# Thread: [SOLVED] Geometry problem from similar triangles section

1. ## [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.

2. We have:

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

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

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