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Topic: Midsegment of a rectangle?
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Subject:   RE: Midsegment of a rectangle?
Author: Mathman
Date: Dec 10 2004
On Dec 10 2004, Alan Cooper wrote:
> On Dec  9 2004, Mathman wrote:
> A quick followup:  I'm thinking of
> a quadrilateral with top and
> bottom of lengths "a" and "b".
> There is a line joining the
> midpoints of the two sides, and its
> length is given as (a+b)/2.  You
> want proof that the top and
> bottom are parallel, [and the midpoint
> divider would then also be
> parallel to them.]  Is this what you
> mean?

Yes.

And I
> think your triangle decomposition (by drawing in a diagonal) leads
> to a very nice proof of this fact.


Very well, but I think then I must be missing something obvious ...it happens.
:-)

For parallelism, I need similarity.  For similarity, I need to determine ratio
of sides [all I can see presently.]

Let's extend sides to the peak of a triangle containing the trapezoid sides, top
"a", bottom "b", with midpoint divider being (a+b)/2.  Then these lengths are
all that we know.  I see three relative triangles, and would have to show the
same similarity simultaneously.  So I have a dilemma.  I say "I", because I
might be missing something.  What I'm thinking of [without working it out yet]
is something similar to the more familiar problem of an isosceles triangle.
Given a base and two equal angle bisectors from that base, it is required to
show that the triangle containing that figure is isosceles.  Of course it is,
but the proof requires indirect reasoning.  I have the same feeling that applies
here also, and so it is a pretty problem.

If you have a more direct proof I'd love to see it, even if it means I'm dead
wrong and need to be eating more fish.  It is easy enough to assume parallelism
for two of the lines and then show the third to be parallel to those as a
consequence, but all we have here is lengths at the start, and need to show
lines parallel [determine angle relationships] with all three lines, the top and
bottom of the trapezoid and the divider, all at the same time.  Again, I could
be dead wrong, and don't mind that at all.  It might be demonstrated with a
tool, but a demonstration is not a proof, being particular rather than general.
I'm thinking of two triangles being similar if the sides are in proportion, and
that can lead to a mess of algebra.

Too bad we can't exchange images here and have to rely on ASCII, but that's high
tech for you.

David.

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