1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.求证:①△ADE≌△ADC ②∠EDF=
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![1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.求证:①△ADE≌△ADC ②∠EDF=](/uploads/image/z/2776583-47-3.jpg?t=1%EF%BC%89%E5%A6%82%E5%9B%BE1%2C%E2%96%B3ABC%E4%B8%AD%2CAB%EF%BC%9EAC%2CAD%E5%B9%B3%E5%88%86%E2%88%A0BAC%E4%BA%A4BC%E4%BA%8E%E7%82%B9D%2C%E5%9C%A8AB%E4%B8%8A%E6%88%AA%E5%8F%96AE%3DAC%2C%E8%BF%87%E7%82%B9E%E4%BD%9CEF%E2%80%96BC%E4%BA%A4AD%E4%BA%8E%E7%82%B9F.1%EF%BC%89%E5%A6%82%E5%9B%BE1%2C%E2%96%B3ABC%E4%B8%AD%2CAB%EF%BC%9EAC%2CAD%E5%B9%B3%E5%88%86%E2%88%A0BAC%E4%BA%A4BC%E4%BA%8E%E7%82%B9D%2C%E5%9C%A8AB%E4%B8%8A%E6%88%AA%E5%8F%96AE%3DAC%2C%E8%BF%87%E7%82%B9E%E4%BD%9CEF%E2%80%96BC%E4%BA%A4AD%E4%BA%8E%E7%82%B9F.%E6%B1%82%E8%AF%81%3A%E2%91%A0%E2%96%B3ADE%E2%89%8C%E2%96%B3ADC+%E2%91%A1%E2%88%A0EDF%3D)
1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.求证:①△ADE≌△ADC ②∠EDF=
1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.
1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.
求证:①△ADE≌△ADC ②∠EDF=∠EFD
(2)如图3,△ABC中,AB>AC,AD平分△的外角∠EAC交BC的延长线于点D,在AB的反向延长线上截取AE=AC,过点E作EF‖BD交AD的反向延长线于点F.求证:四边形CDEF是菱形;
(3)如图2,在(2)的条件下,当四边形CDEF是正方形是,设EF=1,FG=x,试问是否有存在x的值,使△ACB为等腰三角形,若存在,请求出x的值,若
不存在,请说明理由.
图虽然挫了点但还看得过去……
1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.1)如图1,△ABC中,AB>AC,AD平分∠BAC交BC于点D,在AB上截取AE=AC,过点E作EF‖BC交AD于点F.求证:①△ADE≌△ADC ②∠EDF=
(3)如图2
AB=AC
∠B=∠BAC=∠FEB
ADE≌△ADC ∠ACD=∠AED ∠FCA=90-∠ACD ∠FEB=90-∠AED
∠FCA=∠FEB=∠B=∠BAC
∠ACD=∠B+∠BAC=2∠FCA=90-∠FCA
∠FCA=30=∠FEB
FG==1÷√3
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