Midpoints of a quadrilateral form a parallelogram

Take any quadrilateral, like this oneQuadrilateral

then mark the midpoints, and connect them up.
Quad with inner quad

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.

This is the kind of result that seems both random and astonishing. You have to draw a few quadrilaterals just to convince yourself that it even seems to hold. How do you go about proving it in general?

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.

Blue Line

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.

Arbitrary triangleIt 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.

Blue Lines are parallel

So we can conclude:
Lemma. The blue lines above are parallel.

Quad with inner quadTheorem. The orange shape above is a parallelogram.

Proof. Draw in that blue line again. Blue Line

We have the same situation as in the triangle picture from above! Can you see it? Let’s erase the bottom half of the picture, and make the lines that are parallel the same color:

Bottom half erased

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.

All blue lines are parallel

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

Orange lines are parallel too

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?Nonconvex shapes too?

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

Here are a few more questions to consider:

  1. How does the area of the parallelogram you get by connecting the midpoints of the quadrilateral relate to the original quadrilateral?
  2. 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?
  3. Can you find a hexagon such that, when you connect the midpoints of its sides, you get a quadrilateral?

Comments 11

  1. Peter Price

    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 such a good discussion and investigation starter.

  2. Fawn Nguyen

    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.

  3. John

    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.

  4. Andrei K

    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 mass M (the average of A, B, C, and D taken as vectors) is the average of the two midpoints of AB and CD; and also the average of the two midpoints of BC and AD. Thus, the two diagonals of the new quadrilateral are bisected by their intersection point (M), Q.E.D.

  5. Vaishnav Dilip

    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.

  7. Vishal Tiwari

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

    1. Post
  8. János

    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!

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