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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 11 2004 
Let the top and bottom edges be t and b
> and the left and right sides be l and r and let the diagonal from tl
> to br be d. Let L be the midpoint of l, and so on. Then the given
> condition is that LR=(t+b)/2 and we want to show that this forces
> tb.
But, as David pointed out earlier, LDb with LD=b/2.
> Similarly, DRt with DR=t/2.
So, by the shortest distance
> property of a straight line, we have LR<=LD+DR=b/2+t/2 with
> equality only if LD and DR are collinear.
So, if LR=(b+t)/2, then
> LD and DR form a single line and are parallel to one another.
But
> then bLDDRt and we are done.
(I guess a "cute hint" for
> this might be "after making the triangle decomposition use the
> triangle inequality")
I am here to tell you that Alan Cooper was right!!! We have been in touch by
enail, and I suddenly realised he was talking about length, not parallelism, and
am left wondering why I didn't think of it. So i've searched for excuses and
found none. But he is absolutely correct. You can use the lengths of the
segments and compare to the given length of the whole.
OK, I thought of an excuse: I've been busy in Santa's workshop building a toy
box, and had my mind on that.
Anyhow, congratulations Alan, and thanks for your patience.
David.
 
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