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. The proof below is a somewhat shortened version of the original Euclidean proof as it appears in Sir Thomas Heath's translation.
First of all, ΔABF = ΔAEC by SAS. This is because,
∠BAF = ∠BAC + ∠CAF = ∠CAB + ∠BAE = ∠CAE. 
ΔABF 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, ΔAEC 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 ΔAEC 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 area 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), 168170 published several proofs based on the following diagrams
Some properties of this configuration has been proved on the Bride's Chair and others at the special Properties of the Figures in Euclid I.47 page.
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 90°, 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. This proof appears in a dynamic incarnation.
(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), 519528]. 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 strangely 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 tessellation.)
Proof #3
Now we start with four copies of the same triangle. Three of these have been rotated 90°, 180°, and 270°, 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

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)² = 4·ab/2 + c² 
simplifying which we get the needed identity.
A proof which combines this with proof #3 is credited to the 12th century Hindu mathematician Bhaskara (Bhaskara II):
Here we add the two identities
c² = (a  b)² + 4·ab/2 and c² = (a + b)²  4·ab/2 
which gives
2c² = 2a² + 2b². 
The latter needs only be divided by 2. This is the algebraic proof # 36 in Loomis' collection. Its variant, specifically applied to the 345 triangle, has featured in the Chinese classic Chou Pei Suan Ching dated somewhere between 300 BC and 200 AD and which Loomis refers to as proof 253.
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 
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 right triangle, now denoted ABC, and need only one additional construct  the altitude AD. The triangles ABC, DBA, and DAC 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

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.
R. M. Mentock has observed that a little trick makes the proof more succinct. In the common notations,
Finally, it must be mentioned that the configuration exploited in this proof is just a specific case of the one from the next proof  Euclid's second and less known proof of the Pythagorean proposition. A separate page is devoted to a proof by the similarity argument.
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 rightangled 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 rightangled 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 rightangled 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 the 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
Vladimir Nikolin from Serbia supplied a beautiful illustration:
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. The proof admits a hinged dissection illustrated by a Java applet.)
The first two pieces may be combined into one. The result appear in a 1830 book Sanpo Shinsyo  New Mathematics  by Chiba Tanehide (17751849), [H. Fukagawa, A. Rothman, Sacred Mathematics: Japanese Temple Geometry, Princeton University Press, 2008, p. 83].
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:
[Loomis, #53] attributes this construction to the great Leibniz, but lengthens the proof about threefold with meandering and misguided derivations.
B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.3, n. 12 (1896), 299300) offer a somewhat different route. Imagine FK is extended to the second intersection F' with the circle. Then, by the Intersecting Chords theorem,
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. 7475)
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: Sincerely, 
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 (or by Perigal, as above). 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.)) An especially nice variant by Olof Hanner appears on a separate page.
Proof #16
This proof is ascribed to Leonardo da Vinci (14521519) [Eves]. Quadrilaterals ABHI, JHBC, ADGC, and EDGF are all equal. (This follows from the observation that the angle ABH is 45°. This is so because ABC is rightangled, thus center O of the square ACJI lies on the circle circumscribing triangle ABC. Obviously, angle ABO is 45°.) Now,
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 rightangled 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
A dynamic illustration is available elsewhere.
Proof #18
This is another generalization that does not require right angles. It's due to Thâbit ibn Qurra (836901) [Eves]. If angles CAB, AC'B and AB'C are equal then
The same diagram is exploited in a different way by E. W. Dijkstra who concentrates on comparison of BC with the sum
Proof #19
This proof is a variation on #6. On the small side AB add a rightangled triangle ABD similar to ABC. Then, naturally, DBC is similar to the other two. From
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,
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. 1112.)
Proof #23
Another proof is based on the Heron's formula. (In passing, with the help of the formula I displayed the areas in the applet that illustrates Proof #7). 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. (A shorter and a more transparent application of Heron's formula is the basis of proof #75.)
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 CDROM.
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.
Scott Brodie pointed to the obvious relation of this proof to # 9. It is the same configuration but short of one triangle.
Proof #25
B.F.Yanney (1903, [Swetz]) gave a proof using the "shearing 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).
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 rearranged in yet another manner.
This dissection is often attributed to the 17^{th} 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 tessellation in proof #15. As a matter of fact, it may be better explained by a different tessellation. (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), 3334) 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 (that since disappeared) to a page by Donald B. Wagner, an expert on history of science and technology in China. Dr. Wagner appeared to have reconstructed a proof by Liu Hui (third century AD). However (see below), there are serious doubts to the authorship of the proof.
Elisha Loomis cites this as the geometric proof #28 with the following comment:
 Benjir von Gutheil, oberlehrer at Nurnberg, Germany, produced the above proof. He died in the trenches in France, 1914. So wrote J. Adams, August 1933.
 Let us call it the B. von Gutheil World War Proof.
Judging by the Sweet Land movie, such forgiving attitude towards a German colleague may not have been common at the time close to the WWI. It might have been even more guarded in the 1930s during the rise to power of the nazis in Germany.
(I thank D. Rogers for bringing the reference to Loomis' collection to my attention. He also expressed a reservation as regard the attribution of the proof to Liu Hui and traced its early appearance to Karl Julius Walther Lietzmann's Geometrische aufgabensamming Ausgabe B: fuer Realanstalten, published in Leipzig by Teubner in 1916. Interestingly, the proof has not been included in Lietzmann's earlier Der Pythagoreische Lehrsatz published in 1912.)
Proof #29
A mechanical proof of the theorem deserves a page of its own.
Pertinent to that proof is a page "Extrageometric" proofs of the Pythagorean Theorem by Scott Brodie
Proof #39
(By J. Barry Sutton, The Math Gazette, v 86, n 505, March 2002, p72.)
Let in ΔABC, angle C = 90°. As usual,
By construction, C lies on the circle with center A and radius b. Angle DCE subtends its diameter and thus is right:
Triangles DBC and EBC share DBC. In addition,
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. However, this is exactly proof 14 from Elisha Loomis' collection. Furthermore, Loomis provides two earlier references from 1925 and 1905. With the circle centered at A drawn, Loomis repeats the proof as 82 (with references from 1887, 1880, 1859, 1792) and also lists (as proof 89) a symmetric version of the above:
For the right triangle ABC, with right angle at C, extend AB in both directions so that
a² = c²  b², and b² = c²  a². 
Instead of using either of the identities directly, Loomis adds the two:
2(a² + b²) = 2c², 
which appears as both graphical and algebraic overkill.
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
y·dy  x·dx = 0, 
which after integration gives y²  x² = const. The value of the constant is determined from the initial condition for
It is easy to take an issue with this proof. What does it mean for a triangle to be
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
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
An amplified, but apparently independent, version of this proof has been published by Mike Staring (Mathematics Magazine, V. 69, n. 1 (Feb., 1996), 4546).
Assuming Δx > 0 and detecting similar triangles,
Δf / Δx = CQ/CD > CP/CD = CA/CB = x/f(x). 
But also,
Δf / Δx = SD/CD < RD/CD = AD/BD = (x + Δx) / (f(x) + Δf) < x/f(x) + Δx/f(x). 
Passing to the limit as Δx tends to 0^{+}, we get
df / dx = x / f(x). 
The case of Δx < 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
Proof #41
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
For additional details and modifications see a separate page.
Proof #42
The proof is based on the same diagram as #33 [Pritchard, p. 226227].
Area of a triangle is obviously rp, where r is the inradius and
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 117118.
Maciej Maderek informed me that the same proof appeared in a Polish 1988 edition of Sladami Pitagorasa by Szczepan Jelenski:
Jelenski attributes the proof to Möllmann without mentioning a source or a date.
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:
(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), 299300.)
John Molokach suggested a modification based on the following diagram:
From the similarity of triangles, a/b = (b + c)/d, so that
bd + ab/2 = (b + c)d/2 
which simplifies to
ab/2 = (c  b)d/2, or ab = (c  b)d. 
Now using the formula for d:
ab = (c  b)d = (c  b)(c + b)b/a. 
Dividing by b and multiplying by a gives a² = c²  b². This variant comes very close to Proof #82, but with a different motivation.
Finally, the argument shows that the area of an annulus (ring) bounded by circles of radii b and
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 intersection 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.) ΔBCD is isosceles. Therefore,
∠ACD = p/2  (p/2  α/2) = α/2. 
Since ∠AFE is exterior to ΔEFG,
∠AGE = ∠FGE = α/2. 
We now have two lines, CD and EG, crossed by CG with two alternate interior angles, ACD and AGE, equal. Therefore, CDEG. 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.
 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 ΔABD in two ways:
Area(ΔABD) = BD·AF/2 = DE·AB/2. 
Using the notations as indicated in the diagram we get
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 HeadRouce 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,

Multiplying by 2 yields the desired result.
Proof #48
(W. J. Dobbs, The Mathematical Gazette, 8 (19151916), 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:

where, as in the proof #47, c·c is the product of the two perpendicular diagonals of the quadrilateral AECD. On the other hand,

Combining the two we get c²/2 = a²/2 + b²/2, or, after multiplication by 2,
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,

which amounts to the same identity as before.
Douglas Rogers who observed the relationship between the proofs 4649 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,

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 4648. See below.
Proof #50
The area of the big square KLMN is b². The square is split into 4 triangles and one quadrilateral:

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 (19131914), p. 168.)
This one comes courtesy of Douglas Rogers from his extensive collection. As in Proof #2, the triangle is rotated 90 degrees 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
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. 438441.)
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
By this construction, triangles ABE and ADE share side AE, have other two sides equal:
It then follows that in right triangles ABC and BEF angles ABC and EBF add up to 90°. 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 #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), 110113) 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 the 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), 169171, #VII)
Let ABC be right angled at C. Produce BC making
AC/BE = CD/EF. 
But CD = BD  BC = AB  BC. Using this

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 #6۷
This proof was sent to me by a 14 year old Sina Shiehyan from Sabzevar, Iran. The circumcircle aside, the combination of triangles is exactly the same as in S. Brodie's subcase of Euclid's VI.31. However, Brodie's approach if made explicit would require argument different from the one employed by Sina. So, I believe that her derivation well qualifies as an individual proof.
From the endpoints of the hypotenuse AB drop perpendiculars AP and BK to the tangent to the circumcircle of ΔABC at point C. Since OC is also perpendicular to the tangent, C is the midpoint of KP. It follows that

Therefore, Area(ABC) is also Area(ABKP)/2. So that
Area(ACP) + Area(BCK) = Area(ABC) 
Now all three triangles are similar (as being right and having equal angles), their areas therefore related as the squares of their hypotenuses, which are b, a, and c respectively. And the theorem follows.