Proof #1
This is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.
The proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.
First of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and
BAF = BAC + CAF = CAB + BAE = CAE.
ABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.
Similarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.
The configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams
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.
(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)
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 [Pappas], 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.
Two copies of the same trapezoid can be combined in two ways by attaching them along the slanted side of the trapezoid. One leads to the proof #4, the other to proof #52.
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.
In a little different form, this proof appeared in the Mathematics Magazine, 33 (March, 1950), p. 210, in the Mathematical Quickies section, see Mathematical Quickies, by C. W. Trigg.
Taking AB = a, AC = b, BC = c and denoting BD = x, we obtain as above
a2 = cx and b2 = c(c - x),
which perhaps more transparently leads to the same identity.
In a private correspondence, Dr. France Dacar, Ljubljana, Slovenia, has suggested that the diagram on the right may serve two purposes. First, it gives an additional graphical representation to the present proof #6. In addition, it highlights the relation of the latter to proof #1.
It must be mentioned that this proof is just a variant of the next one - Euclid's second and less known proof of the Pythagorean proposition.
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 (II.5) 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.
Confession
I got a real appreciation of this proof only after reading the book by Polya I mentioned above. I hope that a Java applet will help you get to the bottom of this remarkable proof. Note that the statement actually proven is much more general than the theorem as it's generally known. (Another discussion looks at VI.31 from a little different angle.)
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
In hindsight, there is a simpler proof. Look at the rectangle (1+3+4). Its long side is, on one hand, plain c, while, on the other hand, 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. It makes the algebraic part of proof #4 completely redundant. There is nothing much one can add to the two pictures.
(My sincere thanks go to Monty Phister for the kind permission to use the graphics.)
There is an interactive simulation to toy with.
(The proof has been published by Rufus Isaac in Mathematics Magazine, Vol. 48 (1975), p. 198.)
Proof #10
This and the next 3 proofs came from [PWW].
The triangles in Proof #3 may be rearranged in yet another way that makes the Pythagorean identity obvious.
(A more elucidating diagram on the right was kindly sent to me by Monty Phister.)
Proof #11
Draw a circle with radius c and a 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 F, G, H located on the circle form another right triangle with the altitude FK of length a. Its hypotenuse GH is split in two pieces: (c + b) and (c - b). So, as in Proof #6, we get a2 = (c + b)(c - b) = c2 - b2.
B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.3, n. 12 (1896), 299-300) offer a somewhat different route. Imagine FK is extended to the second intersection F' with the circle. Then, by the Intersecting Chords theorem, FK·KF' = GK·KH, with the same implication.
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.
(This proof can be found in H. Eves, In Mathematical Circles, MAA, 2002, pp. 74-75)
Proof #13
In the diagram there is several similar triangles (abc, a'b'c', a'x, and b'y.) 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).
The proof has a dynamic version.
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. (Roger Nelsen ascribes [PWWII, p 3] this proof to Annairizi of Arabia (ca. 900 A.D.))
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 ABC. 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.
David King modifies the argument somewhat
The side lengths of the hexagons are identical. The angles at P (right angle + angle between a & c) are identical. The angles at Q (right angle + angle between b & c) are identical. Therefore all four hexagons are identical.
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, with 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 Thâbit ibn Qurra (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 proposition and proof #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 in the diagram. By construction, ABC = A'BC. In addition, triangles ABB' and ABC' are also equal. Thus we conclude that Area(A'BC) + Area(AB'C) = Area(ABC'). From the similarity of triangles we get as before B'C = AC2/BC and BC' = AC·AB/BC. Putting it all together yields AC·BC + (AC2/BC)·AC = AB·(AC·AB/BC) which is the same as
BC2 + AC2 = AB2.
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 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.
(This proof has been published as number XXIV in a collection of proofs by B. F. Yanney and J. A. Calderhead in Am Math Monthly, v. 4, n. 1 (1897), pp. 11-12.)
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 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).
This is an "unfolded" variant of the above proof. Two pentagonal regions - the red and the blue - are obviously equal and leave the same area upon removal of three equal triangles from each.
The proof is popularized by Monty Phister, author of the inimitable Gnarly Math CD-ROM.
Floor van Lamoen has gracefully pointed me to an earlier source. Eduard Douwes Dekker, one of the most famous Dutch authors, published in 1888 under the pseudonym of Multatuli a proof accompanied by the following diagram.
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 is 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 is 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.
This dissection is often attributed to the 17th century Dutch mathematician Frans van Schooten. [Frederickson, p. 35] considers it as a hinged variant of one by ibn Qurra, see the note in parentheses following proof #2. Dr. France Dacar from Slovenia has pointed out that this same diagram is easily explained with a tesselation in proof #15. As a matter of fact, it may be better explained by a different tesselation. (I thank Douglas Rogers for setting this straight for me.)
The configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v. 6, n. 2 (1899), 33-34) published several proofs based on the following diagrams (multiple proofs per diagram at that)
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
Proof #30
This proof I found in R. Nelsen's sequel Proofs Without Words II. (It's due to Poo-sung Park and was originally published in Mathematics Magazine, Dec 1999). Starting with one of the sides of a right triangle, construct 4 congruent right isosceles triangles with hypotenuses of any subsequent two perpendicular and apices away from the given triangle. The hypotenuse of the first of these triangles (in red in the diagram) should coincide with one of the sides.
The apices of the isosceles triangles form a square with the side equal to the hypotenuse of the given triangle. The hypotenuses of those triangles cut the sides of the square at their midpoints. So that there appear to be 4 pairs of equal triangles (one of the pairs is in green). One of the triangles in the pair is inside the square, the other is outside. Let the sides of the original triangle be a, b, c (hypotenuse). If the first isosceles triangle was built on side b, then each has area b2/4. We obtain
a2 + 4b2/4 = c2
There's a dynamic illustration and another diagram that shows how to dissect two smaller squares and rearrange them into the big one.
Proof #31
Given right ABC, let, as usual, denote the lengths of sides BC, AC and that of the hypotenuse as a, b, and c, respectively. Erect squares on sides BC and AC as on the diagram. According to SAS, triangles ABC and PCQ are equal, so that QPC = A. Let M be the midpoint of the hypotenuse. Denote the intersection of MC and PQ as R. Let's show that MR PQ.
The median to the hypotenuse equals half of the latter. Therefore, CMB is isosceles and MBC = MCB. But we also have PCR = MCB. From here and QPC = A it follows that angle CRP is right, or MR PQ.
With these preliminaries we turn to triangles MCP and MCQ. We evaluate their areas in two different ways:
One one hand, the altitude from M to PC equals AC/2 = b/2. But also PC = b. Therefore, Area(MCP) = b2/4. On the other hand, Area(MCP) = CM·PR/2 = c·PR/4. Similarly, Area(MCQ) = a2/4 and also Area(MCQ) = CM·RQ/2 = c·RQ/4.
We may sum up the two identities: a2/4 + b2/4 = c·PR/4 + c·RQ/4, or a2/4 + b2/4 = c·c/4.
(My gratitude goes to Floor van Lamoen who brought this proof to my attention. It appeared in Pythagoras - a dutch math magazine for schoolkids - in the December 1998 issue, in an article by Bruno Ernst. The proof is attributed to an American High School student from 1938 by the name of Ann Condit.)
Proof #32
Let ABC and DEF be two congruent right triangles such that B lies on DE and A, F, C, E are collinear. BC = EF = a, AC = DF = b, AB = DE = c. Obviously, AB DE. Compute the area of ADE in two different ways.
Area(ADE) = AB·DE/2 = c2/2 and also Area(ADE) = DF·AE/2 = b·AE/2. AE = AC + CE = b + CE. CE can be found from similar triangles BCE and DFE: CE = BC·FE/DF = a·a/b. Putting things together we obtain
c2/2 = b(b + a2/b)/2
(This proof is a simplification of one of the proofs by Michelle Watkins, a student at the University of North Florida, that appeared in Math Spectrum 1997/98, v30, n3, 53-54.)
Douglas Rogers observed that the same diagram can be treated differently:
Proof 32 can be tidied up a bit further, along the lines of the later proofs added more recently, and so avoiding similar triangles.
Of course, ADE is a triangle on base DE with height AB, so of area cc/2.
But it can be dissected into the triangle FEB and the quadrilateral ADBF. The former has base FE and height BC, so area aa/2. The latter in turn consists of two triangles back to back on base DF with combined heights AC, so area bb/2. An alternative dissection sees triangle ADE as consisting of triangle ADC and triangle CDE, which, in turn, consists of two triangles back to back on base BC, with combined heights EF.
The next two proofs have accompanied the following message from Shai Simonson, Professor at Stonehill College in Cambridge, MA:
Greetings,
I was enjoying looking through your site, and stumbled on the long list of Pyth Theorem Proofs.
In my course "The History of Mathematical Ingenuity" I use two proofs that use an inscribed circle in a right triangle. Each proof uses two diagrams, and each is a different geometric view of a single algebraic proof that I discovered many years ago and published in a letter to Mathematics Teacher.
The two geometric proofs require no words, but do require a little thought.
Best wishes,
Shai
Proof #33
Proof #34
Proof #35
Cracked Domino - a proof by Mario Pacek (aka Pakoslaw Gwizdalski) - also requires some thought.
The proof sent via email was accompanied by the following message:
This new, extraordinary and extremely elegant proof of quite probably the most fundamental theorem in mathematics (hands down winner with respect to the # of proofs 367?) is superior to all known to science including the Chinese and James A. Garfield's (20th US president), because it is direct, does not involve any formulas and even preschoolers can get it. Quite probably it is identical to the lost original one - but who can prove that? Not in the Guinness Book of Records yet!
The manner in which the pieces are combined may well be original. The dissection itself is well known (see Proofs 26 and 27) and is described in Frederickson's book, p. 29. It's remarked there that B. Brodie (1884) observed that the dissection like that also applies to similar rectangles. The dissection is also a particular instance of the superposition proof by K.O.Friedrichs.
Proof #36
This proof is due to J. E. Böttcher and has been quoted by Nelsen (Proofs Without Words II, p. 6).
I think cracking this proof without words is a good exercise for middle or high school geometry class.
S. K. Stein, (Mathematics: The Man-Made Universe, Dover, 1999, p. 74) gives a slightly different dissection.
Both variants have a dynamic version.
Proof #37
An applet by David King that demonstrates this proof has been placed on a separate page.
Proof #38
This proof was also communicated to me by David King. Squares and 2 triangles combine to produce two hexagon of equal area, which might have been established as in Proof #9. However, both hexagons tessellate the plane.
For every hexagon in the left tessellation there is a hexagon in the right tessellation. Both tessellations have the same lattice structure which is demonstrated by an applet. The Pythagorean theorem is proven after two triangles are removed from each of the hexagons.
Proof #39
(By J. Barry Sutton, The Math Gazette, v 86, n 505, March 2002, p72.)
Let in ABC, angle C = 90o. As usual, AB = c, AC = b, BC = a. Define points D and E on AB so that AD = AE = b.
By construction, C lies on the circle with center A and radius b. Angle DCE subtends its diameter and thus is right: DCE = 90o. It follows that BCD = ACE. Since ACE is isosceles, CEA = ACE.
Triangles DBC and EBC share DBC. In addition, BCD = BEC. Therefore, triangles DBC and EBC are similar. We have BC/BE = BD/BC, or
a / (c + b) = (c - b) / a.
And finally
a2 = c2 - b2,
a2 + b2 = c2.
The diagram reminds one of Thâbit ibn Qurra's proof. But the two are quite different.
Proof #40
This one is by Michael Hardy from University of Toledo and was published in The Mathematical Intelligencer in 1988. It must be taken with a grain of salt.
Let ABC be a right triangle with hypotenuse BC. Denote AC = x and BC = y. Then, as C moves along the line AC, x changes and so does y. Assume x changed by a small amount dx. Then y changed by a small amount dy. The triangle CDE may be approximately considered right. Assuming it is, it shares one angle (D) with triangle ABD, and is therefore similar to the latter. This leads to the proportion x/y = dy/dx, or a (separable) differential equation
y·dy - x·dx = 0,
which after integration gives y2 - x2 = const. The value of the constant is determined from the initial condition for x = 0. Since y(0) = a, y2 = x2 + a2 for all x.
It is easy to take an issue with this proof. What does it mean for a triangle to be approximately right? I can offer the following explanation. Triangles ABC and ABD are right by construction. We have, AB2 + AC2 = BC2 and also AB2 + AD2 = BD2, by the Pythagorean theorem. In terms of x and y, the theorem appears as
x2 + a2 = y2
(x + dx)2 + a2 = (y + dy)2
which, after subtraction, gives
y·dy - x·dx = (dx2 - dy2)/2.
For small dx and dy, dx2 and dy2 are even smaller and might be neglected, leading to the approximate y·dy - x·dx = 0.
The trick in Michael's vignette is in skipping the issue of approximation. But can one really justify the derivation without relying on the Pythagorean theorem in the first place? Regardless, I find it very much to my enjoyment to have the ubiquitous equation y·dy - x·dx = 0 placed in that geometric context.
An amplified, but apparently independent, version of this proof has been published by Mike Staring (Mathematics Magazine, V. 69, n. 1 (Feb., 1996), 45-46).
Assuming Dx > 0 and detecting similar triangles,
Df / Dx = CQ/CD > CP/CD = CA/CB = x/f(x).
But also,
Df / Dx = SD/CD < RD/CD = AD/BD = (x + Dx) / (f(x) + Df) < x/f(x) + Dx/f(x).
Passing to the limit as Dx tends to 0+, we get
df / dx = x / f(x).
The case of Dx < 0 is treated similarly. Now, solving the differential equation we get
f 2(x) = x2 + c.
The constant c is found from the boundary condition f(0) = b: c = b2. And the proof is complete.
Proof #41
This one was sent to me by Geoffrey Margrave from Lucent Technologies. It looks very much as #8, but is arrived at in a different way. Create 3 scaled copies of the triangle with sides a, b, c by multiplying it by a, b, and c in turn. Put together, the three similar triangles thus obtained to form a rectangle whose upper side is a2 + b2, whereas the lower side is c2. (Which also shows that #8 might have been concluded in a shorter way.)
Also, picking just two triangles leads to a variant of Proofs #6 and #19:
In this form the proof appears in [Birkhoff, p. 92].
Yet another variant that could be related to #8 has been sent by James F.:
The latter has a twin with a and b swapping their roles.
Proof #42
The proof is based on the same diagram as #33 [Pritchard, p. 226-227].
Area of a triangle is obviously rp, where r is the incircle and p = (a + b + c)/2 the semiperimeter of the triangle. From the diagram, the hypothenuse c = (a - r) + (b - r), or r = p - c. The area of the triangle then is computed in two ways:
p(p - c) = ab/2,
which is equivalent to
(a + b + c)(a + b - c) = 2ab,
or
(a + b)2 - c2 = 2ab.
And finally
a2 + b2 - c2 = 0.
(The proof is due to Jack Oliver, and was originally published in Mathematical Gazette 81 (March 1997), p 117-118.)
Proof #43
By Larry Hoehn [Pritchard, p. 229, and Math Gazette].
Apply the Power of a Point theorem to the diagram above where the side a serves as a tangent to a circle of radius b: (c - b)(c + b) = a2. The result follows immediately.
(The configuration here is essentially the same as in proof #39. The invocation of the Power of a Point theorem may be regarded as a shortcut to the argument in proof #39. Also, this is exactly proof XVI by B. F. Yanney and J. A. Calderhead, Am Math Monthly, v.3, n. 12 (1896), 299-300.)
Proof #44
The following proof related to #39, have been submitted by Adam Rose (Sept. 23, 2004.)
Start with two identical right triangles: ABC and AFE, A the midpoint of BE and CF. Mark D on AB and G on extension of AF, such that
BC = BD = FG (= EF).
(For further notations refer to the above diagram.) BCD is isosceles. Therefore, BCD = p/2 - a/2. Since angle C is right,
ACD = p/2 - (p/2 - a/2) = a/2.
Since AFE is exterior to EFG, AFE = FEG + FGE. But EFG is also isosceles. Thus
AGE = FGE = a/2.
We now have two lines, CD and EG, crossed by CG with two alternate interior angles, ACD and AGE, equal. Therefore, CD||EG. Triangles ACD and AGE are similar, and AD/AC = AE/AG:
b/(c - a) = (c + a)/b,
and the Pythagorean theorem follows.
Proof #45
This proof is due to Douglas Rogers who came upon it in the course of his investigation into the history of Chinese mathematics. Two of his articles have online versions:
D. G. Rogers, Pythagoras framed, cut up by Liu Hui
D. G. Rogers, Beyond serendipity: how the Pythagorean proposition turns on the inscribed circle
The proof is a variation on #33, #34, and #42. The proof proceeds in two steps. First, as it may be observed from
a Liu Hui identity (see also Mathematics in China)
a + b = c + d,
where d is the diameter of the circle inscribed into a right triangle with sides a and b and hypotenuse c. Based on that and rearranging the pieces in two ways supplies another proof without words of the Pythagorean theorem:
Proof #46
This proof is due to Tao Tong (Mathematics Teacher, Feb., 1994, Reader Reflections). I learned of it through the good services of Douglas Rogers who also brought to my attention Proofs #47, #48 and #49. In spirit, the proof resembles the proof #32.
Let ABC and BED be equal right triangles, with E on AB. We are going to evaluate the area of ABD in two ways:
Area(ABD) = BD·AF/2 = DE·AB/2.
Using the notations as indicated in the diagram we get c(c - x)/2 = b·b/2. x = CF can be found by noting the similarity (BD AC) of triangles BFC and ABC:
x = a2/c.
The two formulas easily combine into the Pythagorean identity.
Proof #47
This proof which is due to a high school student John Kawamura was report by Chris Davis, his geometry teacher at Head-Rouce School, Oakland, CA (Mathematics Teacher, Apr., 2005, p. 518.)
The configuration is virtually identical to that of Proof #46, but this time we are interested in the area of the quadrilateral ABCD. Both of its perpendicular diagonals have length c, so that its area equals c2/2. On the other hand,
c2/2 = Area(ABCD)
= Area(BCD) + Area(ABD)
= a·a/2 + b·b/2
Multiplying by 2 yields the desired result.
Proof #48
(W. J. Dobbs, The Mathematical Gazette, 8 (1915-1916), p. 268.)
In the diagram, two right triangles - ABC and ADE - are equal and E is located on AB. As in President Garfield's proof, we evaluate the area of a trapezoid ABCD in two ways:
Area(ABCD) = Area(AECD) + Area(BCE)
= c·c/2 + a(b - a)/2,
where, as in the proof #47, c·c is the product of the two perpendicular diagonals of the quadrilateral AECD. On the other hand,
Area(ABCD) = AB·(BC + AD)/2
= b(a + b)/2.
Combining the two we get c2/2 = a2/2 + b2/2, or, after multiplication by 2, c2 = a2 + b2.
Proof #49
In the previous proof we may proceed a little differently. Complete a square on sides AB and AD of the two triangles. Its area is, on one hand, b2 and, on the other,
b2 = Area(ABMD)
= Area(AECD) + Area(CMD) + Area(BCE)
= c2/2 + b(b - a)/2 + a(b - a)/2
= c2/2 + b2/2 - a2/2,
which amounts to the same identity as before.
Douglas Rogers who observed the relationship between the proofs 46-49 also remarked that a square could have been drawn on the smaller legs of the two triangles if the second triangle is drawn in the "bottom" position as in proofs 46 and 47. In this case, we will again evaluate the area of the quadrilateral ABCD in two ways. With a reference to the second of the diagrams above,
c2/2 = Area(ABCD)
= Area(EBCG) + Area(CDG) + Area(AED)
= a2 + a(b - a)/2 + b(b - a)/2
= a2/2 + b2/2,
as was desired.
He also pointed out that it is possible to think of one of the right triangles as sliding from its position in proof #46 to its position in proof #48 so that its short leg glides along the long leg of the other triangle. At any intermediate position there is present a quadrilateral with equal and perpendicular diagonals, so that for all positions it is possible to construct proofs analogous to the above. The triangle always remains inside a square of side b - the length of the long leg of the two triangles. Now, we can also imagine the triangle ABC slide inside that square. Which leads to a proof that directly generalizes #49 and includes configurations of proofs 46-48. See below.
Proof #50
The area of the big square KLMN is b2. The square is split into 4 triangles and one quadrilateral:
b2 = Area(KLMN)
= Area(AKF) + Area(FLC) + Area(CMD) + Area(DNA) + Area(AFCD)
= y(a+x)/2 + (b-a-x)(a+y)/2 + (b-a-y)(b-x)/2 + x(b-y)/2 + c2/2
= [y(a+x) + b(a+y) - y(a+x) - x(b-y) - a·a + (b-a-y)b + x(b-y) + c2]/2
= [b(a+y) - a·a + b·b - (a+y)b + c2]/2
= b2/2 - a2/2 + c2/2.
It's not an interesting derivation, but it shows that, when confronted with a task of simplifying algebraic expressions, multiplying through all terms as to remove all parentheses may not be the best strategy. In this case, however, there is even a better strategy that avoids lengthy computations altogether. On Douglas Rogers' suggestion, complete each of the four triangles to an appropriate rectangle:
The four rectangles always cut off a square of size a, so that their total area is b2 - a2. Thus we can finish the proof as in the other proofs of this series:
b2 = c2/2 + (b2 - a2)/2.
Proof #51
(W. J. Dobbs, The Mathematical Gazette, 7 (1913-1914), p. 168.)
This one comes courtesy of Douglas Rogers from his extensive collection. As in Proof #2, the triangle is rotated 90o around one of its corners, such that the angle between the hypotenuses in two positions is right. The resulting shape of area b2 is then dissected into two right triangles with side lengths (c, c) and (b-a, a+b) and areas c2/2 and (b-a)(a+b)/2 = (b2 - a2)/2:
b2 = c2/2 + (b2 - a2)/2.
J. Elliott adds a wrinkle to the proof by turning around one of the triangles:
Again, the area can be computed in two ways:
ab/2 + ab/2 + b(b - a) = c2/2 + (b - a)(b + a)/2,
which reduces to
b2 = c2/2 + (b2 - a2)/2,
and ultimately to the Pythagorean identity.
Proof #52
This proof, discovered by a high school student, Jamie deLemos (The Mathematics Teacher, 88 (1995), p. 79.), has been quoted by Larry Hoehn (The Mathematics Teacher, 90 (1997), pp. 438-441.)
On one hand, the area of the trapezoid equals
(2a + 2b)/2·(a + b)
and on the other,
2a·b/2 + 2b·a/2 + 2·c2/2.
Equating the two gives a2 + b2 = c2.
The proof is closely related to President Garfield's proof.
Proof #53
Larry Hoehn also published the following proof (The Mathematics Teacher, 88 (1995), p. 168.):
Extend the leg AC of the right triangle ABC to D so that AD = AB = c, as in the diagram. At D draw a perpendicular to CD. At A draw a bisector of the angle BAD. Let the two lines meet in E. Finally, let EF be perpendicular to CF.
By this construction, triangles ABE and ADE share side AE, have other two sides equal: AD = AB, as well as the angles formed by those sides: BAE = DAE. Therefore, triangles ABE and ADE are congruent by SAS. From here, angle ABE is right.
It then follows that in right triangles ABC and BEF angles ABC and EBF add up to 90o. Thus
ABC = BEF and BAC = EBF.
The two triangles are similar, so that
x/a = u/b = y/c.
But, EF = CD, or x = b + c, which in combination with the above proportion gives
u = b(b + c)/a and y = c(b + c)/a.
On the other hand, y = u + a, which leads to
c(b + c)/a = b(b + c)/a + a,
which is easily simplified to c2 = a2 + b2.
Proof #54k
Later (The Mathematics Teacher, 90 (1997), pp. 438-441.) Larry Hoehn took a second look at his proof and produced a generic one, or rather a whole 1-parameter family of proofs, which, for various values of the parameter, included his older proof as well as #41. Below I offer a simplified variant inspired by Larry's work.
To reproduce the essential point of proof #53, i.e. having a right angled triangle ABE and another BEF, the latter being similar to ABC, we may simply place BEF with sides ka, kb, kc, for some k, as shown in the diagram. For the diagram to make sense we should restrict k so that kab. (This insures that D does not go below A.)
Now, the area of the rectangle CDEF can be computed directly as the product of its sides ka and (kb + a), or as the sum of areas of triangles BEF, ABE, ABC, and ADE. Thus we get
ka·(kb + a) = ka·kb/2 + kc·c/2 + ab/2 + (kb + a)·(ka - b)/2,
which after simplification reduces to
a2 = c2/2 + a2/2 - b2/2,
which is just one step short of the Pythagorean proposition.
The proof works for any value of k satisfying kb/a. In particular, for k = b/a we get proof #41. Further, k = (b + c)/a leads to proof #53. Of course, we would get the same result by representing the area of the trapezoid AEFB in two ways. For k = 1, this would lead to President Garfield's proof.
Obviously, dealing with a trapezoid is less restrictive and works for any positive value of k.
Proof #55
The following generalization of the Pythagorean theorem is due to W. J. Hazard (Am Math Monthly, v 36, n 1, 1929, 32-34). The proof is a slight simplification of the published one.
Let parallelogram ABCD inscribed into parallelogram MNPQ is shown on the left. Draw BK||MQ and AS||MN. Let the two intersect in Y. Then
Area(ABCD) = Area(QAYK) + Area(BNSY).
A reference to Proof #9 shows that this is a true generalization of the Pythagorean theorem. The diagram of Proof #9 is obtained when both parallelograms become squares.
The proof proceeds in 4 steps. First, extend the lines as shown below.
Then, the first step is to note that parallelograms ABCD and ABFX have equal bases and altitudes, hence equal areas (Euclid I.35 In fact, they are nicely equidecomposable.) For the same reason, parallelograms ABFX and YBFW also have equal areas. This is step 2. On step 3 observe that parallelograms SNFW and DTSP have equal areas. (This is because parallelograms DUCP and TENS are equal and points E, S, H are collinear. Euclid I.43 then implies equal areas of parallelograms SNFW and DTSP) Finally, parallelograms DTSP and QAYK are outright equal.
(There is a dynamic version of the proof.)
Proof #56
More than a hundred years ago The American Mathematical Monthly published a series of short notes listing great many proofs of the Pythagorean theorem. The authors, B. F. Yanney and J. A. Calderhead, went an extra mile counting and classifying proofs of various flavors. This and the next proof which are numbers V and VI from their collection (Am Math Monthly, v.3, n. 4 (1896), 110-113) give a sample of their thoroughness. Based on the diagram below they counted as many as 4864 different proofs. I placed a sample of their work on a separate page.
Proof #57
Treating triangle a little differently, now extending its sides instead of crossing them, B. F. Yanney and J. A. Calderhead came up with essentially the same diagram:
Following the method they employed in the previous proof, they again counted 4864 distinct proofs of the Pythagorean proposition.
Proof #58
(B. F. Yanney and J. A. Calderhead, Am Math Monthly, v.3, n. 6/7 (1896), 169-171, #VII)
Let ABC be right angled at C. Produce BC making BD = AB. Join AD. From E, the midpoint of CD, draw a perpendicular meeting AD at F. Join BF. DADC is similar to DBFE. Hence.
AC/BE = CD/EF.
But CD = BD - BC = AB - BC. Using this
BE = BC + CD/2
BE = BC + (AB - BC)/2
= (AB + BC)/2
and EF = AC/2. So that
AC·AC/2 = (AB - BC)·(AB + BC)/2,
which of course leads to AB2 = AC2 + BC2.
(As we've seen in proof 56, Yanney and Calderhead are fond of exploiting a configuration in as many ways as possible. Concerning the diagram of the present proof, they note that triangles BDF, BFE, and FDE are similar, which allows them to derive a multitude of proportions between various elements of the configuration. They refer to their approach in proof 56 to suggest that here too there are great many proofs based on the same diagram. They leave the actual counting to the reader.)
Proof #59
(B. F. Yanney and J. A. Calderhead, Am Math Monthly, v.3, n. 12 (1896), 299-300, #XVII)
Let ABC be right angled at C and let BC = a be the shortest of the two legs. With C as a center and a as a radius describe a circle. Let D be the intersection of AC with the circle, and H the other one obtained by producing AC beyond C, E the intersection of AB with the circle. Draw CL perpendicular to AB. L is the midpoint of BE.
By the Intersecting Chords theorem,
AH·AD = AB·AE.
In other words,
(b + a)(b - a) = c(c - 2·BL).
Now, the right triangles ABC and BCL share an angle at B and are, therefore, similar, wherefrom
BL/BC = BC/AB,
so that BL = a2/c. Combining all together we see that
b2 - a2 = c(c - 2a2/c)
and ultimately the Pythagorean identity.
Remark
Note that the proof fails for an esosceles right triangle. To accommodate this case, the authors suggest to make use of the usual method of the theory of limits. I am not at all certain what is the "usual method" that the authors had in mind. Perhaps, it is best to subject this case to Socratic reasoning which is simple and does not require the theory of limits. If the case is exceptional anyway, why not to treat it as such.
Proof #60
(B. F. Yanney and J. A. Calderhead, Am Math Monthly, v.3, n. 12 (1896), 299-300, #XVIII)
The idea is the same as before (proof #59), but now the circle has the radius b, the length of the longer leg. Having the sides produced as in the diagram, we get
AB·BK = BJ·BF,
or
c·BK = (b - a)(b + a).
BK, which is AK - c, can be found from the similarity of triangles ABC and AKH: AK = 2b2/c.
Note that, similar to the previous proof, this one, too, dos not work in case of the isosceles triangle.
Proof #61
(B. F. Yanney and J. A. Calderhead, Am Math Monthly, v.3, n. 12 (1896), 299-300, #XIX)
This is a third in the family of proofs that invoke the Intersecting Chords theorem. The radius of the circle equals now the altitude from the right angle C. Unlike in the other two proofs, there are now no exceptional cases. Refering to the diagram,
AD2 = AH·AE = b2 - CD2,
BD2 = BK·BL = a2 - CD2,
2AD·BD = 2CD2.
Adding the three yields the Pythagorean identity.
Proof #62
This proof, which is due to Floor van Lamoen, makes use of some of the many properties of the symmedian point. First of all, it is known that in any triangle ABC the symmedian point K has the barycentric coordinates proportional to the squares of the triangle's side lengths. This implies a relationship between the areas of triangles ABK, BCK and ACK:
Area(BCK) : Area(ACK) : Area(ABK) = a2 : b2 : c2.
Next, in a right triangle, the symmedian point is the midpoint of the altitude to the hypotenuse. If, therefore, the angle at C is right and CH is the altitude (and also the symmedian) in question, AK serves as a median of DACH and BK as a median of DBCH. Recollect now that a median cuts a triangle into two of equal areas. Thus,
Area(ACK) = Area(AKH) and
Area(BCK) = Area(BKH).
But
Area(ABK) = Area(AKH) + Area(BKH)
= Area(ACK) + Area(BCK),
so that indeed k·c2 = k·a2 + k·b2, for some k > 0; and the Pythagorean identity follows.
Floor also suggested a different approach to exploiting the properties of the symmedian point. Note that the symmedian point is the center of gravity of three weights on A, B and C of magnitudes a2, b2 and c2 respectively. In the right triangle, the foot of the altitude from C is the center of gravity of the weights on B and C. The fact that the symmedian point is the midpoint of this altitude now shows that a2 + b2 = c2.
Proof #63
This is another proof by Floor van Lamoen; Floor has been led to the proof via Bottema's theorem. However, the theorem is not actually needed to carry out the proof.
In the figure, M is the center of square ABA'B'. Triangle AB'C' is a rotation of triangle ABC. So we see that B' lies on C'B''. Similarly, A' lies on A''C''. Both AA'' and BB'' equal a + b. Thus the distance from M to AC' as well as to B'C' is equal to (a + b)/2. This gives
Area(AMB'C') = Area(MAC') + Area(MBC')
= (a + b)/2 · b/2 + (a + b)/2 · a/2
= a2/4 + ab/2 + b2/4.
But also:
Area(AMB'C') = Area(AMB') + Area(ABC')
= c2/4 + ab/2.
This yields a2/4 + b2/4 = c2/4 and the Pythagorean theorem.
The basic configuration has been exploited by B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 10, (1987), 250-251) to produce several proofs based on the following diagrams
None of their proofs made use of the centrality of point M.
Proof #64
And yet one more proof by Floor van Lamoen; in a quintessentially mathematical spirit, this time around Floor reduces the general statement to a particular case, that of a right asosceles triangle. The latter has been treated by Socrates and is shown independently of the general theorem.
FH devides the square ABCD of side a + b into two equal quadrilaterals, ABFH and CDHF. The former consists of two equal triangles with area ab/2, and an isosceles right triangle with area c2/2. The latter is composed of two isosceles right triangles: one of area a2/2, the other b2/2, and a right triangle whose area (by the introductory remark) equals ab! Removing equal areas from the two quadrlaterals, we are left with the identity of areas: a2/2 + b2/2 = c2/2.
Proof #65
This and the following proof are also due to Floor van Lamoen. Both a based on the following lemma, which appears to generalize the Pythagorean theorem: Form squares on the sides of the orthodiagonal quadrilateral. The squares fall into two pairs of opposite squares. Then the sum of the areas of the squares in two pairs are equal.
The proof is based on the friendly relationship between a triangles and its flank triangles: the altitude of a triangle through the right angle extended beyond the vertex is the median of the flank triangles at the right angle. With this in mind, note that the two parallelograms in the left figure not only share the base but also have equal altitudes. Therefore they have equal areas. Using shearing, we see that the squares at hand split into pairs of rectangles of equal areas, which can be combined in two ways proving the lemma.
For the proof now imagine two adjacent vertices of the quadrilateral closing in towards the point of intersection of the diagonals. In the limit, the quadrilateral will become a right triangle and one of the squares shrink to a point. Of the remaining three squares two will add up to the third.
Proof #66
(Floor van Lamoen). The lemma from Proof 65 can be used in a different way:
Let there be two squares: APBMc and C1McC2Q with a common vertex Mc. Rotation through 90o in the positive direction around Mc moves C1Mc into C2Mc and BMc into AMc. This implies that DBMcC1 rotates into DAMcC2 so that AC2 and BC1 are orthogonal. Quadrilateral ABC2C1 is thus orthodiagonal and the lemma applies: the red and blue squares add up to the same area. The important point to note is that the sum of the areas of the original squares APBMc and C1McC2Q is half this quantity.
Now assume the configurations is such that Mc coincides with the point of intersection of the diagonals. Because of the resulting symmetry, the red squares are equal. Therefore, the areas of APBMc and C1McC2Q add up to that of a red square!