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Image

Based on the attached, if $AB$ and $DC$ are parallel, then $$\frac{|OD|}{|OA|}=\frac{|OC|}{|OB|}=\frac{|CD|}{|AB|}$$ Using the theorem, how can I show the following when the image is dilated by a dilation factor of $r$:

The line segments go to parallel line segments

I am not sure what this question is asking. I thought it was saying to show that $rAB$ and $rDC$ are parallel (so kind of the theorem in reverse). If so, can I show that the lines are parallel proceeding by contradiction and assuming they're not? So then, if the lines aren't parallel, then they must intersect at some point, $p$? Any insight is appreciated.

Thanks

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If we dilate everything by a factor of $r$, then the ratios $$\frac{|OD|}{|OA|}=\frac{|OC|}{|OB|}=\frac{|CD|}{|AB|}$$ will continue to hold, right? Have you learned that that is a sufficient condition for $AB$ and $CD$ to be parallel?

In particular, you can show that $OAB$ and $ODC$ are similar triangles (try Side angle side), and therefore angle $OAB$ and angle $ODC$ are the same.

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  • $\begingroup$ Someone asked me this question, so I'm not sure if I can assume that. I did show the first part of your statement though. If I can't assume your second statement, how can I prove it? $\endgroup$ – Alti Sep 23 '15 at 1:52
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    $\begingroup$ @Alti I added some. $\endgroup$ – Eric Auld Sep 23 '15 at 1:53
  • $\begingroup$ Thank you. So, it wouldn't be correct to prove by showing they don't intersect? $\endgroup$ – Alti Sep 23 '15 at 1:55
  • $\begingroup$ @Alti Not to say that that is an incorrect method, but I don't see how to do it that way. $\endgroup$ – Eric Auld Sep 23 '15 at 2:45
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    $\begingroup$ @Alti That seems like it would work. $\endgroup$ – Eric Auld Sep 23 '15 at 3:38

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