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ABCD is a rectangle, in which BC=2AB. A point E lies on ray CD, such that CE=2BC. Prove that BE perpindicular AC

  • mathematics -

    let AB = x
    then BC = 2x

    then CE = 4x

    angle B and angle C are right
    AB/BC = BC/CE = 1/2
    triangle ABC similar to triangle BCE
    Call F intersection of BE and AC
    Now angle EBC = FAB (similar triasngles)
    Angle FCB = BEC (also same similar triangles)
    BUT those angles add to 90 degrees (two angles in right triangle)
    so
    angle BFC = 90 degrees

  • mathematics -

    thanks for ur help..

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