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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.


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:



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   
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry

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