Most of the proofs of Pythagorean Theorem that I see all seem to involve the concept of area, which to me does not seem "trivial" to prove.

Others show proof for a particular triangle but it does not seem clear to me if it works for all right triangles or just specific variants.

Is there a proof that is purely algebraic based on algebraic triangle constraints? Or one that does not rely on area at least and works for any arbitrary right triangle?

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    $\begingroup$ "Algebraic Triangle Constraints" Do you mean the triangle inequality, where a+b>c? If this is all, I'm not sure this gets you the pythagorean theorem. The theorem is married to Euclidean space and fails in metric spaces that aren't Euclidean (e.g. spherical triangles). $\endgroup$ Apr 1, 2018 at 22:19
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    $\begingroup$ If you're not happy about area, are you happy about length? If you are, what is the distinction? $\endgroup$ Apr 1, 2018 at 22:19
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    $\begingroup$ Take your pick from the list here. $\endgroup$
    – dxiv
    Apr 1, 2018 at 22:19
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    $\begingroup$ @user525966 I do not hold out much hope for your question. Note that the correspondence between the arithmetical numbers and geometric quantities like lengths and areas does not come for free. You have to specify what the correspondence is, and what algebraic quantities correspond to which geometrical quantities, as you have implicitly done already with length and real numbers (in your constraints). In order to be able to state Pythagoras, you must assign a geometric meaning to the algebraic quantity $a^{2}$. If not area, then what? $\endgroup$ Apr 1, 2018 at 22:31
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    $\begingroup$ I am disturbed by the vector proof, because the definition of the length of a vector presumes Pythagoras. $\endgroup$ Apr 2, 2018 at 6:52

4 Answers 4


I realize this question is old, but I wonder if the OP would have been OK with the following:

Let $\triangle ABC$ be a right triangle with $\angle ACB$ the right angle. Drop an altitude from $C$ to $\overline{AB}$ at $D$. Then $\triangle DCA$ and $\triangle DBC$ are both right triangles and similar to $\triangle ABC$.

enter image description here

By similarity,

$$ \frac{AD}{AC} = \frac{AC}{AB} $$

and hence

$$ AD = \frac{AC^2}{AB} $$

Similarly (!),

$$ \frac{DB}{BC} = \frac{BC}{AB} $$

and therefore

$$ DB = \frac{BC^2}{AB} $$


$$ AB = AD+DB = \frac{AC^2}{AB}+\frac{BC^2}{AB} $$

leading directly to

$$ AB^2 = AC^2 + BC^2 $$

as desired.

I'm not sure this avoids things that are fundamentally equivalent to assuming area, but perhaps it would have been satisfactory to the OP?

  • $\begingroup$ Similarity use Thales theorem (Intercept theorem), Thales theorem use Areas. $\endgroup$ Jan 17, 2021 at 3:08
  • $\begingroup$ @MikhailIvanov: I suspect if you dig deep enough, there's no way around it. That doesn't mean it wouldn't be acceptable to the OP, though. $\endgroup$
    – Brian Tung
    Jan 17, 2021 at 18:26
  • $\begingroup$ @MikhailIvanov i disagree. you just need the Parallel Postulate, for this similarity $\endgroup$
    – symbiont
    Jul 11, 2021 at 9:20
  • $\begingroup$ @symbiont How Parallel Postulate can say something about ratios of sides? $\endgroup$ Jul 13, 2021 at 10:18
  • $\begingroup$ @MikhailIvanov maybe it's not the Parallel Postulate. it is possible to extend a few lines to find all angles (not sure what this is called). i expect that the angles, can then be related to ratios of sides. so, without using area $\endgroup$
    – symbiont
    Jul 20, 2021 at 20:07

Actually there are many concepts of area, some of them just involving additivity, some of them involving $\sigma$-additivity and completeness. Taking as a reference this recent answer of mine, all of them agree on the following facts:

  1. the area/measure of a rectangle in the plane is $\text{base}\times\text{height}$;
  2. isometric measurable sets have the same measure;
  3. if $A,B$ are measurable and almost-disjoint (meaning that $A\cap B$ is empty or it is just a polygonal path) the measure of $A\cup B$ is the sum of the measures of $A$ and $B$.

In particular, all of them agree on the fact that the area of a right triangle (i.e. half a rectangle) is half the product of the lengths of the legs. So there is no issue in using any naive concept of area for proving the Pythagorean theorem, which is usually done by decomposing a square in a smaller square and four isometric right triangles, or by similar approaches by dissection (they just exploit 3.).

Anyway, if the unusual appeals to you, you may just prove that the classical definitions of $\sin$ and $\cos$ match with the definition of $\sin$ and $\cos$ as the imaginary/real parts of the complex exponential function (Euler-De Moivre's formula), then prove the Pythagorean theorem in the form $\sin^2\theta+\cos^2\theta=1$ through $e^{z}\cdot e^{w}=e^{w+z}$, see here.

On the other hand, as already pointed out in the comments, you already need completeness to define what a length actually is, so it is kind of artificial to want to avoid completeness for dealing with measures in geometry.

  • $\begingroup$ I had forgotten that the Euler formula could be used there... that makes this a lot easier since that doesn't require me to use area, although I think it does require me to have defined real exponentiation $\endgroup$
    – user525966
    Apr 2, 2018 at 0:04
  • $\begingroup$ @user525966: it requires to prove that $e^z$ (for $z\in\mathbb{C}$) defined through $$ e^z=\sum_{n\geq 0}\frac{z^n}{n!}$$ fulfills $e^{z}\cdot e^{w}=e^{z+w}$, which is a nice and classical exercise about the Cauchy product of series. $\endgroup$ Apr 2, 2018 at 0:08
  • $\begingroup$ For that one could go through Taylor series correct? $\endgroup$
    – user525966
    Apr 2, 2018 at 0:10
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    $\begingroup$ @user525966: or just through differential equations. Given the series definition we have $\frac{d}{dz}e^z=e^z$ and $e^0=1$, so both sides of $e^z\cdot e^w=e^{z+w}$ are solutions of the same Cauchy problem, and they actually are the same function. $\endgroup$ Apr 2, 2018 at 0:15
  • $\begingroup$ Unfortunately I know absolutely 0 about diffeqs but I'll keep that in mind when I get there -- thanks! $\endgroup$
    – user525966
    Apr 2, 2018 at 0:21

Use differential calculus as according to Wikipedia:

The triangle $ABC$ is a right triangle, as shown in the upper part of the diagram, with $BC$ the hypotenuse. At the same time the triangle lengths are measured as shown, with the hypotenuse of length $y$, the side $AC$ of length $x$ and the side $AB$ of length $a$, as seen in the lower diagram part: https://en.wikipedia.org/wiki/File:Pythag_differential_proof.

If $x$ is increased by a small amount $dx$ by extending the side $AC$ slightly to $D$, then $y$ also increases by $dy$. These form two sides of a triangle, $CDE$, which (with $E$ chosen so $CE$ is perpendicular to the hypotenuse) is a right triangle approximately similar to $ABC$. Therefore, the ratios of their sides must be the same, that is:

${\displaystyle {\frac {dy}{dx}}={\frac {x}{y}}.} {\displaystyle {\frac {dy}{dx}}={\frac {x}{y}}.}$ This can be rewritten as ${\displaystyle y\,dy=x\,dx} {\displaystyle y\,dy=x\,dx}$ , which is a differential equation that can be solved by direct integration:

${\displaystyle \int y\,dy=\int x\,dx\,,} {\displaystyle \int y\,dy=\int x\,dx\,,}$ giving

${\displaystyle y^{2}=x^{2}+C}$ ${\displaystyle y^{2}=x^{2}+C.}$ The constant can be deduced from $x = 0$, $y = a$ to give the equation

${\displaystyle y^{2}=x^{2}+a^{2}.}$


I got this from https://en.wikipedia.org/wiki/Pythagorean_theorem#Proof_using_differentials.

  • $\begingroup$ why was I downvoted? $\endgroup$
    – KingLogic
    Apr 1, 2018 at 22:38
  • $\begingroup$ I'm not the one who downvoted, but I guess that the reason was that the OP explicitly said he doesn't feel comfortable with the concept of area, so how do you think (s)he feels about integrals and differentials? $\endgroup$ Apr 1, 2018 at 22:40
  • $\begingroup$ The concept of integrals involves limits of Riemann sums, which is an area estimate. $\endgroup$ Apr 1, 2018 at 23:01

I think the best answer that adds something to the already existing answers is pretty much the same as this answer so I will write a similar answer here.

Without any assumptions about what properties distance follows, you can't prove anything about what the distance formula is. We could make the assumptions that distance is a binary function from $\mathbb{R}^2$ to $\mathbb{R}$, in otherwords, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfying the following properties

  1. For any points $(x, y)$ and $(z, w)$ in $\mathbb{R}^2$, $d((0, 0), (z, w)) = d((x, y), (x + z, y + w))$
  2. $\forall x \in \mathbb{R}\forall z \in \mathbb{R}^+d((0, 0), (z\cos(x) ,z\sin(x))) = z$

Then we could derive the distance formula as follows. $\cos$ and $\sin$ are defined by the following differential equations.

  • $\cos(0) = 1$
  • $\sin(0) = 0$
  • $\cos' = -\sin$
  • $\sin' = \cos$

$\frac{d}{dx}\cos^2(x) + \sin^2(x) = \frac{d}{dx}\cos^2(x) + \frac{d}{dx}\sin^2(x) = 2\cos(x)(-\sin(x)) + 2\sin(x)\cos(x) = 0$ so $\cos^2(x) + \sin^2(x)$ is constant. Also $\cos^2(0) + \sin^2(0) = 1$ so $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$. From this and those two assumptions, we can show that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$. This just shows that for any right angle triangle whose legs are parallel to the axes, the Pythagoren theorem holds. To prove the Pythagorean theorem holds for all right angle triangles, we have to show that distance also satisfies the following property.

  • For any points $(x, y)$ and $(z, w)$ in $\mathbb{R}^2$, $d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

That can be done as follows. $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$

Now how do we show that there actually exists a way of defining distance that satisfies the assumptions I made? Because it's trivial to show that the function $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ actually satisfies those properties.

Some people make other assumptions about what properties distance follows. Here are some assumptions about distance in $\mathbb{R}$ each of which some people make.

  1. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  2. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (z, w))$ is nonnegative
  3. $\forall \text{ nonnegative } x \in \mathbb{R}d((0, 0), (x, 0)) = x$
  4. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, -y)) = d((0, 0), (x, y))$
  5. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$
  6. The area of any square is the square of the length of its edges
  7. $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$

How do we know there exists a way of defining distance that satisfies all 7 properties? Because it has been proven in this answer that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying the first 5 properties from this list and it also satisfies properties 6 and 7 from this list.

The second assumption I made earlier does not appear as one of them. That's because using properties 3, 5, and 7, we can deduce that that definition of distance satisfies the second assumption I made earlier.


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