Yesterday in geometry students were analyzing proofs of the Pythagorean Theorem using this lesson by The Shell Foundation. It ties in nicely with my favorite proof which they discuss while watching this video (there’s something about doing proofs to classical music which delights me).

At the end of class, a student came up to me and asked if there was a theorem which says the following:

I responded that I had no idea if it was a theorem, but if he could prove it, then it would become a theorem and be named after him.

It sure SEEMS like it’s true.

But then again, there’s this:

And the above diagram actually satisfies his initial question which was if you have a square, and you draw 4 congruent segments inside the square which intersect the sides of the square, will the resulting figure also be a square?

I can’t wait for the day I get to name a theorem in my class after a kid :)
Btw, in your drawing, ABCD is most certainly not a square! I think I figured out what you meant, though — Square ABCD with point E on AB, F on BC, G on CD, and H on DA?
Then yes, EFGH will be a square. We already have that the segments are congruent, so now we just need to see that the inner angles are right. Take a look at the angles with vertex at (e.g.) point G. Since HDG and GCF form right triangles, and the three angles adjacent at point G are supplementary, we get the angle HBF is right.
The same argument holds for each angle, so yes we do have a square here!

One thing I’ve noticed is that in some way I’m not sure I understand fully, being aware of how you know things, or why you know things, is at least as importing as being aware of what you think you know.

In other words, proof make you question what you know and think about how you know it. That quality is noticeably lacking in many people who walk around believing false “knowledge”.

In response to Andrew, I think you somehow would need to prove that HDG and GCF are *congruent* right triangles to know that the relevant angles add up to 90 degrees.

Sybilla: I am so excited! My student wasn’t able to solve it (and nor was I), so we were letting it hang out there in hopes someone in the class or I would come up with some new approaches to it. Very cool! My class has just recently started looking at properties of special quadrilaterals, so this will make for some exciting conversations next week. Thanks!

I have not read Sybilla’s solution. Here’s what I see in your and your student’s diagrams…

Your student had four vertices: H, B, D and F. You have five. If you force (1) four vertices (and thus a quadrilateral), and (2) congruent segments, it’s going to be a square. I feel that symmetry is going to be important to use, but I haven’t worked on it. Lovely question; thanks for sharing it.

I recommend you this free video on the Pythagorean Theorem, especially in the end when different geometric proofs are shown: http://www.youtube.com/watch?v=Pqy9TlNWxFg (German language)

I can’t wait for the day I get to name a theorem in my class after a kid :)

Btw, in your drawing, ABCD is most certainly not a square! I think I figured out what you meant, though — Square ABCD with point E on AB, F on BC, G on CD, and H on DA?

Then yes, EFGH will be a square. We already have that the segments are congruent, so now we just need to see that the inner angles are right. Take a look at the angles with vertex at (e.g.) point G. Since HDG and GCF form right triangles, and the three angles adjacent at point G are supplementary, we get the angle HBF is right.

The same argument holds for each angle, so yes we do have a square here!

One thing I’ve noticed is that in some way I’m not sure I understand fully, being aware of how you know things, or why you know things, is at least as importing as being aware of what you think you know.

In other words, proof make you question what you know and think about how you know it. That quality is noticeably lacking in many people who walk around believing false “knowledge”.

In response to Andrew, I think you somehow would need to prove that HDG and GCF are *congruent* right triangles to know that the relevant angles add up to 90 degrees.

Wow, what a great question! I agree with what Joshua Zucker said, that Andrew’s argument is not complete.

I posted about this question at the Mathematics Teaching Community at https://mathematicsteachingcommunity.math.uga.edu

I posted a solution to the question here https://mathematicsteachingcommunity.math.uga.edu/index.php/282/is-a-rhombus-inscribed-in-a-square-necessarily-a-square at the Mathematics Teaching Community

Sybilla: I am so excited! My student wasn’t able to solve it (and nor was I), so we were letting it hang out there in hopes someone in the class or I would come up with some new approaches to it. Very cool! My class has just recently started looking at properties of special quadrilaterals, so this will make for some exciting conversations next week. Thanks!

I have not read Sybilla’s solution. Here’s what I see in your and your student’s diagrams…

Your student had four vertices: H, B, D and F. You have five. If you force (1) four vertices (and thus a quadrilateral), and (2) congruent segments, it’s going to be a square. I feel that symmetry is going to be important to use, but I haven’t worked on it. Lovely question; thanks for sharing it.

I recommend you this free video on the Pythagorean Theorem, especially in the end when different geometric proofs are shown: http://www.youtube.com/watch?v=Pqy9TlNWxFg (German language)

The programs used in the video can be found here: http://www.echteinfach.tv/trigonometrie/rechtwinklige-dreiecke-und-satz-des-pythagoras#p