The proof usually called Einstein's uses the altitude from the right angle to split a right triangle into two smaller triangles with the same shape as the whole. The area of each such triangle is one fixed number times the square of its hypotenuse, so the three areas are ka2, kb2 and kc2. The two small triangles fill the big one, so a2 + b2 = c2.
Albert Einstein recalled proving the theorem with similar triangles as a boy, but he never wrote his proof down. The argument on this page is the best-known reconstruction of it, and the first section explains exactly how much we know.
What Einstein actually said
In his "Autobiographical Notes", written in 1946 and published in 1949 in the volume Albert Einstein: Philosopher-Scientist, Einstein describes the excitement of receiving a small book on plane geometry at the age of 12, a book he called the "holy geometry booklet". Then he adds:
I remember that an uncle told me the Pythagorean theorem before the holy geometry booklet had come into my hands. After much effort I succeeded in "proving" this theorem on the basis of the similarity of triangles.
He goes on to say that it seemed "evident" to him that the ratios of the sides of a right triangle must be completely fixed by one of its acute angles. That is all he says. The uncle is usually identified as his uncle Jakob, an engineer.
So which proof did he find? Biographers have generally assumed he rediscovered a standard similar triangles proof. In his 1991 book Fractals, Chaos, Power Laws, the physicist Manfred Schroeder presented the area argument below and attributed it to Einstein. Steven Strogatz retold it in The New Yorker in 2015, stressing that no one can be sure it is the proof the young Einstein found. It fits his description well: it rests entirely on similar triangles and on the idea that a right triangle's shape is fixed by one angle.
The key fact: area grows with the square of size
Enlarge any shape by a scale factor and its area grows by the square of that factor. Double every length and the area becomes 4 times as large; triple every length and it becomes 9 times as large. For triangles this is easy to see, because area is 1 over 2 × base × height and both the base and the height are multiplied by the scale factor.
So for right triangles of one fixed shape, the area is always the same number times the square of the hypotenuse. Call that number k. It depends on the shape but not on the size:
area = k · (hypotenuse)2For the shape of the 3-4-5 triangle, the area is 6 and the hypotenuse is 5, so k = 6/25 = 0.24. Check it on the 6-8-10 triangle, which has the same shape: its area is 24, and 0.24 × 102 = 24.
The proof in three steps
- Split the triangle. In a right triangle ABC with the right angle at C, draw the altitude from C to the hypotenuse, meeting it at D. This cuts ABC into two triangles, ACD and CBD. Each shares an acute angle with ABC and has a right angle at D, so both are similar to ABC: all three triangles have the same shape.
- Write each area as k times a hypotenuse squared. Because the three triangles have the same shape, they share the same k. The big triangle's hypotenuse is c. Triangle CBD's hypotenuse is the leg BC, which is a. Triangle ACD's hypotenuse is the leg AC, which is b. So their areas are kc2, ka2 and kb2.
- Add the pieces. The two small triangles together make up the big one, so their areas add up. Then divide by k, which is not zero:
Key ideaA right triangle is the only kind of triangle that splits into two smaller copies of itself. Since area scales with the square of size, that self-similarity turns directly into a2 + b2 = c2.
The complete proof, written out
| Statement | Reason |
|---|---|
| 1. In △ABC, ∠ACB = 90°; CD ⊥ AB, with D between A and B. | Given; D lies on segment AB because angles A and B are acute |
| 2. △ACD ∼ △ABC and △CBD ∼ △ABC | AA: each shares an acute angle with ABC and has a right angle at D |
| 3. If two triangles are similar with scale factor r, their areas are in the ratio r2. | Area is 1 over 2 base × height, and both are multiplied by r |
| 4. Let k = area(ABC)/c2. Then area(CBD) = ka2 and area(ACD) = kb2. | Step 3 with scale factors a/c and b/c |
| 5. area(ABC) = area(ACD) + area(CBD) | The altitude divides ABC into these two triangles |
| 6. kc2 = ka2 + kb2, so c2 = a2 + b2 | Substitute 4 into 5 and divide by k > 0 |
Step 4 is where the area scaling does its work: triangle CBD is triangle ABC shrunk by the factor a/c, so its area is kc2 · (a/c)2 = ka2. The similar triangles proof uses exactly the same figure but compares side lengths instead of areas, giving a2 = cp and b2 = cq.
The dimensional-analysis version
Physicists like to tell the same proof in the language of units. A right triangle is completely fixed by its hypotenuse c and one acute angle θ, which is Einstein's own remark that the angle determines the ratios of the sides. So its area depends only on c and θ.
Area is measured in square units, and the only length available is c. An angle has no units. The only way to build square units out of c is c2, so the area must be c2 times a pure number that depends on θ alone:
area = c2f(θ)The three triangles in the proof all have the same angle θ at matching corners, so the same f(θ) appears in all three areas:
c2f(θ) = a2f(θ) + b2f(θ) ⇒ c2 = a2 + b2This version is a favorite example in physics courses, where checking units is an everyday tool (see the theorem in physics). As a proof, it needs one more ingredient. The claim "area = c2f(θ)" is really the scaling fact from earlier, and that fact comes from similar triangles. With that justification the argument is rigorous; units alone would only make it a very good guess.
NoteThe argument quietly relies on flat geometry. On a sphere, a bigger triangle has bigger angles, so there are no similar triangles of different sizes, and the scaling step breaks down. Fittingly, the Pythagorean theorem itself is false on a sphere; see the theorem in non-Euclidean geometry.
Why this proof is special
It uses self-similarity. The right triangle is the only triangle that one straight cut can split into two triangles similar to itself. Such a cut must run from a corner to the opposite side, and the two angles it makes there add up to 180°. Both are angles of pieces similar to the original, so both are angles of the original triangle, and two different angles of a triangle cannot add up to 180°. So the two angles are equal, each 90°: the cut is an altitude, and the original triangle must contain a right angle.
It works for any shape on the sides. Nothing in the argument is special to triangles except that triangles happen to fit together. Put any three similar shapes on the sides of a right triangle, such as squares, semicircles or regular hexagons, and their areas are ka2, kb2 and kc2 for one shape constant k. If the relationship holds for one shape, it holds for all of them. This is Euclid's Book VI, Proposition 31, explored on the page on the theorem for other shapes. Euclid's better-known Book I proof works with squares directly.
For more ways to prove the theorem, from cutting and sliding to trigonometry, see the proofs overview, and for the wider story behind the theorem, the history of the Pythagorean theorem.
Frequently asked questions
Did Einstein really prove the Pythagorean theorem?
By his own account, yes: as a boy he proved it "on the basis of the similarity of triangles" after an uncle told him the statement. He never recorded the details, so the area argument on this page is a plausible reconstruction, popularized by Manfred Schroeder and Steven Strogatz, not a document in Einstein's hand.
How old was Einstein when he proved it?
He says the geometry booklet reached him at the age of 12 and that his uncle had told him the theorem before that. So he was about 12, or a little younger.
Is the Pythagorean theorem related to E = mc²?
Not directly; the resemblance between kc2 and mc2 is a coincidence. There is a genuine link elsewhere in relativity, though: the energy of a moving particle satisfies E2 = (mc2)2 + (pc)2, where p is its momentum. That has the form a2 + b2 = c2 and is often drawn as a right triangle.
What is the dimensional analysis proof of the Pythagorean theorem?
It argues that a right triangle's area depends only on its hypotenuse c and an acute angle θ, so it must equal c2f(θ). Applying this to the big triangle and the two pieces cut off by the altitude gives c2f(θ) = a2f(θ) + b2f(θ), and dividing by f(θ) gives the theorem.
Is Einstein's proof the same as the similar triangles proof?
They use the same figure: the altitude and three similar right triangles. The usual similar triangles proof compares sides and gets a2 = cp and b2 = cq. Einstein's version compares areas, which avoids naming the pieces of the hypotenuse at all.
Sources and further reading
- Albert Einstein, "Autobiographical Notes", in Paul Arthur Schilpp (ed.), Albert Einstein: Philosopher-Scientist, The Library of Living Philosophers, 1949.
- Steven Strogatz, Einstein's First Proof, The New Yorker, November 2015.
- Manfred Schroeder, Fractals, Chaos, Power Laws: Minutes from an Infinite Paradise, W. H. Freeman, 1991.
- Euclid, Elements, Book VI, Proposition 31, ed. D. E. Joyce, Clark University.
Keep exploring
- ProofsSimilar Triangles Proof of the Pythagorean TheoremProve the Pythagorean theorem with similar triangles: the altitude to the hypotenuse makes three similar right triangles, so a² = cp, b² = cq and a² + b² = c².
- LearnThe Pythagorean Theorem with Semicircles and Other ShapesEuclid VI.31 extends the Pythagorean theorem from squares to semicircles, equilateral triangles and any similar shapes. See why it works, with examples.
- ProofsEuclid's Proof of the Pythagorean TheoremEuclid's windmill proof of the Pythagorean theorem, Book I Proposition 47, step by step: congruent triangles, an animated shear and the converse, I.48.
- ProofsProofs of the Pythagorean TheoremProofs of the Pythagorean theorem: an animated proof, how a proof differs from checking examples, all 12 proofs on this site by type, and where to start.
- Real-life usesThe Pythagorean Theorem in PhysicsThe Pythagorean theorem in physics: resultant vectors, vector magnitude in 2D and 3D, projectile speed, AC impedance and the spacetime interval, with units.
- HistoryHistory of the Pythagorean TheoremThe history of the Pythagorean theorem from Babylon (1800 BCE) to a 2023 proof: who discovered it, who proved it first, and why it bears Pythagoras' name.