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Math Help - DE Parallel to BC?

  1. #1
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    DE Parallel to BC?

    Given triangle ABC and segment DE, if AD =2, DB =3,
    AE =4 and EC =4, is DE parallel to BC?

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    This is converse of Basic Proportionality Theorem
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    A riddle wrapped in an enigma
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    Quote Originally Posted by magentarita View Post
    Given triangle ABC and segment DE, if AD =2, DB =3,
    AE =4 and EC =4, is DE parallel to BC?
    See diagram.

    The "Triangle Proportionality Theroem" states: If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments of proportional length.

    So, if \overline{DE} \parallel \overline{BC}, then

    \frac{AD}{DB}=\frac{AE}{EC}

    \frac{2}{3} \neq \frac{4}{4}

    Therefore, \overline{DE} \nparallel \overline{BC}
    Attached Thumbnails Attached Thumbnails DE Parallel to BC?-tri.jpg  
    Last edited by masters; December 16th 2008 at 10:01 AM.
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  4. #4
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    ok...

    Quote Originally Posted by masters View Post
    See diagram.

    The "Triangle Proportionality Theroem" states: If a line is parallel to one side of a triangle and intersects the other two sides in two distinct points, then it separates these sides into segments of proportional length.

    So, if \overline{DE} \parallel \overline{BC}, then

    \frac{AD}{DB}=\frac{AE}{EC}

    \frac{2}{3} \neq \frac{4}{4}

    Therefore, \overline{DE} \nparallel \overline{BC}
    I thank both replies. I want to thank masters for always answering my questions quickly and professionally.
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