On the original proof of the Pythagorean theorem in Euclid's "Elements"


Without any doubt the Pythagorean theorem is the most famous theorem in geometry! Nowadays we think about it as an algebraic formula, from which the length of one side of a right triangle can be calculated given the length of the other two sides. But that is not how it appears in Euclides' "Elements". There it's a relation among the squares constructed on the sides of a right (or right-angled) triangle and the approach is puraly geometrical, without any algebraic formula.

In his monumental book "The thirteen books of Euclid's Elements, volume I: Books I-II" (Dover Publications, second edition, 1956), augmented with many mathematical and historical annotations, Thomas L. Heath states proposition I47, better known as the Pythagorean Theorem as follows:

Proposition I47

In right-angled triangles the square on the side subtending the right angle is equal to the squares on the sides containing the right angle.



As the approach is purely geometrical, no algebraic expressions are used. "Equality" between figures (line segments, triangles, parallelograms, rectangles, triangles... ) is first expressed in the sense of congruence. However, in some propositions preceeding I47 the notion of equality is extended to some figures that do not need to be congruent! So, for the proof of I47 proposition I38 is important. I38 extends the equality of triangles to some triangles that aren't congruent.

Proposition I38

Triangles which are on equal bases and in the same parallels are equal to one another.




Besides congruence and equality of figures in accordance to I38, the proof of the Pythagorean theorem I47 in the Elements also uses some of the so-called "Common Notions":
1. Things which are equal to the same thing are also equal to one another.
2. If equals are added to equals, the wholes are equal.
3. If equals are subtracted from equals, the remainders are equal.
4. Things which coincide with one another are equal to one another.
5. The whole is greater than the part.

Hundreds of proofs of the Pyhagorean theorem exist. Some of these proofs use algebraic formulas. Other proofs, such as the original proof in the Elements, are purely geometrical. They compare areas without using algebraic formulas.

At the first sight the original proof looks rather complicated. The reason is probably the number of auxiliary constructions. In reality only congruence, equality of figures in the sense of proposition I38 and some of the Common Notions are used.


As the triangles ACH and HCF are congruent (figure on the left), triangle ACH is half the square HACF. In the central figure, according to I38 the triangle ACH is equal to the triangle HAB (same base HA and C and B on the same parallel. And in the figure on the right triangle HAB is equal to triangle NAC as both triangles are congruent.


In the figure on the left, according to I38 the triangle NAC is equal to the triangle NAP (same base NA and C and P on the same parallel). So the triangle ACH we started with is equal to triangle NAP (next figure). As triangle ACH is half the square HACF and triangle NAP is half the rectangle NAPQ it follows that the square HACF is equal to the rectangle NAPQ (figure on the right).

In the same way we can prove that the square CBLJ is equal to the rectangle PQMB. From Common Notion 2 then follows that the square ANMB is equal to the combination of the squares HACF and CBLJ. And this is what we had to prove.



This approach is purely geometrical. If we allow the use of algebraic expressions for the length of a line segment and the area of a square, it's clear that the Pythagorean theorem for a right triangle with a and b as the lengths of the rectangular sides and c as the length of the hypothenosa is expressed in the well-known algebraic formula:

a2 + b2 = c2



For another nice, simple and also purely geometrical proof of the Pythagorean theorem using a rearrangement of triangles within a square click here

On YouTube I posted an animation about the Pythagorean Theorem, accompanied by a song... click here
Herman Serras (December 2025)
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