≌△DEF,∴AC=DF,∵∠ACB=∠F,∴AC∥DF,∴四边形ACFD是平行四边形,∴AD∥CF,AD=CF,∵EC=CF,∴AD∥EC,AD=CE,∴四边形AECD是平行四边形11解:1∵四边形ABCD为平行四边形,∴AD∥BC,∴∠EAO=∠FCO,又∵OA=OC,∠AOE=∠COF,∴△OAE≌△OCFASA,∴OE=OF,同理OG=OH,∴四边形EGFH是平行四边形2GBCH,ABFE,EFCD,EGFH12解:1∵△ABC是等边三角形,∴∠ABC=60°,又∵∠EFB=60°,∴∠ABC=∠EFB,∴EF∥BC,又∵DC=EF,∴四边形EFCD是平行四边形2连接BE,∵∠EFB=60°,BF=EF,∴△BEF为等边三角形,∴BE=BF=EF,∠ABE=60°,∵CD=EF,∴BE=CD,又∵△ABC为等边三角形,∴AB=AC,∠ACD=60°,∴∠ABE=∠ACD,∴△ABE≌△ACDSAS,∴AE=AD
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