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Math Help - trisecting diagonal

  1. #1
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    trisecting diagonal

    ABCD is a cyclic quadrilateral such that AB is 7 , BC is 24 , CD is 15 and DA is 20 units respectively. points E and F lie on AC such that AE=EF=FC . what is the area of the quadrilateral BEDF.
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  2. #2
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    Re: trisecting diagonal

    Quote Originally Posted by geniusgarvil View Post
    ABCD is a cyclic quadrilateral such that AB is 7 , BC is 24 , CD is 15 and DA is 20 units respectively. points E and F lie on AC such that AE=EF=FC . what is the area of the quadrilateral BEDF.
    1. By first inspection:

    AD^2+CD^2 = AB^2+BC^2

    That means AC = 25 and ABCD consists of 2 right triangles with a common hypotenuse.

    2. The long way: In a cyclic quadrilateral the interior angles of opposite vertices add up to 180 (or pi).

    Use Cosine rule:

    AC^2 = AD^2 + DC^2 - 2 \cdot AD \cdot DC \cdot \cos(\angle(ADC))

    AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(-\angle(ADC))

    Solve this system of equations for x and AC.

    3. Let a denote the area. Then

    a_{BEDF} = \frac13 \cdot a_{ABCD}
    Last edited by earboth; February 13th 2013 at 12:50 AM.
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