Monday Math 156

For a trapezoid, the median (also known as the midline or mid-segment) is the segment connecting the midpoints of the legs of the trapezoid. This post presents a proof that:

  1. the median of a trapezoid is parallel to the bases;
  2. the length of the median is half the sum of the lengths of the bases;
  3. the midpoints of the diagonals of a trapezoid also lie on its midline.

The key to all of these is the triangle midline theorem.

Consider the trapezoid ABCD with bases AB and CD. Let M and N be the midpoints of the legs AD and BC, respectively, and let M′ be the midpoint of the diagonal BD.


Then, by the midline theorem with regards to ∆ABD, \overline{MM\prime}\parallel\overline{AB}, and MM′AB. Applying the midline theorem to ∆BCD, \overline{M\prime{N}}\parallel\overline{CD}, and M′NCD.

Now, since \overline{CD}\parallel\overline{AB}, we see \overline{M\prime{N}}\parallel\overline{AB}.

Since \stackrel{\longleftrightarrow}{MM'} and \stackrel{\longleftrightarrow}{M'N} are both parallel to the same line \stackrel{\longleftrightarrow}{AB}, and pass through the same point M′, then we see that by Playfair’s axiom, they must be the same line; M, M′ and N are collinear, and so the median \overline{MN}\parallel\overline{AB}\parallel\overline{CD}. Further, MN=MM′+M′NABCD =½(AB+CD).

Noting that we could perform the same procedure using diagonal AC and its midpoint N′, we see that the midpoints of both diagonals must lie on the median MN.


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