在矩形ABCD中DE垂直AC，EF//AB，AB=6 D=3求EF

在矩形ABCD中DE垂直AC，EF//AB，AB=6 D=3求EF

解：矩形ABCD的对角线  AC=3√5

    易证△DEA ∽△ABC

    ∴ AE/CB=DA/AC

     ∴ AE=3×3/（3√5）=3/√5

      由△CEF ∽△CAB得

      EF/AB=CE/CA

    EF=AB×CE/(3√5)

    =[6×(3√5-3/√5]/( 3√5)

    =6[3√5×(1/3√5)-3/√5×(1/3√5) ]

    =6(1-1/5)

    =24/5

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证明：∵∠CEF=90°∴∠AEF+∠AFE=∠AEF+∠CED=90°∴∠AFE=∠CED∵∠A=∠D∴△AEF∽△DCE∴EF/CE =AF /DE∵AE =DE ∴EF/CE =AF /AE∵∠A=∠FEC∴△AEF∽ECF(两边成比例，夹角相等)

AC=3√5,DE*AC=18,DE=6√5/5,CE=12√5/5,

EF/6=CE/AC,EF=24/5

解：在直角三角形ADC中，CD = AB = 6

    AD = 3

    所以，AC = 3倍根号5

    易证明，CDE与DEA相似

    所以，CE/DE = DE/AE = CD/AB = 2

    所以，CE/DE*DE/AE = 4

    即：CE/EA = 4

    CE/CA = 4/5

    易得，CEF与CAB相似

    所以，EF/AB = CE/CA = 4/5

    因此，EF = 4/5*AB = 24/5