证明:如图,连接BD、AE,
∵DA⊥AB,FC⊥AB,
∴AD∥CF,∠DAB=∠BCF=90°,
又∵DA=BC,FC=AB,
∴△DAB≌△BCF(SAS),
∴BD=BF,
∴∠BDF=∠BFD,
又∵AD∥CF,
∴∠ADF=∠CFD,
∴∠ABF=∠DFB+∠ADF=∠BFC+2∠CFD,
同理可得,∠BAF=∠AFC+2∠CFE,
又∵∠AFB=51°,
∴∠ABF+∠BAF=129°,
∴∠BFC+2∠CFD+∠AFC+2∠CFE=51°+2∠DFE=129°,
∴∠DFE=39°.
答:∠DFE度数是39°.
如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
数学人气:142 ℃时间:2019-11-21 19:37:00
优质解答
我来回答
类似推荐
- 如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
- 如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
- 如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
- 如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
- 如图,点C在线段AB上,DA⊥AB,EB⊥AB,FC⊥AB,且DA=BC,EB=AC,FC=AB,∠AFB=51°,求∠DFE度数.
猜你喜欢
- 1From what number can one take half and leave nothing?
- 2(a strong wind blows his scart off his
- 3Who is your teacher?
- 4We have to find a hotel to live __.A.\ B.in
- 5大自然中的哲理~8条~
- 6碳酸钠,碳酸氢钠的混合物与200ml1.5mol/L的盐酸恰好完全反应,产生的气体通入澄清石灰水产生20g沉淀
- 7"嗯,现在我明白了,谢谢你"这句话英语怎么说.
- 8如果连续三个偶数的和是2010,那么这3个数分别是多少啊?
- 9谁有好的开头结尾 {是一片文章上的}
- 10-42分之1/(6分之1+3分之2-7分之2-14分之3)=