Midline in Trapezoid

In a trapezoid, a midline (or a midsegment) is the line joining the midpoints of the sides.

In a trapezoid, the midline is parallel to the bases and its length is half their sum. Conversely, the line joining points on the two sides of a rapezoid, parallel to its bases and half as long is their sum is the midline.

A. Bogomolny, 19 January 2015, Created with GeoGebra

Proof

Assume $AD\,$ is the smaller of the two bases. Let $AB\,$ and $CD\,$ meet at $E.\,$ In $\Delta BCE,\,$ $AD\parallel BC\,$ so that, by Thales' Theorem,

$\displaystyle\frac{AE}{DE} = \frac{AB}{CD} = \frac{AB/2}{CD/2}.\,$

If $M'\,$ is the midpoint of $AB\,$ and $N'\,$ that of $CD\,$, then the above implies $\displaystyle\frac{AM'}{AE} = \frac{DN'}{DE},\,$ from which, in turn, it follows that $\displaystyle \frac{EM'}{AE} = \frac{EN'}{DE}.\,$ With another reference to Thales' Theorem, $M'N'\parallel AD.$

Thus there are three parallel lines and three similar triangles which supply several proportions. Of interest for us are the following:

$\displaystyle\frac{EB}{EM'} = \frac{BC}{M'N'}\,$

and

$\displaystyle\frac{EA}{EM'} = \frac{AD}{M'N'}\,$,

which add up to

$\displaystyle\frac{EB + EA}{EM'} = \frac{BC + AD}{M'N'}\,$

but $EB+EA=2EM'\,$ such that finally $2M'N'=BC+AD.\,$

Reversing the steps yields the converse. Indeed, $2M'N'=BC+AD\,$ implies $EB+EA=2EM',\,$ meaning that $M'\,$ is the midpoint of $AB.\,$ $N'\,$ is similarly shown to be the midpoint of $CD.\,$


The midline of a trapezoid is directly related to the midlines of triangles formed by its diagonals. In part, if $M\,$ and $N\,$ are the points of intersection of $M'N'\,$ with $AC\,$ and $BD,\,$ then $M\,$ and $N\,$ are the midpoints of $AC\,$ and $BD,\,$ respectively. In addition,

$\displaystyle M'N = MN' = \frac{AD}{2}\,$

and

$\displaystyle MM' = NN' = \frac{BC}{2}\,$.

In a trapezoid, the line joining the midpoints $M\,$ and $N\,$ of the diagonals is the midline because the latter passes through these points. I mention that with a problem for young mathematicians offered at a Moscow Math Olympiad in 1957 in mind. It is considered on a separate page.


Related material
Read more...

  • Thales' Theorem
  • Midline in Triangle
  • Midline in Quadrilateral
  • Bimedians in a Quadrilateral
  • Midline in Similar Triangles
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