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Math Help - Ratios of line segments in a triangle

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    Ratios of line segments in a triangle

    Not sure if this belongs here, it's pretty tough to solve (and I have the solution from the book!) I thought I would share it here in case others wanted to be challenged, or if anyone can provide the reasoning behind the solution.

    We have a triangle ABC as shown in the figure, with point E selected on side AB in such a way that AE : EB = 1:3, in other words, the ratio of the length of segment AE to the length of segment EB is 1 to 3. Draw the line segment CE.

    We then have a point D selected on side BC so that CD : DB = 1:2. Then draw a line segment from A to D. This gives a point F, which is the intersection of segments AD and CE.

    Solve for (EF/FC) + (AF/FD).
    Attached Thumbnails Attached Thumbnails Ratios of line segments in a triangle-contest-problem-book-2-page-44.jpg  
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