According to Van der Waerden, it is very likely that the ratio in general form was known in Babylon around the 18th century BC. e.
Around 400 BC. BC, according to Proclus, Plato gave a method for finding Pythagorean triplets, combining algebra and geometry. Around 300 BC. e. The oldest axiomatic proof of the Pythagorean theorem appeared in Euclid's Elements.
Formulations
The basic formulation contains algebraic operations - in a right triangle, the lengths of which are equal a (\displaystyle a) And b (\displaystyle b), and the length of the hypotenuse is c (\displaystyle c), the following relation is satisfied:
.An equivalent geometric formulation is also possible, resorting to the concept of area of a figure: in a right triangle, the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs. The theorem is formulated in this form in Euclid’s Elements.
Converse Pythagorean theorem- a statement about the rectangularity of any triangle, the lengths of the sides of which are related by the relation a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)). As a consequence, for every triple of positive numbers a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c), such that a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), there is a right triangle with legs a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c).
Proof
There are at least 400 proofs of the Pythagorean theorem recorded in the scientific literature, which is explained by both its fundamental significance for geometry and the elementary nature of the result. The main directions of proofs are: algebraic use of relations between the elements of a triangle (for example, the popular method of similarity), the method of areas, there are also various exotic proofs (for example, using differential equations).
Through similar triangles
The classical proof of Euclid is aimed at establishing the equality of areas between rectangles formed by dissecting the square above the hypotenuse by the height of the right angle with the squares above the legs.
The construction used for the proof is as follows: for a right triangle with a right angle C (\displaystyle C), squares over the legs and and squares over the hypotenuse A B I K (\displaystyle ABIK) height is being built CH and the ray that continues it s (\displaystyle s), dividing the square above the hypotenuse into two rectangles and . The proof aims to establish the equality of the areas of the rectangle A H J K (\displaystyle AHJK) with a square over the leg A C (\displaystyle AC); the equality of the areas of the second rectangle, constituting the square above the hypotenuse, and the rectangle above the other leg is established in a similar way.
Equality of areas of a rectangle A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) is established through the congruence of triangles △ A C K (\displaystyle \triangle ACK) And △ A B D (\displaystyle \triangle ABD), the area of each of which is equal to half the area of the squares A H J K (\displaystyle AHJK) And A C E D (\displaystyle ACED) accordingly, in connection with the following property: the area of a triangle is equal to half the area of a rectangle if the figures have a common side, and the height of the triangle to the common side is the other side of the rectangle. The congruence of triangles follows from the equality of two sides (sides of squares) and the angle between them (composed of a right angle and an angle at A (\displaystyle A).
Thus, the proof establishes that the area of a square above the hypotenuse, composed of rectangles A H J K (\displaystyle AHJK) And B H J I (\displaystyle BHJI), is equal to the sum of the areas of the squares over the legs.
Proof of Leonardo da Vinci
The area method also includes a proof found by Leonardo da Vinci. Let a right triangle be given △ A B C (\displaystyle \triangle ABC) with right angle C (\displaystyle C) and squares A C E D (\displaystyle ACED), B C F G (\displaystyle BCFG) And A B H J (\displaystyle ABHJ)(see picture). In this proof on the side HJ (\displaystyle HJ) of the latter, a triangle is constructed on the outer side, congruent △ A B C (\displaystyle \triangle ABC), moreover, reflected both relative to the hypotenuse and relative to the height to it (that is, J I = B C (\displaystyle JI=BC) And H I = A C (\displaystyle HI=AC)). Straight C I (\displaystyle CI) splits the square built on the hypotenuse into two equal parts, since triangles △ A B C (\displaystyle \triangle ABC) And △ J H I (\displaystyle \triangle JHI) equal in construction. The proof establishes the congruence of quadrilaterals C A J I (\displaystyle CAJI) And D A B G (\displaystyle DABG), the area of each of which turns out to be, on the one hand, equal to the sum of half the areas of the squares on the legs and the area of the original triangle, on the other hand, half the area of the square on the hypotenuse plus the area of the original triangle. In total, half the sum of the areas of the squares over the legs is equal to half the area of the square over the hypotenuse, which is equivalent to the geometric formulation of the Pythagorean theorem.
Proof by the infinitesimal method
There are several proofs using the technique of differential equations. In particular, Hardy is credited with a proof using infinitesimal increments of legs a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c), and preserving similarity with the original rectangle, that is, ensuring the fulfillment of the following differential relations:
d a d c = c a (\displaystyle (\frac (da)(dc))=(\frac (c)(a))), d b d c = c b (\displaystyle (\frac (db)(dc))=(\frac (c)(b))).Using the method of separating variables, a differential equation is derived from them c d c = a d a + b d b (\displaystyle c\ dc=a\,da+b\,db), whose integration gives the relation c 2 = a 2 + b 2 + C o n s t (\displaystyle c^(2)=a^(2)+b^(2)+\mathrm (Const) ). Application of initial conditions a = b = c = 0 (\displaystyle a=b=c=0) defines the constant as 0, which results in the statement of the theorem.
The quadratic dependence in the final formula appears due to the linear proportionality between the sides of the triangle and the increments, while the sum is associated with independent contributions from the increment of different legs.
Variations and generalizations
Similar geometric shapes on three sides
An important geometric generalization of the Pythagorean theorem was given by Euclid in the Elements, moving from the areas of squares on the sides to the areas of arbitrary similar geometric figures: the sum of the areas of such figures built on the legs will be equal to the area of a similar figure built on the hypotenuse.
The main idea of this generalization is that the area of such a geometric figure is proportional to the square of any of its linear dimensions and, in particular, to the square of the length of any side. Therefore, for similar figures with areas A (\displaystyle A), B (\displaystyle B) And C (\displaystyle C), built on legs with lengths a (\displaystyle a) And b (\displaystyle b) and hypotenuse c (\displaystyle c) Accordingly, the following relation holds:
A a 2 = B b 2 = C c 2 ⇒ A + B = a 2 c 2 C + b 2 c 2 C (\displaystyle (\frac (A)(a^(2)))=(\frac (B )(b^(2)))=(\frac (C)(c^(2)))\,\Rightarrow \,A+B=(\frac (a^(2))(c^(2) ))C+(\frac (b^(2))(c^(2)))C).Since according to the Pythagorean theorem a 2 + b 2 = c 2 (\displaystyle a^(2)+b^(2)=c^(2)), then done.
In addition, if it is possible to prove without invoking the Pythagorean theorem that the areas of three similar geometric figures on the sides of a right triangle satisfy the relation A + B = C (\displaystyle A+B=C), then using the reverse of the proof of Euclid's generalization, one can derive a proof of the Pythagorean theorem. For example, if on the hypotenuse we construct a right triangle congruent with the initial one with an area C (\displaystyle C), and on the sides - two similar right-angled triangles with areas A (\displaystyle A) And B (\displaystyle B), then it turns out that triangles on the sides are formed as a result of dividing the initial triangle by its height, that is, the sum of the two smaller areas of the triangles is equal to the area of the third, thus A + B = C (\displaystyle A+B=C) and, applying the relation for similar figures, the Pythagorean theorem is derived.
Cosine theorem
The Pythagorean theorem is a special case of the more general cosine theorem, which relates the lengths of the sides in an arbitrary triangle:
a 2 + b 2 − 2 a b cos θ = c 2 (\displaystyle a^(2)+b^(2)-2ab\cos (\theta )=c^(2)),where is the angle between the sides a (\displaystyle a) And b (\displaystyle b). If the angle is 90°, then cos θ = 0 (\displaystyle \cos \theta =0), and the formula simplifies to the usual Pythagorean theorem.
Free Triangle
There is a generalization of the Pythagorean theorem to an arbitrary triangle, operating solely on the ratio of the lengths of the sides, it is believed that it was first established by the Sabian astronomer Thabit ibn Qurra. In it, for an arbitrary triangle with sides, an isosceles triangle with a base on the side fits into it c (\displaystyle c), the vertex coinciding with the vertex of the original triangle, opposite the side c (\displaystyle c) and angles at the base equal to the angle θ (\displaystyle \theta ), opposite side c (\displaystyle c). As a result, two triangles are formed, similar to the original one: the first - with sides a (\displaystyle a), the side farthest from it of the inscribed isosceles triangle, and r (\displaystyle r)- side parts c (\displaystyle c); the second - symmetrically to it from the side b (\displaystyle b) with the side s (\displaystyle s)- the corresponding part of the side c (\displaystyle c). As a result, the following relation is satisfied:
a 2 + b 2 = c (r + s) (\displaystyle a^(2)+b^(2)=c(r+s)),degenerating into the Pythagorean theorem at θ = π / 2 (\displaystyle \theta =\pi /2). The relationship is a consequence of the similarity of the formed triangles:
c a = a r , c b = b s ⇒ c r + c s = a 2 + b 2 (\displaystyle (\frac (c)(a))=(\frac (a)(r)),\,(\frac (c) (b))=(\frac (b)(s))\,\Rightarrow \,cr+cs=a^(2)+b^(2)).Pappus's theorem on areas
Non-Euclidean geometry
The Pythagorean theorem is derived from the axioms of Euclidean geometry and is not valid for non-Euclidean geometry - the fulfillment of the Pythagorean theorem is equivalent to the Euclidian parallelism postulate.
In non-Euclidean geometry, the relationship between the sides of a right triangle will necessarily be in a form different from the Pythagorean theorem. For example, in spherical geometry, all three sides of a right triangle, which bound the octant of the unit sphere, have a length π / 2 (\displaystyle \pi /2), which contradicts the Pythagorean theorem.
Moreover, the Pythagorean theorem is valid in hyperbolic and elliptic geometry if the requirement that the triangle is rectangular is replaced by the condition that the sum of two angles of the triangle must be equal to the third.
Spherical geometry
For any right triangle on a sphere with radius R (\displaystyle R)(for example, if the angle in a triangle is right) with sides a , b , c (\displaystyle a,b,c) the relationship between the sides is:
cos (c R) = cos (a R) ⋅ cos (b R) (\displaystyle \cos \left((\frac (c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)).This equality can be derived as a special case of the spherical cosine theorem, which is valid for all spherical triangles:
cos (c R) = cos (a R) ⋅ cos (b R) + sin (a R) ⋅ sin (b R) ⋅ cos γ (\displaystyle \cos \left((\frac ( c)(R))\right)=\cos \left((\frac (a)(R))\right)\cdot \cos \left((\frac (b)(R))\right)+\ sin \left((\frac (a)(R))\right)\cdot \sin \left((\frac (b)(R))\right)\cdot \cos \gamma ). ch c = ch a ⋅ ch b (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b),Where ch (\displaystyle \operatorname (ch) )- hyperbolic cosine. This formula is a special case of the hyperbolic cosine theorem, which is valid for all triangles:
ch c = ch a ⋅ ch b − sh a ⋅ sh b ⋅ cos γ (\displaystyle \operatorname (ch) c=\operatorname (ch) a\cdot \operatorname (ch) b-\operatorname (sh) a\cdot \operatorname (sh) b\cdot \cos \gamma ),Where γ (\displaystyle \gamma )- an angle whose vertex is opposite to the side c (\displaystyle c).
Using the Taylor series for the hyperbolic cosine ( ch x ≈ 1 + x 2 / 2 (\displaystyle \operatorname (ch) x\approx 1+x^(2)/2)) it can be shown that if a hyperbolic triangle decreases (that is, when a (\displaystyle a), b (\displaystyle b) And c (\displaystyle c) tend to zero), then the hyperbolic relations in a right triangle approach the relation of the classical Pythagorean theorem.
Application
Distance in two-dimensional rectangular systems
The most important application of the Pythagorean theorem is determining the distance between two points in a rectangular coordinate system: distance s (\displaystyle s) between points with coordinates (a , b) (\displaystyle (a,b)) And (c , d) (\displaystyle (c,d)) equals:
s = (a − c) 2 + (b − d) 2 (\displaystyle s=(\sqrt ((a-c)^(2)+(b-d)^(2)))).For complex numbers, the Pythagorean theorem gives a natural formula for finding the modulus of a complex number - for z = x + y i (\displaystyle z=x+yi) it is equal to the length
One thing you can be one hundred percent sure of is that when asked what the square of the hypotenuse is, any adult will boldly answer: “The sum of the squares of the legs.” This theorem is firmly ingrained in the minds of every educated person, but you just need to ask someone to prove it, and difficulties can arise. Therefore, let's remember and consider different ways to prove the Pythagorean theorem.
Brief biography
The Pythagorean theorem is familiar to almost everyone, but for some reason the biography of the person who brought it into the world is not so popular. This can be fixed. Therefore, before exploring the different ways to prove Pythagoras’ theorem, you need to briefly get to know his personality.
Pythagoras - philosopher, mathematician, thinker originally from Today it is very difficult to distinguish his biography from the legends that have developed in memory of this great man. But as follows from the works of his followers, Pythagoras of Samos was born on the island of Samos. His father was an ordinary stone cutter, but his mother came from a noble family.
Judging by the legend, the birth of Pythagoras was predicted by a woman named Pythia, in whose honor the boy was named. According to her prediction, the born boy was supposed to bring a lot of benefit and good to humanity. Which is exactly what he did.
Birth of the theorem
In his youth, Pythagoras moved to Egypt to meet famous Egyptian sages there. After meeting with them, he was allowed to study, where he learned all the great achievements of Egyptian philosophy, mathematics and medicine.
It was probably in Egypt that Pythagoras was inspired by the majesty and beauty of the pyramids and created his great theory. This may shock readers, but modern historians believe that Pythagoras did not prove his theory. But he only passed on his knowledge to his followers, who later completed all the necessary mathematical calculations.
Be that as it may, today not one method of proving this theorem is known, but several at once. Today we can only guess how exactly the ancient Greeks performed their calculations, so here we will look at different ways to prove the Pythagorean theorem.
Pythagorean theorem
Before you begin any calculations, you need to figure out what theory you want to prove. The Pythagorean theorem goes like this: “In a triangle in which one of the angles is 90°, the sum of the squares of the legs is equal to the square of the hypotenuse.”
There are a total of 15 different ways to prove the Pythagorean theorem. This is a fairly large number, so we will pay attention to the most popular of them.
Method one
First, let's define what we've been given. These data will also apply to other methods of proving the Pythagorean theorem, so it is worth immediately remembering all the available notations.
Suppose we are given a right triangle with legs a, b and a hypotenuse equal to c. The first method of proof is based on the fact that you need to draw a square from a right triangle.
To do this, you need to add a segment equal to leg b to leg length a, and vice versa. This should result in two equal sides of the square. All that remains is to draw two parallel lines, and the square is ready.
Inside the resulting figure, you need to draw another square with a side equal to the hypotenuse of the original triangle. To do this, from the vertices ас and св you need to draw two parallel segments equal to с. Thus, we get three sides of the square, one of which is the hypotenuse of the original right triangle. All that remains is to draw the fourth segment.
Based on the resulting figure, we can conclude that the area of the outer square is (a + b) 2. If you look inside the figure, you can see that in addition to the inner square, there are four right triangles. The area of each is 0.5av.
Therefore, the area is equal to: 4 * 0.5ab + c 2 = 2av + c 2
Hence (a+c) 2 =2ab+c 2
And, therefore, c 2 =a 2 +b 2
The theorem has been proven.
Method two: similar triangles
This formula for proving the Pythagorean theorem was derived based on a statement from the section of geometry about similar triangles. It states that the leg of a right triangle is the average proportional to its hypotenuse and the segment of the hypotenuse emanating from the vertex of the 90° angle.
The initial data remains the same, so let's start right away with the proof. Let us draw a segment CD perpendicular to side AB. Based on the above statement, the sides of the triangles are equal:
AC=√AB*AD, SV=√AB*DV.
To answer the question of how to prove the Pythagorean theorem, the proof must be completed by squaring both inequalities.
AC 2 = AB * AD and CB 2 = AB * DV
Now we need to add up the resulting inequalities.
AC 2 + CB 2 = AB * (AD * DV), where AD + DV = AB
It turns out that:
AC 2 + CB 2 =AB*AB
And therefore:
AC 2 + CB 2 = AB 2
The proof of the Pythagorean theorem and various methods for solving it require a versatile approach to this problem. However, this option is one of the simplest.
Another calculation method
Descriptions of different methods of proving the Pythagorean theorem may not mean anything until you start practicing on your own. Many techniques involve not only mathematical calculations, but also the construction of new figures from the original triangle.
In this case, it is necessary to complete another right triangle VSD from the side BC. Thus, now there are two triangles with a common leg BC.
Knowing that the areas of similar figures have a ratio as the squares of their similar linear dimensions, then:
S avs * c 2 - S avd * in 2 = S avd * a 2 - S vsd * a 2
S avs *(from 2 - to 2) = a 2 *(S avd -S vsd)
from 2 - to 2 =a 2
c 2 =a 2 +b 2
Since out of the various methods of proving the Pythagorean theorem for grade 8, this option is hardly suitable, you can use the following method.
The easiest way to prove the Pythagorean Theorem. Reviews
According to historians, this method was first used to prove the theorem in ancient Greece. It is the simplest, as it does not require absolutely any calculations. If you draw the picture correctly, then the proof of the statement that a 2 + b 2 = c 2 will be clearly visible.
The conditions for this method will be slightly different from the previous one. To prove the theorem, assume that right triangle ABC is isosceles.
We take the hypotenuse AC as the side of the square and draw its three sides. In addition, it is necessary to draw two diagonal lines in the resulting square. So that inside it you get four isosceles triangles.
You also need to draw a square to the legs AB and CB and draw one diagonal straight line in each of them. We draw the first line from vertex A, the second from C.
Now you need to carefully look at the resulting drawing. Since on the hypotenuse AC there are four triangles equal to the original one, and on the sides there are two, this indicates the veracity of this theorem.
By the way, thanks to this method of proving the Pythagorean theorem, the famous phrase was born: “Pythagorean pants are equal in all directions.”
Proof by J. Garfield
James Garfield is the twentieth President of the United States of America. In addition to making his mark on history as the ruler of the United States, he was also a gifted autodidact.
At the beginning of his career, he was an ordinary teacher in a public school, but soon became the director of one of the higher educational institutions. The desire for self-development allowed him to propose a new theory for proving the Pythagorean theorem. The theorem and an example of its solution are as follows.
First you need to draw two right triangles on a piece of paper so that the leg of one of them is a continuation of the second. The vertices of these triangles need to be connected to ultimately form a trapezoid.
As you know, the area of a trapezoid is equal to the product of half the sum of its bases and its height.
S=a+b/2 * (a+b)
If we consider the resulting trapezoid as a figure consisting of three triangles, then its area can be found as follows:
S=av/2 *2 + s 2 /2
Now we need to equalize the two original expressions
2ab/2 + c/2=(a+b) 2 /2
c 2 =a 2 +b 2
More than one volume of textbooks could be written about the Pythagorean theorem and methods of proving it. But is there any point in it when this knowledge cannot be applied in practice?
Practical application of the Pythagorean theorem
Unfortunately, modern school curricula provide for the use of this theorem only in geometric problems. Graduates will soon leave school without knowing how they can apply their knowledge and skills in practice.
In fact, anyone can use the Pythagorean theorem in their daily life. And not only in professional activities, but also in ordinary household chores. Let's consider several cases when the Pythagorean theorem and methods of proving it may be extremely necessary.
Relationship between the theorem and astronomy
It would seem how stars and triangles on paper can be connected. In fact, astronomy is a scientific field in which the Pythagorean theorem is widely used.
For example, consider the movement of a light beam in space. It is known that light moves in both directions at the same speed. Let's call the trajectory AB along which the light ray moves l. And let's call half the time it takes light to get from point A to point B t. And the speed of the beam - c. It turns out that: c*t=l
If you look at this same ray from another plane, for example, from a space liner that moves with speed v, then when observing bodies in this way, their speed will change. In this case, even stationary elements will begin to move with speed v in the opposite direction.
Let's say the comic liner is sailing to the right. Then points A and B, between which the beam rushes, will begin to move to the left. Moreover, when the beam moves from point A to point B, point A has time to move and, accordingly, the light will already arrive at a new point C. To find half the distance by which point A has moved, you need to multiply the speed of the liner by half the travel time of the beam (t ").
And to find how far a ray of light could travel during this time, you need to mark half the path with a new letter s and get the following expression:
If we imagine that points of light C and B, as well as the space liner, are the vertices of an isosceles triangle, then the segment from point A to the liner will divide it into two right triangles. Therefore, thanks to the Pythagorean theorem, you can find the distance that a ray of light could travel.
This example, of course, is not the most successful, since only a few can be lucky enough to try it in practice. Therefore, let's consider more mundane applications of this theorem.
Mobile signal transmission range
Modern life can no longer be imagined without the existence of smartphones. But how much use would they be if they couldn’t connect subscribers via mobile communications?!
The quality of mobile communications directly depends on the height at which the mobile operator’s antenna is located. In order to calculate how far from a mobile tower a phone can receive a signal, you can apply the Pythagorean theorem.
Let's say you need to find the approximate height of a stationary tower so that it can distribute a signal within a radius of 200 kilometers.
AB (tower height) = x;
BC (signal transmission radius) = 200 km;
OS (radius of the globe) = 6380 km;
OB=OA+ABOB=r+x
Applying the Pythagorean theorem, we find out that the minimum height of the tower should be 2.3 kilometers.
Pythagorean theorem in everyday life
Oddly enough, the Pythagorean theorem can be useful even in everyday matters, such as determining the height of a wardrobe, for example. At first glance, there is no need to use such complex calculations, because you can simply take measurements using a tape measure. But many people wonder why certain problems arise during the assembly process if all measurements were taken more than accurately.
The fact is that the wardrobe is assembled in a horizontal position and only then raised and installed against the wall. Therefore, during the process of lifting the structure, the side of the cabinet must move freely both along the height and diagonally of the room.
Let's assume there is a wardrobe with a depth of 800 mm. Distance from floor to ceiling - 2600 mm. An experienced furniture maker will say that the height of the cabinet should be 126 mm less than the height of the room. But why exactly 126 mm? Let's look at an example.
With ideal cabinet dimensions, let’s check the operation of the Pythagorean theorem:
AC =√AB 2 +√BC 2
AC=√2474 2 +800 2 =2600 mm - everything fits.
Let's say the height of the cabinet is not 2474 mm, but 2505 mm. Then:
AC=√2505 2 +√800 2 =2629 mm.
Therefore, this cabinet is not suitable for installation in this room. Because lifting it into a vertical position can cause damage to its body.
Perhaps, having considered different ways of proving the Pythagorean theorem by different scientists, we can conclude that it is more than true. Now you can use the information received in your daily life and be completely confident that all calculations will be not only useful, but also correct.
The potential for creativity is usually attributed to the humanities, leaving the natural science to analysis, a practical approach and the dry language of formulas and numbers. Mathematics cannot be classified as a humanities subject. But without creativity you won’t go far in the “queen of all sciences” - people have known this for a long time. Since the time of Pythagoras, for example.
School textbooks, unfortunately, usually do not explain that in mathematics it is important not only to cram theorems, axioms and formulas. It is important to understand and feel its fundamental principles. And at the same time, try to free your mind from cliches and elementary truths - only in such conditions are all great discoveries born.
Such discoveries include what we know today as the Pythagorean theorem. With its help, we will try to show that mathematics not only can, but should be exciting. And that this adventure is suitable not only for nerds with thick glasses, but for everyone who is strong in mind and strong in spirit.
From the history of the issue
Strictly speaking, although the theorem is called the “Pythagorean theorem,” Pythagoras himself did not discover it. The right triangle and its special properties were studied long before it. There are two polar points of view on this issue. According to one version, Pythagoras was the first to find a complete proof of the theorem. According to another, the proof does not belong to the authorship of Pythagoras.
Today you can no longer check who is right and who is wrong. What is known is that the proof of Pythagoras, if it ever existed, has not survived. However, there are suggestions that the famous proof from Euclid’s Elements may belong to Pythagoras, and Euclid only recorded it.
It is also known today that problems about a right triangle are found in Egyptian sources from the time of Pharaoh Amenemhat I, on Babylonian clay tablets from the reign of King Hammurabi, in the ancient Indian treatise “Sulva Sutra” and the ancient Chinese work “Zhou-bi suan jin”.
As you can see, the Pythagorean theorem has occupied the minds of mathematicians since ancient times. This is confirmed by about 367 different pieces of evidence that exist today. In this, no other theorem can compete with it. Among the famous authors of proofs we can recall Leonardo da Vinci and the twentieth US President James Garfield. All this speaks of the extreme importance of this theorem for mathematics: most of the theorems of geometry are derived from it or are somehow connected with it.
Proofs of the Pythagorean theorem
School textbooks mostly give algebraic proofs. But the essence of the theorem is in geometry, so let’s first consider those proofs of the famous theorem that are based on this science.
Evidence 1
For the simplest proof of the Pythagorean theorem for a right triangle, you need to set ideal conditions: let the triangle be not only right-angled, but also isosceles. There is reason to believe that it was precisely this kind of triangle that ancient mathematicians initially considered.
Statement “a square built on the hypotenuse of a right triangle is equal to the sum of the squares built on its legs” can be illustrated with the following drawing:
Look at the isosceles right triangle ABC: On the hypotenuse AC, you can construct a square consisting of four triangles equal to the original ABC. And on sides AB and BC a square is built, each of which contains two similar triangles.
By the way, this drawing formed the basis of numerous jokes and cartoons dedicated to the Pythagorean theorem. The most famous is probably "Pythagorean pants are equal in all directions":
Evidence 2
This method combines algebra and geometry and can be considered a variant of the ancient Indian proof of the mathematician Bhaskari.
Construct a right triangle with sides a, b and c(Fig. 1). Then construct two squares with sides equal to the sum of the lengths of the two legs - (a+b). In each of the squares, make constructions as in Figures 2 and 3.
In the first square, build four triangles similar to those in Figure 1. The result is two squares: one with side a, the second with side b.
In the second square, four similar triangles constructed form a square with a side equal to the hypotenuse c.
The sum of the areas of the constructed squares in Fig. 2 is equal to the area of the square we constructed with side c in Fig. 3. This can be easily checked by calculating the area of the squares in Fig. 2 according to the formula. And the area of the inscribed square in Figure 3. by subtracting the areas of four equal right triangles inscribed in the square from the area of a large square with a side (a+b).
Writing all this down, we have: a 2 +b 2 =(a+b) 2 – 2ab. Open the brackets, carry out all the necessary algebraic calculations and get that a 2 +b 2 = a 2 +b 2. In this case, the area inscribed in Fig. 3. square can also be calculated using the traditional formula S=c 2. Those. a 2 +b 2 =c 2– you have proven the Pythagorean theorem.
Evidence 3
The ancient Indian proof itself was described in the 12th century in the treatise “The Crown of Knowledge” (“Siddhanta Shiromani”) and as the main argument the author uses an appeal addressed to the mathematical talents and observation skills of students and followers: “Look!”
But we will analyze this proof in more detail:
Inside the square, build four right triangles as indicated in the drawing. Let us denote the side of the large square, also known as the hypotenuse, With. Let's call the legs of the triangle A And b. According to the drawing, the side of the inner square is (a-b).
Use the formula for the area of a square S=c 2 to calculate the area of the outer square. And at the same time calculate the same value by adding the area of the inner square and the areas of all four right triangles: (a-b) 2 2+4*1\2*a*b.
You can use both options for calculating the area of a square to make sure that they give the same result. And this gives you the right to write down that c 2 =(a-b) 2 +4*1\2*a*b. As a result of the solution, you will receive the formula of the Pythagorean theorem c 2 =a 2 +b 2. The theorem has been proven.
Proof 4
This curious ancient Chinese proof was called the “Bride’s Chair” - because of the chair-like figure that results from all the constructions:
It uses the drawing that we have already seen in Fig. 3 in the second proof. And the inner square with side c is constructed in the same way as in the ancient Indian proof given above.
If you mentally cut off two green rectangular triangles from the drawing in Fig. 1, move them to opposite sides of the square with side c and attach the hypotenuses to the hypotenuses of the lilac triangles, you will get a figure called “bride’s chair” (Fig. 2). For clarity, you can do the same with paper squares and triangles. You will make sure that the “bride’s chair” is formed by two squares: small ones with a side b and big with a side a.
These constructions allowed the ancient Chinese mathematicians and us, following them, to come to the conclusion that c 2 =a 2 +b 2.
Evidence 5
This is another way to find a solution to the Pythagorean theorem using geometry. It's called the Garfield Method.
Construct a right triangle ABC. We need to prove that BC 2 = AC 2 + AB 2.
To do this, continue the leg AC and construct a segment CD, which is equal to the leg AB. Lower the perpendicular AD line segment ED. Segments ED And AC are equal. Connect the dots E And IN, and E And WITH and get a drawing like the picture below:
To prove the tower, we again resort to the method we have already tried: we find the area of the resulting figure in two ways and equate the expressions to each other.
Find the area of a polygon ABED can be done by adding up the areas of the three triangles that form it. And one of them, ERU, is not only rectangular, but also isosceles. Let's also not forget that AB=CD, AC=ED And BC=SE– this will allow us to simplify the recording and not overload it. So, S ABED =2*1/2(AB*AC)+1/2ВС 2.
At the same time, it is obvious that ABED- This is a trapezoid. Therefore, we calculate its area using the formula: S ABED =(DE+AB)*1/2AD. For our calculations, it is more convenient and clearer to represent the segment AD as the sum of segments AC And CD.
Let's write down both ways to calculate the area of a figure, putting an equal sign between them: AB*AC+1/2BC 2 =(DE+AB)*1/2(AC+CD). We use the equality of segments already known to us and described above to simplify the right side of the notation: AB*AC+1/2BC 2 =1/2(AB+AC) 2. Now let’s open the brackets and transform the equality: AB*AC+1/2BC 2 =1/2AC 2 +2*1/2(AB*AC)+1/2AB 2. Having completed all the transformations, we get exactly what we need: BC 2 = AC 2 + AB 2. We have proven the theorem.
Of course, this list of evidence is far from complete. The Pythagorean theorem can also be proven using vectors, complex numbers, differential equations, stereometry, etc. And even physicists: if, for example, liquid is poured into square and triangular volumes similar to those shown in the drawings. By pouring liquid, you can prove the equality of areas and the theorem itself as a result.
A few words about Pythagorean triplets
This issue is little or not studied at all in the school curriculum. Meanwhile, it is very interesting and is of great importance in geometry. Pythagorean triples are used to solve many mathematical problems. Understanding them may be useful to you in further education.
So what are Pythagorean triplets? This is the name for natural numbers collected in groups of three, the sum of the squares of two of which is equal to the third number squared.
Pythagorean triples can be:
- primitive (all three numbers are relatively prime);
- not primitive (if each number of a triple is multiplied by the same number, you get a new triple, which is not primitive).
Even before our era, the ancient Egyptians were fascinated by the mania for numbers of Pythagorean triplets: in problems they considered a right triangle with sides of 3, 4 and 5 units. By the way, any triangle whose sides are equal to the numbers from the Pythagorean triple is rectangular by default.
Examples of Pythagorean triplets: (3, 4, 5), (6, 8, 10), (5, 12, 13), (9, 12, 15), (8, 15, 17), (12, 16, 20 ), (15, 20, 25), (7, 24, 25), (10, 24, 26), (20, 21, 29), (18, 24, 30), (10, 30, 34), (21, 28, 35), (12, 35, 37), (15, 36, 39), (24, 32, 40), (9, 40, 41), (27, 36, 45), (14 , 48, 50), (30, 40, 50), etc.
Practical application of the theorem
The Pythagorean theorem is used not only in mathematics, but also in architecture and construction, astronomy and even literature.
First, about construction: the Pythagorean theorem is widely used in problems of various levels of complexity. For example, look at a Romanesque window:
Let us denote the width of the window as b, then the radius of the major semicircle can be denoted as R and express through b: R=b/2. The radius of smaller semicircles can also be expressed through b: r=b/4. In this problem we are interested in the radius of the inner circle of the window (let's call it p).
The Pythagorean theorem is just useful to calculate R. To do this, we use a right triangle, which is indicated by a dotted line in the figure. The hypotenuse of a triangle consists of two radii: b/4+p. One leg represents the radius b/4, another b/2-p. Using the Pythagorean theorem, we write: (b/4+p) 2 =(b/4) 2 +(b/2-p) 2. Next, we open the brackets and get b 2 /16+ bp/2+p 2 =b 2 /16+b 2 /4-bp+p 2. Let's transform this expression into bp/2=b 2 /4-bp. And then we divide all terms by b, we present similar ones to get 3/2*p=b/4. And in the end we find that p=b/6- which is what we needed.
Using the theorem, you can calculate the length of the rafters for a gable roof. Determine how high a mobile communications tower is needed for the signal to reach a certain populated area. And even install a Christmas tree sustainably in the town square. As you can see, this theorem lives not only on the pages of textbooks, but is often useful in real life.
In literature, the Pythagorean theorem has inspired writers since antiquity and continues to do so in our time. For example, the nineteenth-century German writer Adelbert von Chamisso was inspired to write a sonnet:
The light of truth will not dissipate soon,
But, having shone, it is unlikely to dissipate
And, like thousands of years ago,
It will not cause doubt or controversy.
The wisest when it touches your gaze
Light of truth, thank the gods;
And a hundred bulls, slaughtered, lie -
A return gift from the lucky Pythagoras.
Since then the bulls have been roaring desperately:
Forever alarmed the bull tribe
Event mentioned here.
It seems to them that the time is about to come,
And they will be sacrificed again
Some great theorem.
(translation by Viktor Toporov)
And in the twentieth century, the Soviet writer Evgeny Veltistov, in his book “The Adventures of Electronics,” devoted an entire chapter to proofs of the Pythagorean theorem. And another half chapter to the story about the two-dimensional world that could exist if the Pythagorean theorem became a fundamental law and even a religion for a single world. Living there would be much easier, but also much more boring: for example, no one there understands the meaning of the words “round” and “fluffy”.
And in the book “The Adventures of Electronics,” the author, through the mouth of mathematics teacher Taratar, says: “The main thing in mathematics is the movement of thought, new ideas.” It is precisely this creative flight of thought that gives rise to the Pythagorean theorem - it is not for nothing that it has so many varied proofs. It helps you go beyond the boundaries of the familiar and look at familiar things in a new way.
Conclusion
This article was created so that you can look beyond the school curriculum in mathematics and learn not only those proofs of the Pythagorean theorem that are given in the textbooks “Geometry 7-9” (L.S. Atanasyan, V.N. Rudenko) and “Geometry 7” -11” (A.V. Pogorelov), but also other interesting ways to prove the famous theorem. And also see examples of how the Pythagorean theorem can be applied in everyday life.
Firstly, this information will allow you to qualify for higher scores in mathematics lessons - information on the subject from additional sources is always highly appreciated.
Secondly, we wanted to help you feel how interesting mathematics is. Confirm with specific examples that there is always room for creativity. We hope that the Pythagorean theorem and this article will inspire you to independently explore and make exciting discoveries in mathematics and other sciences.
Tell us in the comments if you found the evidence presented in the article interesting. Did you find this information useful in your studies? Write to us what you think about the Pythagorean theorem and this article - we will be happy to discuss all this with you.
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Theorem
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs (Fig. 1):
$c^(2)=a^(2)+b^(2)$
Proof of the Pythagorean Theorem
Let triangle $A B C$ be a right triangle with right angle $C$ (Fig. 2).
Let us draw the height from the vertex $C$ to the hypotenuse $A B$, and denote the base of the height as $H$.
Right triangle $A C H$ is similar to triangle $A B C$ at two angles ($\angle A C B=\angle C H A=90^(\circ)$, $\angle A$ is common). Likewise, triangle $C B H$ is similar to $A B C$ .
By introducing the notation
$$B C=a, A C=b, A B=c$$
from the similarity of triangles we get that
$$\frac(a)(c)=\frac(H B)(a), \frac(b)(c)=\frac(A H)(b)$$
From here we have that
$$a^(2)=c \cdot H B, b^(2)=c \cdot A H$$
Adding the resulting equalities, we get
$$a^(2)+b^(2)=c \cdot H B+c \cdot A H$$
$$a^(2)+b^(2)=c \cdot(H B+A H)$$
$$a^(2)+b^(2)=c \cdot A B$$
$$a^(2)+b^(2)=c \cdot c$$
$$a^(2)+b^(2)=c^(2)$$
Q.E.D.
Geometric formulation of the Pythagorean theorem
Theorem
In a right triangle, the area of the square built on the hypotenuse is equal to the sum of the areas of the squares built on the legs (Fig. 2):
Examples of problem solving
Example
Exercise. Given a right triangle $A B C$, the sides of which are 6 cm and 8 cm. Find the hypotenuse of this triangle.
Solution. According to the leg condition $a=6$ cm, $b=8$ cm. Then, according to the Pythagorean theorem, the square of the hypotenuse
$c^(2)=a^(2)+b^(2)=6^(2)+8^(2)=36+64=100$
From this we obtain that the desired hypotenuse
$c=\sqrt(100)=10$ (cm)
Answer. 10 cm
Example
Exercise. Find the area of a right triangle if it is known that one of its legs is 5 cm larger than the other and the hypotenuse is 25 cm.
Solution. Let $x$ cm be the length of the smaller leg, then $(x+5)$ cm be the length of the larger one. Then, according to the Pythagorean theorem, we have:
$$x^(2)+(x+5)^(2)=25^(2)$$
We open the brackets, reduce similar ones and solve the resulting quadratic equation:
$x^(2)+5 x-300=0$
According to Vieta's theorem, we obtain that
$x_(1)=15$ (cm) , $x_(2)=-20$ (cm)
The value $x_(2)$ does not satisfy the conditions of the problem, which means that the smaller leg is 15 cm, and the larger leg is 20 cm.
The area of a right triangle is equal to half the product of the lengths of its legs, that is
$$S=\frac(15 \cdot 20)(2)=15 \cdot 10=150\left(\mathrm(cm)^(2)\right)$$
Answer.$S=150\left(\mathrm(cm)^(2)\right)$
Historical reference
Pythagorean theorem- one of the fundamental theorems of Euclidean geometry, establishing the relationship between the sides of a right triangle.
The ancient Chinese book "Zhou Bi Xuan Jing" talks about a Pythagorean triangle with sides 3, 4 and 5. The leading German historian of mathematics, Moritz Cantor (1829 - 1920), believes that the equality $3^(2)+4^(2)=5^ (2) $ was already known to the Egyptians around 2300 BC. According to the scientist, builders then built right angles using right triangles with sides 3, 4 and 5. Somewhat more is known about the Pythagorean theorem among the Babylonians. One text gives an approximate calculation of the hypotenuse of an isosceles right triangle.
Currently, 367 proofs of this theorem have been recorded in the scientific literature. Probably, the Pythagorean theorem is the only theorem with such an impressive number of proofs. Such diversity can only be explained by the fundamental significance of the theorem for geometry.
Pythagorean theorem- one of the fundamental theorems of Euclidean geometry, establishing the relation
between the sides of a right triangle.
It is believed that it was proven by the Greek mathematician Pythagoras, after whom it was named.
Geometric formulation of the Pythagorean theorem.
The theorem was originally formulated as follows:
In a right triangle, the area of the square built on the hypotenuse is equal to the sum of the areas of the squares,
built on legs.
Algebraic formulation of the Pythagorean theorem.
In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs.
That is, denoting the length of the hypotenuse of the triangle by c, and the lengths of the legs through a And b:
Both formulations Pythagorean theorem are equivalent, but the second formulation is more elementary, it does not
requires the concept of area. That is, the second statement can be verified without knowing anything about the area and
by measuring only the lengths of the sides of a right triangle.
Converse Pythagorean theorem.
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then
right triangle.
Or, in other words:
For every triple of positive numbers a, b And c, such that
there is a right triangle with legs a And b and hypotenuse c.
Pythagorean theorem for an isosceles triangle.
Pythagorean theorem for an equilateral triangle.
Proofs of the Pythagorean theorem.
Currently, 367 proofs of this theorem have been recorded in the scientific literature. Probably the theorem
Pythagoras is the only theorem with such an impressive number of proofs. Such diversity
can only be explained by the fundamental significance of the theorem for geometry.
Of course, conceptually all of them can be divided into a small number of classes. The most famous of them:
proof area method, axiomatic And exotic evidence(For example,
by using differential equations).
1. Proof of the Pythagorean theorem using similar triangles.
The following proof of the algebraic formulation is the simplest of the proofs constructed
directly from the axioms. In particular, it does not use the concept of area of a figure.
Let ABC there is a right triangle with a right angle C. Let's draw the height from C and denote
its foundation through H.
Triangle ACH similar to a triangle AB C at two corners. Likewise, triangle CBH similar ABC.
By introducing the notation:
we get:
,
which corresponds to -
Folded a 2 and b 2, we get:
or , which is what needed to be proven.
2. Proof of the Pythagorean theorem using the area method.
The proofs below, despite their apparent simplicity, are not so simple at all. All of them
use properties of area, the proofs of which are more complex than the proof of the Pythagorean theorem itself.
- Proof through equicomplementarity.
Let's arrange four equal rectangular
triangle as shown in the figure
on right.
Quadrangle with sides c- square,
since the sum of two acute angles is 90°, and
unfolded angle - 180°.
The area of the entire figure is equal, on the one hand,
area of a square with side ( a+b), and on the other hand, the sum of the areas of four triangles and
Q.E.D.
3. Proof of the Pythagorean theorem by the infinitesimal method.
Looking at the drawing shown in the figure and
watching the side changea, we can
write the following relation for infinitely
small side incrementsWith And a(using similarity
triangles):
Using the variable separation method, we find:
A more general expression for the change in the hypotenuse in the case of increments on both sides:
Integrating this equation and using the initial conditions, we obtain:
Thus we arrive at the desired answer:
As is easy to see, the quadratic dependence in the final formula appears due to the linear
proportionality between the sides of the triangle and the increments, while the sum is related to the independent
contributions from the increment of different legs.
A simpler proof can be obtained if we assume that one of the legs does not experience an increase
(in this case the leg b). Then for the integration constant we obtain: