# Varignon’s theorem proved in one line with vectors

Today I was reading some math book when the author mentions Varignon’s theorem, and gives a proof. The proof was not very long, but it was somewhat confusing. On Wikipedia, several more short proofs were given, but they were all more confusing than need be.

I remembered seeing the theorem proven using vector geometry before, but I couldn’t find the text (nor any other page / book that proves it this way) —

[image shamelessly taken from Wikipedia]

Varignon’s theorem states that in any quadrilateral, if we join the midpoints of the sides, then we get a parallelogram.

In the diagram, it suffices to prove that vector HG is equal to vector EF — vectors must have both the same orientation and length to be equal. This works since any method that proves HG = EF can also prove HE = GF. The proof goes as follows —

$\vec {HG} = \vec{HD} + \vec{DG} = \frac{1}{2} (\vec{AD} + \vec{DC}) = \frac{1}{2} \vec{AC} = \vec{EF}$

And we’re done. (the last step is due to symmetry of HG and EF)

## 7 thoughts on “Varignon’s theorem proved in one line with vectors”

1. Patrick says:

The prove is not correct, since:

HD+DG > HG.

This is the basic rule for triangle.

The easiest way to prove it is as follows:

1. Connect AC, we have triangle ACD. Since H is the middle point of AD and G is the middle point of CD, then:

HG // AC;

Similarly, we can prove:

EF // AC.

Thus: HG // EF

2. Connect BD, we can prove HG // CG in the same way.

So the EFHG must be a parallelogram.

1. Patrick says:

correction( my typing error):

2. Connect BD, we can prove HE // GF in the same way.

The other option will be to prove EF=HG and EH=GF.

still using triangle theorem.

2. Oh yea I suppose I should’ve been more clear on that. The proof deals with _vectors_, not line segments — so HD + DG would be the vector sum of the vectors HD and DG. (I’m also used to not putting arrows atop vectors xD)

2. Robin says:

This reminds me of the random discovery I found back in middle school – if you start with some random polygon (I’d say most polygons would work), and connect the midpoints of the sides to form a new polygon, and connect its midpoints to form yet a new one, and so on, and the shape of the polygon will approach an ellipse. I don’t know of a proof though.

3. Anonymous says:

what do u mean when u say connect, connect how?

4. Zzurrat says:

Where is the proof that 1/2 ac = ef?
You say that the last step is due to symmetry but that doesn’t mean much when you assume the thing it is asking you to prove. Hence the proof is invalid.

1. We proved HG = 1/2 AC. By the exact same argument you can prove EF = 1/2 AC.