AB=CD,DF⊥AC于F,BE⊥AC于E,DF=BE.试说明:AF=CE
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/29 20:32:04
xMN@ǯBHAgcz
3iiUh.H\hbX-DqQ ;`E{ }/YYv=~h_wX"Owlu?~>u%z`I++sZM~W+y/ո JdqOsED(Agq"y
NTׄIK"D*apUF.1:ljzC#Ma
ʛh1R%܊,FBBM@.'a`r!#=
qjlRd-kO]Z9ϧMer[R7Vdy>XNF_7Q/l`K2g;5
AB=CD,DF⊥AC于F,BE⊥AC于E,DF=BE.试说明:AF=CE
AB=CD,DF⊥AC于F,BE⊥AC于E,DF=BE.试说明:AF=CE
AB=CD,DF⊥AC于F,BE⊥AC于E,DF=BE.试说明:AF=CE
∵∠DFC=∠AEB=90°
且AB=CD,DF=EB
∴根据勾股定理 AE=CF
∵AE-EF=CF-FE
∴AF=CE
已知AB=CD,AF=CE,BE⊥AC于E,DF⊥AC于F.求证:AB//CD
AB=CD,DF⊥AC于F,BE⊥AC于E,DF=BE.试说明:AF=CE
如图,已知AD=BC,BE⊥AC于E,DF⊥AC于F,且BE=DF.求证AB‖CD
如图,AB‖CD,AD‖BC,BE⊥AC于E,DF‖AC于F,求证BE=DF谢.
如图,CD⊥AB,BE⊥AC,垂足分别为D、E,BE交CD于F,且AD=DF,求证:AC=BF
梯形ABCD中,AB//CD,AB>CD,AC= BC ,过B做BE⊥AD,交AC于F,连接DF,求证∠BFC=∠DFC
如图,AB=CD,AF=CE,BE⊥AC于E,DF⊥AC于F,试说明△ABE≌△CDF.
如图,AB=CD,AF=CE,BE⊥AC于E,DF⊥AC于F,求证△ABE≌△CDF
如图 AB=CD DF垂直于AC于F BE垂直于AC于E DF=BE 求AF=CE
如图,AB\CD,AD\BC,BE垂直于AC于E,DF垂直于AC于F,求证BE=DF.明天要交了急!
已知AB=CD,BE⊥AC于E,DF⊥AC于F,且BE=DF,求证:AB‖CD已知∠1=∠2,∠3=70度,求∠4的度数
已知:AD=BC,BE⊥AC于E,DF⊥AC于F,且BE=DF,求证:1.△ABE全等与△CDF;2.AB=CD;3.AB平行CD,
如图,CD⊥AB,BE⊥AC,垂足分别为D、E,交于CD于F,且AD=DF,求证∶AC=BF.
已知E、F在BC边上,AE⊥BC于E,DF⊥BC于F,且AC=BD,BE=CF,求证AB=CD
△ABC中,AB=AC,AD是高,DE⊥AB于E,DF⊥DF于E,DF⊥于F.求证:BE=CF
如图,AB=AC,BD=CD,DE⊥AB于E,DF⊥AC于F,求证DE=DF,
如图,AB=AC,BD=CD,DE⊥AB于点E,DF⊥AC于点F.求证:DE=DF
如图,AB=AC,BD=CD,DE⊥AB于点E,DF⊥AC于点F.求证DE=DF