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Thread: triangles

  1. #1
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    triangles

    In the figure , AFD and BFE are straight lines and AB is parallel to ED . THe point C is such that AB=AC and FC bisects $\displaystyle \angle ACD $. Prove that DC=DE
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  2. #2
    MHF Contributor red_dog's Avatar
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    $\displaystyle \Delta FAB\sim\Delta FDE\Rightarrow\frac{AB}{DE}=\frac{AF}{FD}$

    In the triangle ACD apply the bisector theorem: $\displaystyle \frac{AF}{FD}=\frac{AC}{CD}$

    Then $\displaystyle \frac{AB}{DE}=\frac{AC}{CD}$.

    But $\displaystyle AB=AC\Rightarrow DE=CD$
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