Discussion:  Dynamic Geometry Exploration: Properties of the Midsegment of a Trapezoid tool 
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 
> ...and then there's the followup question ("part (b)" if you like):
> Is it possible to find *any* nontrapezoidal example of a
> quadrilateral for which the distance between the midpoints of two
> nonadjacent sides is equal to the average of the lengths of the
> other two sides?
Alan
P.S. I am firmly in the camp of those
> who consider parallelograms to be trapezoidal so I wouldn't consider
> a parallelogram or rectangle to be examples of what I am asking for.
In a word, "No." If it's a general quadrilateral, it does not work in general.
That is; it does not work for *all* quadrilaterals. The only distinction
between the general quadrilateral and any other quadrilateral is to do with
parallelism. Either one pair or two pairs are parallel ...and there are only
four sides, so that' it. That is, a quadrilateral is a trapezoid, or it is
not.
To show that it is not true in general, consider the reverse theorem of the
divided triangle as a simpler example: If, in triangle ABC, DE is such that D
is the midpoint of AB, and E the midpoint of AC, then DE is parallel and half of
BC. ...Otherwise, it is not. The quadrilateral can be divided into two
triangles and then looked at from that point of view. The thing is that similar
figures [equal angles/parallel lines] are necessary for the relationship to
hold, and joining midpoints does not do that in general.
An interesting sidetrack: If you join the midpoints of any quadrilateral to
make a quadrilateral, it will be a parallelogram.
David.
 
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