Theorems from theorems
I often describe mathematics to people as the process of building up a collection of tools, which can then be used by others to do mathematics, build up more tools, and eventually solve real-world problems (although I tend to leave that last bit to applied mathematicians). It’s lovely to see an example of when a neat result can be connected to an existing fact, and I recently spotted a nice example I thought I’d share.
Varignon’s theorem is a result about quadrilaterals named after mathematician Pierre Varignon, which he proved in 1731 – the proof was discovered after his death, among his notes and published as a part of Elemens de Mathematique, a book collecting Varignon’s work.
The theorem states that if you find the midpoints of each side of any quadrilateral, and join them to make a four-sided shape, that shape will be a parallelogram. That is, opposite pairs of sides on the shape will be the same length as each other, and parallel.
You can probably picture this, and look at the example here to convince yourself it’s possible for such a parallelogram to exist.
But the theorem states that this will work for absolutely any quadrilateral, no matter how irregular – not just convex ones, but also concave quadrilaterals (where part of the shape points inwards), crossed quadrilaterals which have a twist across the middle, and even skew (non-planar) quadrilaterals in 3D space will still have a nice flat parallelogram sitting across their edges’ midpoints.
If you’d like to play with this a little, there’s an interactive Desmos sheet you can use, in which the internal shape changes as you drag the corners around – but it’s always a parallelogram. If you’re still not satisfied, building your own version of this in Desmos or Geogebra might be a fun diversion.
This theorem is so neat and intriguing that it was named by maths teacher Patrick Honner as his own personal favourite, when interviewed for the My Favorite theorem podcast. As part of the discussion, he explains how he uses the theorem in class with his high school students, since there are some really accessible ways to prove it, and doing so can demonstrate to students the satisfaction and elegance of a simple proof.
How to prove it
The proof given on the Wikipedia page for the theorem uses the similarity of triangles – the small triangle in the top left must be similar (having equal angles and sides in proportion) to the larger one that covers half the quadrilateral. This is because the triangles share a common angle in the top left, and two sides of the smaller one are by definition half the length of the equivalent sides on the larger one, as they’re formed from their midpoints.
This can then be used to show that the two marked angles are equal, and hence that the edges of the parallelogram are each parallel to the corresponding diagonal of the quadrilateral. If this is repeated using different sets of triangles within the shape, the same method can be used to show the other pair of sides are parallel too.
While it seems fairly self-contained, this proof actually relies on several other known mathematical results – the side-angle-side criterion used to show that the two triangles are similar, and it could also be considered to use the property of corresponding angles, also known as F-angles – that two lines branching off a third at the same angle must be parallel (and vice versa).
Another way to prove the theorem, shared in a blog post by Patrick Honner, is to use the existing known fact, sometimes called the triangle midsegment theorem, that the segment between the midpoints of two sides of a triangle is parallel to the third side (and half its length, although this fact isn’t needed). Considering the quadrilateral as two back-to-back triangles gives the same result as before.
There’s even a proof using vectors, shared by mathematician Steven Strogatz on Twitter, which considers each side of the quadrilateral as a vector. Adding together two adjacent sides of the quadrilateral will give a vector that points to the far corner, so the sum of these two must be equal to the sum of the other two.
Then the midpoints of the sides can be found by considering half of each of the vectors, and the proof follows from there – since a + b = c + d, then ½(a+b) = ½(c+d), which means the vectors between the midpoints must be parallel. Even though this proof seems to stand alone, it relies on properties of vector addition that have been stated and proved previously by others.
You might be wondering how Varignon proved this theorem. The original proof is available (in French) and follows along similar lines to the ones we’ve seen – still using an existing piece of maths, and this time referring all the way back to the beginnings of geometry, quoting Euclid’s Elements, proposition VI.2.
This is a version of the triangle midsegment theorem and states that if a line intersects the sides of a triangle proportionally – that is, the same proportion of the way along each side – then it must be parallel to the base of the triangle. Varignon uses this to show that opposite pairs of sides of the internal shape are parallel to the diagonals of the quadrilateral, and hence to each other.
Even though much of this feels like stating the same obvious facts in very slightly different ways, it shows that there are many approaches to proving any given statement in maths, and it pleases me deeply that each relies on other pieces of maths – sometimes stretching as far back as ancient Greece – to complete the result. Each tiny step taken by a mathematician adds to our knowledge in ways that will allow future mathematicians to prove new and exciting results – by standing on the (parallel) shoulders of giants.