Bhāskara's Proof of the Pythagorean Theorem

One square, four triangles, one small square and, according to the famous story, a single word of explanation. Here is the figure in motion, the reasoning it leaves out, and the second proof Bhāskara also gave.

By the PythagoreanTheorem.net teamUpdated 8 min read

Bhāskara's proof cuts the square on the hypotenuse into four copies of the right triangle and a small square of side b − a, then rearranges the same five pieces into two squares of areas a2 and b2. The 12th-century Indian mathematician Bhāskara II is said to have drawn the figure and written a single word: "Behold!"

The figure

1 2 3 4 1 2 3 4 c c 5 5 b a b a c² = 4 triangles + square 5 same 5 pieces = a² + b²
Left: the square on the hypotenuse, cut into triangles 1 to 4 and a small square 5. Right: the same five pieces, rearranged.

Take a right triangle with legs a and b, where b is the longer leg, and hypotenuse c. On the left, four copies of it sit inside the square of side c, each with its hypotenuse along one side of the square, each a quarter turn from the next. In the middle they leave a small square, piece 5.

On the right, triangles 1 and 2 and square 5 have stayed where they were, and triangles 3 and 4 have moved to the bottom corners. The outline is now an L-shaped region. The dashed line splits it into a square of side b on the left and a square of side a on the right. Same five pieces, so same area: c2 = a2 + b2.

Watch the pieces move

c²cc b²a²ba THE SQUARE ON THE HYPOTENUSE4 triangles + 1 small square SWING TWO TRIANGLES DOWNthe same 5 pieces make a² + b² Behold: c² = a² + b²
The two white triangles each swing a quarter turn about one corner of the tilted square. The grey triangles and the small square never move.

Each moving triangle turns 90° about a corner of the square on the hypotenuse, like a door on a hinge: the top-left triangle about the left corner, the top-right triangle about the right corner. Turning a shape doesn't change its area, so the area of the L-shaped region is still c2. That region is made of a b × b square and an a × a square, so its area is also a2 + b2.

Why the pieces fit exactly

A picture can hide small gaps or overlaps, so it is worth checking that everything fits. It is easiest to run the animation backwards, starting from the L shape.

  1. Build the L. Place a square of side b and a square of side a side by side with their bottoms on one line. The bottom edge has length a + b. Mark the point P on it at distance a from the left end, so it is at distance b from the right end.
  2. Make two cuts. Cut from P to the top-left corner T of the big square, and from P to the top-right corner Q of the small square. The first cut is the hypotenuse of a triangle with legs a (along the bottom) and b (up the left side); the second, of a triangle with legs b (along the bottom) and a (up the right side). By SAS both triangles are congruent to the original, so TP = PQ = c.
  3. The angle at P is 90°. Along the bottom edge at P, the left triangle shows its angle opposite b, call it β, and the right triangle its angle opposite a, call it α. Since α + β = 90°, the angle TPQ between the cuts is 180° − 90° = 90°.
  4. Swing the triangles up. Turn the left triangle 90° about T and the right triangle 90° about Q. The turn about T carries P to the point at distance c from T at right angles to TP; the turn about Q carries P to the point at distance c from Q at right angles to QP. Because TP and PQ are equal and perpendicular, both turns send P to the same point R, the fourth corner of a square TPQR.
  5. Nothing is left over. The swung triangles have legs lying along the top edges of the two squares, so they fill exactly the two notches between the L shape and the square TPQR. The pieces now form the square on the hypotenuse.

Since the five pieces were only moved, never stretched, the square TPQR has the same area as the L shape: c2 = a2 + b2.

The algebra version

Many textbooks present Bhāskara's figure with algebra instead of scissors. The left figure alone is enough: the square of side c is made of four triangles, each of area 1 over 2ab, and a small square of side b − a.

c2= 4 · 1 over 2ab + (b − a)2= 2ab + b2 − 2ab + a2= a2 + b2

For the 3-4-5 triangle that reads 4 × 6 + 12 = 25 = 52. For the 8-15-17 Pythagorean triple it reads 4 × 60 + 72 = 240 + 49 = 289 = 172. Why the middle hole is a square of side b − a is explained on the algebraic proof page, which also shows the related (a + b)2 version.

A complete written proof

Theorem. In a right triangle with legs a ≤ b and hypotenuse c, a2 + b2 = c2.

Proof. Let α and β be the acute angles opposite a and b, so α + β = 90°.

  1. Place a square OUVT of side b and a square UWQX of side a so that they share the corner U and their bottom sides OU and UW lie on one line. Their union L has area a2 + b2.
  2. Let P be the point on OW with OP = a, so PW = b. Triangle TOP has legs OT = b and OP = a, and triangle PWQ has legs PW = b and WQ = a, with right angles at O and W. By SAS both are congruent to the original triangle, so TP = PQ = c.
  3. The angles of these triangles at P are β and α, and with ∠TPQ they fill the straight angle at P. Hence ∠TPQ = 90°.
  4. Let R be the point such that TPQR is a square. The rotation by 90° about T maps P to R and O to V, so it maps triangle TOP onto triangle TVR. The rotation by 90° about Q maps P to R and W to X, so it maps triangle PWQ onto triangle RXQ.
  5. The square TPQR is the union of L without the triangles TOP and PWQ, together with the triangles TVR and RXQ, and these pieces do not overlap. Rotations preserve area, so area(TPQR) = area(L), that is, c2 = a2 + b2. ∎

When a = b the small square vanishes and the two squares in the L shape are equal; the argument still works.

Bhāskara's second proof: similar triangles

Bhāskara gave another demonstration that uses no rearranging at all. Drop a perpendicular from the right angle to the hypotenuse.

A B C D b a q p h c = p + q
The altitude CD splits triangle ABC into two smaller triangles, each similar to the whole.

Triangle CBD has a right angle at D and shares angle B with triangle ABC, so the two are similar (AA). Triangle ACD has a right angle at D and shares angle A with triangle ABC, so they are similar too. Matching sides in proportion:

p over a = a over c⇒  a2 = cpq over b = b over c⇒  b2 = cq

Adding the two equations, and using p + q = c:

a2 + b2 = cp + cq = c(p + q) = c2

For the 3-4-5 triangle, p = 9 over 5 = 1.8 and q = 16 over 5 = 3.2, and indeed 5 × 1.8 + 5 × 3.2 = 9 + 16 = 25. According to Florian Cajori's History of Mathematics (1893), this proof was unknown in Europe until John Wallis rediscovered it in the 17th century. Today it is the standard proof in many geometry courses; the similar triangles proof page treats it in full, and the same figure gives the geometric mean theorem, h2 = pq.

Bhāskara and his world

Bhāskara II (1114–c. 1185), also called Bhāskarācārya, "Bhāskara the teacher", was the leading Indian mathematician and astronomer of the 12th century. The numeral distinguishes him from an earlier astronomer, Bhāskara I, of the 7th century. His best-known works are the Līlāvatī, on arithmetic and measurement, the Bījagaṇita, on algebra, and the astronomical Siddhānta Śiromaṇi, completed around 1150. The dissection figure is usually traced to the Bījagaṇita.

The famous story, that the whole proof consisted of the figure and the word "Behold!", is told in Cajori's 1893 history and in many books since. Some modern writers call it a legend, so read it as a description of the proof's spirit rather than of the manuscript: the figure really does speak for itself.

Bhāskara was working in a long tradition. Indian geometers had stated the theorem in the Śulba Sūtras many centuries earlier, and essentially the same four-triangle figure appears in a 3rd-century Chinese commentary, the hypotenuse diagram. Whether Bhāskara knew the Chinese figure is not known. For the full picture, see the history of the theorem, and for other ways to prove it, the guide to all the proofs.

Frequently asked questions

What does "Behold!" mean in Bhāskara's proof?

It refers to the story that Bhāskara drew the figure and, instead of an explanation, wrote only "Behold!", inviting the reader to see the proof. The word stands for the idea of a proof without words. Some historians regard the one-word version as a legend.

Who was Bhāskara?

Bhāskara II (1114–c. 1185) was an Indian mathematician and astronomer, author of the Līlāvatī, the Bījagaṇita and the Siddhānta Śiromaṇi. He is known for work on algebra, on equations such as Pell's equation, and on astronomy.

Why is the small square's side b − a?

Along each long leg b of a triangle lies the short leg a of its neighbor, because the two acute angles at each corner of the big square add to 90°. What remains of the long leg, b − a, is one side of the small square.

Is Bhāskara's proof the same as the Chinese proof?

The figure is essentially the same: four right triangles around a central square inside the square on the hypotenuse. The Chinese version, in Zhao Shuang's 3rd-century commentary, is about nine centuries older and was usually drawn on a grid for the 3-4-5 triangle. See the Chinese proof.

What is Bhāskara's second proof?

It drops the altitude from the right angle to the hypotenuse, creating two triangles similar to the original. The proportions give a2 = cp and b2 = cq, and adding them gives a2 + b2 = c2.

Sources and further reading

  1. Florian Cajori, A History of Mathematics (Macmillan, 1893), section on the Hindus; the "Behold!" passage and the second proof are on Wikisource.
  2. J. J. O'Connor and E. F. Robertson, Bhaskara II, MacTutor History of Mathematics, University of St Andrews.
  3. H. T. Colebrooke (translator), Algebra, with Arithmetic and Mensuration, from the Sanscrit of Brahmegupta and Bháscara (London, 1817), which contains English translations of the Līlāvatī and Bījagaṇita.
  4. Eli Maor, The Pythagorean Theorem: A 4,000-Year History (Princeton University Press, 2007).
  5. Xuan tu, Wikipedia, on the shared figure and its appearance in the Bījagaṇita.

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