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Chord parallel to AB and 2CD=AB

  1. Feb 9, 2004 #1
    AB is the diameter of a circle CD is a chord parallel to AB and 2CD=AB. The tangent at B meets the line AC produced at E. Prove that AE=2AB.

    I'm finding no way to solve this ?

    Still i thought of applying AE*CE=BE2 but that is not enough to slove the pro ???

    Any other Hint to solve
     
  2. jcsd
  3. Feb 9, 2004 #2

    matt grime

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    So A and C are on the same end of the chords?

    Let O be the centre of the circle

    the triangle OAC is equilateral, because the triangle OCD is and everything is symmetric.. well, convince yourself somehow

    so angle BAE is 60,

    so AB/AE = cos60 = 1/2; AE=d. QED

    Opposite ends of te chords gives you...?
     
  4. Feb 10, 2004 #3

    MathematicalPhysicist

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    i think you should bulid a trapozoid with CD and AB (with two sides equals to each other and to the small base-CD) and to see that we have THE TRAINGLE BAE as a right triangle because the angle between radius and a tangent is 90.
    here are my computations that brought to the answer:
    BE^2=CE*AE
    AB^2+BE^2=AE^2 (PYTHAGORAS THEOREM)
    AB^2+CE*AE=AE^2
    AB^2=AE(AE-CE)=AE*CE
    CE=CD=AB/2
    AB^2/CE=AE
    AB^2/(AB/2)=AE
    2AB=AE

    so you were half right himanshu :wink:
     
  5. Feb 11, 2004 #4

    MathematicalPhysicist

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    i have an illustration of the triangle and trapaoid the problem is my scanner doesnt work so i cant upload it here, sorry.
     
  6. Feb 12, 2004 #5
    Ya i got it
    I was half way write upto formation of equation and it just required manipulations & rearrangement

    Ya i convinced myself for equilateral triangle
    Thnxs
     
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