Theorem of Pythagore

The theorem of Pythagore is a Théorème of Euclidean Géométrie which states that in a Right-angled triangle (which has a Right angle) the square of the Hypoténuse (side opposed to the Right angle) is equal to the sum of the squares on the two other sides. This theorem is named according to Pythagore de Samos which was a Mathématicien, Philosophe and Astronome of the ancient Greece.

Theorem The most known form of the theorem of Pythagore is the following one:

In a right-angled triangle ABC out of C, AB being the hypotenuse, where AB = C, AC = B and BC = has (cf appears opposite), one will thus have:

$BC^2\; +\; AC^2\; =\; AB^2\; \backslash ,\; \backslash !$

or:

$a^2+b^2\; =\; c^2$

The theorem of Pythagore thus makes it possible to calculate the Length of one on the sides of a Right-angled triangle if the two others are known. Example: with the notations above, that is to say the right-angled triangle on sides has = 3 and B = 4; then the length on the third side, C, is given by:

+ B has = 3 + 4 = 25 = C

from where   C = 5.

A triplet of integers such as (3, 4,5), representing the length on the sides of a right-angled triangle is called a Triplet pythagorician.

Reciprocal

The Réciproque of the theorem of Pythagore (proposal 47 of the first book of the Elements of Euclide) is also true:

The theorem of Pythagore is thus an index property of the right-angled triangles.

Another formulation: “So in a triangle ABC there is AC + BC = AB, then this triangle is right-angled out of C.”

This can be proven by using the law of the cosine (or Théorème of Al-Kashi, already known of Euclide in its Éléments : proposals 12 and 13 of the book II) which is a generalization of the theorem of Pythagore applied to all the triangles (Euclidean).

History

That the property of Pythagore is known since the Antiquité is a fact which one can find trace in the history. It is enough for that to observe the Corde with thirteen nodes of which the Arpenteur S Egyptians were useful. This cord made it possible to measure distances but also to build, without square, a right angle since the 13 nodes (and twelve intervals) made it possible to build a triangle whose dimensions were (3 - 4 - 5), triangle which proves to be right-angled. This cord will remain a tool of geometrician during still all the Moyen-âge.

The theorem of Pythagore was used by the magicians, the gnostic ones and the sects esoteric. It is a question of building oppositions between the material and the spiritual one, the sky and the ground, the human one and the divine one, etc Ainsi, Albert Pike affirms that the theorem of Pythagore constitutes the large secrecy of freemasonry. This theorem reveals at the same time as several of these sects make a worship secret towards Isis and Osiris.

The oldest representation of triplets pythagoricians (right-angled triangle whose sides are whole) is on Mégalithe S (towards 2500 av. J. - C., Great Britain). One finds also the trace of triplets pythagoricians on Babylonian shelves (shelf of Plimpton 322 worms 1800 av. J. - C.) which prove that, more than 1000 years before Pythagore, geometricians knew the existence of triplets pythagoricians.

But between the discovery of a Property: “it is observed that certain right-angled triangles check this property”, its generalization: “it seems that all the right-angled triangles check this property” and its demonstration: “it is true that all the right-angled triangles (and they only) in an Euclidean plan check this property”, several centuries often should be waited.

The historical evidence of the life of Pythagore is already so rare that it is not astonishing that one cannot allot to him with certainty the paternity of the demonstration. The first hard copy figure in the Elements of Euclide in the following form:

“With the right-angled triangles, the square on the side which supports the right angle, is equal to the squares on the two other sides. ”
(Book I, proposal XLVII)

With its reciprocal:

“If the square of the one on the sides of a triangle is equal to the squares on the two other sides, the angle supported by these east coasts right. ”
(Book I, proposal XLVIII)

However, the comments of Proclos of the Éléments of Euclide (approximately 400 a. J. - C.) seem to indicate that Euclide would have done nothing but retranscribe one demonstration older than Proclos allots to Pythagore.

It is thus between the {{VIe}} and third century BC that one can go back the demonstration to this property. It is told that it is on this occasion that the existence of irrational Nombre would have been discovered. Indeed, it is easy to build an isosceles right-angled triangle on side 1. Then the square on the hypotenuse would be worth 2. However a simple demonstration accessible from the time of Pythagore proves that no rational has a square equal to 2. It is told that this discovery was held secret by the school pythagorician under penalty of death.

Parallel to these discoveries, it seems that in China also the property is known. One finds trace of the existence of this theorem in one of the oldest Chinese mathematical works the Zhoubi suanjing . This work, probably writes during the Dynastie Han (206 av. J. - C. - 220 a. J. - C.), gathers techniques of calculation dating from the Dynastie Zhou (- 256 av. J. - C.). A demonstration of the theorem, which bears to China the name of theorem of Gougu (bases and altitude), figure in the Jiuzhang suanshu (nine chapters on mathematical art, 100 av. J. - C. - 50 a. J. - C.), demonstration which resembles of nothing that Euclide and which proves the originality of the Chinese step.

In India, towards 300 av. J. - C., one finds the trace of a numerical demonstration of the property (proof carried out on particular numbers but which can spread easily).

Of a geometrical property, the theorem of Pythagore takes also an arithmetic development with the search for all the triplets of entireties associated with the three sides with a right-angled triangle: they are the triplets pythagoricians. This research will open the door with another: the search for triplets checking the equality $a^n\; +\; b^n\; =\; c^n$, research which leads to the Conjecture of Fermat solved in 1994 by Andrew Wiles.

There actually exists of many demonstrations of this theorem, that of Euclide to that of the Chinese, while passing by that of India, that using of the similarities, that of Léonard de Vinci and even that of the US president James Garfield. One cannot overlook Al Kashi which gives for an unspecified triangle a relation whose formula of Pythagore becomes then the particular case of the right-angled triangle: the Theorem of Al-Kashi .

Demonstrations

It is undoubtedly the theorem which has the greatest number of known evidence (the quadratic Loi of reciprocity is also distinguished in this field). Here are three:

The proof according to Euclide

For making the Demonstration, it is necessary to prove two proposals. The first proposal that it is necessary for us to prove (proposal XXXV in the 1st book of the Éléments ) is the equivalence of two Parallélogramme S of the same bases and of the same Height:

“The parallelograms made up on the same basis, and between same parallel S, are equal between them. ”

Let us consider the two parallelograms ABCD and BCFE, both on the same basis, BC, and between the same parallels, BC and AF. Observe that AD is equal to BC (because they are the two bases of parallelogram ABCD), and BC is equal to EFF (because they are the two bases of parallelogram BCFE), then AD is equal to EFF.

However, there are only three possibilities (shown in the image) for the position of the point E relative to D; E can be with the left of D, at the point D, or with the right-hand side of D. Examinons each case:

1. If E fall to the left from D, ED is the common part of AD and EFF, then it is possible to check that AD and EFF are equal. But note that sides AB and cd. are equal, because they are sides opposite of parallelogram ABCD. Also, because the points has, E, D and F are colinéaires, angles BAE and CDF are equal. Consequently, the Triangle S BAE and CDF are equal, because two sides of the one are equal to two sides of the other, and an angle is common. Thus parallelograms ABCD and CBEF are only various arrangements of the Trapèze BEDC and triangle BAE (or CDF). CQFD

2. If E fall at the point D, one finds in a way similar to 1 that triangles BAE and CDF are equal, and whereas it is possible to obtain parallelograms ABCD and BCFE by adding to part common BCD triangle BAE (or CDF). CQFD
3. If E fall to the right-hand side from D, note that, because the segment S AD and EFF are equal, by adding to each one the line OF, we find that AE and DF are equal. By an argument similar to those used in cases 1 and 2, it is possible to prove that the triangles BAE and CDF, and consequently trapezoids BADG and CGEF, are equal. Then, it is obvious that parallelograms ABCD and CBEF are obtained by adding to triangle common BCG trapezoid BADG (or CGEF). CQFD

The replacement of a parallelogram by another of the same base and even height, justified by this proposal, is known in mathematics under the name of Cisaillement . Shearing will be very important in the proof of the following proposal:

“If a parallelogram, and a triangle have the same base, and are between same parallels; the parallelogram will be double triangle. ”

Let us consider a parallelogram ABCD, and is E a point on the extension of AD. We want to show that the surface of ABCD is twice the surface of NOZZLE. Tracing the diagonal AC, we see that the surface of ABCD is twice the surface of ABC. But, the surface of the triangle ABC is equal to the surface of the triangle NOZZLE, because they have the same base. Then, twice the surface of NOZZLE equalizes the surface of ABC twice, i.e. the surface of ABCD. We showed that ABCD (which is double ABC) is double of NOZZLE. CQFD

Now, we can continue the Démonstration.

Let us consider the right-angled triangle ABC in A. Are BCED, ABFG and ACIH the squares on the sides BC, AB and AC respectively. That is to say J the intersection of AK and BC. What we want to show is that the surface of BCED is equal to sum of the surfaces of ABFG and ACIH. We prove this fact by showing that the surface of square ABFG is equal to the surface of the Rectangle BJKD and that the surface of square ACIH is equal to the surface of rectangle CEKJ.

Let us show the first equality, note that sides BFR and BC are equal to sides AB and data base, respectively. Because angles ABF and CBD are equal, angles FBC (FBA + ABC) and ABD (ABC + CBD) are equal. Consequently, triangles FBC and ABD are equal too. However, note that, by proposal XLI, the surface of square ABFG is double of that of triangle FBC and that the surface of rectangle BJKD is double of that of triangle ABD. As FBC and ABD are equal, the surface of ABFG is quite equal to that of BJKD.

The second equality is proven in a similar way: observing that IC and CB equalize AC and EC, respectively, and that angle ICB equalizes angle ACE, we conclude that triangles ICB and ACE are equal. Then, knowing that the surface of square ACIH is double of that of ICB and that the surface of rectangle CEKJ is double of that of ACE, and that triangle ICB is equal to triangle ACE, the surface of ACIH is thus equal to the surface of CEKJ.

Consequently, the surface of BCED, equal to the sum of the surface of BJKD and that of CEKJ, is quite equal to the sum surface of ABFG and that of ACIH. CQFD

In this form, the theorem of Pythagore is a particular case of the theorem of Clairaut.

A proof of the theorem of Guogu (China)

Note: The theorem of Guogu is reconstituted according to the comments of the Chinese mathematician Liu Hui ({{IIIe}} century a. J. - C.) on the JiuZhang SuanShu 九章算術 “nine chapters of Arithmetic” (206 av. - 220 a. J. - C.), and the Zhoubi Suanjian 周髀算經, “the shade of the cycles, delivers calculations” (a book of astronomy). This proof uses the principle of the Puzzle: two equal surfaces after Cutting finished and recombining have even surface. It is it should be noted that Euclide, in its property of shearing, uses same the Principe. In the figure opposite, the right-angled triangle is traced in fat, the square on the large side was traced outside the triangle, the square on the small side and that of the hypotenuse is turned towards the triangle. The parts of the squares on the sides of the right angle which exceed square on the hypotenuse were cut out and replaced inside this square.

The red triangle is equal to the starting triangle. The yellow triangle has for large side of the right angle the small side of the starting triangle and has same angles as the initial triangle. The blue triangle has for large side of the right angle, the difference on the sides of the initial triangle and has same angles as the initial triangle.

A modern proof

Let us consider a right-angled triangle whose sides are lengths has , B and C . Then let us recopy this triangle three times and place the triangle and its copies so as to have the side has each one aligned at the side B of another, and so that the legs of the triangles form a square of which the east coast $a+b$, as in the image. Then, we try to find the surface of the square formed by the sides C . Obviously, it is $c^2$, but it is also equal to the difference between the surface of the square external and the sum of the surfaces of the triangles. The surface of the square is $(a+b)^2$ (because its east coast $a+b$) and the total surface of the triangles is four times the surface of only one, i.e. $4\; (ab/2)$, therefore the difference is $(a+b)^2\; -\; 4\; (ab/2)$, which one can simplify like $a^2\; +\; 2ab\; +\; b^2\; -\; 2ab$, or $a^2\; +\; b^2$. We showed that the surface of the square on side C is equal to $a^2+b^2$; indeed, $c^2\; =\; a^2+b^2$. CQFD

There exist many other demonstrations of the theorem of Pythagore; the twentieth president of the United States of America, James Abram Garfield developed one of them itself, very close to the preceding one. One of most interesting is the calculative proof based on the Formule of Euler. (See the external bonds below for a presentation of various evidence of the theorem of Pythagore).

Variations on the theorem

Contraposée

The contraposée of the theorem affirms this:

“If the lengths on the sides of a triangle ABC check $AB^2\; \backslash \; AC^2+CB^2\; \backslash ,\; \backslash !$, then the triangle is not right-angled out of C.”

Let us note that contraposée is logically equivalent to the direct theorem, it does not have on the other hand the same use in demonstration since the theorem is used to calculate the third side lacking a right-angled triangle whereas contraposée is used to show that a triangle which one knows the lengths on the three sides is not right-angled.

Contraposée of the reciprocal one

Lastly, contraposée of reciprocal of the theorem of Pythagore this stipulates:

If the triangle ABC is not right-angled out of C then $AB^2\; \backslash \; AC^2+CB^2\; \backslash ,\; \backslash !$

Generalization with other figures that squares

Another generalization of the theorem of Pythagore was already stated by Euclide in its Éléments (Proposal 31 of book VI):

“In the right-angled triangles, the figure built on the side which underlies the right angle, is equal to the similar figures and similarly described on the sides which include/understand the right angle. ”

In other words:

“If one sets up similar figures (see Géométrie) on the sides of a right triangle, then the sum of the surfaces of the two smaller figures equalizes the surface of largest. ”

This property makes it possible to show that the surface of the right-angled triangle is equal to the sum of the surfaces of the Lunule S drawn on each side of the right angle (see the Théorème of the two lunules).

Uses

• In Coordinated Cartesian in a Reference mark orthonormé, the theorem of Pythagore makes it possible to express the distance between two points of the plan: thus, if $A\; (x\_a,\; y\_a)$ and $B\; (x\_b,\; y\_b)$ are points of the Euclidean plan, the distance separating them is given by:

$\backslash \; sqrt\; \{(x\_b-x\_a)\; ^2\; +\; (y\_b\; there\; \_a)\; ^2\}.$

Indeed, if C is the point of coordinates $(x\_b,\; y\_a)$, the triangle ACB is right-angled in C , the distances CA and CB are given by CA= |x_b - x_a| and CB = |y_b - y_a| and outdistances it AB represents the hypotenuse of the right-angled triangle ACB .

• more generally, in a Euclidean Space (or in a space refines Euclidean) of finished size, the distance from $(x\_1,\; \backslash \; dowries,\; x\_n)$ with $(y\_1,\; \backslash \; dowries,\; y\_n)$ is written

$\backslash \; sqrt\; \{\backslash \; sum\_\; \{k=1\}\; ^\; \{k=n\}\; \{(x\_k\; there\; \_k)\; ^2\}\}.$

• the identity of Parseval can be seen like a generalization of the theorem of Pythagore to the infinite families of vectors of a Espace préhilbertien.

• the theorem of Pythagore also spreads in the Simplexe S of higher dimension. If a Tétraèdre has a formed corner of right angle (a corner of Cube), then the square of the surface of the face opposed to the corner is the sum of the squares of the surfaces of the three others faces. This theorem is also known under the name of theorem of Gua .

Theorem of Pythagore in other spaces

Vectorial writing

While utilizing the concept of Vector, one can reformulate the theorem as follows:

“Being given two vectors $\backslash \; vec\; \{U\}$ and $\backslash \; vec\; \{v\}$, $\backslash \; Vert\; \backslash \; vec\; \{U\}\; +\; \backslash \; vec\; \{v\}\; \backslash \; Vert^2\; =\; \backslash \; Vert\; \backslash \; vec\; \{U\}\; \backslash \; Vert^2\; +\; \backslash \; Vert\; \backslash \; vec\; \{v\}\; \backslash \; Vert^2$ if and only if $\backslash \; vec\; u$ and $\backslash \; vec\; v$ is orthogonal. ”

In a general way, there is simply the triangular Inégalité:

$||\backslash \; vec\; \{U\}\; +\; \backslash \; vec\; \{v\}||^2\; \backslash \; it\; ||\backslash \; vec\; \{U\}||^2\; +\; ||\backslash \; vec\; \{v\}||^2\; +\; 2||\backslash \; vec\; \{U\}||\; \backslash \; cdot\; ||\backslash \; vec\; \{v\}||$

that one writes

$in\; general||\backslash \; vec\; \{U\}\; +\; \backslash \; vec\; \{v\}||\; \backslash \; it\; ||\backslash \; vec\; \{U\}||\; +\; ||\backslash \; vec\; \{v\}||.$

In a space préhilbertien

The theorem of Pythagore rises in fact directly from the definition from the scalar Produit, and spreads with all Espace préhilbertien. Within this general framework, he affirms that if $u$ and $v$ are two orthogonal vectors, then:

$\backslash \; left\; \backslash \; Green\; U\; \backslash \; right\; \backslash \; Vert^2\; +\; \backslash \; Green\; left\; \backslash \; v\; \backslash \; right\; \backslash \; Vert^2\; =\; \backslash \; Green\; left\; \backslash \; u+v\; \backslash \; right\; \backslash \; Vert^2$

The reciprocal one is true in the real case.

Moreover, this formula spreads with a family of orthogonal vectors. For it, the sum of square of the Norme S is equal to the square of the standard of the sum. This very general result in particular makes it possible to show the Inégalité of Bessel, and the equality of Parseval.

In nonEuclidean geometry

This property resists badly the transfer in other geometries because of their Courbure:

• if the curve is positive (spherical geometry), one obtains: C < has + B;
• if the curve is negative (hyperbolic Géométrie), one obtains: C > has + B;
• if the curve is null (plane or cylindrical geometry), one preserves: C = has + B.

More precisely, for any right-angled triangle on a Sphere of ray R , the theorem of Pythagore takes the following form:

$\backslash \; cos\; \backslash \; left\; (\backslash \; frac\; \{C\}\; \{R\}\; \backslash \; right)\; =\; \backslash \; cos\; \backslash \; left\; (\backslash \; frac\; \{has\}\; \{R\}\; \backslash \; right)\; \backslash ,\; \backslash \; cos\; \backslash \; left\; (\backslash \; frac\; \{B\}\; \{R\}\; \backslash \; right).$

By using a Development limited of order 2 of the function cosine, one finds well, for great values of R , the traditional formula of the theorem of Pythagore.

• For any right-angled triangle in hyperbolic geometry, with a curve of -1, the theorem of Pythagore takes the following form

$\backslash \; cosh\; c=\; \backslash \; cosh\; has\; \backslash ,\; \backslash \; cosh\; b$

where cosh is the hyperbolic cosine. By using the Development limited of order 2 of this function, one finds well, for small values on the sides, the traditional form of the theorem of Pythagore.

Space physical

As the theorem of Pythagore is derived from Axiome S of the Euclidean Géométrie, and that physical spaces are not always Euclidean, it should not be valid for the triangles in physical spaces. One of the first mathematicians to carry out this was Carl Friedrich Gauss, which thus measured attentively large right-angled triangles within the framework of its geographical study in order to check this theorem. It did not find any counterexample with its measuring accuracy. The theory of the General relativity supports that the Matière and the energy lead the space to be not-Euclidean and the theorem thus does not apply strictly in the presence of energy. However, the deviation compared to Euclidean space is weak except near imposing gravitational sources like the black holes. To determine if the theorem is enfreint on important cosmological scales, i.e. to measure the curve of the Universe, is an open problem for the Cosmologie.

See too

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