Take any quadrilateral, like this one

It sure looks like we’ve built a parallelogram, doesn’t it? The amazing fact here is that *no matter what quadrilateral you start with, you always get a parallelogram when you connect the midpoints*.

Some students asked me why this was true the other day. I had totally forgotten how to approach the problem, so I got the chance to play around with it fresh. I had two ideas of how to start. The first was to draw another line in the drawing and see if that helped.

Doesn’t it look like the blue line is parallel to the orange lines above and below it? If that were true, that would give us a powerful way forward. It also presages my second idea: try connecting the midpoints of a triangle rather than a quadrilateral.

Here’s what it looks like for an arbitrary triangle.

It sure looks like connecting those midpoints creates four congruent triangles, doesn’t it? In fact, that’s not too hard to prove. Once we know that, we can see that any pair of touching triangles forms a parallelogram. That means that we have the two blue lines below are parallel.

So we can conclude:

Lemma. The blue lines above are parallel.

Theorem. The orange shape above is a parallelogram.

See that the blue lines are parallel? The top line connects the midpoints of a triangle, so we can apply our lemma!

But the same holds true for the bottom line and the middle line as well! So all the blue lines below must be parallel.

The same holds true for the orange lines, by the same argument.

So the quadrilateral is a parallelogram after all!

I found this quite a pretty line of argument: drawing in the lines from opposite corners turns the unfathomable into the (hopefully) obvious. That resolution from confusion to clarity is, for me, one of the greatest joys of doing math.

The next question is whether we can break the result by pushing back on the initial setup. Does our result hold, for example, when the quadrilateral isn’t convex?

Looks like it will still hold. I’ll leave that one to you.

Here are a few more questions to consider:

- How does the area of the parallelogram you get by connecting the midpoints of the quadrilateral relate to the original quadrilateral?
- There is a hexagon where, when you connect the midpoints of its sides, you get a hexagon with a larger area than you started with. Can you find a hexagon with this property?
- Can you find a hexagon such that, when you connect the midpoints of its sides, you get a quadrilateral?

## Comments 10

Well explained, Dan.

This really looks like a fairly advanced concept, including as it does formal geometry and proofs. My focus is in elementary-middle years, where we want students to explore and discover without having to get to the really formal side of math. Your post does that really nicely: You can intuitively feel that the new shape is a parallelogram, and go a step further to thinking logically about it, without going the whole hog and proving it. This is

sucha good discussion and investigation starter.What Peter said :-). My geometry kids “discover” this on Geometer’s Sketchpad, but they are done exploring too quickly. So what Peter said is so important, that we can expose kids to a lot of rich problems at an early age, without expecting the “formal” conjectures. Thanks for the great post, Dan.

Wonderfully elegant and simple. A student and I were perplexed for the last week on this problem and finally we had to give in and look it up. She was a bit disappointed that it wasn’t complex and confusing, and upset with herself that she could figure it out. I told her some of the most amazing things in life are right in front of us and they should be cherished for their simplicity.

Another simple proof (though, admittedly, appealing to somewhat more advanced concepts) is via the center of mass. Place unit masses in the vertices of the original quadrilateral

ABCD. Then its center of massM(the average ofA,B,C, andDtaken as vectors) is the average of the two midpoints ofABandCD; and also the average of the two midpoints ofBCandAD. Thus, the two diagonals of the new quadrilateral are bisected by their intersection point (M), Q.E.D.Nice! My first thought was just to bash it out with vectors, but this is much nicer.

can you prove that the area remaining after the formation of parallelogram is equal to the area of the parallelogram?Please send the proof to my email.

Did you also know that the four triangles formed by connecting the four midpoints of the quadrilateral fit into the quadrilateral. I don’t know the proof of this but if you should take a piece of cardboard about a sixteenth of an inch thick, the four triangles will fit into the parallelogram if you are very accurate.

But sir, it may be rectangle because of here it is not clear that angle formed by sides is not 90 degree?

Author

In this context, we’re considering a rectangle as a special kind of parallelogram.

I have a really irrelevant math teacher who literally can’t teach, and math especially is the kind of subject that has to be explained. I finally understand this proof! I’m glad I found this website. Thank you!