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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.

Thanks.
```

```
Date: 03/04/2002 at 10:04:07
From: Doctor Floor
Subject: Re: Geometry

Hi, Leonard,

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

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