Pythagorean Theorem

Let's build up squares on the sides of a right triangle. The Pythagoras' Theorem then claims that the sum of (areas of) the two small squares equals (the area of) the large one.

In algebraic terms, a2+b2=c2 where c is the hypotenuse while a and b are the sides of the triangle.

The theorem is of fundamental importance in the Euclidean Geometry where it serves as a basis for the definition of distance between two points. It's so basic and well known that, I believe, anyone who took geometry classes in high school couldn't fail to remember it long after other math notions got solidly forgotten.

I plan to present several geometric proofs of the Pythagorean Theorem. An impetus for this page was provided by a remarkable Java applet written by Jim Morey. This constitutes the first proof on this page. There is nothing like learning while doing and, as an exercise in Java programming, I'll later offer an original Java applet. But, for now, let consider several plain HTML proofs.

Remark

  1. The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. Jim Morey's applet follows the Proposition I.47 (First Book, Proposition 47), mine VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a2+b2=c2 is right angled. Euclid was the first (I.48) to mention and prove this fact.
     

     

  2. W.Dunham [Ref 1] cites a book The Pythagorean Proposition by an early 20th century professor Elisha Scott Loomis. The book is a collection of 367 proofs of the Pythagorean Theorem and has been republished by NCTM in 1968.
     

     

  3. Pythagorean Theorem generalizes to spaces of higher dimensions. Some of the generalizations are far from obvious.
     

     

  4. The Theorem whose formulation leads to the notion of Euclidean distance and Euclidean and Hilbert spaces, plays an important role in Mathematics as a whole. I began collecting math facts whose proof may be based on the Pythagorean Theorem.
  5. Wherever all three sides of a right triangle are integers, their lengths form a Pythagorean triple (or Pythagorean numbers). There is a general formula for obtaining all such numbers.

Proof #2

We start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.
The construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.
As a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.

Proof #3

Now we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.

The square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4*ab/2), we get

c2 = (a-b)2+2ab = a2-2ab+b2+2ab = a2+b2



 
 

Proof #4

The fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus

(a+b)2=4*ab/2+c2

simplifying which we get the needed identity.
 
 


Proof #5

This proof, discovered by President J.A. Garfield in 1876 (T. Pappas, The Joy of Mathematics, p200, Wide World Publishing/Tetra, 1989), is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2 * (a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c*c/2. As before, simplifications yield a2+b2=c2.
 
 


Proof #6

We start with the original triangle, now denoted ABC, and need only one additional construct - the altitude AD. The triangles ABC, BDA and ADC are similar which leads to two ratios:

AB/BC=BD/AB and AC/BC=DC/AC.

Written another way these become

AB*AB=BD*BC and AC*AC=DC*BC

Summing up we get

AB*AB+AC*AC=BD*BC+DC*BC=(BD+DC)*BC=BC*BC.

Proof #7

The next proof is taken verbatim from Euclid VI.31 in translation by Sir Thomas L. Heath. The great G. Polya analyzes it in his Induction and Analogy in Mathematics which is a recommended reading to students and teachers of Mathematics.

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

Let ABC be a right-angled triangle having the angle BAC right; I say that the figure on BC is equal to the similar and similarly described figures on BA, AC.

Let AD be drawn perpendicular. Then since, in the right-angled triangle ABC, AD has been drawn from the right angle at A perpendicular to the base BC, the triangles ABD, ADC adjoining the perpendicular are similar both to the whole ABC and to one another [VI.8].

And, since ABC is similar to ABD, therefore, as CB is to BA so is AB to BD [VI.Def.1].

And, since three straight lines are proportional, as the first is to the third, so is the figure on the first to the similar and similarly described figure on the second [VI.19]. Therefore, as CB is to BD, so is the figure on CB to the similar and similarly described figure on BA.

For the same reason also, as BC is to CD, so is the figure on BC to that on CA; so that, in addition, as BC is to BD, DC, so is the figure on BC to the similar and similarly described figures on BA, AC.

But BC is equal to BD, DC; therefore the figure on BC is also equal to the similar and similarly described figures on BA, AC. 

Therefore etc. Q.E.D.

Proof #8

Playing with the applet that demonstrates the Euclid's proof (#7), I have discovered another one which, although ugly, serves the purpose nonetheless.

Thus starting with the triangle 1 we add three more in the way suggested in proof #7: similar and similarly described triangles 2, 3, and 4. Deriving a couple of ratios as was done in proof #6 we arrive at the side lengths as depicted on the diagram. Now, it's possible to look at the final shape in two ways: 

  • as a union of the rectangle (1+3+4) and the triangle 2, or
  • as a union of the rectangle (1+2) and two triangles 3 and 4.

Equating areas leads to

ab/c * (a2+b2)/c + ab/2 = ab + (ab/c * a2/c + ab/c * b2/c)/2

Simplifying we get

ab/c * (a2+b2)/c/2 = ab/2, or (a2+b2)/c2 = 1

Remark

On a second look at the diagram, there is a simpler proof. Viz., look at the rectangle (1+3+4). Its long side is, on the one hand, plain c while, on the other, it's a2/c+b2/c and we again have the same identity.
 
 

Proof #9

Another proof stems from a rearrangement of rigid pieces, much like proof #2. There is nothing much one can add to the two pictures.
 
 




Proof #10

This and the next 3 proofs came from [R.B.Nelsen].

The triangles in Proof #3 may be rearranged in yet another way that makes the Pythagorean identity obvious.
 
 

Proof #11

Draw a circle with radius c and a the right triangle with sides a and b as shown. In this situation, one may apply any of a few well known facts. For example, in the diagram three points located on the circle form another right triangle with the height of length a. Its hypotenuse is split in the ratio (c+b)/(c-b). So, as in Proof #6, we get a2 = (c+b)(c-b) = c2 - b2.

Proof #12

This proof is a variation on #1, one of the original Euclid's proofs. In parts 1,2, and 3, the two small squares are sheared towards each other such that the total shaded area remains unchanged (and equal to a2+b2.) In part 3, the length of the vertical portion of the shaded area's border is exactly c because the two leftover triangles are copies of the original one. This means one may slide down the shaded area as in part 4. From here the Pythagorean Theorem follows easily.

Proof #13

In the diagram there is several similar triangles (abc, a'b'c', b'x, and a'c'.) We successively have

y/b = b'/c, x/a = a'/c, cy + cx = aa' + bb'. 

And, finally, cc' = aa' + bb'. This is very much like Proof #6 but the result is more general.

Proof #14

This proof by H.E.Dudeney (1917) starts by cutting the square on the larger side into four parts that are then combined with the smaller one to form the square built on the hypotenuse.

Greg Frederickson from Purdue University, the author of a truly illuminating book, Dissections: Plane & Fancy (Cambridge University Press, 1997), pointed out the historical inaccuracy:
 
You attributed proof #14 to H.E. Dudeney (1917), but it was actually published earlier (1873) by Henry Perigal, a London stockbroker. A different dissection proof appeared much earlier, given by the Arabian mathematician/astronomer Thabit in the tenth century. I have included details about these and other dissections proofs (including proofs of the Law of Cosines) in my recent book "Dissections: Plane & Fancy", Cambridge University Press, 1997. You might enjoy the web page for the book:

http://www.cs.purdue.edu/homes/gnf/book.html

Sincerely,
Greg Frederickson

Bill Casselman from the University of British Columbia seconds Greg's information. Mine came from Proofs Without Words by R.B.Nelsen (MAA, 1993).

Proof #15

This remarkable proof by K.O.Friedrichs is a generalization of the previous one by Dudeney. It's indeed general. It's general in the sense that an infinite variety of specific geometric proofs may be derived from it. 

Proof #16

This proof is ascribed to Leonardo da Vinci (1452-1519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle abH is 45o. This is so because ABC is right-angled, thus center O of the square ACJI lies on the circle circumscribing triangle vC. Obviously, angle ABO is 45o.) Now, area(ABHI)+area(JHBC)=area(ADGC)+area(EDGF). Each sum contains two areas of triangles equal to ABC (IJH or BEF) removing which one obtains the Pythagorean Theorem.

Proof #17

This proof appears in the Book IV of Mathematical Collection by Pappus of Alexandria (ca A.D. 300) [Eves, Pappas]. It generalizes the Pythagorean Theorem in two ways: the triangle ABC is not required to be right-angled and the shapes built on its sides are arbitrary parallelograms instead of squares. Thus build parallelograms CADE and CBFG on sides AC and, respectively, BC. Let DE and FG meet in H and draw AL and BM parallel and equal to HC. Then area(ABML)=area(CADE)+area(CBFG). Indeed, the sheering transformation already used in proofs #1 and #12, area(CADE)=area(CAUH)=area(SLAR) and also area(CBFG)=area(CBVH)=area(SMBR). Now, just add up what's equal.

Proof #18

This is another generalization that does not require right angles. It's due to Tabit ibn Qorra (836-901). [Eves]. If angles CAB, AC'B and AB'C are equal then AC2+AB2=BC(CB'+BC'). Indeed, triangles ABC, AC'B and AB'C are similar. Thus we have AB/BC'=BC/AB and AC/CB'=BC/AC which immediately leads to the required identity. In case the angle A is right, the theorem reduces to the Pythagorean and the proof to the #6.

Proof #19

This proof is a variation on #6. On the small side AB add a right-angled triangle abD similar to abC. Then, naturally, DBC is similar to the other two. From area(ABD) + area(abC) = area(DBC), AD = AB2/AC and BD = AB*BC/AC we derive (ab2/AC)*AB + AB*AC = (AB*BC/AC)*BC. Dividing by AB/AC leads to AB2 + AC2 = BC2.

Proof #20

This one is a cross between #7 and #19. Construct triangles ABC', BCA', and ACB' similar to ABC, as on the diagram. By construction, ABC = ACB'. In addition, triangles BCC' and BCA' are also equal. Thus we conclude that area(ACB') + area(ABC') = area(BCA'). From the similarity of triangles we get as before AC' = AB2/AC and CA' = AB*BC/AC. Putting all together yields (AB2/AC)*AB + AB*AC = BC*(AB*BC/AC) which is the same as in #19.

Proof #21

The following is an excerpt from a letter by Dr. Scott Brodie from the Mount Sinai School of Medicine, NY who sent me a couple of proofs of the theorem proper and its generalization to the Law of Cosines:

The first proof I merely pass on from the excellent discussion in the Project Mathematics series, based on Ptolemy's theorem on quadrilaterals inscribed in a circle: for such quadrilaterals, the sum of the products of the lengths of the opposite sides, taken in pairs equals the product of the lengths of the two diagonals. For the case of a rectangle, this reduces immediately to a2 + b2 = c2.

Proof #22

Here is the second proof from the Dr. Scott Brodie's letter.

We take as known a "power of the point" theorems: If a point is taken exterior to a circle, and from the point a segment is drawn tangent to the circle and another segment (a secant) is drawn which cuts the circle in two distinct points, then the square of the length of the tangent is equal to the product of the distance along the secant from the external point to the nearer point of intersection with the circle and the distance along the secant to the farther point of intersection with the circle.

Let abC be a right triangle, with the right angle at C. Draw the altitude from C to the hypotenuse; let P denote the foot of this altitude. Then since CPB is right, the point P lies on the circle with diameter BC; and since CPA is right, the point P lies on the circle with diameter AC. Therefore the intersection of the two circles on the legs BC, CA of the original right triangle coincides with P, and in particular, lies on ab. Denote by x and y the lengths of segments BP and PA, respectively, and, as usual let a, b, c denote the lengths of the sides of abC opposite the angles A, B, C respectively. Then, x + y = c.

Since angle C is right, BC is tangent to the circle with diameter CA, and the power theorem states that a2 = xc; similarly, AC is tangent to the circle with diameter BC, and b2 = yc. Adding, we find a2 + b2 = xc + yc = c2, Q.E.D.

Dr. Brodie also created a Geometer's SketchPad file to illustrate this proof.

Proof #23

Another proof is based on the Heron's formula which I already used in Proof #7 to display triangle areas. This is a rather convoluted way to prove the Pythagorean Theorem that, nonetheless reflects on the centrality of the Theorem in the geometry of the plane.

Proof #24

[Swetz] ascribes this proof to abu' l'Hasan Thâbit ibn Qurra Marwân al'Harrani (826-901). It's the second of the proofs given by Thâbit ibn Qurra. The first one is essentially the #2 above.

The proof resembles part 3 from proof #12.abC =FLC =FMC =BED =AGH =FGE. On the one hand, the area of the shape abDFH equals AC2 + BC2 + area(abC + FMC + FLC). On the other hand, area(abDFH) = ab2 + area(BED + FGE + AGH).

Proof #25

B.F.Yanney (1903, [Swetz]) gave a proof using the "sliding argument" also employed in the Proofs #1 and #12. Successively, areas of LMOA, LKCA, and ACDE (which is AC2) are equal as are the areas of HMOB, HKCB, and HKDF (which BC2). BC = DF. Thus AC2 + BC2 = area(LMOA) + area(HMOB) = area(abHL) = ab2.
 

Proof #26

This proof I discovered at the site maintained by Bill Casselman where it presented by a Java applet.

With all the above proofs, this one must be simple. Similar triangles like in proofs #6 or #13.

Proof #27

The same pieces as in proof #26 may be rearrangened in yet another manner.
 

Proof #28

Melissa Running from MathForum has kindly sent me a link to A proof of the Pythagorean Theorem by Liu Hui (third century AD). The page is maintained by Donald B. Wagner, an expert on history of science and technology in China. The diagram is a reconstruction from a written description of an algorithm by Liu Hui (third century AD). For details you are referred to the original page.
 
 



Proof #29

A mechanical proof of the theorem deserves a page of its own.

Pertinent to that proof is a page "Extra-geometric" proofs of the Pythagorean Theorem by Scott Brodie

References

  1. W.Dunham, The Mathematical Universe, John Wiley & Sons, NY, 1994.
  2. W.Dunham, Journey through Genius, Penguin Books, 1991
  3. H.Eves, Great Moments in Mathematics Before 1650, MAA, 1983
  4. R.B.Nelsen, Proofs Without Words, MAA, 1993
  5. J.A.Paulos, Beyond Numeracy, Vintage Books, 1992
  6. T.Pappas, The Joy of Mathematics, Wide World Publishing, 1989
  7. F.J.Swetz, From Five Fingers to Infinity, Open Court, 1996, third printing

On Internet

  1. The Mathematical Association of America offers a video tape. Tell me if it's good.
  2. One more proof which appears to be a cross between #2 and #3.
  3. Pythagoras, biography
  4. Another biography
  5. The birthplace; all about it in a language much different from English
  6. Ask Dr. Math
  7. Eric's Treasure Trove features more than 10 proofs.
  8. Pythagoras' Theorem by Bill Casselman
    A huge collection of proofs with Java demonstations. The proofs are grouped by the basic geometric transformations. The site also offers delightful historical information.
  9. A proof of the Pythagorean Theorem by Liu Hui (third century AD)
    An interesting page from which I borrowed Proof #28

Copyright © 1996-2000 Alexander Bogomolny