## 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.