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Topic:  Midsegment of a rectangle? 
Related Item:  http://mathforum.org/mathtools/tool/15621/ 
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Subject:  RE: Midsegment of a rectangle? 
Author:  Mathman 
Date:  Dec 9 2004 
> Hi David,
Unless I have misread you, I think you have answered
> Cynthia's original question rather than my "followup question".
> Cynthia asked whether the average length property holds for *all*
> quadrilaterals, and my question was whether there *exist* any (other
> than trapezoids). In other words, "part(b)" could be rephrased as
> "If the midsegment satisfies the average length property, *must* the
> edges that it separates be parallel?"
Interestingly, your "two
> triangles" construction does lead (after a couple more lines of
> argument)to a proof that the answer to part (b) is "Yes" (in terms
> of the second phrasing)  and that proof is much nicer than the
> coordinatebased argument I had in mind.
Perhaps my misinterpretation then Alan. I had thought you were asking if there
were quadrilaterals other than the trapezoid that had this property. I've found
I'm prone to error after a long day, so I'll get back to you on that. Euclid is
my favourite for pastime math. Other material is deeper, but I like it
relatively simple. Getting too old for the heavy stuff. Again intuitively, and
I could be dead wrong, but I'd look to trig for a possible answer. Parallelism
allows a deadon approach, but it might take a workaround for the reverse
condition ...As I said, it's late here, and I'll get back to it.
I had included the "triangles" as an intuitive approach and hadn't really
followed up on that either. Glad it worked for you. Busy painting and spending
time in Santa's workshop [my hobby]. Don't ever retire. You're available
24/7.
David.
 
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