The story

The Pythagorean Theorem: The Way of Truth

Pythagoras (569-475 BC) is recognized as the world's first mathematician. He was born on the island of Samos and was thought to study with Thales and Anaximander (recognized as the first western philosophers). Pythagoras believed that numbers were not only the way to truth, but truth itself. Through mathematics, one could attain harmony and live an easier life. He is said to have proposed a number of mathematical theorems to this end but, of all these, only the famous Pythagorean Theorem remains (Allen, 1966).

The historian Robinson writes, “The statement that `Pythagoras worked very hard at the arithmetical side of geometry' is further borne out by the tradition that he investigated the arithmetical problem of finding triangles having the square on one side equal to the sum of the squares on the other two” and did so, early on, by using stones in rows to understand the truths he was trying to convey (1968). The Pythagorean Theorem states that a² + b² = c². This is used when we are given a triangle in which we only know the length of two of the three sides. C is the longest side of the angle known as the hypotenuse. If a is the adjacent angle then b is the opposite side. If b is the adjacent angle then a is the opposite side. If a = 3, and b = 4, we could then solve for c. 32 + 42 = c². 9 + 16 = c². 25 = c². c = 5. This is one of the prime uses of the Pythagorean Theorem.

There are many proofs of the Pythagorean Theorem, the best known being Euclid's proof from Book I of his Elements.

Proposition: In right-angled triangles the square on the hypotenuse is equal to the sum of the squares on the legs.

Euclid started with a Pythagorean configuration and then drew a line through a diagram illustrating the equalities of the areas. He concluded that AB/AC = AC/HA, therefore (AC)² = (HA)(AB). Since AB=AJ, the area of the rectangle HAJG corresponds to the area of the square on side AC. Similarly, AB/BC = BC/BH also written as (BC)² = (BH)(AB) = (BH)(BD) and since AB=BD. Thus we see that the sum of the areas of the rectangles is the area of the square on the hypotenuse. In the words of Stephanie Morris, “This completes the proof” (Morris, 2011).

Another proof, which is easier for people to understand, starts off with a rectangle divided into three triangles, all with right angles.

Triangle BEA and triangle BCE overlap triangle ACD. Comparing triangle BCE and triangle ACD, and looking at their corresponding sides, we see that AC/BC = AD/EC. Since AD = BC, AC/AD = AD/EC. Through multiplication this equation is rendered (AD)² = (AC)(AE). From triangles ABC and ABE, noting that AB = CD, comparing the right angles of these two figures we render the equation AC/AB = CD/AE. From the original rectangle shape we had AB = CD also given as AC/CD = CD/AE, which is written as a multiplication problem as (CD)² = (AC)(AE) and by adding the equations we have so far, we obtain two new formulas which are (CD)² + (AD)² = (AC)(AE) + (AC) (EC) and (CD)² + (AD)² = (AC)(AE + EC). Since AC = AE + EC, we obtain (CD)² + (AD)² = (AC)². As with the earlier proof, this shows the validity of the Pythagorean Theorem (Morris, 2011).

In the Pythagorean Theorem every side/angle is a critical piece of information that helps us determine other angles/sides. Pythagoras believed in an objective truth which was number. The Pythagorean Theorem allows for truths to be known through the mathematical equations above which means that there does exist an objective truth, outside of any personal opinion, which can actually be proven; and this, finally, is what Pythagoras wanted to prove through his work.

Love History?

Sign up for our free weekly email newsletter!

The Pythagorean Theorem: The Way of Truth - History

This essay was inspired by a class that I am taking this quarter. The class is the History of Mathematics . In this class, we are learning how to include the history of mathematics in teaching a mathematics. One way to include the history of mathematics in your classroom is to incorporate ancient mathematics problems in your instruction. Another way is to introduce a new topic with some history of the topic. Hopefully, this essay will give you some ideas of how to include the history of the Pythagorean Theorem in the teaching and learning of it.

We have been discussing different topics that were developed in ancient civilizations. The Pythagorean Theorem is one of these topics. This theorem is one of the earliest know theorems to ancient civilizations. It was named after Pythagoras, a Greek mathematician and philosopher. The theorem bears his name although we have evidence that the Babylonians knew this relationship some 1000 years earlier. Plimpton 322 , a Babylonian mathematical tablet dated back to 1900 B.C., contains a table of Pythagorean triples. The Chou-pei , an ancient Chinese text, also gives us evidence that the Chinese knew about the Pythagorean theorem many years before Pythagoras or one of his colleagues in the Pythagorean society discovered and proved it. This is the reason why the theorem is named after Pythagoras.

Pythagoras lived in the sixth or fifth century B.C. He founded the Pythagorean School in Crotona. This school was an academy for the study of mathematics, philosophy, and natural science. The Pythagorean School was more than a school it was "a closely knit brotherhood with secret rites and observances" (Eves 75). Because of this, the school was destroyed by democratic forces of Italy. Although the brotherhood was scattered, it continued to exist for two more centuries. Pythagoras and his colleagues are credited with many contributions to mathematics.

The following is an investigation of how the Pythagorean theorem has been proved over the years.

"The square on the hypotenuse of a right triangle is equal to the sum of the squares on the two legs" (Eves 80-81).

This theorem is talking about the area of the squares that are built on each side of the right triangle.

Accordingly, we obtain the following areas for the squares, where the green and blue squares are on the legs of the right triangle and the red square is on the hypotenuse.

area of the green square is
area of the blue square is
area of the red square is

From our theorem, we have the following relationship:

area of green square + area of blue square = area of red square or

As I stated earlier, this theorem was named after Pythagoras because he was the first to prove it. He probably used a dissection type of proof similar to the following in proving this theorem.

"Let a, b, c denote the legs and the hypotenuse of the given right triangle, and consider the two squares in the accompanying figure, each having a+b as its side. The first square is dissected into six pieces-namely, the two squares on the legs and four right triangles congruent to the given triangle. The second square is dissected into five pieces-namely, the square on the hypotenuse and four right triangles congruent to the given triangle. By subtracting equals from equals, it now follows that the square on the hypotenuse is equal to the sum of the squares on the legs" (Eves 81).

Consider the following figure.

The area of the first square is given by (a+b)^2 or 4(1/2ab)+ a^2 + b^2.
The area of the second square is given by (a+b)^2 or 4(1/2ab) + c^2.
Since the squares have equal areas we can set them equal to another and subtract equals. The case (a+b)^2=(a+b)^2 is not interesting. Let's do the other case.
4(1/2ab) + a^2 + b^2 = 4(1/2ab)+ c^2
Subtracting equals from both sides we have

concluding Pythagoras' proof.
Over the years there have been many mathematicians and non-mathematicians to give various proofs of the Pythagorean Theorem. Following are proofs from Bhaskara and one of our former presidents, President James Garfield. I have chosen these proofs because any of them would be appropriate to use in any classroom.

Bhaskara's First Proof

Bhaskara's proof is also a dissection proof. It is similar to the proof provided by Pythagoras. Bhaskara was born in India. He was one of the most important Hindu mathematicians of the second century AD. He used the following diagrams in proving the Pythagorean Theorem.

In the above diagrams, the blue triangles are all congruent and the yellow squares are congruent. First we need to find the area of the big square two different ways. First let's find the area using the area formula for a square.
Thus, A=c^2.
Now, lets find the area by finding the area of each of the components and then sum the areas.
Area of the blue triangles = 4(1/2)ab
Area of the yellow square = (b-a)^2
Area of the big square = 4(1/2)ab + (b-a)^2
= 2ab + b^2 - 2ab + a^2
= b^2 + a^2

Since, the square has the same area no matter how you find it
A = c^2 = a^2 + b^2,
concluding the proof.

Bhaskara's Second Proof of the Pythagorean Theorem

In this proof, Bhaskara began with a right triangle and then he drew an altitude on the hypotenuse. From here, he used the properties of similarity to prove the theorem.

Now prove that triangles ABC and CBE are similar.
It follows from the AA postulate that triangle ABC is similar to triangle CBE, since angle B is congruent to angle B and angle C is congruent to angle E. Thus, since internal ratios are equal s/a=a/c.
Multiplying both sides by ac we get

Now show that triangles ABC and ACE are similar.
As before, it follows from the AA postulate that these two triangles are similar. Angle A is congruent to angle A and angle C is congruent to angle E. Thus, r/b=b/c. Multiplying both sides by bc we get

Now when we add the two results we get
sc + rc = a^2 + b^2.
c(s+r) = a^2 + b^2
c^2 = a^2 + b^2,
concluding the proof of the Pythagorean Theorem.

Garfield's Proof

The twentieth president of the United States gave the following proof to the Pythagorean Theorem. He discovered this proof five years before he become President. He hit upon this proof in 1876 during a mathematics discussion with some of the members of Congress. It was later published in the New England Journal of Education .. The proof depends on calculating the area of a right trapezoid two different ways. The first way is by using the area formula of a trapezoid and the second is by summing up the areas of the three right triangles that can be constructed in the trapezoid. He used the following trapezoid in developing his proof.

First, we need to find the area of the trapezoid by using the area formula of the trapezoid.
A=(1/2)h(b1+b2) area of a trapezoid

In the above diagram, h=a+b, b1=a, and b2=b.

Now, let's find the area of the trapezoid by summing the area of the three right triangles.
The area of the yellow triangle is

The area of the red triangle is

The area of the blue triangle is
A= 1/2(ab).

The sum of the area of the triangles is
1/2(ba) + 1/2(c^2) + 1/2(ab) = 1/2(ba + c^2 + ab) = 1/2(2ab + c^2).

Since, this area is equal to the area of the trapezoid we have the following relation:
(1/2)(a^2 + 2ab + b^2) = (1/2)(2ab + c^2).

Pythagorean Theorem

Why is mathematics different (in a good way) from every other subject you learned in school?

Two words: Pythagorean Theorem.

Let me explain. The Pythagorean Theorem in itself isn’t really the reason math is unique it is merely an example I wish to use to illustrate my point. I chose this Theorem for an example because it has been my experience that it is one of the few things everyone remembers from math class, regardless of how much they enjoyed math or how well they did in the course. But just in case the P.T. slipped your mind, here is a recap:

For any right triangle, the square of the hypotenuse (side opposite the right (90 degree) angle), is equal to the sum of the square of the other two sides.

This result is attributed to the Greek mathematician and philosopher Pythagoras (hence the creative name for the theorem). Pythagoras lived between the 5th and 6th century B.C. and while he is ultimately the one credited with proving the theorem, there is evidence that the result of the theorem was known to the Babylonians 1000 years before Pythagoras was born. Notice this old tablet:

Wow, that is old. Here you can read more about the Babylonians and the Pythagorean Theorem.

My point is that in what other class are you performing the same operations as people were performing 3000 years ago? Certainly in history class you learn about earlier civilizations, but you are not being taught how to do history in the same manner as those civilizations. The precision that modern history requires was largely unknown to those ancient people. Perhaps in literature you read Homer’s Iliad and Odyssey, but again, you aren’t being taught to write in the same style of epic poetry.

So then why is it that in math class, while advancements have been made and technology certainly has come a long way, we still find it beneficial to perform calculations the way they were performed thousands of years ago?

My answer: there is nothing to perfect, nothing ot improve upon, when you come across truth. Real truth.

To all of us who hold the Christian belief that God is truth, anything that is true is a fact about God, and mathematics is a branch of theology.

"The chief aim of all investigations of the external world should be to discover the rational order and harmony which has been imposed on it by God and which He revealed to us in the language of mathematics."

Pythagoras's theorem in Babylonian mathematics

In this article we examine four Babylonian tablets which all have some connection with Pythagoras's theorem. Certainly the Babylonians were familiar with Pythagoras's theorem. A translation of a Babylonian tablet which is preserved in the British museum goes as follows:-

All the tablets we wish to consider in detail come from roughly the same period, namely that of the Old Babylonian Empire which flourished in Mesopotamia between 1900 BC and 1600 BC.

Here is a map of the region where the Babylonian civilisation flourished.

The article Babylonian mathematics gives some background to how the civilisation came about and the mathematical background which they inherited.

The four tablets which interest us here we will call the Yale tablet YBC 7289 , Plimpton 322 ( shown below ) , the Susa tablet, and the Tell Dhibayi tablet. Let us say a little about these tablets before describing the mathematics which they contain.

The Yale tablet YBC 7289 which we describe is one of a large collection of tablets held in the Yale Babylonian collection of Yale University. It consists of a tablet on which a diagram appears. The diagram is a square of side 30 with the diagonals drawn in. The tablet and its significance was first discussed in [ 5 ] and recently in [ 18 ] .

Plimpton 322 is the tablet numbered 322 in the collection of G A Plimpton housed in Columbia University.

You can see from the picture that the top left hand corner of the tablet is damaged as and there is a large chip out of the tablet around the middle of the right hand side. Its date is not known accurately but it is put at between 1800 BC and 1650 BC. It is thought to be only part of a larger tablet, the remainder of which has been destroyed, and at first it was thought, as many such tablets are, to be a record of commercial transactions. However in [ 5 ] Neugebauer and Sachs gave a new interpretation and since then it has been the subject of a huge amount of interest.

The Susa tablet was discovered at the present town of Shush in the Khuzistan region of Iran. The town is about 350 km from the ancient city of Babylon. W K Loftus identified this as an important archaeological site as early as 1850 but excavations were not carried out until much later. The particular tablet which interests us here investigates how to calculate the radius of a circle through the vertices of an isosceles triangle.

Finally the Tell Dhibayi tablet was one of about 500 tablets found near Baghdad by archaeologists in 1962 . Most relate to the administration of an ancient city which flourished in the time of Ibalpiel II of Eshunna and date from around 1750 . The particular tablet which concerns us is not one relating to administration but one which presents a geometrical problem which asks for the dimensions of a rectangle whose area and diagonal are known.

Before looking at the mathematics contained in these four tablets we should say a little about their significance in understanding the scope of Babylonian mathematics. Firstly we should be careful not to read into early mathematics ideas which we can see clearly today yet which were never in the mind of the author. Conversely we must be careful not to underestimate the significance of the mathematics just because it has been produced by mathematicians who thought very differently from today's mathematicians. As a final comment on what these four tablets tell us of Babylonian mathematics we must be careful to realise that almost all of the mathematical achievements of the Babylonians, even if they were all recorded on clay tablets, will have been lost and even if these four may be seen as especially important among those surviving they may not represent the best of Babylonian mathematics.

There is no problem understanding what the Yale tablet YBC 7289 is about.

Here is a Diagram of Yale tablet

It has on it a diagram of a square with 30 on one side, the diagonals are drawn in and near the centre is written 1 , 24 , 51 , 10 and 42 , 25 , 35 . Of course these numbers are written in Babylonian numerals to base 60 . See our article on Babylonian numerals. Now the Babylonian numbers are always ambiguous and no indication occurs as to where the integer part ends and the fractional part begins. Assuming that the first number is 1 24 , 51 , 10 then converting this to a decimal gives 1 . 414212963 while √ 2 = 1 . 414213562 . Calculating 30 × [ 1 24 , 51 , 10 ] gives 42 25 , 35 which is the second number. The diagonal of a square of side 30 is found by multiplying 30 by the approximation to √ 2 .

This shows a nice understanding of Pythagoras's theorem. However, even more significant is the question how the Babylonians found this remarkably good approximation to √ 2 . Several authors, for example see [ 2 ] and [ 4 ] , conjecture that the Babylonians used a method equivalent to Heron's method. The suggestion is that they started with a guess, say x x x . They then found e = x 2 − 2 e = x^ <2>- 2 e = x 2 − 2 which is the error. Then

This is certainly possible and the Babylonians' understanding of quadratics adds some weight to the claim. However there is no evidence of the algorithm being used in any other cases and its use here must remain no more than a fairly remote possibility. May I [ EFR ] suggest an alternative. The Babylonians produced tables of squares, in fact their whole understanding of multiplication was built round squares, so perhaps a more obvious approach for them would have been to make two guesses, one high and one low say a a a and b b b . Take their average a + b 2 Largefrac 2 2 a + b ​ and square it. If the square is greater than 2 then replace b b b by this better bound, while if the square is less than 2 then replace a a a by a + b 2 Largefrac 2 2 a + b ​ . Continue with the algorithm.

Now this certainly takes many more steps to reach the sexagesimal approximation 1 24 , 51 , 10 . In fact starting with a = 1 a = 1 a = 1 and b = 2 b = 2 b = 2 it takes 19 steps as the table below shows: However, the Babylonians were not frightened of computing and they may have been prepared to continue this straightforward calculation until the answer was correct to the third sexagesimal place.

Next we look again at Plimpton 322

The tablet has four columns with 15 rows. The last column is the simplest to understand for it gives the row number and so contains 1 , 2 , 3 , . , 15 . The remarkable fact which Neugebauer and Sachs pointed out in [ 5 ] is that in every row the square of the number c c c in column 3 minus the square of the number b b b in column 2 is a perfect square, say h h h .

So the table is a list of Pythagorean integer triples. Now this is not quite true since Neugebauer and Sachs believe that the scribe made four transcription errors, two in each column and this interpretation is required to make the rule work. The errors are readily seen to be genuine errors, however, for example 8 , 1 has been copied by the scribe as 9 , 1 .

Several historians ( see for example [ 2 ] ) have suggested that column 1 is connected with the secant function. However, as Joseph comments [ 4 ] :-

Zeeman has made a fascinating observation. He has pointed out that if the Babylonians used the formulas h = 2 m n , b = m 2 − n 2 , c = m 2 + n 2 h = 2mn, b = m^<2>-n^<2>, c = m^<2>+n^ <2>h = 2 m n , b = m 2 − n 2 , c = m 2 + n 2 to generate Pythagorean triples then there are exactly 16 triples satisfying n ≤ 60 , 30 ° ≤ t ≤ 45 ° n ≤ 60, 30° ≤ t ≤ 45° n ≤ 6 0 , 3 0 ° ≤ t ≤ 4 5 ° , and tan ⁡ 2 t = h 2 / b 2 an^<2>t = h^<2>/b^ <2>tan 2 t = h 2 / b 2 having a finite sexagesimal expansion ( which is equivalent to m , n , b m, n, b m , n , b having 2 , 3 , and 5 as their only prime divisors ) . Now 15 of the 16 Pythagorean triples satisfying Zeeman's conditions appear in Plimpton 322 . Is it the earliest known mathematical classification theorem? Although I cannot believe that Zeeman has it quite right, I do feel that his explanation must be on the right track.

To give a fair discussion of Plimpton 322 we should add that not all historians agree that this tablet concerns Pythagorean triples. For example Exarchakos, in [ 17 ] , claims that the tablet is connected with the solution of quadratic equations and has nothing to do with Pythagorean triples:-

The Susa tablet sets out a problem about an isosceles triangle with sides 50 , 50 and 60 . The problem is to find the radius of the circle through the three vertices.

Pythagoras' Other Theorem: A Short History of Vegetarianism

Recently, on her weekly Heritage Radio Network program, "A Taste of the Past," Linda Pellacio interviewed Rynn Berry, an author and historical advisor to the North American Vegetarian Society.

Berry has been a vegetarian since learning, as a teenager, that animals experience anxiety before slaughter. His vegetarianism has since evolved into a vegan lifestyle, which means he excludes all animal products, including honey, not just from his diet but also from his clothing.

With Pellacio, Berry discussed the trajectory of vegetarianism, which has been a documented part of history since the sixth century B.C. According to Berry, the first vegetarian society was founded by the ancient Greek mathematician, Pythagoras (a key player in ninth grade geometry). Not only did Pythagoras demystify triangles, he also spread the gospel of the Buddha, a contemporary of Pythagoras who inspired him personally to practice non-violent vegetarianism. For Pythagoras, abstaining from meat was rooted in his spiritual values nutrition would not become a factor in the diet until much later in history. In fact, a diet void of any animal products was actually called a "Pythagorean" diet until 1944, when Donald Watson, founder of the Vegan Society, coined the word vegan. Vegetarianism was first documented in 1848, most likely by an Oxford Scholar.

Berry has written several books on vegetarianism, including Famous Vegetarians. Notable meat-abstainers include Benjamin Franklin, whom Berry described as "the only founding father to have a fling with vegetarianism," as well as George Bernard Shaw, who was famously told by a team of doctors that he needed to eat meat or starve. Not only did he not starve, he lived until the age of 94.

Other vegetarians of the 19th century have had the legacy of their names enter the industrial food lexicon of today. John Harvey Kellogg, a Seventh-Day Adventist and the inventor of corn flakes, created the cereal as an alternative meatless breakfast option. Sylvester Graham, a Presbyterian minister who preached for temperance, whole grains, and vegetarian diets, created a cracker he believed was a nutritionally superior product. S'mores enthusiasts can rest assured that the modern version of the beloved campfire treat, the graham cracker, bears little resemblance to the original prototype.

The trajectory of vegetarianism is particularly interesting, especially in America, where history has recorded its renaissance on several occasions. The early vegetarians who I have named in this article were all inspired by their respective religions to cling to a meat-free diet. Their goals may have varied, but the common impetus was a spiritual sense of clarity that was thought to be achieved by eating a diet void of flesh. It wasn't until the 20th century that America embraced vegetarianism in a secular fashion. The baby-boomer generation, spurred by the violence of the 1960's and disparaged by threats of imminent ecological disasters, largely embraced a diet inspired by ecology and a desire to get closer to the Earth. By the time Frances Moore Lappé's iconic book, Diet for a Small Planet (1971), was published, vegetarianism had found its way into mainstream America's collective consciousness.

Today we are seeing a vegetarian redux. On one hand, nutrition is worshipped in society and a meat-free diet has become an acceptable entry point into a healthy lifestyle. Even extreme versions of vegetarianism, such as veganism and raw-food diets, have begun to shed their stigma. William Jefferson Clinton, who was not a founding father but is a beloved former president, has been outspoken about his drastic transition from a fast-food fueled diet to a strict vegan one. Rynn Berry would refer to Clinton as a "coronary vegetarian," someone who moves to a plant-based diet on the recommendation of their doctor after having a heart attack or a major procedure. Possibly inspired by their former president, or perhaps just riding the current trend wave, the American people have listened to a litany of testaments from celebrities who swear by their new meat-free diets. Rarely are ethics the impetus, and the nutrition-focused impulse has created a cross section of players on the corner of "I want my meat" and "I want to feel good about it too." This new business of wanting to stay healthy without sacrificing taste cravings has inspired movements such as "Meatless Mondays," which encourages a commitment to eating lower on the food chain without having to go cold turkey. Or cold tofurkey, as the case may be.

The ethical devotion to a cruelty-free diet, we can see, has been tempered and popularized by a re-shift of focus. Yes, we still care about animals, but now that we know that we can consume creatures that have lived healthy and happy lives, we no longer have to stress about their blood staying on our hands. It is important to note that only five percent of Americans identify themselves as vegetarians and for the majority of the population who are meat eaters, there are other ways to impact the environment and one's own health in a powerful and positive way. Pushing for breed diversity in our meat supply and purchasing only sustainably raised livestock are effective and important choices that meat-eaters should consider. Vegetarianism in itself is inherently complicated the deep relationships between the livestock and dairy industries make for considerable debate when choosing to exclude meat from a diet but not cheese and milk. Regardless of the choices you make in your diet, the more the dots are connected between health, compassion and ecology, the more nourishing your diet will become for your mind and your body.

Listen to the original interview between Linda Pellacio and Rynn Berry here.

To learn more about Rynn Berry and his books on vegetarianism, click here.

The Pythagorean Theorem: The Way of Truth - History

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

In algebraic terms, a 2 + b 2 = c 2 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. One of my first Java applets was written to illustrate another Euclidean proof. Presently, there are several Java illustrations of various proofs, but the majority have been rendered in plain HTML with simple graphic diagrams.


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 a 2 +b 2 =c 2 is right angled. Euclid was the first (I.48) to mention and prove this fact.

W. Dunham [Mathematical Universe] 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.

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

Larry Hoehn came up with a plane generalization which is related to the law of cosines but is shorther and looks nicer.

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.

(EWD) sign( a + b - g ) = sign(a 2 + b 2 - c 2 ),

where sign(t) is the signum function:

The theorem this page is devoted to is treated as "If then Dijkstra deservedly finds (EWD) more symmetric and more informative. Absence of transcendental quantities ( p ) is judged to be an additional advantage.

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 a 2 +b 2 .

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 o , 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 c 2 .

(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 90 o , 180 o , and 270 o , 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 Summing up its area and 2ab, the area of the four triangles (4·ab/2), we get

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 + c 2

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 a 2 +b 2 =c 2 .

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


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.

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.


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.

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.

ab/c · (a 2 +b 2 )/c + ab/2 = ab + (ab/c · a 2 /c + ab/c · b 2 /c)/2

ab/c · (a 2 +b 2 )/c/2 = ab/2, or (a 2 +b 2 )/c 2 = 1


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 a 2 /c+b 2 /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.)

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 the ratio (c+b)/(c-b). So, as in Proof #6, we get a 2 = (c+b)(c-b) = c 2 - b 2 .

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 a 2 +b 2 .) 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:

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. (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 45 o . 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 45 o .) 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 Indeed, triangles ABC, AC'B and AB'C are similar. Thus we have and 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 AD = AB 2 /AC and BD = AB·BC/AC we derive Dividing by AB/AC leads to

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, In addition, triangles ABB' and ABC' are also equal. Thus we conclude that From the similarity of triangles we get as before B'C = AC 2 /BC and BC' = AC·AB/BC. Putting it all together yields which is the same as

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 a 2 + b 2 = c 2 .

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 a 2 = xc similarly, AC is tangent to the circle with diameter BC, and b 2 = yc. Adding, we find a 2 + b 2 = xc + yc = c 2 , 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 one hand, the area of the shape ABDFH equals AC 2 + BC 2 + area(ABC + FMC + FLC). On the other hand, area(ABDFH) = AB 2 + 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.

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 AC 2 ) are equal as are the areas of HMOB, HKCB, and HKDF (which BC 2 ). BC = DF. Thus AC 2 + BC 2 = area(LMOA) + area(HMOB) = area(ABHL) = AB 2 .

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 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 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.)

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 b 2 /4. We obtain

Here'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 Let M be the midpoint of the hypotenuse. Denote the intersection of MC and PQ as R. Let's show that

The median to the hypotenuse equals half of the latter. Therefore, CMB is isosceles and But we also have From here and it follows that angle CRP is right, or

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 Therefore, On the other hand, Similarly, and also

We may sum up the two identities: or

(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. , , . Obviously, Compute the area of ADE in two different ways.

Area(ADE) = AB·DE/2 = c 2 /2 and also CE can be found from similar triangles BCE and DFE: Putting things together we obtain

(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:

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.

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.

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 = 90 o . As usual, Define points D and E on AB so that

By construction, C lies on the circle with center A and radius b. Angle DCE subtends its diameter and thus is right: It follows that Since ACE is isosceles,

Triangles DBC and EBC share DBC. In addition, Therefore, triangles DBC and EBC are similar. We have or

a 2 = c 2 - b 2 ,
a 2 + b 2 = c 2 .

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 and 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 or a (separable) differential equation

which after integration gives y 2 - x 2 = const. The value of the constant is determined from the initial condition for Since for all x.

It is easy to take an issue with this proof. What does it mean for a triangle to be ? I can offer the following explanation. Triangles ABC and ABD are right by construction. We have, and also by the Pythagorean theorem. In terms of x and y, the theorem appears as

x 2 + a 2 = y 2
(x + dx) 2 + a 2 = (y + dy) 2

which, after subtraction, gives

For small dx and dy, dx 2 and dy 2 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 placed in that geometric context.

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 form a rectangle whose upper side is , whereas the lower side is c 2 . (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 the semiperimeter of the triangle. From the diagram, the hypothenuse or The area of the triangle then is computed in two ways:

(The proof is due to Jack Oliver, and was originally published in Mathematical Gazette 81 (March 1997), p 117-118.)

Proof #43

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 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.)

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

(For further notations refer to the above diagram.) BCD is isosceles. Therefore, Since angle C is right,

Since AFE is exterior to EFG, But EFG is also isosceles. Thus

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:

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. The two have also online versions:

The proof is a variation on #33, #34, and #42. The proof proceeds in two steps. First, as it may be observed from

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:

Using the notations as indicated in the diagram we get can be found by noting the similarity of triangles BFC and ABC:

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 c 2 /2. On the other hand,

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:

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 c 2 /2 = a 2 /2 + b 2 /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, b 2 and, on the other,

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,

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 b 2 . 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 b 2 - a 2 . Thus we can finish the proof as in the other proofs of this series:

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 90 o around one of its corners, such that the angle between the hypotenuses in two positions is right. The resulting shape of area b 2 is then dissected into two right triangles with side lengths and and areas c 2 /2 and

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

Equating the two gives a 2 + b 2 = c 2 .

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 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: as well as the angles formed by those sides: 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 90 o . Thus

The two triangles are similar, so that

But, EF = CD, or x = b + c, which in combination with the above proportion gives

On the other hand, y = u + a, which leads to

which is easily simplified to c 2 = a 2 + b 2 .

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 (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

which after simplification reduces to

which is just one step short of the Pythagorean proposition.

The proof works for any value of k satisfying kb/a. In particular, for we get proof #41. Further, leads to proof #53. Of course, we would get the same result by representing the area of the trapezoid AEFB in two ways. For this would lead to President Garfield's proof.

Obviously, dealing with a trapezoid is less restrictive and works for any positive value of k.

The Pythagorean Theorem: The Way of Truth - History

Department of Mathematics Education
J. Wilson, EMT 669

The Pythagorean Theorem

The Pythagorean Theorem was one of the earliest theorems known to ancient civilizations. This famous theorem is named for the Greek mathematician and philosopher, Pythagoras. Pythagoras founded the Pythagorean School of Mathematics in Cortona, a Greek seaport in Southern Italy. He is credited with many contributions to mathematics although some of them may have actually been the work of his students.

The Pythagorean Theorem is Pythagoras' most famous mathematical contribution. According to legend, Pythagoras was so happy when he discovered the theorem that he offered a sacrifice of oxen. The later discovery that the square root of 2 is irrational and therefore, cannot be expressed as a ratio of two integers, greatly troubled Pythagoras and his followers. They were devout in their belief that any two lengths were integral multiples of some unit length. Many attempts were made to suppress the knowledge that the square root of 2 is irrational. It is even said that the man who divulged the secret was drowned at sea.

The Pythagorean Theorem is a statement about triangles containing a right angle. The Pythagorean Theorem states that:

"The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides."

According to the Pythagorean Theorem, the sum of the areas of the two red squares, squares A and B, is equal to the area of the blue square, square C.

Thus, the Pythagorean Theorem stated algebraically is:

for a right triangle with sides of lengths a, b, and c, where c is the length of the hypotenuse.

Although Pythagoras is credited with the famous theorem, it is likely that the Babylonians knew the result for certain specific triangles at least a millennium earlier than Pythagoras. It is not known how the Greeks originally demonstrated the proof of the Pythagorean Theorem. If the methods of Book II of Euclid's Elements were used, it is likely that it was a dissection type of proof similar to the following:

"A large square of side a+b is divided into two smaller squares of sides a and b respectively, and two equal rectangles with sides a and b each of these two rectangles can be split into two equal right triangles by drawing the diagonal c. The four triangles can be arranged within another square of side a+b as shown in the figures.

The area of the square can be shown in two different ways:

1. As the sum of the area of the two rectangles and the squares:

2. As the sum of the areas of a square and the four triangles:

Now, setting the two right hand side expressions in these equations equal, gives

Therefore, the square on c is equal to the sum of the squares on a and b. (Burton 1991)

There are many other proofs of the Pythagorean Theorem. One came from the contemporary Chinese civilization found in the oldest extant Chinese text containing formal mathematical theories, the Arithmetic Classic of the Gnoman and the Circular Paths of Heaven.

The proof of the Pythagorean Theorem that was inspired by a figure in this book was included in the book Vijaganita, (Root Calculations), by the Hindu mathematician Bhaskara. Bhaskara's only explanation of his proof was, simply, "Behold" .

These proofs and the geometrical discovery surrounding the Pythagorean Theorem led to one of the earliest problems in the theory of numbers known as the Pythgorean problem.

Find all right triangles whose sides are of integral length, thus finding all solutions in the positive integers of the Pythagorean equation:

The three integers (x, y, z) that satisfy this equation is called a Pythagorean triple.

The formula that will generate all Pythagorean triples first appeared in Book X of Euclid's Elements :

where n and m are positive integers of opposite parity and m>n.

In his book Arithmetica , Diophantus confirmed that he could get right triangles using this formula although he arrived at it under a different line of reasoning.

The Pythagorean Theorem can be introduced to students during the middle school years. This theorem becomes increasingly important during the high school years. It is not enough to merely state the algebraic formula for the Pythagorean Theorem. Students need to see the geometric connections as well. The teaching and learning of the Pythagorean Theorem can be enriched and enhanced through the use of dot paper, geoboards, paper folding, and computer technology, as well as many other instructional materials. Through the use of manipulatives and other educational resources, the Pythagorean Theorem can mean much more to students than just

and plugging numbers into the formula.

The following is a variety of proofs of the Pythagorean Theorem including one by Euclid. These proofs, along with manipulatives and technology, can greatly improve students' understanding of the Pythagorean Theorem.

The following is a summation of the proof by Euclid, one of the most famous mathematicians. This proof can be found in Book I of Euclid's Elements .

Proposition: In right-angled triangles the square on the hypotenuse is equal to the sum of the squares on the legs.

Euclid began with the Pythagorean configuration shown above in Figure 2. Then, he constructed a perpendicular line from C to the segment DJ on the square on the hypotenuse. The points H and G are the intersections of this perpendicular with the sides of the square on the hypotenuse. It lies along the altitude to the right triangle ABC. See Figure 3.

Next, Euclid showed that the area of rectangle HBDG is equal to the area of square on BC and that the are of the rectangle HAJG is equal to the area of the square on AC. He proved these equalities using the concept of similarity. Triangles ABC, AHC, and CHB are similar. The area of rectangle HAJG is (HA)(AJ) and since AJ = AB, the area is also (HA)(AB). The similarity of triangles ABC and AHC means

or, as to be proved, the area of the rectangle HAJG is the same as the areaof the square on side AC. In the same way, triangles ABC and CHG are similar. So

Since the sum of the areas of the two rectangles is the area of the square on the hypotenuse, this completes the proof.

Euclid was anxious to place this result in his work as soon as possible. However, since his work on similarity was not to be until Books V and VI, it was necessary for him to come up with another way to prove the Pythagorean Theorem. Thus, he used the result that parallelograms are double the triangles with the same base and between the same parallels. Draw CJ and BE.

The area of the rectangle AHGJ is double the area of triangle JAC, and the area of square ACLE is double triangle BAE. The two triangles are congruent by SAS. The same result follows in a similar manner for the other rectangle and square. (Katz, 1993)

Click here for a GSP animation to illustrate this proof.
The next three proofs are more easily seen proofs of the Pythagorean Theorem and would be ideal for high school mathematics students. In fact, these are proofs that students could be able to construct themselves at some point.
The first proof begins with a rectangle divided up into three triangles, each of which contains a right angle. This proof can be seen through the use of computer technology, or with something as simple as a 3x5 index card cut up into right triangles.

It can be seen that triangles 2 (in green) and 1 (in red), will completely overlap triangle 3 (in blue). Now, we can give a proof of the Pythagorean Theorem using these same triangles.

I. Compare triangles 1 and 3.

Angles E and D, respectively, are the right angles in these triangles. By comparing their similarities, we have

and from Figure 6, BC = AD. So,

By cross-multiplication, we get :

II. Compare triangles 2 and 3:

By comparing the similarities of triangles 2 and 3 we get:

From Figure 4, AB = CD. By substitution,

Finally, by adding equations 1 and 2, we get:

We have proved the Pythagorean Theorem.

The next proof is another proof of the Pythagorean Theorem that begins with a rectangle. It begins by constructing rectangle CADE with BA = DA. Next, we construct the angle bisector of <BAD and let it intersect ED at point F. Thus, <BAF is congruent to <DAF, AF = AF, and BA = DA. So, by SAS, triangle BAF = triangle DAF. Since <ADF is a right angle, <ABF is also a right angle.

Next, since m<EBF + m<ABC + m<ABF = 180 degrees and m<ABF = 90 degrees, <EBF and <ABC are complementary. Thus, m<EBF + m<ABC = 90 degrees. We also know that
m<BAC + m<ABC + m<ACB = 180 degrees. Since m<ACB = 90 degrees, m<BAC + m<ABC = 90 degrees. Therefore, m<EBF + m<ABC = m<BAC + m<ABC and m<BAC = m<EBF.

By the AA similarity theorem, triangle EBF is similar to triangle CAB.

Now, let k be the similarity ratio between triangles EBF and CAB.

Thus, triangle EBF has sides with lengths ka, kb, and kc. Since FB = FD, FD = kc. Also, since the opposite sides of a rectangle are congruent, b = ka + kc and c = a + kb. By solving for k, we have

and we have completed the proof.

The next proof of the Pythagorean Theorem that will be presented is one that begins with a right triangle. In the next figure, triangle ABC is a right triangle. Its right angle is angle C.

Next, draw CD perpendicular to AB as shown in the next figure.

Compare triangles 1 and 3 :

Triangle 1 (green) is the right triangle that we began with prior to constructing CD. Triangle 3 (red) is one of the two triangles formed by the construction of CD.

Figure 13
Triangle 1. Triangle 3.

By comparing these two triangles, we can see that

Compare triangles 1 and 2 :

Triangle 1 (green) is the same as above. Triangle 2 (blue) is the other triangle formed by constructing CD. Its right angle is angle D.

Figure 14
Triangle 1. Triangle 2.

By comparing these two triangles, we see that

By adding equations 3 and 4 we get:

From Figures 11 and 12, with CD, we have that (p + q) = c. By substitution, we get

The next proof of the Pythagorean Theorem that will be presented is one in which a trapezoid will be used.

By the construction that was used to form this trapezoid, all 6 of the triangles contained in this trapezoid are right triangles. Thus,

Area of Trapezoid = The Sum of the areas of the 6 Triangles

And by using the respective formulas for area, we get:

We have completed the proof of the Pythagorean Theorem using the trapezoid.

The next proof of the Pythagorean Theorem that I will present is one that can be taught and proved using puzzles. These puzzles can be constructed using the Pythagorean configuration and then, dissecting it into different shapes.

Before the proof is presented, it is important that the next figure is explored since it directly relates to the proof.

In this Pythagorean configuration, the square on the hypotenuse has been divided into 4 right triangles and 1 square, MNPQ, in the center. Since MN = AN - AM = a - b. Each side of square MNPQ has length of a - b. This gives the following:

Area of Square on the hypotenuse = Sum of the Areas of the 4 triangles and the Area of Square MNPQ

As mentioned above, this proof of the Pythagorean Theorem can be further explored and proved using puzzles that are made from the Pythagorean configuration. Students can make these puzzles and then use the pieces from squares on the legs of the right triangle to cover the square on the hypotenuse. This can be a great connection because it is a "hands-on" activity. Students can then use the puzzle to prove the Pythagorean Theorem on their own.

To create this puzzle, copy the square on BC twice, once placed below the square on AC and once to the right of the square on AC as shown in Figure 17.

Triangle CDE is congruent to triangle ACB by leg-leg.

In triangle ACB, m<ACB =90 and the sides have lengths a, b, c.

In triangle CDE, m<CDE =90 and the sides have lengths a,b, c.

Triangle EGH is congruent to triangle ACB by leg-leg. The m<EGH =90 and its sides have lengths a and c. Since EF=b-a=AI, EG=b. Thus the diagonals CE and EH are both equal to c.


Pythagoras is a Greek mathematician at the same time ancient a philosopher for the 6th century. He is very influence for science especially in mathematics. One of his famous omission is Pythagoras Theorem which almost people have ever heard it. Pythagoras Theorem said that hypotenuse of right triangle is sum of square of 2nd other side from the right triangle. Because of his omissions in mathematics, he also called as “The Father of Number”.

One of the pupils which so called Hippasus said that 𕔆 which is hypotenuse of isosceles triangle which having length each feet is 1 is the irrational number. However, Hippasus then murdered because Pythagoras cannot argue evidence raised by Hippasus.

Hippasus is a pupil of Pythagoras coming from Metapontum. He also a mathematician at the same time ancient Greek philosopher about the 6th century. He considered to be inventor of irrational number, especially prove that square root of 2 𕔆 is irrational number. Ironically, the invention exactly cause the death. Pythagoras argue existence of irrational number. Pythagoras and the other pupils assumed that all nmber are rational number and there is no irrational number. Hippasus prove this theorem by using reductio ad absurdum (prove by contradiction) proving number that it is irrational number. Pythagoras cannot argue this statement and assume that Hippasus is errant teaching follower so that he set mind on to engulf Hippasus.

Irrational number is a real number which cannot divided (result for its have never desisted). In this case, irrational number cannot expressed as a/b, with a and b as integer, and b unlike null. So irrational number is not rational number. The example for irrational number such as π, 𕔆, and e number. Phi number (π) which during the time we recognizing, actually imprecise 3,14 but 3,1415926535897932…. Also 𕔆 number which if we formulate becoming 1,41421356237309504880….And e number that is 2,71828182….

The irrational number can be proved with using reductio ad absurdum or in english called proof by contradiction. It is logic argument started with an assumption, then from the assumption found an absurd result, illogical, or contradictive, so the conclusion of the assumption will to be wrong valuable and the deny will become correct valuable. A mathematical statement sometime be proved by reductio ad absurdum that is by assuming the deny (negation) from the statement which will be proved, then from the assumption degraded a contradiction. When contradiction reachable logically, then the assumption have proven fault, so that the statement is correct.

Proved by contradiction or reductio ad absurdum is not wrong argument, but if done truly will be valid argument. If prove by contradiction yield a mistake, the mistake lay at the process degradation of contradiction, not at may of the prove.

The classical example for the prove by contradiction at ancient Greek era is prove that square root of two is irrational number (cannot expressed as comparison of integer). This statement is provable by the way of assuming on the contrary that 2 is rational number, so that can expressed as comparison of integer a/b in simplest fraction. But if a/b = 𕔆, so a2 = 2b2.It means a2 is even number. Because square from odd number is not possibly even, then a is even number. Because a/b is simplest fraction, b surely anomalous (because fraction of even/even number still can be made moderate). But because a is even number (assume 2r = a, mean a2 = 4r2) is fold number of 4, and b2 is fold number of 2 (even). This mean b is also even number, and this is contradiction to conclusion before all that b surely anomalous. Because assumption early that 2 is rational number result the contradiction, the assumption surely wrong, and the deny (that 2 is irrational) is correct statement.

One of the Pythagoras omission that very popular is Pythagoras Theorem. The theorem called as the ancient Greek mathematician and philosopher, he is Pythagoras. Pythagoras is not the inventor of the theorem but he is the first people who proved the truth of the theorem so he given appreciation with give name the theorem like his name.

This theorem express that summing up wide squares at foots a right triangles equal broadly squares in hypotenuses. Right triangle is triangle having a right angle (90o0) the foots are two sides which the bevels angular shapes, and hypotenuse is third side dealing with the right angle. The formula of this theorem is a2+ b2 = c2, where a and b is the sides of the right triangle, and c is the hypotenuse.

Useful Application: Try Any Shape

We used triangles in our diagram, the simplest 2-D shape. But the line segment can belong to any shape. Take circles, for example:

Now what happens when we add them together?

You guessed it: Circle of radius 5 = Circle of radius 4 + Circle of radius 3.

Pretty wild, eh? We can multiply the Pythagorean Theorem by our area factor (pi, in this case) and come up with a relationship for any shape.

Remember, the line segment can be any portion of the shape. We could have picked the circle's radius, diameter, or circumference -- there would be a different area factor, but the 3-4-5 relationship would still hold.

So, whether you're adding up pizzas or Richard Nixon masks, the Pythagorean theorem helps you relate the areas of any similar shapes. Now that's something they didn't teach you in grade school.

The Pythagorean Theorem Makes Construction and GPS Possible

OK, time for a pop quiz. You've got a right-angled triangle — that is, one where two of the sides come together to form a 90-degree angle. You know the length of those two sides. How do you figure out the length of the remaining side?

That's easy, provided that you took geometry in high school and know the Pythagorean theorem, a mathematical statement that's thousands of years old.

The Pythagorean theorem states that with a right-angled triangle, the sum of the squares of the two sides that form the right angle is equal to the square of the third, longer side, which is called the hypotenuse. As a result, you can determine the length of the hypotenuse with the equation a 2 + b 2 = c 2 , in which a and b represent the two sides of the right angle and c is the long side.

Who Was Pythagoras?

A pretty slick trick, huh? But the man whom this math trick is named for is nearly as fascinating. Pythagoras, an ancient Greek thinker who was born on the island of Samos and lived from 570 to 490 B.C.E, was kind of a trippy character — equal parts philosopher, mathematician and mystical cult leader. In his lifetime, Pythagoras wasn't known as much for solving for the length of the hypotenuse as he was for his belief in reincarnation and adherence to an ascetic lifestyle that emphasized a strict vegetarian diet, adherence to religious rituals and plenty of self-discipline that he taught to his followers.

Pythagoras biographer Christoph Riedweg describes him as a tall, handsome and charismatic figure, whose aura was enhanced by his eccentric attire — a white robe, trousers and a golden wreath on his head. Odd rumors swirled around him — that he could perform miracles, that he had a golden artificial leg concealed beneath his clothes and that he possessed the power to be in two places at one time.

Pythagoras founded a school near what is now the port city of Crotone in southern Italy, which was named the Semicircle of Pythagoras. Followers, who were sworn to a code of secrecy, learned to contemplate numbers in a fashion similar to the Jewish mysticism of Kaballah. In Pythagoras' philosophy, each number had a divine meaning, and their combination revealed a greater truth.

With a hyperbolic reputation like that, it's little wonder that Pythagoras was credited with devising one of the most famous theorems of all time, even though he wasn't actually the first to come up with the concept. Chinese and Babylonian mathematicians beat him to it by a millennium.

"What we have is evidence they knew the Pythagorean relationship through specific examples," writes G. Donald Allen, a math professor and director of the Center for Technology-Mediated Instruction in Mathematics at Texas A&M University, in an email. "An entire Babylonian tablet was found that shows various triples of numbers that meet the condition: a 2 + b 2 = c 2 ."

How Is the Pythagorean Theorem Useful Today?

The Pythagorean theorem isn't just an intriguing mathematical exercise. It's utilized in a wide range of fields, from construction and manufacturing to navigation.

As Allen explains, one of the classic uses of the Pythagorean theorem is in laying the foundations of buildings. "You see, to make a rectangular foundation for, say, a temple, you need to make right angles. But how can you do that? By eyeballing it? This wouldn't work for a large structure. But, when you have the length and width, you can use the Pythagorean theorem to make a precise right angle to any precision."

Beyond that, "This theorem and those related to it have given us our entire system of measurement," Allen says. "It allows pilots to navigate in windy skies, and ships to set their course. All GPS measurements are possible because of this theorem."

In navigation, the Pythagorean theorem provides a ship's navigator with a way of calculating the distance to a point in the ocean that's, say, 300 miles north and 400 miles west (480 kilometers north and 640 kilometers west). It's also useful to cartographers, who use it to calculate the steepness of hills and mountains.

"This theorem is important in all of geometry, including solid geometry," Allen continues. "It is also foundational in other branches of mathematics, much of physics, geology, all of mechanical and aeronautical engineering. Carpenters use it and so do machinists. When you have angles, and you need measurements, you need this theorem."

One of the formative experiences in the life of Albert Einstein was writing his own mathematical proof of the Pythagorean theorem at age 12. Einstein's fascination with geometry eventually played a role in his development of the theories of special and general relativity.

List of site sources >>>

Watch the video: Does Maths Prove God? Alexey Burov vs James Fodor (January 2022).