I’m a little bit puzzled by geometrical proofs, like the common algebraic proof for the Pythagorean theorem listed Wikipedia's "Pythagorean theorem" entry.

I understand the idea of arranging the right triangles and the area $c^2$ in a neat way to form another square and writing the area of the new square in different terms and going from there.

But I’m a little confused on how you actually know that the right triangle you draw isn’t just the one special right triangle with which you can actually form such a square.

How do we know this way of rearranging the pieces works for any other right triangle?

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    $\begingroup$ You got two excellent answers. Note too that we take as axiomatic that all right angles are equal, and that these "presto" kinds of visual proofs are just one of many ways to show/prove this theorem. You can do entirely synthetic geometric proofs that have nothing to do with rearrangements, and which specify nothing about the sizes of the triangles. $\endgroup$ Feb 28 at 11:53
  • $\begingroup$ Interesting, I'll check those out too. Thanks for the info. $\endgroup$
    – Name
    Feb 28 at 22:00
  • $\begingroup$ Here's an animated diagram I created using Python, a decade or so ago newtonexcelbach.files.wordpress.com/2012/02/pythanim1a.gif $\endgroup$
    – PM 2Ring
    Feb 28 at 23:12
  • $\begingroup$ How about finding a right angled triangle such that the rearrangement proofs do not apply to that triangle? $\endgroup$ Feb 29 at 7:29
  • $\begingroup$ These proofs-by-pictures are meant to suggest an idea, not to be completely rigorous themselves. They suggest a construction that can then be carried out rigorously using the axioms of geometry. For the details, you might see for example John Lee's "Axiomatic Geometry". $\endgroup$ Mar 1 at 16:30

4 Answers 4


But I’m a little confused on how you actually know that the right triangle you draw isn’t just the one special right triangle with which you can actually form such a square.

If I am not mistaken, this is the figure you are referring to.

enter image description here

Ask yourself, what is it about this right triangle that is not applicable to every right triangle possible? We are not making any unwarranted assumptions about its side lengths or angles, right? We are not saying something like - one side has a length $4$ units or one acute angle measures $40^{\circ}$. So undoubtedly, this isn't "just one special right triangle".

As for being able to rearrange the triangles into a square, again, imagine any right triangle. Can you make four copies of it? Of course, you can. Can you arrange the four triangles as shown in the figures below? sure, why not?

Perhaps you are not sure if the corners would make right angles. Well, they have to. If you observe any corner made by the triangles, it's either the right angle itself, or the sum of the two acute angles of the right triangle. And remember, the sum of the acute angles in a right triangle is $90^{\circ}$.

Also, the quadrilaterals formed have to be squares. Each angle is $90^{\circ}$ and the sides are all equal in length ($a, b, a+b, \text{ or } c$depending on the square).

Response to OP's comment

How can we know that we can place the $c^2$ square in the middle for any right triangles. It seems a bit hand-wavy due to the square fitting in there while being all slanted like that.

We know the quadrilateral formed in the center (right figure) is a square because the sides are all equal ($c$) and the angles are $90^{\circ}$ (straight angle minus the two acute angles = $180^{\circ} - 90^{\circ}$). Again, this is independent of the shape and size of the right triangle.

Also, as ilkkachu says in the comments the squares are all formed as a result of us arranging the triangles. We are not trying to fit any squares.

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    $\begingroup$ Ok so now I see I should have been a bit more specific. I’m referring to the right side of the figure. How can we know that we can place the $c^2$ square in the middle for any right triangles. It seems a bit hand-wavy due to the square fitting in there while being all slanted like that. Since we’re drawing this, aren't we making assumptions about how whether or not we could do this for any right triangle we draw? $\endgroup$
    – Name
    Feb 28 at 8:24
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    $\begingroup$ @Name, note that in both cases, the sides of the big square are a+b, and c is just the hypotenuse of the triangles. It's not as much $c^2$ "fitting in there", but that putting four copies of the triangle in that arrangement in the corners of the big square just forms the $c^2$ square in the middle. $\endgroup$
    – ilkkachu
    Feb 28 at 17:53
  • $\begingroup$ @Name I believe your doubts have already been addressed. Still, edited the answer to include my response to your comment. $\endgroup$
    – Haris
    Feb 28 at 20:36
  • $\begingroup$ Thanks for the additional answers, but I want to clarify I wasn't really thinking about fitting the bigger square in there. I was trying to ask about how we know the resulting square in the middle would always be a square with area $c^2$ even if we made $a$ bigger and wouldn't, for example, become a parallelogram or rhombus or something. $\endgroup$
    – Name
    Feb 28 at 21:48
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    $\begingroup$ @Name There are several reasons. For example, it is clearly has four sides (so a quadrilateral), with each side equal to $c$ (so a rhombus) and by rotational symmetry each angle equal (so these must be right angles) so it is a square. $\endgroup$
    – Henry
    Feb 29 at 18:22

Somewhat more briefly, look at the detail of the angles in the second picture. I've labelled the angles as alpha and beta.

Pythagorean figure angles detail

In the light blue triangle, since a triangle has 180 degrees and it has a top-left right angle, the sum of alpha and beta must be 90 degrees. Now looking at the joint on top, since the symmetric alpha and beta also add up to 90 degrees, and the straight line segment represents 180, then the internal white angle must be 90. So the white figure has right angles all around and it's a square.

  • $\begingroup$ That’s also a good point. Thank you for this concise answer. $\endgroup$
    – Name
    Feb 28 at 17:00

Consider a right triangle with legs of arbitrary length $a,b$. Let $c$ denote the length of the hypotenuse. Without loss of generality, suppose $b \geq a$. Then orient this triangle such that its vertices are $V_1 = \{(-a,0), (0,0), (0,b)\}$. Note that a second triangle with vertices $V_2 = \{ (-a,0),(b-a,0),(b-a,-a)\}$ would be congruent (as can be seen by checking the difference in corresponding coordinates). Similarly, $V_3 = \{ (b-a,-a),(b-a,b-a),(b,b-a)\}$ and $V_4 = \{ (b,b-a),(0,b-a),(0,b)\}$ are also congruent. This arrangement of the four congruent triangles ensures that the square of side length $b-a$ with vertices $S_{int}=\{(0,0),(b-a,0),(b-a,b-a),(0,b-a)\}$, which degenerates to a point in the case that $b=a$, does not overlap with any triangle. Moreover, all four triangles fall within the square of side length $c$ with vertices $S_{ext}=\{(-a,0),(b-a,-a),(b,b-a),(0,b)\}$. You can look at the coordinates of the vertices to convince yourself of this. Critically, this construction holds for arbitrary $a,b$. Notice how the interior square of side length $b-a$ is guaranteed to exist (unless it degenerates to a point when $b=a$) by the very way we laid out the triangles. It is the construction itself that gives this guarantee. I included a diagram for when $a=\sqrt{2},b=\pi$ below as example. Construction with <span class=$a=\sqrt{2}, b= \pi$" />

  • $\begingroup$ Alright, that answers my question. I Thank you both for helping me out. $\endgroup$
    – Name
    Feb 28 at 8:50
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    $\begingroup$ This answer contains a small mistake. We cannot assume that $b > a$ without loss of generality, as it excludes the very possible case $a = b$. This is fixed by assuming $b \geq a$, which can actually be done without loss of generality. In this case, the square $S_{int}$ vanishes. $\endgroup$ Feb 29 at 11:46
  • $\begingroup$ @HermanRohrbach, thank you for reminding me to address this case. I have edited my answer to include what happens when $b=a$. $\endgroup$ Feb 29 at 17:06

Others have explained why it is a square, but I feel like there's a deeper confusion in the question.

When someone gives a "proof by picture" such as the ones shown here, the point is not to say

Look at the picture, the thing in the middle looks like a square so it must be a square.


Look at the picture, I say this is a square so it must be a square, trust me.

A proof by picture is not intended to be a formal proof. The point is to quickly communicate the idea of the proof to someone who is assumed to recognize some basic geometry concepts that the proof uses and then be able to go throught the necessary steps to convince themselves of the theorem that is being proven.

The same rearrangement proof could be given in just words and mathematical symbols without any pictures, it would just be waaaaaaay more cumbersome to read and understand.

For example, to give the same proof with basic geometry concepts, one could write something like: (No need to read this carefully, I didn't even check closely whether all my notation is right because I'm demonstrating this is hard to read. And I'm skipping some details on the angle arrangements.)

Take a square $ABCD$ with side length $a+b$.

  • Let $E$ be the point in $AB$ such that $|AE|=a$ and $|EB|=b$.
  • Let $F$ be the point in $BC$ such that $|BF|=a$ and $|FC|=b$.
  • Let $G$ be the point in $CD$ such that $|CG|=a$ and $|GD|=b$.
  • Let $H$ be the point in $DA$ such that $|DH|=a$ and $|HA|=b$.

Now $AEH$, $BFE$, $CGF$, and $DHG$ are all congruent: The side $AE$ is of length $a$, the side $AH$ is of length $b$ and the angle $\angle EAH$ is a right angle. (Repeat for the 3 other triangles.) So we can say that $|HE|=|EF|=|FG|=|GH|=c$

Now we can prove that $\angle HEF$ is a right angle as follows:

  • Angles $\angle AEH$ and $\angle BEF$ are the two non-right angles of a right-angled triangle, so their sum is $90^\circ$.
  • But also $\angle AEH + \angle HEF + \angle BEF = \angle AEB = 180^\circ $ (because $E$ is in the segment $AB$)
  • So $\angle HEF = 90 ^\circ$.

With similar arguments, we can show that $\angle HEF$, $\angle EFG$, $\angle FGH$ and $\angle GHE$ are all right angles, and we know that $|HE|=|EF|=|FG|=|GH|$ because the congruence of the triangles we showed above. So $EFGH$ is in fact a square.

So in essence the idea of the "proof by picture" is to make the reader think something like:

Ahh so we have four congruent triangles like that. And two non-right angles of the triangle meet at those points so this angle is therefore a right angle. So all of the four angles are right angles, ahh so that is indeed a square.

But such a thought process is hard to communicate exactly without using so many symbols that it becomes a chore to read. When someone who is well-versed in the basics of geometry sees something like the image, they will quickly understand the concept of the proof - they might even have a thought process similar to what I described above but without the need to go to details of so many symbols and written arguments.

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    $\begingroup$ So this is actually the level of rigor I was wondering about initially. But I didn’t know whether or not such a proof exists since I’ve never seen it before. Now I see why. $\endgroup$
    – Name
    Mar 1 at 7:12

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