## 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)

### 5 Responses to 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.

• 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.

• luckytoilet says:

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?