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Have I created the proportion correctly? Some of my friends say to use the side splitter theorem but could I do this?

https://i.sstatic.net/SxP09.jpg

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    $\begingroup$ yeah, you are right. $\endgroup$
    – Aang
    Jan 25, 2013 at 2:45
  • $\begingroup$ @Avatar , what is the side splitter theorem ? $\endgroup$
    – FNH
    Jan 25, 2013 at 2:49
  • $\begingroup$ @MathsLover see e.g. this. In particular, the "side splitter theorem" is not applicable here, since it deals with a triangle cut by a line parallel to one of its sides. $\endgroup$ Jan 25, 2013 at 2:51
  • $\begingroup$ Side splitter theorem sounds funny. $\endgroup$ Jan 25, 2013 at 3:09
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    $\begingroup$ This reminds me of the joke where the student circled the "X" in the figure and wrote "here it is"! $\endgroup$ Jan 25, 2013 at 3:25

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Yes, the way you've set it up is correct.

The "side splitter theorem" is not applicable here, since it deals with a triangle cut by a line parallel to one of its sides, while the cut here is not parallel to any of the sides. What you've done is essentially what the side splitter theorem does, finding the similar triangles and setting up the inequalities. Well done.

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