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Thread: [Trigonometric Ratios] Dividing a triangle with equal area

  1. #1
    Junior Member Cthul's Avatar
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    [Trigonometric Ratios] Dividing a triangle with equal area


    The area must be equal.
    How do I do this?
    Last edited by Cthul; Jul 10th 2010 at 10:20 PM. Reason: Missing information.
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Use median from point C to AB (why?)
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  3. #3
    Junior Member Cthul's Avatar
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    Oops, I forgot to mention the triangle has to have equal area.
    Will the area be the same if I just divide it like that?
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  4. #4
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by Cthul View Post
    Oops, I forgot to mention the triangle has to have equal area.
    I have considered that already...
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  5. #5
    Junior Member Cthul's Avatar
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    Thanks. I didn't know finding the median will make the area still equivalent.
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  6. #6
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by Cthul View Post
    Thanks. I didn't know finding the median will make the area still equivalent.
    There's a proof here. (The proof is for a stronger result.) This is related to the centroid being centre of mass for a thin triangular slab with uniform density.
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  7. #7
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    Hello, Cthul!

    "Also Sprach" is absolutely correct.


    We have $\displaystyle \Delta ABC$ with median $\displaystyle AM$ drawn to side $\displaystyle BC.$
    . . Hence: .$\displaystyle BM \,=\,MC.$

    Draw altitude $\displaystyle h$ from vertex $\displaystyle A$ to side $\displaystyle BC.$

    Code:
                    A
                    o
                   *:* *
                  * : *   *
                 *  :  *     *
                *   :   *       *
               *    :h   *         *
              *     :     *           *
             *      :      *             *
            *       :       *               *
           *        :        *                 *
        B o * * * * + * * * * o * * * * * * * * * o C
                              M

    Triangles $\displaystyle ABM$ and $\displaystyle AMC$ have equal bases and the same altitude.
    . . Hence, they have equal areas.


    A median always divides a triangle into two equal areas.

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