The Varignon Quadrilateral

The Varignon quadrilateral is a very interesting figure.

Take an arbitrary quadrilateral $ABCD$ (assume it is convex for now) and denote the midpoints of $AB$, $BC$, $CD$, and $AD$ as $K$, $L$, $M$, and $N$, respectively.

Surprisingly, the quadrilateral $KLMN$ is a parallelogram, no matter what quadrilateral $ABCD$ is! How cool is that??

Obviously, you're not just going to take my word for it (I hope), so we'll prove it here!

I'll give you some time to think about how to prove that $KLMN$ is a parallelogram.

Hint: You have to draw auxiliary lines. Think you know which one?

Draw $AC$. This is a really big giveaway, so see if you can finish it up yourself.

Recall this theorem:

If a line passes through the midpoints of two sides of a triangle, then it is parallel to the third side.

We can apply this theorem to triangles $ABC$ and $ADC$! Directly applying the theorem gives us $LK \parallel AC$ and $MN \parallel AC$. But since parallelism is transitive, this gives $LK \parallel MN$!

Now how can we finish up our problem?

You can now literally draw $BD$ and do the same exact thing to triangles $ABD$ and $BCD$, and you'll get that $LM \parallel KN$.

But two pairs of opposite parallel sides determine a parallelogram, so we're done! $\Box$

Another property of the Varignon parallelogram is that it has one-half the area of the big quadrilateral! Try to prove this yourself (note: you will have to draw in altitudes and auxiliary lines.)

For more information on the Varignon quadrilateral, see this link!

The Varignon quadrilateral of a given quadrilateral is the quadrilateral formed by the midpoints of the sides of the given quadrilateral. This quadrilateral has some notable properties.

Regardless of whether $ABCD$ is a planar or a skew quadrilateral, the Varignon quadrilateral is a parallelogram.

We may prove this using vectors. Let $M_1$, $M_2$, $M_3$, and $M_4$ be the midpoints of $AB$, $BC$, $CD$, and $DA$, respectively. Also, let $\vec{P}$ denote the position vector of a point $P$.

Using the midpoint formula for vectors, we have the following: $$\vec{M_1}=\frac{1}{2}(\vec{A}+\vec{B})$$ $$\vec{M_2}=\frac{1}{2}(\vec{B}+\vec{C})$$ $$\vec{M_3}=\frac{1}{2}(\vec{C}+\vec{D})$$ $$\vec{M_4}=\frac{1}{2}(\vec{D}+\vec{A})$$

Notice the following: $$\vec{M_1M_2}=\frac{1}{2}(\vec{C}-\vec{A})=\vec{M_4M_3}$$ $$\vec{M_1M_4}=\frac{1}{2}(\vec{D}-\vec{B})=\vec{M_2M_3}$$

This means that $M_1M_2\parallel M_4M_3$ and $M_1M_4\parallel M_2M_3$. Thus, $M_1M_2M_3M_4$ is a parallelogram, and the proof is complete.

The second notable property is that the Varignon quadrilateral has area equal to one half of the area of the original quadrilateral. Unlike the above property, this only holds if the quadrilateral is planar and simple, meaning that it does not cross itself.

For now, assume that the quadrilateral is convex. The concave case is essentially the same in most respects.

Consider the triangles $AM_4M_1$ and $ADB$. We know: $$AM_4=\frac{1}{2}AD$$ $$AM_1=\frac{1}{2}AB$$ Thus, we have the following area relation: $$[AM_4M_1]=\frac{1}{4}[ADB]$$ By applying a similar process at the other vertices, $$[BM_1M_2]=\frac{1}{4}[BAC]$$ $$[CM_2M_3]=\frac{1}{4}[CBD]$$ $$[DM_3M_4]=\frac{1}{4}[DCA]$$ Adding the four triangle areas together, $$[AM_4M_1]+[BM_1M_2]+[CM_2M_3]+[DM_3M_4]\\=\frac{1}{4}([ADB]+[BAC]+[CBD]+[DCA])$$ $$[AM_4M_1]+[BM_1M_2]+[CM_2M_3]+[DM_3M_4]\\=\frac{1}{4}([ABCD]+[ABCD])$$ $$[AM_4M_1]+[BM_1M_2]+[CM_2M_3]+[DM_3M_4]=\frac{1}{2}[ABCD]$$ The remaining area in the quadrilateral is the area of the Varignon quadrilateral. Therefore, we have the following: $$[M_1M_2M_3M_4]=[ABCD]-([AM_4M_1]+[BM_1M_2]+[CM_2M_3]+[DM_3M_4])$$ $$[M_1M_2M_3M_4]=\frac{1}{2}[ABCD]$$ Thus, we have proven that the area of the Varignon quadrilateral is one half that of the original quadrilateral.