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Math Help - [SOLVED] Geometry problem from similar triangles section

  1. #1
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    [SOLVED] Geometry problem from similar triangles section

    In triangle ABC, draw line segments AQ and BP, where P and Q lie on sides AC and BC, respectively. Now draw PY parallel to AQ and QX parallel to BP, where X and Y line on AC and BC. Show that XY is parallel to AB.

    I realize this is harder without a picture. Sorry.

    I know that showing ABC and XYC are similar triangles would prove XY parallel to AB since they both share vertex C, but I don't know how. I know that the intersection of the sets of parallel lines for a parallelogram, but I don't know if that plays in or not.
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  2. #2
    MHF Contributor red_dog's Avatar
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    We have:

    \frac{XC}{PX}=\frac{QC}{BQ}\Rightarrow\frac{XC}{PC  }=\frac{QC}{BC}\Rightarrow PC\cdot QC=XC\cdot BC (1)

    \frac{PC}{PA}=\frac{YC}{YQ}\Rightarrow\frac{PC}{AC  }=\frac{CY}{QC}\Rightarrow PV\cdot QC=AC\cdot CY (2)

    From (1) and (2) XC\cdot BC=CY\cdot AC\Rightarrow\frac{XC}{AC}=\frac{CY}{BC}\Rightarro  w XY\parallel AB
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