Trapezoid Diagonals and Midpoints of Parallel Sides
Date: 03/04/2002 at 09:13:54 From: Leonard Subject: Geometry Why, in a trapezoid, are the midpoints of the parallel sides collinear with the intersection of the diagonals? It seems so easy, but I don't see it. Thanks.
Date: 03/04/2002 at 10:04:07 From: Doctor Floor Subject: Re: Geometry Hi, Leonard, Thanks for your question. Consider the following trapezoid in which AB and CD are parallel and diagonals AC and BD intersect in E. M is the midpoint of AB, and ME intersects CD in X: D---------X----------C E A------M-------B We will show that X is the midpoint of CD. We start by noting that <MAE = <XCE and <AEM = < CEX. This shows that triangles AEM and CEX are similar, and that r = XE/ME is their ratio. In the same way we see that MEB and DEX are similar, with the same ratio r. But since AM = MB, this shows that r = DX/MB and also r = CX/AM = CX/MB, and thus DX = CX. Hence X must be the midpoint of DC, as desired. If you have more questions, just write back. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/
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