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Math Help - Help with proving an identity.

  1. #1
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    Help with proving an identity.

    I recently found this problem in my maths book, and I don't know how I should solve it:

    Help with proving an identity.-namnloes.gif Show that L^2=(ab+cd)(ac+bd)/(bc+ad)

    I looked in the answers section of the book, and there it said that I should begin by using the law of cosines on the triangles ABD and BDC, and then use the fact that \cos C= -\cos A, as A+C=180 degrees. Why is that? Have I forgotten about some basic geometric rule or something?

    And still, even after using the fact above, I couldn't complete the task. So how do I complete this task?

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  2. #2
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    Quote Originally Posted by scounged View Post
    I recently found this problem in my maths book, and I don't know how I should solve it:

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ID:	20790 Show that L^2=(ab+cd)(ac+bd)/(bc+ad)

    I looked in the answers section of the book, and there it said that I should begin by using the law of cosines on the triangles ABD and BDC, and then use the fact that \cos C= -\cos A, as A+C=180 degrees. Why is that? Have I forgotten about some basic geometric rule or something?

    opposite angles of a quadrilateral inscribed in a circle are supplementary.
    law of cosines ...

    L^2 = a^2 + d^2 - 2ad \cos{C}<br />

    L^2 = b^2 + c^2 - 2bc \cos{A} = b^2 + c^2 + 2bc \cos{C}


    a^2 + d^2 - 2ad \cos{C} = b^2 + c^2 + 2bc \cos{C}

    a^2 + d^2 - b^2 - c^2 = 2bc \cos{C} + 2ad \cos{C}<br />

    a^2 + d^2 - b^2 - c^2 = 2\cos{C}(bc + ad)

    \dfrac{a^2 + d^2 - b^2 - c^2}{bc+ad} = 2\cos{C}

    sub the left side of the above equation into the equation ...

    L^2 = b^2 + c^2 + 2bc \cos{C}

    L^2 = b^2 + c^2 + \dfrac{bc(a^2 + d^2 - b^2 - c^2)}{bc+ad}

    common denominator ...

    L^2 = \dfrac{(b^2+c^2)(bc+ad) + bc(a^2 + d^2 - b^2 - c^2)}{bc+ad}

    L^2 = \dfrac{b^3c+ab^2d + bc^3 + adc^2 + a^2bc + bcd^2 - b^3c - bc^3}{bc+ad}

    L^2 = \dfrac{ab^2d + adc^2 + a^2bc + bcd^2}{bc+ad}

    L^2 = \dfrac{a^2bc + ab^2d + adc^2 + bcd^2}{bc+ad}


    L^2 = \dfrac{(ab+cd)(ac+bd)}{bc+ad}
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  3. #3
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    Thanks alot for the quick reply!
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