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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 
> The first proof I thought of was based on analytic geometry  giving
> names to the coordinates of the corners and algebraically converting
> the average length condition to deduce the equality of slopes. Even
> with some simplification by scaling (or equivalently using vector
> language with basis vectors along two edges) this was still rather
> messy.
But David's triangle decomposition leads to a much
> simpler proof as follows:
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.
Sorry, but my suggestion was in reference to my previous assumption on what you
had meant. In this case, it does not hold so far as I can see form what
follows. For, all that we know are lengths. We do not know (yet) that LDb
for example. If it is, then, yes, the rest follows. It is only on this
assumption that the rest follows, and it is assumed here, not proven. Again,
since all that we know are lengths, we need to find either similarity form equal
ratio of sufficient sides [all three at once!] in order to establish that the
triangles are similar. Then, and only then do we know that angles are equal
from that similarity. Then it follows that lines are parallel. That is the
dilemma I was referring to earlier.
Consider: You have a quadrilateral, yet to be determined as a trapezoid. The
initial information is that you have a line joining the two sides, so you have
distances as in your diagram of l/2, l/2, and r/2, r/2 that may be indicated.
You have the distance across the divider (t+b)/2, having been given a top of "t"
and bottom of "b". That is ALL that is known. Parallelism has to be
established from that, not assumed. You would need to show, for example, that
the diagonal is divided in two to yeild a parallel portion of the divider, or
establish [not assume] that the divider is parallel and then divides the
diagonal in two. Both can not be assumed simultaneously. As I said, a pretty
problem, which is still how to show that the divider [or part of it] is parallel
from the dimensions given.
In other words, it remains as much of the problem to show initially that the
divider is parallel to one of the top or bottom lines as it is to show that the
top and bottom are parallel.
David.
 
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