如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2=(3)若D2D3=1/3D2B,E2E3=1/3E2C,
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![如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2=(3)若D2D3=1/3D2B,E2E3=1/3E2C,](/uploads/image/z/3976919-71-9.jpg?t=%E5%A6%82%E5%9B%BE%EF%BC%9A%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CBC%3D1%2C%EF%BC%881%EF%BC%89%E8%8B%A5AD1%3D1%2F3AB%2CAE1%3D1%2F3AC%2C%E5%88%99D1E1%3D+%EF%BC%882%EF%BC%89%E8%8B%A5D1D2%3D1%2F3D1B%2CE1E2%3D1%2F3E1C%2C%E5%88%99D2E2%E5%A6%82%E5%9B%BE%EF%BC%9A%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CBC%3D1%2C%EF%BC%881%EF%BC%89%E8%8B%A5AD1%3D1%2F3AB%2CAE1%3D1%2F3AC%2C%E5%88%99D1E1%3D+%EF%BC%882%EF%BC%89%E8%8B%A5D1D2%3D1%2F3D1B%2CE1E2%3D1%2F3E1C%2C%E5%88%99D2E2%3D%EF%BC%883%EF%BC%89%E8%8B%A5D2D3%3D1%2F3D2B%2CE2E3%3D1%2F3E2C%2C)
如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2=(3)若D2D3=1/3D2B,E2E3=1/3E2C,
如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2
如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1=
(2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2=
(3)若D2D3=1/3D2B,E2E3=1/3E2C,则D3E3=
(4)若Dn-1Dn=1/3Dn-1B,En-1En=1/3En-1C,则DnEn=
如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2如图:在△ABC中,BC=1,(1)若AD1=1/3AB,AE1=1/3AC,则D1E1= (2)若D1D2=1/3D1B,E1E2=1/3E1C,则D2E2=(3)若D2D3=1/3D2B,E2E3=1/3E2C,
(1):在△ABC和△AD1E1中,共同角A,AD1/AB=AE1/AC=1/3,所以△ABC和△AD1E1相似,所以D1E1/BC=AD1/AB=1/3;
(2):AD1=1/3AB,D1D2=1/3D1B,所以AD2=5/9AB,同理,AE2=5/9AC,又因共同角A,所以△ABC和△AD2E2相似,所以D2E2=5/9BC=5/9;
(3):同理可推,D3E3=19/27;
(4):同理可推,DnEn=1-(2/3)^n
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我晕图呢?