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Math Help - Triangles

  1. #1
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    Triangles

    Hello,

    I need the solution of the following question:

    Consider a triangle \triangle ABC and let X be a point on the side AB such that AX=\frac{1}{3}AB and let Y be a point on the side AC such that CY=\frac{1}{3}AC. Prove that the area of the triangle \triangle AXC equals the area of the triangle \triangle BYC.


    Best Regards.
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  2. #2
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    Hello, raed!

    \text{Consider }\Delta ABC\text{; let }X\text{ be a point on side }AB\text{ such that: }AX\,=\,\frac{1}{3}AB
    \text{and let }Y\text{ be a point on the side }AC\text{ such that }CY\,=\,\frac{1}{3}AC.

    \text{Prove that the area of }\Delta AXC\text{ equals the area of }\Delta BYC.

    Side \,AB is divided in the ratio 1:2

    Compare \Delta AXC and \Delta ABC.

    They have the same height \,h.

    Code:
                        C
                        o
                      **|*
                    * * | *
                  *  *  |  *
                *   *   |h  *
              *    *    |    *
            *     *     |     *
          *      *      |      *
      A o * * * o * * * * * * * o B
        : - 1 - X - - - 2 - - - :

    The base of \Delta AXC is one-third the base of \Delta ABC.

    Hence: . (\text{area }\Delta AXC) \;=\;\frac{1}{3}(\text{area }\Delta ABC)


    In a similar fashion, we prove that: . \text{(Area }\Delta BYC)} \;=\;\frac{1}{3}(\text{area }\Delta ABC})


    And we're done!

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  3. #3
    Member
    Joined
    Sep 2009
    Posts
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    Quote Originally Posted by Soroban View Post
    Hello, raed!


    Side \,AB is divided in the ratio 1:2

    Compare \Delta AXC and \Delta ABC.

    They have the same height \,h.

    Code:
                        C
                        o
                      **|*
                    * * | *
                  *  *  |  *
                *   *   |h  *
              *    *    |    *
            *     *     |     *
          *      *      |      *
      A o * * * o * * * * * * * o B
        : - 1 - X - - - 2 - - - :

    The base of \Delta AXC is one-third the base of \Delta ABC.

    Hence: . (\text{area }\Delta AXC) \;=\;\frac{1}{3}(\text{area }\Delta ABC)


    In a similar fashion, we prove that: . \text{(Area }\Delta BYC)} \;=\;\frac{1}{3}(\text{area }\Delta ABC})


    And we're done!

    Thank you very much for your reply.

    Best Regards,
    Follow Math Help Forum on Facebook and Google+

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