Today we learned about Pythagoras' theorem. Sadly, I can't understand the logic behind it.

$A^{2} + B^{2} = C^{2}$

enter image description here

$C^{2} = (5 \text{ cm})^2 + (7 \text{ cm})^2$
$C^{2} = 25 \text{ cm}^2 + 49 \text{ cm}^2$
$C^{2} = 74 \text{ cm}^2$

${x} = +\sqrt{74} \text{ cm}$

Why does the area of a square with a side of $5$ cm + the area of a square with a side of $7$ cm always equal to the missing side's length squared?

I asked my teacher but she's clueless and said Pythagoras' theorem had nothing to do with squares.

However, I know it does because this formula has to somehow make sense. Otherwise, it wouldn't exist.

  • 65
    $\begingroup$ Your teacher is wrong to make that generalization. The square of a positive number is geometrically the area of a square with that corresponding side length. $\endgroup$
    – David P
    Commented Feb 13, 2014 at 20:46
  • 165
    $\begingroup$ If at all possible, find a better teacher. The theorem is "A squared plus B squared equals C squared" and that has nothing to do with squares? Where does your teacher believe the expression "squared" came from? $\endgroup$ Commented Feb 14, 2014 at 0:17
  • 13
    $\begingroup$ ViHart explains it beautifully as usual. youtube.com/watch?v=z6lL83wl31E After watching this I suggest you watch everything she's ever made. You will look at mathematics through a whole new light. $\endgroup$
    – Cruncher
    Commented Feb 14, 2014 at 15:17
  • 43
    $\begingroup$ I like the animation at i.sstatic.net/VqNSK.gif $\endgroup$
    – j08691
    Commented Feb 14, 2014 at 16:27
  • 58
    $\begingroup$ "If at all possible, find a better teacher" - You shouldn't reinforce the OP's disrespect of his/her teacher based on a single quote you didn't hear directly. $\endgroup$
    – Stefan
    Commented Feb 15, 2014 at 21:56

16 Answers 16


Ponder this image and you will see why your intuition about what is going on is correct:

enter image description here

  • 43
    $\begingroup$ This is all correct and this is how one can prove PT, with one caveat: This proof assumes that we have a well-developed notion of area. Proving (from Euclid's axioms) that area for convex polygons satisfies the property that area of subdivided polygon is the sum of areas of pieces (so that the area of rectangle is the product of its sides) is harder than it looks. There is a more elementary but less intuitive proof of PT using similarity of triangles. $\endgroup$ Commented Feb 13, 2014 at 20:53
  • 11
    $\begingroup$ I really like the simplicity of this answer $\endgroup$
    – nl-x
    Commented Feb 13, 2014 at 21:12
  • 1
    $\begingroup$ @studiosus Indeed! Euclid proves this by splitting the squares on the legs into parts that fill the square on the hypothenuse. So no “sums of areas”, just equidecomposability. $\endgroup$
    – egreg
    Commented Feb 13, 2014 at 22:12
  • 2
    $\begingroup$ I have a moving version at se16.info/hgb/pyth.htm $\endgroup$
    – Henry
    Commented Feb 14, 2014 at 13:45
  • 5
    $\begingroup$ @fish.frog: This is an excellent question. There are several ways of doing this: Few involve "cheating" in the sense that they first introduce coordinates on the Euclidean plane (which is done using Pythagorean formula!). An alterantive way to define area and to verify that it has all the required properties (without using coordinates) can be found in the book by E. Moise "Elementary geometry from an advanced standpoint". It takes Moise about 10 pages! This book also discusses several proofs of Pythagorean formula (some using areas and one using similarities as mentioned above). $\endgroup$ Commented Feb 14, 2014 at 19:56

It is not quite important that the shapes of the figures you put on the edges of your right triangle are squares. They can be any figure, since the area of a figure goes with the square of its side. So you can put pentagons, stars, or a camel's shape... In the following picture the area of the large pentagon is the sum of the areas of the two smaller ones:

I'm am not able to rescale Camels with geogebra... so here are pentagons

But you can also put a triangle similar to the original one, and you can put it inside instead of outside. So here comes the proof of Pitagora's Theorem.

enter image description here

Let $ABC$ be your triangle with a right angle in $C$. Let $H$ be the projection of $C$ onto $AB$. You can easily notice that $ABC$, $ACH$ and $CBH$ are similar. And they are the triangles constructed inside the edges of $ABC$. Clearly the area of $ABC$ is the sum of the areas of $ACH$ and $CBH$, so the theorem is proven.

In my opionion this is the proof which gives the essence of the theorem.

  • 6
    $\begingroup$ @rschwieb : Since the three pentagons are similar to each other, the area of each pentagon is proportional to the square of its side. $\endgroup$ Commented Feb 14, 2014 at 17:49
  • 43
    $\begingroup$ Can you show an example with a Camel's shape? $\endgroup$
    – typ1232
    Commented Feb 14, 2014 at 21:10
  • 10
    $\begingroup$ Given how the image of the Pythagorean triangle with squares attached to its edges is seen as the quintessential symbol of the universality of mathematics, this first image of yours is pretty mind-blowing. I even once saw a suggestion of planting trees in the shape of a Pythagorean triangle to signal our intelligence to aliens. Maybe they'd just be like "WTF? LOL!" and scram ;) $\endgroup$
    – Christian
    Commented Feb 14, 2014 at 22:55
  • 17
    $\begingroup$ I wish I could upvote this twice. It's perfect. $\endgroup$
    – MJD
    Commented Feb 15, 2014 at 2:50
  • 7
    $\begingroup$ @typ1232 I had difficulties in copying and rescaling a camel's shape with geogebra. But maybe the picture with pentagons is better in view of Christian comment. $\endgroup$ Commented Feb 15, 2014 at 7:45

Bulberage's diagram is a common way to visualize it. I like this one too. Make four copies of your right triangle and arrange them like this: enter image description here

The total area of the square is $C^{2} $

The area of the white square is $(A-B)^{2} = A^{2}-2AB+B^{2}$. (Note that it doesn't matter if $A-B$ is negative because we're squaring it.)

The area of the four triangles is $2AB$.

The total area of the outer square is equal to the four triangles plus the inner square:

$C^{2} = A^{2} - 2AB + B^{2} + 2AB $

and we're done.


This is an extended addendum to all the neat proofs listed in the other answers to the post (and prompted by a reference request from @fish.frog):

There is no free lunch: All the proofs which use the notion of area (unlike the one via similarities) depend on the following

Theorem. There exists a function $A$ (usually called "area") on the set of all planar polygonal regions which takes values in ${\mathbb R}_+$, and which satisfies the following properties:

  1. The function is invariant under congruences, that is, if $P, Q$ are congruent polygonal regions then $A(P)=A(Q)$.

  2. The function $A$ is additive in the sense that if we subdivide a polygonal region $P$ in finitely many non-overlapping polygonal regions $P_1,...,P_n$ then $$ A(P)=\sum_{i=1}^n A(P_i). $$

  3. If $P$ is a rectangle with the base $a$ and height $b$, then $A(P)=ab$.

  4. Edit: the following is not needed, it follows from 1, 2 and 3 as noted by Hagen von Eitzen: $A(kP)=k^2A(P)$ where $kP$ is the dilation if $P$ by $k$.

Furthermore, if one has a well-developed notion of a limit (which Euclid did not have) then 3 can be replaced by a single normalization condition:

3'. $A(Q)=1$, where $Q$ is a unit square.

Existence of such function is not at all obvious (to a mathematician; anybody else would say that it is completely obvious, but this site is dedicated to mathematics). The shortest proof I know (using only axioms of Euclidean geometry) can be found in the book

E. Moise, "Elementary Geometry from an Advanced Standpoint" (3rd Edition).

The proof is long (about 10 pages) and rather tedious. (Somebody has to pay for a free lunch!)

Edit.: Proofs of existence of the area function for planar polygonal regions that one might see in a typical textbook or a class go as follows:

a. Compute area of triangle from the area of rectangles (Property 3). No attempt is made to show that different way to do so result in the same value of area of triangles.

b. Triangulate a polygonal region $P$ (in the convex case). No attempt is made to do this in the nonconvex case. (That's OK if one reduces attention only to convex polygons.) Proof of existence of a triangulation in the nonconvex case requires few clever but not difficult ideas. Archimedes could have done it (and maybe did!).

c. End of the story: No attempt is made to show that the result is independent of a triangulation, i.e., that the constructed area function is well-defined.

(This is not meant as a criticism of the common treatments of the notion of area, since there is only so much time and space one can devote to this topic, but to show where the problems arise and why a rigorous treatment is needed.)

Moise does a very thorough job here. His proof of c is very modern, it depends in the set of ideas used to prove "Schoenflies theorem" for simple closed polygonal curves in the plane: that every such curve bounds a planar polygon (that's Jordan's separation theorem in this context) which is homeomorphic to a convex polygon via a piecewise-linear homeomorphism. (Moise was a topologist.) I do not believe anybody could have done a proof that area is well-defined rigorously before, say, middle of the 19th century (Riemann integration or, alternatively, appearance of topology as a field).

By the way, Moise also gives several proofs of Pythagorean formula: Two using areas and one using similarities.

There are alternative ways to prove the theorem on existence of the area function, but, from what I know, they all require introduction of Cartesian coordinates on the Euclidean plane, which, in turn, depend on the Pythagorean formula. (One such approach is via Riemann or Lebesgue integration, developing which takes much more than 10 pages; another proof uses determinants.)

  • 7
    $\begingroup$ Well, Euclid assumed property 1 in his Common Notion 4 (things which coincide with one another equal one another), and he took property 2 for granted, but he gave quite a rigorous proof of property 3 in Book VI Proposition 14, which states that if you have two rectangles with dimensions $l_1,w_1$ and $l_2,w_2$ respectively, then the areas are equal if and only if the ratio of $l_1$ to $w_2$ is equal to the ratio of $l_2$ to $w_1$ (i.e $ l_1w_1=l_2w_2$ in modern notation). This is done by using the Eudoxian theory of proportions, which inspired the Dedekind cut construction of the real numbers. $\endgroup$ Commented Feb 15, 2014 at 5:06
  • 4
    $\begingroup$ And the remarkable thing was that Euclid did this without even having the modern notion of real numbers. He just used ratios of magnitudes, with equality of ratios defined by Eudoxus' theory of proportions: in modern language, two ratios are equal if and only if the same rational numbers are less than, greater than, and equal to them. $\endgroup$ Commented Feb 15, 2014 at 5:10
  • $\begingroup$ @studiosus You gave me an answer I was badly waiting for: math.stackexchange.com/questions/653264/… . Perhaps you can rewrite it for my question, and I will accept it. $\endgroup$
    – Sawarnik
    Commented Feb 15, 2014 at 5:13
  • 2
    $\begingroup$ @Keshav: Property 2 is the really hard one, this is where the bulk of the work is. $\endgroup$ Commented Feb 15, 2014 at 5:29
  • 3
    $\begingroup$ Here is the proof from Moise's book, if anyone's interested: tinyurl.com/moiseproof $\endgroup$ Commented Feb 15, 2014 at 16:37

Surely no set of answers to this question is complete without mentioning Euclid's proof?

Diagram for Euclid's proof

In this figure, the right angle is at A, and AL is perpendicular to BC.

Euclid shows a little more than just the Pythagorean theorem; namely, that the two pink regions have equal areas, as do the two blue regions.

First, observe that triangle FBC can be rotated about B to become triangle ABD. This follows from side-angle-side (FB=AB, angle FBC = angle ABD, BC = BD). So these two triangles are congruent.

(Aside: That rotation is a perfect 90 degrees, so we have just shown in passing that segments CF and DA are (a) equal and (b) perpendicular. Strictly speaking, this diversion is unnecessary... But getting sidetracked is one of the joys of Euclidean geometry, in my opinion. I only wish had I learned this in school. End aside.)

Next, observe that both the pink square (FBAG) and the triangle FBC have base FB and height BA. So the pink square has twice the area of the triangle.

Next, observe that both the pink rectangle (BDLK) and the triangle ABD have base BD and height DL. So the pink rectangle also has twice the area of the triangle.

Since the triangles are equal, the pink square has the same area as the pink rectangle. A symmetric argument holds for the blue square and blue rectangle, which completes the proof.

Of course, as @studiosus points out, this does assume that the blue and pink rectangles add up to the square of BC...


I like Garfield's proof the best. By the way, do you know that he became the president of the US five years after giving this proof! I like because it uses simple geometry and no rearrangement stuff, that others here have resorted to. Here it goes:

enter image description here

What I have done here is constructed a trapezium $ADCE$, such that $AD=a+b$, $AC = a$ and $DE = b$. Now we place a point $B$ on $AD$, such that $AB= b$ and $BD = a$. So, we can see by $SAS$ criteria, that triangles $ABC$ and $BDE$ are congruent.

This gives us $BC = BE = c$. Similarly we can show that, $\angle CBE = 180^{\circ} -(\angle ABC + \angle DBE) = 90^{\circ}$. Now we come to the crux of the proof, equating the area of the trapezoid and the individual triangles.

We know the area of the trapezoid, using the area formula of the trapezoid, is $\frac{(a+b)^2}{2}$. You can visualize it by extending the sides [as shown in the diagram], and observing that it becomes a square of side $a+b$, and seeing that the two trapezoids formed are congruent. Now:

$$\begin{align} (ACDE) & = (ABC) + (BCE) + (BDE) \\ \frac{(a+b)^2}{2} & = \frac{1}{2}ab + \frac{1}{2}ab + \frac{1}{2}c^2 \\ (a+b)^2 & = ab + ab +c^2 \\ a^2+b^2+2ab & = 2ab + c^2 \\ a^2+b^2 & = c^2 \end{align}$$

Got it! And Bhaskara's proof, as suggested by Free, is worth as look as well. You might be surprised to learn that this theorem has the most number of proofs for any theorem in mathematics! You can see 100 of them here: http://www.cut-the-knot.org/pythagoras/index.shtml .

  • 5
    $\begingroup$ How easy is it to find the area of the trapzoid without pythagoras theorem though ? $\endgroup$
    – C4stor
    Commented Feb 14, 2014 at 7:26
  • 1
    $\begingroup$ @C4stor: can't you use (area) = (base) $\times$ (height)? A trapezoid consists of a rectangle and a right triangle (which is half a rectangle). $\endgroup$ Commented Feb 14, 2014 at 10:10
  • $\begingroup$ @C4stor Have you thought about it? You just need area of a triangle. In this case, drop a perpendicular from C to DE, and sum the area of the two parts. $\endgroup$
    – Sawarnik
    Commented Feb 14, 2014 at 10:11
  • $\begingroup$ @C4stor And if you wonder about how to get area of the triangle ($\frac{1}{2}bh$), that follows directly from the area of a square. Then derive the area of rectangle, then the parallelogram, and then the triangle. $\endgroup$
    – Sawarnik
    Commented Feb 14, 2014 at 10:35
  • 4
    $\begingroup$ I used to like this proof before I realized it was the top proof with the square cut in half. $\endgroup$ Commented Feb 14, 2014 at 22:57

Similar Triangle Approach

Here is the simplest proof I have seen that doesn't use areas. It just uses similar triangles. However, what is most intuitive is a subjective decision.

By similar triangles in the diagram below, we have $$ \large\frac{c_a}{a}=\frac ac\qquad\normalsize\text{and}\large\qquad\frac{c_b}{b}=\frac bc $$

$\hspace{3.5cm}$enter image description here

Thus, $$ \large c=c_a+c_b=a\frac ac+b\frac bc $$ Therefore, $$ \large c^2=a^2+b^2 $$

Ratio of Areas Approach

Consider a right triangle whose hypotenuse is coincident with the side of a square. Let the legs of the triangle be $a$ and $b$ and the hypotenuse, $c$. The ratio of the area of the triangle to that of the square is $$ \frac{ab}{2c^2}=\frac12\frac ac\frac bc $$ Since $\frac ac$ and $\frac bc$ are the same for all similar triangles, and the red, green, and purple triangles below are similar, the ratios of the areas of each triangle to the areas of each corresponding square are equal.

$\hspace{3.5cm}$enter image description here

since the area of the purple triangle is the sum of the areas of the exterior red and green triangles, the area of the purple square is the sum of the areas of the red and green squares.

  • $\begingroup$ Hi Rob. This proof certainly does avoid the concept of area, but I suppose that behind the scenes, one must first prove that the axioms of Euclidean geometry imply the existence of a length function on the set of line segments, so that congruent line segments are assigned the same length. After this is done, one can proceed with the similarity-based argument above to obtain the Pythagorean Theorem. $\endgroup$ Commented Sep 1, 2016 at 1:48
  • $\begingroup$ I am saying this because many people I know take the Pythagorean Distance Formula for granted and so think that the Pythagorean Theorem is self-evident: “As the length of $ c $ is $ \sqrt{a^{2} + b^{2}} $ by definition, then isn’t this already the Pythagorean Theorem?” The point is, one must be careful to explain how a length is assigned to a line segment. $\endgroup$ Commented Sep 1, 2016 at 1:59
  • $\begingroup$ I like this proof so I upvoted it. I know people sometimes write mixed fractions where you add the integer and the fraction part instead of multiplying them but I could figure out that if you instead mean multiplication, then your answer gives a proof of the Pythagoren theorem assuming certain properties of distance. I'm not sure the extra confusing bit at the end is needed after you already gave the proof, but I also realize that some people are not satisfied with just seeing a proof because they don't find the proof very intuitive and want to know more about what's going on. For example, $\endgroup$
    – Timothy
    Commented Jun 27, 2019 at 20:01
  • $\begingroup$ some people might find some proofs of a statement in Naive set theory intuitive and other proofs in Naive set theory not very intuitive because that way of thinking is what ends up working to have them not accept proofs in Naive set theory of an absurd result. I wrote an answer at math.stackexchange.com/questions/895076/… and it did more than just give a proof of the existence of irrational numbers and also explained more of what was going on and then ended up with an upvote and maybe that's the reason. That answer was slightly different at $\endgroup$
    – Timothy
    Commented Jun 27, 2019 at 20:06
  • $\begingroup$ the time it got the upvote. Maybe you also wanted to show using the assumption that shapes scaled identically have the area get multiplied by the same number that the the properties of distance you assumed when writing the proof by similar triangles are consistent, that those assumptions are also consistent with the additional assumption the area of any shape varies as the square of its size. I was able to follow through with understanding that proof. I don't see a problem. Those who like the first part of your answer a lot don't have to finish reading the rest of the answer if they $\endgroup$
    – Timothy
    Commented Jun 27, 2019 at 20:15

I would have posted this as a comment, but I can't yet; so please forgive me!

I strongly suggest following and watching the user Vi Hart on YouTube, she has many fascinating videos on math, and a couple about Geometry with specific relevance to squaring and why the Greeks invented a lot a math based on the shapes of objects; they used the areas of shapes to work out the answers.

In a very literal sense, whilst we know Pythagoras's theory as $A^2+B^2=C^2$; he wouldn't have squared a number on a page; he would have drawn a triangle and extend each of the faces of the triangle into a square and add up their respective areas.

I hope this helps!

Do check out Vi Hart though, she's fascinating and your teacher might learn something too. :)

  • $\begingroup$ Her actual youtube username is 'vihart' but I'm sure Google will work either way. $\endgroup$
    – Cryogen
    Commented Feb 14, 2014 at 2:27
  • 2
    $\begingroup$ is there a specific video of hers about the pitagorean theorem ? $\endgroup$
    – josinalvo
    Commented Feb 14, 2014 at 16:25
  • $\begingroup$ a link would be nice =P $\endgroup$
    – josinalvo
    Commented Feb 14, 2014 at 16:26
  • 3
    $\begingroup$ youtube.com/watch?v=z6lL83wl31E $\endgroup$
    – josinalvo
    Commented Feb 14, 2014 at 16:34

You can profitably use Eli Maor, The Pythagorean theorem A 4,000-year history (2007) about the

well over four hundred proofs of it [today] known.

It is interesting to read the Ch.2 : Pythagoras [page 17-on]: it is not very clear what is the real contribution of Pythagoas itself to the question, due to the paucity of information rlated to his historical personality, but we can surely assert that the Pythagorean theorem is a milestone of ancient Greek mathematics and geometry.

One of greek "reflection" about it has the following aspect :

Of particular interest to the Pythagoreans were figurative numbers [... they are obtained as patterns of physical objects, like pebbles, that show special properties; among them square numbers, like $4=2 \times 2$ and $9 = 3 \times 3$]. Such explorations led the Pythagoreans to develop a kind of primitive algebra, based on the interrelation of various figures. For example, the familiar formula $(a + b)^2 = a^2 + 2ab + b^2$ can be proved geometrically by considering a square of side $(a + b)$, as shown in figure [see Bulberage's figure]. We can dissect this square into two smaller squares of areas $a^2$ and $b^2$ and two rectangles of areas $a \times b$ and $b \times a$. But the two rectangles are congruent, so their areas are equal. Thus the total area of the dissected parts is $a^2 + 2ab + b^2$, and this is equal to the area of the original square, $(a + b)^2$.

We may inspect the figures [see Bulberage's answer] and notes that the LH is a big square (call it $SQ$) with inside two squares (of sides $a$ and $b$ respectively) and four congruent triangles (of sides $a$ and $b$).

With the algebraic expression of the area we have that :

$SQ = a^2 + b^2 +4 \times ab/2$.

The RH figure has the same square $SQ$ withe the four triangles "rearranged": what remains is an inside square (call it $sq$).

Again we have .

$SQ = sq + 4 \times ab/2$.

With algebraic manipulations, we get :

$a^2 + b^2 = sq$

where the square $sq$ is precisely the square constructed on the hypotenuse of the triangle of sides $a$ and $b$, i.e. a formulation of the Pythagorean theorem.

Of course, ancient Greeks did not know about algebraic manipulations; but a "proof by inspection" on the two figures shows that, starting from two equal squares, and subtracting from them the same "quantities" (the four triangles), by an easy application of Euclid's Elements Common notion 3 : "If equals are subtracted from equals, then the remainders are equal", what remains must be equal; i.e.

the square constructed on the hypotenuse of the triangle of sides $a$ and $b$ is equal to the sum of the squares constructed on the sides $a$ and $b$ respectively.

In conclusion, starting from squares (numbers), we ended with squares (geometrical shapes) and we have historical support to the statement that the Pythagorean theorem :

was also about squares, and not only about triangles.

  • 4
    $\begingroup$ Although Answers that mainly rely on references are certain allowed, it might be difficult for a casual Reader to follow up with the little you've provided here. Perhaps, if you have the time, you could elaborate on at least one Proof that differs from what others have already detailed? $\endgroup$
    – hardmath
    Commented Feb 15, 2014 at 14:33

You may not realize it but you have two very different questions here. One is about proof, a lock-tight convincing argument removing all doubt. And then there is one about intuition, what gives you a quick good idea why.

The first, about proof, can be as dull and uninformative as following all the details of long division (e.g. prove that there is a remainder of 1 when dividing 1111 by 37). The second, about intuition, is more informative, but not necessarily lock-tight, say comparing lengths in two pictures.

For the proof of pythagorean's theorem itself, there are many examples of that around:online, Euclid's I.47... in fact at least 101 proofs. So I won't try to repeat them.

One might be able to get 'intuition' by looking at one proof over and over, following each step, understanding not just the truth of each step but also the direction it is going in (one might call this 'why each step is made'). Or you might think 'intuition' is a really easy proof (many of which have been given here).

Also, 'intuition' is just a good guess (looking at pictures and guessing that because two lengths are the same in the picture they're the same for any change in the picture), and it only becomes a reliable guess when backed by understanding well the mechanical details (that's a encouragement not to rely solely on intuition because proofs are too 'hard'). Intuition is not always geometric or visual (e.g. in algebra, your intuition may tell you that certain things will cancel out easily without having to expand a multiplication fully).

Now to actual content about the intuition of Pythagorean's theorem (where it will become more clear what that means). Really, the point of the theorem is that something is special about right triangles. One knows (intuitively) that the legs can be any lengths, but whatever they are the hypotenuse is then defined (that might seem obvious) and there is a simple formula for it.

The weird part is that the formula involves squaring and square roots, right? And it's also weird that it involves (relatedly) going from distances to areas (actual squares) which are very different things (despite the fact that we use numbers for both).

The closest to a 'why' proof is one that is actually more general; that is, that similar (in the technical sense) figures on the feet of a right triangle add to the similar figure on the hypotenuse.

And finally, after a lot of talk, the closest to an actual not-in-quotes intuition for the traditional Pythagorean theorem is to look at 3 examples. One example is the isosceles right triangle. If you construct some examples like the one here, you simple add tiles together and see that the the squares on the feet obviously add up to that on the hypotenuse. The second example is a bit degenerate, suppose one foot is extremely small, then the square on the other foot is almost identical to the square on the hypotenuse (and so as the small foot goes to zero, the sum must approach equality (if they weren't already equal).

Now you take -any- example in between (not equal feet, not one foot zero). If the sum of squares equals the square of the hypotenuse it means the as you move from isosceles to a sliver, at three points the equation holds. Then intuitively it would just be strange for it not to hold at every point in between. (it wouldn't have been strange for it to hold at the two extremes but in between to be off). And since there do exist Pythagorean triples ($3^2 + 4^2 = 5^2$) most likely the Pythagorean theorem holds for all right triangles. Note that this isn't an air-tight proof because maybe it could hold on just those Pythagorean triple but maybe nowhere else).

So that's the intuition: the idea holds for two extremes and then at least one in between, so it probably holds for everything in between.

  • $\begingroup$ Your "degenerate" case is very unsatisfying, IMO, since it applies to any exponent and not just 2. $\endgroup$
    – Nemo
    Commented Aug 5, 2019 at 3:13
  • $\begingroup$ @Nemo Good point. Then the 'sliver' example would work as an anchor (where the equality holds) for a proof for any exponent. It's just that the examples on the other side (isosceles) don't work. The intuition still needs something on the sliver side. $\endgroup$
    – Mitch
    Commented Aug 5, 2019 at 12:29

I think this excellent sequence of answers to the OP's interesting question needs one more observation: the Pythagorean Theorem is logically equivalent to Euclid's Parallel Postulate.

All the proofs above depend on that postulate, either explicitly (e.g. drawing parallel lines with useful properties) or implicitly ( e.g. relying on arguments about similar figures).

The theorem is false in spherical or hyperbolic geometry.

@TomCollinge said as much in a comment above.

See http://www.cut-the-knot.org/triangle/pythpar/PTimpliesPP.shtml


A great question, and lots of interesting answers! Mauro ALLEGRANZA's answer that the ancient Greeks played with pebbles (we get our word "calculation" from the Greek word for a pebble) is a good one - that gave them a feeling for the way squares could add up. Like for instance the way that 12² plus 5² would equal 13²: two squares sum to a squareEmanuele Paolini's thoughts about how any shape will do, not just a square, and that way of dividing a triangle he shows, and which robjon elaborates - those seem the likely routes to me. More thoughts here.

  • $\begingroup$ Actually the way Greek mathematicians dealt with quantities like areas and lengths has little or nothing to see with pebbles. Btw "calculus" is a Latin word, not Greek $\endgroup$ Commented Mar 2, 2017 at 21:53
  • $\begingroup$ Aristotle describes the Pythagorean Eurytus as one who "decided what was the number of a thing... by imitating its form with pebbles. as some bring numbers into the shapes of triangle and square." Metaphysics 1092b10-14 Calculate, from from Latin calculus "small stone, pebble" (used in reckoning), dim. of calx, calc- "limestone," from Greekk. khalix "small pebble," kakhlex "round pebble" $\endgroup$
    – Simon G
    Commented Mar 7, 2017 at 20:58
  • $\begingroup$ Yes, but I was not referring to Aristotle's authority, which is almost zero in mathematics and in physics; rather, to the great Greek scientific tradition (consider instead Euclid's or Archimedes' geometric proofs). Also, and maybe this is a minor point, I think it is more safe to say the Greek χάλιξ is affine to the Latin calx, in the sense they share a common root, rather than saying one derives from the other. Besides, χάλιξ has no connection with computations, AFAIK. Note that e.g. the English word chalk has also the same root (and we even use chalk for computations :) ) $\endgroup$ Commented Mar 9, 2017 at 18:45
  • $\begingroup$ Well, @Pietro , an ancient Greek would make use of an abax, a sand tray. en.wikipedia.org/wiki/Sand_table Aristophanes' comedy Wasps, staged in 422 BC mentions this kind of counting: "And not with pebbles precisely ranged, but roughly thus on your fingers count the tribute paid by the subject states, and just consider its whole amount; and then, in addition to this, compute the many taxes and one-per-cents, the fees and the fines, and the silver mines, the markets and harbours and sales and rents." $\endgroup$
    – Simon G
    Commented Mar 10, 2017 at 20:45
  • $\begingroup$ Aristotle was probably referring to the Pythagoreans interest in figurate numbers, such as the triangle numbers, including the tetractys. $\endgroup$
    – Simon G
    Commented Mar 10, 2017 at 20:49

Many people seem to confuse proof with explanation. Proof proves; it does not explain. If you think about it you can see that regardless of the shape of a triangle, the lengths of any two sides together with the angle between them determine the length of the third side, i.e., the side opposite the angle. Draw a triangle and you'll see for yourself that this is true. Now it happens that the relationship between these four quantities is known. When you study trigonometry you'll see it. It's called the law of cosines. It is a peculiarity of that relationship that when the angle is ninety degrees (pi over 2 radians), the term in that relationship that involves the angle becomes zero, i.e., the opposite side is ONLY related to the lengths of the other two sides. That is the relationship called the Pythagorean theorem. (Pythagorus didn't know about the law of cosines. He (or whoever) came by the discovery a different way.) This is what explains it to me. Does this help? Of course this leaves the law of cosines to be explained but that's a different question. Mathematics is loaded to the gills with questions. Good for you for asking. Keep it up. We need much more explanation and much less proof. Proof is important to specialists (professional mathematicians) and most of us are not specialists.


The accepted answer provides no explanation for the diagrams: enter image description here

Each diagram shows an outer square, $\color{red}{(a+b)}$ units on a side, divided into other figures. The diagrams suggest different expressions for the area of the outer square. Equating these expressions leads to the equation $a^2+b^2=c^2$.

Total area of $\color{red}{outer}$ square = Area not covered by triangles + Area covered by triangles

$\color{red}{(a+b)^2} = c^2 + 4(\dfrac1{2}ab)$ ________$\color{red}{(a+b)^2} = a^2 +b^2 + 4(\dfrac1{2}ab)$

$c^2 + 4(\dfrac1{2}ab)= a^2 +b^2 + 4(\dfrac1{2}ab)$

$\therefore c^2 = a^2+b^2$

source: Algebra Structure and Method Book $1$ by Brown, Dolciani et al.

Here is a visual aid to help the new learner move from one diagram to the other: enter image description here


enter image description here

  • 4
    $\begingroup$ That's well more complicated than necessary. Isn't it obvious that the green areas are the same on both sides, and the orange areas are the same on both sides, and so the grey areas must also be equal? $\endgroup$
    – user856
    Commented Jul 28, 2018 at 5:33
  • $\begingroup$ You do have a good point. I once independently thought of the same concern as this answer addresses but it doesn't add anything to the answer at math.stackexchange.com/questions/675522/… so I downvoted it. I actually independently thought of a proof like the one in that answer because it seemed a bit more straight forward. $\endgroup$
    – Timothy
    Commented Jun 27, 2019 at 20:27

Consider a right triangle ABC with hypotenuse AB (which I'll call c) and legs CB (which I'll call a) and AC (which I'll call b). If we accept that the right angle at C means that vectors $\vec{CA}$ and $\vec{CB}$ are perpendicular and that $\vec{CA}$ and $\vec{CB}$ are perpendicular implies that $\vec{CA}\cdot\vec{CB}=0$ (so $\vec{CB}\cdot\vec{CA}=0$ by symmetry of the inner product on $\mathbb{R}^n$ with the standard metric) and that for any vector $\vec{v}$, the norm squared of $\vec{v}$ is given by $|\vec{v}|^2=\vec{v}\cdot\vec{v}$, then since vectors add tail to heads, $\vec{CA}=\vec{CB}+\vec{BA}$ implies

$\vec{CA}\cdot\vec{CA}=(\vec{CB}+\vec{BA})\cdot(\vec{CB}+\vec{BA})$ $=\vec{CB}\cdot\vec{CB}+\vec{CB}\cdot\vec{BA}+\vec{BA}\cdot\vec{CB}+\vec{BA}\cdot\vec{BA}$ by linearity of the inner product

gives $c^2 = |\vec{CA}|^2 = |\vec{CB}|^2 + 0 + 0 + |\vec{BA}|^2 = a^2 + b^2$.

I say all of this because to me, the intuition behind Pythagoras' theorem is "forces which act in perpendicular directions can be considered separately in physics," or "normal forces don't interact."


I had the same struggle, so I tried to animate the theorem:
Pythagoras with Parallelograms.

The only thing you need to understand is that shearing squares and rectangles into parallelograms, preserves their area. Hope you like it.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .