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 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: Alan Cooper 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 t||b.
But, as David pointed out earlier, LD||b with |LD|=b/2.
Similarly, DR||t 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 b||LD||DR||t and we are done.

(I guess a "cute hint" for this might be "after making the triangle
decomposition use the triangle inequality")