History of Mathematics Project | Ibn Qurra's Pythagorean Theorem Proofs

History of
Mathematics

Project

around 850

Ibn Qurra's Pythagorean Theorem Proofs

Pythagorean theorem proofs by dissection

Arab polymath Thābit ibn Qurra revised an earlier translation of the Elements and translated several other Greek works from the times of Euclid to Ptolemy. He also produced two original proofs of the Pythagorean theorem as well a generalization of the theorem (which, while valid, was stated without proof).

Thābit ibn Qurra presented two original proofs of the Pythagorean theorem using a method of dissection which he called the method of reduction to triangles and rearrangement by juxtaposition. These proofs appeared in a letter to a friend written around around 850 CE and are preserved in the library of the Hagia Sophia Museum in Istanbul, Turkey.

Artifact origin

Baghdad, Iraq

Current artifact location

Hagia Sophia, Istanbul

Catalog number

4832

Timeline

Interactive Content

Computational Explanation

In a letter to a friend, ibn Qurra gave two proofs of the Pythagorean theorem using a dissection method he called the method of reduction and composition, or the method of reduction to triangles and rearrangement by juxtaposition. Both proofs are valid for any right triangle. He also gave a generalization of the theorem that, however, was stated without proof.

First Proof

Consider a right triangle ΔABC and draw a figure as follows:

Out[]=

Here, AA'B'B and DFB'D' are squares on the sides AB and BC, respectively, but with the former reflected across AB and the latter shifted to the left by a distance |BB'| and then reflected across the line FB' extending BC. The square on the hypotenuse AC is drawn as ACDE.

The first two squares can be obtained by combining the shaded region with the triangles numbered 1 and 2:

Out[]=

Similarly, the square on the hypotenuse can be obtained by combining the same shaded region with the triangles numbered 3 and 4:

Out[]=

Since triangles 1–4 are all equal by construction, the sum of the areas of the first two squares must equal the area of the third. The Pythagorean theorem has therefore been proved.

In 1873, H. Perigal published an independent rediscovery of this proof.

Second Proof

Again consider a right triangle ΔABC and draw a figure as follows:

Out[]=

Here, BCF'B' and ACFA' are squares on the sides BC and AC, respectively, and the square on the hypotenuse is drawn reflected across the line AB as ABED.

Removing the three congruent shaded triangles ΔABC, ΔCFG and ΔGF'C gives the squares on the two sides AC and BC of the original right triangle:

Out[]=

However, those triangles are also congruent to the triangles ΔAA'D, ΔDGF' and ΔEB'B by construction:

Out[]=

… but removing those three triangles gives the square on the hypotenuse.

Therefore, the sum of the areas of the first two squares must be equal to the area of the third, and the Pythagorean theorem has again been proved.

In 1910, A. R. Colburn published an amazingly similar proof of the Pythagorean theorem using this method of dissection.

Generalization of the Pythagorean Theorem

Ibn Qurra stated the following beautiful theorem with proof. It is applicable to all triangles, whether right, isosceles, equilateral or scalene. Consider an arbitrary triangle ΔABC and draw lines AB' and AC' intersecting the side BC opposite A such that angles ∠AB'B and ∠AC'C are equal to the angle at vertex A, i.e. ∠BAC:

Out[]=

Then

BC(

′

)

This result can be proved using the Wolfram Language after constructing the points in question with the aid of a little trigonometry and simplifying the result assuming the triangle inequality (which holds for any triangle and states that the sum of the lengths of any two sides is greater than the length of the remaining side):

In[]:=

Assuming[a+b>c&&b+c>a&&a+c>b,FullSimplify[Module[{A,B,C,ptA,ptB,ptBp,ptC,ptCp,tri=SSSTriangle[b,c,a]},{ptB,ptC,ptA}=tri〚1〛;{B,C,A}=PolygonAngle[tri];ptBp={ptA〚1〛+cSin[B]Cot[A],0};ptCp={ptA〚1〛-cSin[B]Cot[A],0};SquaredEuclideanDistance[ptA,ptB]+SquaredEuclideanDistance[ptA,ptC]EuclideanDistance[ptB,ptC](EuclideanDistance[ptB,ptBp]+EuclideanDistance[ptC,ptCp])]]]

Out[]=

True

A short and elegant geometric proof follows from similarity of ΔBAC, ΔAC'C and ΔBB'A, which implies equality of the distance ratios as follows:

BC:AC:ABAC:CC':AB'AB:AC':BB'.

It therefore follows that:

BB'



⇒

(BC)(BB')

CC'



⇒

(BC)(CC')

Adding and factoring the above two equations then establishes the theorem:

BC(BB'+CC').

Numerical evidence for the generalization can be obtained using computational geometry functionality built into the Wolfram Language:

In[]:=

gs=RandomInstance[GeometricScene[{{"A","B","C","B'","C'"},{"AB","AC","BC","BB'","CC'"}},{Triangle[{"A","B","C"}],Line[{"B","B'","C'","C"}],PlanarAngle[{"B","A","C"}]PlanarAngle[{"A","B'","B"}]PlanarAngle[{"A","C'","C"}],"AB"EuclideanDistance["A","B"],"AC"EuclideanDistance["A","C"],"BC"EuclideanDistance["B","C"],"BB'"EuclideanDistance["B","B'"],"CC'"EuclideanDistance["C","C'"]}]]

Out[]=

Plugging in the values corresponding to the preceding random scene shows the stated equality is indeed satisfied:

In[]:=

{"AB"^2+"AC"^2,"BC"("BB'"+"CC'")}/.gs["Quantities"]

Out[]=

{7.61593,7.61593}

A more visual proof of ibn Qurra's theorem can be obtained for triangles ΔABC such that AB/BC and AC/BC are rational. (This proof can be extended to general triangles by using arbitrarily accurate rational approximations.) Consider the following figure:

Out[]=

Note that ΔABC is similar to

ΔB'BA

, and draw copies

Δ(B1)B(A1)

and

Δ(B2)B(A2)

ΔB'BA

, with BC and B(B1) forming adjacent sides of rectangle BC(C*)(B1), and AB and B(A2) forming adjacent sides of square AB(A2)(A*). Let BC/AB  p/q for positive integers p and q, and partition square AB(A2)(A*) into a p × q grid of rectangles each of size AB/p × AB/q. Copy that grid onto side BC of ΔABC after rotating each individual rectangle sideways in place to get a p × q grid of rectangles each of size AB/q × AB/p. Altogether, this grid creates a rectangle of size BC × BB' since AB p/q  BC and AB q/p  AB(B'B)/AB  B'B  (B1)B:

Out[]=

The preceding argument establishes the identity

BC(BB')

, a proof that

BC(CC')

similarly follows.

Other Resources

Wolfram Demonstration
Pythagorean Theorem

Additional Reading

Beggren, J. L. "Mathematics in Medieval Islam." Ch. 5 in The Mathematics of Egypt, Mesopotamia, China, India, and Islam: A Sourcebook (Ed. V. J. Katz). Princeton, NJ: Princeton University Press, pp. 517 and 547, 2007.
Colburn, A. R. "The Pons Asinorum.—II. New Solutions of the Pythagorean Theorem." Scientific American Supplement, Vol. 70, No. 1823, p. 382–383, 1910.
Eves, H. An Introduction to the History of Mathematics, 6th ed. Cengage Learning, p. 73, 1990.
Maor, E. The Pythagorean Theorem: A 4000-Year History. Princeton, NJ: Princeton University Press, pp. 69–71, 2007.
Merzbach, U. C. and Boyer, C. B. A History of Mathematics, 3rd ed. Hoboken, NJ: Wiley, pp. 213–214, 2011.
Perigal, H. "On Geometric Dissections and Transformations." Messenger of Mathematics, Vol. 2, pp. 103–106, 1873.
Sayili, A. "Thabat ibn Qurra's Generalization of the Pythagorean Theorem." Isis, Vol, 51, 35–37, 1960.
Scribs, C. J. "John Wallis' Treament of Angular Sections and Thābit ibn Qurra's Generalization of the Pythagorean Theorem." Isis, Vol. 56, pp. 56–67, XXX.
Shloming, S. "Thabit ibn Qurra and the Pythagorean Theorem." The Mathematics Teacher, Vol. 63, pp. 519–528, 1970.
Suter, H. Die Mathematiker und Astronomen der Araber und Ihre Werke. Leipzig, Germany: Teubner, p. 37, 1900.