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Thread: Triangles

  1. #1
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    Triangles

    Hello,

    I need the solution of the following question:

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


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  2. #2
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    Hello, raed!

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

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

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

    Compare $\displaystyle \Delta AXC$ and $\displaystyle \Delta ABC.$

    They have the same height $\displaystyle \,h.$

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

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

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


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


    And we're done!

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, raed!


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

    Compare $\displaystyle \Delta AXC$ and $\displaystyle \Delta ABC.$

    They have the same height $\displaystyle \,h.$

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

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

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


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


    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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