Midpoints, midpoints, everywhere!

I didn’t encounter the Quadrilateral Midpoint Theorem (QMT) until I had been teaching a few years.  Following is a minor variation on my approach to the QMT this year plus a fun way I leveraged the result to introduce similarity.

In case you haven’t heard of it, the surprisingly lovely QMT says that if you connect, in order, the midpoints of the four sides of a quadrilateral–any quadrilateral–even if the quadrilateral is concave or if its sides cross–the resulting figure will always be a parallelogram.

Parallel1

Parallel2

Parallel3

This is a cool and easy property to explore on any dynamic geometry software package (GeoGebra, TI-Nspire, Cabri, …).

SKETCH OF THE TRADITIONAL PROOF:  The proof is often established through triangle similarity:  Whenever you connect the midpoints of two sides of a triangle, the resulting segment will be parallel to and half the length of the triangle’s third side.  Draw either diagonal in the quadrilateral to create two triangles.  Connecting the midpoints of the other two sides of each triangle creates two congruent parallel sides, so the quadrilateral connecting all four midpoints must be a parallelogram.

NEW APPROACH THIS YEAR:  I hadn’t yet led my class into similarity, but having just introduced coordinate proofs, I tried an approach I’d never used before.  I assigned a coordinate proof of the QMT.  I knew the traditional approach existed, but I wanted them to practice their new technique.  From a lab in December, they already knew the result of the QMT, but they hadn’t proved it.

PART I:  Let quadrilateral ABCD be defined by the points , A=(a,b), B=(c,d), C=(e,f),  and D=(g,h).  There are several ways to prove that the midpoints of ABCD are the vertices of a parallelogram.  Provide one such coordinate proof.

All groups quickly established the midpoints of the four sides:  AB_{mid}=\left( \frac{a+c}{2},\frac{b+d}{2} \right)BC_{mid}=\left( \frac{c+e}{2},\frac{d+f}{2} \right)CD_{mid}=\left( \frac{e+g}{2},\frac{f+h}{2} \right), and DA_{mid}=\left( \frac{g+a}{2},\frac{h+b}{2} \right).  From there, my students took three approaches to the final proof, each relying on a different sufficiency condition for parallelograms.

The most common was to show that opposite sides were parallel.  \displaystyle slope \left( AB_{mid} \text{ to } BC_{mid} \right) = \frac{\frac{a-e}{2}}{\frac{b-f}{2}}=\frac{a-e}{b-f} and \displaystyle slope \left( CD_{mid} \text{ to } DA_{mid} \right) =\frac{a-e}{b-f}, making those two midpoint segments parallel.  Likewise, \displaystyle slope \left( BC_{mid} \text{ to } CD_{mid} \right) = \displaystyle slope \left( DA_{mid} \text{ to } AB_{mid} \right) = \frac{c-g}{d-h}, proving the other opposite side pair also was parallel.  With both pairs of opposite sides parallel, the midpoint quadrilateral was necessarily a parallelogram.

I had two groups leverage the fact that the diagonals of parallelograms were mutually bisecting.    \displaystyle midpoint \left( AB_{mid} \text{ to } CD_{mid} \right) = \left( \frac{a+c+e+g}{4},\frac{b+d+f+h}{4}\right) = = midpoint \left( BC_{mid} \text{ to } DA_{mid} \right).  QED.

One student even proved that opposite sides were congruent.

While it was not readily available for my students this year, I can imagine allowing CAS for these manipulations if I use this activity in the future.

EXTENDING THE QMT TO SIMILARITY:  For the next stage, I asked my students to explains what happens when the QMT is applied to degenerate quadrilaterals.

PART II:  You could think of triangles as being degenerate quadrilaterals when two quadrilateral vertices coincide to make one side of the quadrilateral have side length 0.  Apply this to generic quadrilateral ABCD from above where points A and D coincide to create triangle BCD.  Use this to explain how the segment connecting the midpoints of any two sides of a triangle is related to the third side of the triangle.

I encourage you to construct this using a dynamic geometry package, but here’s the result.

Parallel4

 

Heres a brief video showing the quadrilateral going degenerate.

Notice the parallelogram still exists and forms two midpoint segments on the triangle (degenerate quadrilateral).  By parallelogram properties, each of these segments is parallel and congruent to the opposite side of the parallelogram, making them parallel to and half the length of the opposite side of the triangle.

CONCLUSION:  I think it critical to teach in a way that draws connections between ideas and units. This exercise made a lovely transition from quadrilaterals through coordinate proofs to the triangle midpoint theorem.

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3 responses to “Midpoints, midpoints, everywhere!

  1. Reblogged this on Singapore Maths Tuition and commented:
    Interesting post about Quadrilateral Midpoint Theorem (QMT), which states that if you connect, in order, the midpoints of the four sides of a quadrilateral–any quadrilateral–even if the quadrilateral is concave or if its sides cross–the resulting figure will always be a parallelogram.

  2. Neat post! Oddly enough, I learned this theorem in 9th grade but was
    never shown a proof of it. I did, however, figure out that the theorem
    could be applied to triangles (degenerate quads) even though that wasn’t
    taught. I jokingly called it the FK Theorem, since I didn’t see it in the
    book. Does it have an actual name?

    • No name at all for thinking of the triangle case as a degenerate quad maintaining all the parallelogram properties.

      As for the case working directly from triangles & similarity, it’s known by a few different names. The one I’ve seen most is the “Side-Splitter Theorem”.

      I encountered the degenerate case this past fall when I was toying around with general quadrilaterals & their midpoints on a dynamic geometry package. Trying to “shake the pattern apart”, I tried typical boundary approaches that usually break stuff like this. I started with convex quads and made them concave. No break. Then I tried crossing the sides of the quad to make a
      non-quadrilateral because polygons are typically defined as non-crossing sides. The QMT still held! (That was actually a pretty cool moment.) Then I tried squeezing one of the quad’s sides close to zero. I wasn’t actually thinking of degenerates at the time, but the closer I got to a zero side length, the live image more closely morphed into the triangle and inspiration struck. It was a great moment of discovery and connection.

      Moments like this convince me that math classes everywhere would definitely benefit from many more opportunities to play.

      As for a name, I’ve never seen this connection in print. From now on in my classes, it will be known as the “FK Theorem” in honor of your prior discovery!

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