Pythagorean theorem in stereometry formula. Right triangle. The Complete Illustrated Guide (2019)

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 segment ED. Segments ED And AC are equal. Connect the dots E And IN, and also 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 studied little or not 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 communication 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 doubts or disputes.

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: 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 Evgeniy Veltistov, in his book “The Adventures of Electronics,” devoted an entire chapter to proofs of the Pythagorean theorem. And another half chapter to a story about a 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.

website, when copying material in full or in part, a link to the source is required.

Different ways to prove Pythagoras' theorem

student of 9th "A" class

Municipal educational institution secondary school No. 8

Scientific supervisor:

math teacher,

Municipal educational institution secondary school No. 8

Art. Novorozhdestvenskaya

Krasnodar region.

Art. Novorozhdestvenskaya

ANNOTATION.

The Pythagorean theorem is rightfully considered the most important in the course of geometry and deserves close attention. It is the basis for solving many geometric problems, the basis for studying theoretical and practical geometry courses in the future. The theorem is surrounded by a wealth of historical material related to its appearance and methods of proof. Studying the history of the development of geometry instills a love for this subject, promotes the development of cognitive interest, general culture and creativity, and also develops research skills.

As a result of the search activity, the goal of the work was achieved, which was to replenish and generalize knowledge on the proof of the Pythagorean theorem. It was possible to find and consider various methods of proof and deepen knowledge on the topic, going beyond the pages of the school textbook.

The collected material further convinces us that the Pythagorean theorem is a great theorem of geometry and has enormous theoretical and practical significance.

Introduction. Historical background 5 Main part 8

3. Conclusion 19

4. Literature used 20
1. INTRODUCTION. HISTORICAL BACKGROUND.

The essence of the truth is that it is for us forever,

When at least once in her insight we see the light,

And the Pythagorean theorem after so many years

For us, as for him, it is undeniable, impeccable.

To rejoice, Pythagoras made a vow to the gods:

For touching infinite wisdom,

He slaughtered a hundred bulls, thanks to the eternal ones;

He offered prayers and praises after the victim.

Since then, when the bulls smell it, they push,

That the trail again leads people to a new truth,

They roar furiously, so there’s no point in listening,

Such Pythagoras instilled terror in them forever.

The bulls, powerless to resist the new truth,

What remains? - Just closing your eyes, roaring, trembling.

It is not known how Pythagoras proved his theorem. What is certain is that he discovered it under the strong influence of Egyptian science. A special case of the Pythagorean theorem - the properties of a triangle with sides 3, 4 and 5 - was known to the builders of the pyramids long before the birth of Pythagoras, and he himself studied with Egyptian priests for more than 20 years. A legend has been preserved that says that, having proven his famous theorem, Pythagoras sacrificed a bull to the gods, and according to other sources, even 100 bulls. This, however, contradicts information about the moral and religious views of Pythagoras. In literary sources you can read that he “forbade even killing animals, much less feeding on them, for animals have souls, just like us.” Pythagoras ate only honey, bread, vegetables and occasionally fish. In connection with all this, the following entry can be considered more plausible: “... and even when he discovered that in a right triangle the hypotenuse corresponds to the legs, he sacrificed a bull made of wheat dough.”

The popularity of the Pythagorean theorem is so great that its proofs are found even in fiction, for example, in the story “Young Archimedes” by the famous English writer Huxley. The same Proof, but for the special case of an isosceles right triangle, is given in Plato’s dialogue “Meno”.

Fairy tale "Home".

“Far, far away, where even planes don’t fly, is the country of Geometry. In this unusual country there was one amazing city - the city of Teorem. One day a beautiful girl named Hypotenuse came to this city. She tried to rent a room, but no matter where she applied, she was turned down. Finally she approached the rickety house and knocked. A man who called himself Right Angle opened the door to her, and he invited Hypotenuse to live with him. The hypotenuse remained in the house in which Right Angle and his two young sons, named Katetes, lived. Since then, life in the Right Angle house has changed in a new way. The hypotenuse planted flowers on the window and planted red roses in the front garden. The house took the shape of a right triangle. Both legs really liked the Hypotenuse and asked her to stay forever in their house. In the evenings, this friendly family gathers at the family table. Sometimes Right Angle plays hide and seek with his kids. Most often he has to look, and the Hypotenuse hides so skillfully that it can be very difficult to find. One day, while playing, Right Angle noticed an interesting property: if he manages to find the legs, then finding the Hypotenuse is not difficult. So the Right Angle uses this pattern, I must say, very successfully. The Pythagorean theorem is based on the property of this right triangle.”

(From the book by A. Okunev “Thank you for the lesson, children”).

A humorous formulation of the theorem:

If we are given a triangle

And, moreover, with a right angle,

That is the square of the hypotenuse

We can always easily find:

We square the legs,

We find the sum of powers -

And in such a simple way

We will come to the result.

While studying algebra and the beginnings of analysis and geometry in the 10th grade, I became convinced that in addition to the method of proving the Pythagorean theorem discussed in the 8th grade, there are other methods of proof. I present them for your consideration.
2. MAIN PART.

Theorem. In a right triangle there is a square

The hypotenuse is equal to the sum of the squares of the legs.

1 METHOD.

Using the properties of the areas of polygons, we will establish a remarkable relationship between the hypotenuse and the legs of a right triangle.

Proof.

a, c and hypotenuse With(Fig. 1, a).

Let's prove that c²=a²+b².

Proof.

Let's complete the triangle to a square with side a + b as shown in Fig. 1, b. The area S of this square is (a + b)². On the other hand, this square is made up of four equal right-angled triangles, each of which has an area of ​​½ aw  , and a square with side With, therefore S = 4 * ½ aw + s² = 2aw + s².

Thus,

(a + b)² = 2 aw + s²,

c²=a²+b².

The theorem has been proven.
2 METHOD.

After studying the topic “Similar triangles”, I found out that you can apply the similarity of triangles to the proof of the Pythagorean theorem. Namely, I used the statement that the leg of a right triangle is the mean proportional to the hypotenuse and the segment of the hypotenuse enclosed between the leg and the altitude drawn from the vertex of the right angle.

Consider a right triangle with right angle C, CD – height (Fig. 2). Let's prove that AC² +NE² = AB² .

Proof.

Based on the statement about the leg of a right triangle:

AC = , SV = .

Let us square and add the resulting equalities:

AC² = AB * AD, CB² = AB * DB;

AC² + CB² = AB * (AD + DB), where AD+DB=AB, then

AC² + CB² = AB * AB,

AC² + CB² = AB².

The proof is complete.
3 METHOD.

To prove the Pythagorean theorem, you can apply the definition of the cosine of an acute angle of a right triangle. Let's look at Fig. 3.

Proof:

Let ABC be a given right triangle with right angle C. Let us draw the altitude CD from the vertex of right angle C.

By definition of cosine of an angle:

cos A = AD/AC = AC/AB. Hence AB * AD = AC²

Likewise,

cos B = ВD/ВС = ВС/АВ.

Hence AB * BD = BC².

Adding the resulting equalities term by term and noting that AD + DB = AB, we obtain:

AC² + sun² = AB (AD + DB) = AB²

The proof is complete.
4 METHOD.

Having studied the topic “Relationships between the sides and angles of a right triangle”, I think that the Pythagorean theorem can be proven in another way.

Consider a right triangle with legs a, c and hypotenuse With. (Fig. 4).

Let's prove that c²=a²+b².

Proof.

sin B= high quality ; cos B= a/c , then, squaring the resulting equalities, we get:

sin² B= in²/s²; cos² IN= a²/c².

Adding them up, we get:

sin² IN+cos² B=в²/с²+ а²/с², where sin² IN+cos² B=1,

1= (в²+ а²) / с², therefore,

c²= a² + b².

The proof is complete.

5 METHOD.

This proof is based on cutting squares built on the legs (Fig. 5) and placing the resulting parts on a square built on the hypotenuse.

6 METHOD.

For proof on the side Sun we are building BCD ABC(Fig. 6). We know that the areas of similar figures are related as the squares of their similar linear dimensions:

Subtracting the second from the first equality, we get

c2 = a2 + b2.

The proof is complete.

7 METHOD.

Given(Fig. 7):

ABC,= 90° , sun= a, AC=b, AB = c.

Prove:c2 = a2 +b2.

Proof.

Let the leg b A. Let's continue the segment NE per point IN and build a triangle BMD so that the points M And A lay on one side of the straight line CD and, in addition, BD =b, BDM= 90°, DM= a, then BMD= ABC on two sides and the angle between them. Points A and M connect with segments AM. We have M.D. CD And A.C. CD, that means it's straight AC parallel to the line M.D. Because M.D.< АС, then straight CD And A.M. not parallel. Therefore, AMDC- rectangular trapezoid.

In right triangles ABC and BMD 1 + 2 = 90° and 3 + 4 = 90°, but since = =, then 3 + 2 = 90°; Then AVM=180° - 90° = 90°. It turned out that the trapezoid AMDC is divided into three non-overlapping right triangles, then by the area axioms

(a+b)(a+b)

Dividing all terms of the inequality by , we get

Ab + c2 + ab = (a +b) , 2 ab+ c2 = a2+ 2ab+ b2,

c2 = a2 + b2.

The proof is complete.

8 METHOD.

This method is based on the hypotenuse and legs of a right triangle ABC. He constructs the corresponding squares and proves that the square built on the hypotenuse is equal to the sum of the squares built on the legs (Fig. 8).

Proof.

1) DBC= FBA= 90°;

DBC+ ABC= FBA+ ABC, Means, FBC = DBA.

Thus, FBC=ABD(on two sides and the angle between them).

2) , where AL DE, since BD is a common base, DL- total height.

3) , since FB is a foundation, AB- total height.

4)

5) Similarly, it can be proven that

6) Adding term by term, we get:

, BC2 = AB2 + AC2 . The proof is complete.

9 METHOD.

Proof.

1) Let ABDE- a square (Fig. 9), the side of which is equal to the hypotenuse of a right triangle ABC= s, BC = a, AC =b).

2) Let DK B.C. And DK = sun, since 1 + 2 = 90° (like the acute angles of a right triangle), 3 + 2 = 90° (like the angle of a square), AB= BD(sides of the square).

Means, ABC= BDK(by hypotenuse and acute angle).

3) Let EL D.K., A.M. E.L. It can be easily proven that ABC = BDK = DEL = EAM (with legs A And b). Then KS= CM= M.L.= L.K.= A -b.

4) SKB = 4S+SKLMC= 2ab+ (a - b),With2 = 2ab + a2 - 2ab + b2,c2 = a2 + b2.

The proof is complete.

10 METHOD.

The proof can be carried out on a figure jokingly called “Pythagorean pants” (Fig. 10). Its idea is to transform squares built on the sides into equal triangles that together make up the square of the hypotenuse.

ABC move it as shown by the arrow, and it takes position KDN. The rest of the figure AKDCB equal area of ​​the square AKDC this is a parallelogram AKNB.

A parallelogram model has been made AKNB. We rearrange the parallelogram as sketched in the contents of the work. To show the transformation of a parallelogram into an equal-area triangle, in front of the students, we cut off a triangle on the model and move it down. Thus, the area of ​​the square AKDC turned out to be equal to the area of ​​the rectangle. Similarly, we convert the area of ​​a square into the area of ​​a rectangle.

Let's make a transformation for a square built on a leg A(Fig. 11,a):

a) the square is transformed into an equal parallelogram (Fig. 11.6):

b) the parallelogram rotates a quarter turn (Fig. 12):

c) the parallelogram is transformed into an equal rectangle (Fig. 13): 11 METHOD.

Proof:

PCL - straight (Fig. 14);

KLOA= ACPF= ACED= a2;

LGBO= SVMR =CBNQ= b 2;

AKGB= AKLO +LGBO= c2;

c2 = a2 + b2.

The proof is over .

12 METHOD.

Rice. Figure 15 illustrates another original proof of the Pythagorean theorem.

Here: triangle ABC with right angle C; segment B.F. perpendicular NE and equal to it, the segment BE perpendicular AB and equal to it, the segment AD perpendicular AC and equal to it; points F, C,D belong to the same line; quadrilaterals ADFB And ASVE equal in size, since ABF = ECB; triangles ADF And ACE equal in size; subtract from both equal quadrilaterals the triangle they share ABC, we get

, c2 = a2 + b2.

The proof is complete.

13 METHOD.

The area of ​​a given right triangle, on one side, is equal to , on the other hand, ,

3. CONCLUSION.

As a result of the search activity, the goal of the work was achieved, which was to replenish and generalize knowledge on the proof of the Pythagorean theorem. It was possible to find and consider various ways to prove it and deepen knowledge on the topic, going beyond the pages of the school textbook.

The material I have collected convinces me even more that the Pythagorean theorem is a great theorem of geometry and has enormous theoretical and practical significance. In conclusion, I would like to say: the reason for the popularity of the Pythagorean triune theorem is its beauty, simplicity and significance!

4. LITERATURE USED.

1. Entertaining algebra. . Moscow "Science", 1978.

2. Weekly educational and methodological supplement to the newspaper “First of September”, 24/2001.

3. Geometry 7-9. etc.

4. Geometry 7-9. etc.

Animated proof of the Pythagorean theorem - one of fundamental theorems of Euclidean geometry establishing the relationship between the sides of a right triangle. It is believed that it was proven by the Greek mathematician Pythagoras, after whom it is named (there are other versions, in particular the alternative opinion that this theorem in general form was formulated by the Pythagorean mathematician Hippasus).
The theorem states:

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.

Determining the length of the hypotenuse of the triangle c, and the lengths of the legs are like a And b, we get the following formula:

Thus, the Pythagorean theorem establishes a relationship that allows you to determine the side of a right triangle, knowing the lengths of the other two. The Pythagorean theorem is a special case of the cosine theorem, which determines the relationship between the sides of an arbitrary triangle.
The converse statement has also been proven (also called the converse of the Pythagorean theorem):

For any three positive numbers a, b and c such that a ? + b ? = c ?, there is a right triangle with legs a and b and hypotenuse c.

Visual evidence for the triangle (3, 4, 5) from the book "Chu Pei" 500-200 BC. The history of the theorem can be divided into four parts: knowledge about Pythagorean numbers, knowledge about the ratio of sides in a right triangle, knowledge about the ratio of adjacent angles, and the proof of the theorem.
Megalithic structures around 2500 BC. in Egypt and Northern Europe, contain right triangles with whole number sides. Bartel Leendert van der Waerden hypothesized that at that time Pythagorean numbers were found algebraically.
Written between 2000 and 1876 BC. papyrus from the Middle Egyptian Kingdom Berlin 6619 contains a problem whose solution is Pythagorean numbers.
During the reign of Hammurabi the Great, Babylonian tablet Plimpton 322, written between 1790 and 1750 BC contains many entries closely related to Pythagorean numbers.
In the Budhayana sutras, which are variously dated to the eighth or second centuries B.C. in India, contains Pythagorean numbers derived algebraically, a statement of the Pythagorean theorem and a geometric proof for a equilateral right triangle.
The Apastamba Sutras (circa 600 BC) contain a numerical proof of the Pythagorean theorem using area calculations. Van der Waerden believes that it was based on the traditions of its predecessors. According to Albert Burco, this is the original proof of the theorem and he suggests that Pythagoras visited Arakon and copied it.
Pythagoras, whose years of life are usually indicated as 569 - 475 BC. uses algebraic methods for calculating Pythagorean numbers, according to Proklov's commentaries on Euclid. Proclus, however, lived between 410 and 485 AD. According to Thomas Guise, there is no indication of the authorship of the theorem until five centuries after Pythagoras. However, when authors such as Plutarch or Cicero attribute the theorem to Pythagoras, they do so as if the authorship was widely known and certain.
Around 400 BC According to Proclus, Plato gave a method for calculating Pythagorean numbers that combined algebra and geometry. Around 300 BC, in Beginnings Euclid we have the oldest axiomatic proof that has survived to this day.
Written sometime between 500 BC. and 200 BC, the Chinese mathematical book "Chu Pei" (? ? ? ?), gives a visual proof of the Pythagorean theorem, called the Gugu theorem (????) in China, for a triangle with sides (3, 4, 5). During the Han Dynasty, from 202 BC. to 220 AD Pythagorean numbers appear in the book "Nine Branches of the Mathematical Art" along with a mention of right triangles.
The first recorded use of the theorem was in China, where it is known as the Gugu (????) theorem, and in India, where it is known as Bhaskar's theorem.
It has been widely debated whether Pythagoras' theorem was discovered once or repeatedly. Boyer (1991) believes that the knowledge found in the Shulba Sutra may be of Mesopotamian origin.
Algebraic proof
Squares are formed from four right triangles. More than a hundred proofs of the Pythagorean theorem are known. Here is a proof based on the existence theorem of the area of ​​a figure:

Let's place four identical right triangles as shown in the figure.
Quadrangle with sides c is a square, since the sum of two acute angles is , and a straight angle is .
The area of ​​the entire figure is equal, on the one hand, to the area of ​​a square with side “a + b”, and on the other, to the sum of the areas of four triangles and the inner square.

Which is what needs to be proven.
By similarity of triangles
Using similar triangles. Let ABC- a right triangle in which the angle C straight as shown in the picture. Let's draw the height from the point C, and let's call H point of intersection with the side AB. A triangle is formed ACH similar to a triangle ABC, since they are both rectangular (by definition of height) and they have a common angle A, Obviously the third angle in these triangles will also be the same. Similar to peace, triangle CBH also similar to a triangle ABC. With similarity of triangles: If

This can be written as

If we add these two equalities, we get

HB + c times AH = c times (HB + AH) = c ^ 2, ! Src = "http://upload.wikimedia.org/math/7/0/9/70922f59b11b561621c245e11be0b61b.png" />

In other words, the Pythagorean theorem:

Euclid's proof
Euclid's proof in Euclidean Elements, the Pythagorean theorem is proven by the method of parallelograms. Let A, B, C vertices of a right triangle, with right angle A. Let's drop a perpendicular from the point A to the side opposite the hypotenuse in a square built on the hypotenuse. The line divides the square into two rectangles, each of which has the same area as the squares built on the sides. The main idea in the proof is that the upper squares turn into parallelograms of the same area, and then return and turn into rectangles in the lower square and again with the same area.

Let's draw segments CF And A.D. we get triangles BCF And B.D.A.
Angles CAB And BAG– straight; respectively points C, A And G– collinear. Also B, A And H.
Angles CBD And FBA– both are straight lines, then the angle ABD equal to angle FBC, since both are the sum of a right angle and an angle ABC.
Triangle ABD And FBC level on two sides and the angle between them.
Since the points A, K And L– collinear, the area of ​​the rectangle BDLK is equal to two areas of the triangle ABD (BDLK = BAGF = AB 2)
Similarly, we obtain CKLE = ACIH = AC 2
On one side the area CBDE equal to the sum of the areas of the rectangles BDLK And CKLE, and on the other side the area of ​​the square BC 2, or AB 2 + AC 2 = BC 2.

Using differentials
Use of differentials. The Pythagorean theorem can be arrived at by studying how the increase in side affects the size of the hypotenuse as shown in the figure on the right and applying a little calculation.
As a result of the increase in side a, of similar triangles for infinitesimal increments

Integrating we get

If a= 0 then c = b, so "constant" is b 2. Then

As can be seen, the squares are due to the proportion between the increments and the sides, while the sum is the result of the independent contribution of the increments of the sides, not obvious from the geometric evidence. In these equations da And dc– correspondingly infinitesimal increments of sides a And c. But what do we use instead? a And? c, then the limit of the ratio if they tend to zero is da / dc, derivative, and is also equal to c / a, the ratio of the lengths of the sides of the triangles, as a result we obtain a differential equation.
In the case of an orthogonal system of vectors, the equality holds, which is also called the Pythagorean theorem:

If – These are projections of the vector onto the coordinate axes, then this formula coincides with the Euclidean distance and means that the length of the vector is equal to the square root of the sum of the squares of its components.
The analogue of this equality in the case of an infinite system of vectors is called the Parseval equality.

Around and around

The history of the Pythagorean theorem goes back centuries and millennia. In this article, we will not dwell in detail on historical topics. For the sake of intrigue, let’s just say that, apparently, this theorem was known to the ancient Egyptian priests who lived more than 2000 years BC. For those who are curious, here is a link to the Wikipedia article.

First of all, for the sake of completeness, I would like to present here the proof of the Pythagorean theorem, which, in my opinion, is the most elegant and obvious. The picture above shows two identical squares: left and right. It can be seen from the figure that on the left and right the areas of the shaded figures are equal, since in each of the large squares there are 4 identical right triangles shaded. This means that the unshaded (white) areas on the left and right are also equal. We note that in the first case the area of ​​the unshaded figure is equal to , and in the second case the area of ​​the unshaded region is equal to . Thus, . The theorem is proven!

How to call these numbers? You can’t call them triangles, because four numbers can’t form a triangle. And here! Like a bolt from the blue

Since there are such quadruples of numbers, it means there must be a geometric object with the same properties reflected in these numbers!

Now all that remains is to select some geometric object for this property, and everything will fall into place! Of course, the assumption was purely hypothetical and had no basis in support. But what if this is so!

The selection of objects has begun. Stars, polygons, regular, irregular, right angle, and so on and so forth. Again nothing fits. What to do? And at this moment Sherlock gets his second lead.

We need to increase the size! Since three corresponds to a triangle on a plane, then four corresponds to something three-dimensional!

Oh no! Too many options again! And in three dimensions there are much, much more different geometric bodies. Try to go through them all! But it's not all bad. There is also a right angle and other clues! What do we have? Egyptian fours of numbers (let them be Egyptian, they need to be called something), a right angle (or angles) and some three-dimensional object. Deduction worked! And... I believe that quick-witted readers have already realized that we are talking about pyramids in which, at one of the vertices, all three angles are right. You can even call them rectangular pyramids similar to a right triangle.

New theorem

So, we have everything we need. Rectangular (!) pyramids, side facets and secant face-hypotenuse. It's time to draw another picture.

The picture shows a pyramid with its apex at the origin of rectangular coordinates (the pyramid seems to be lying on its side). The pyramid is formed by three mutually perpendicular vectors plotted from the origin along the coordinate axes. That is, each side face of the pyramid is a right triangle with a right angle at the origin. The ends of the vectors define the cutting plane and form the base face of the pyramid.

Theorem

Let there be a rectangular pyramid formed by three mutually perpendicular vectors, the areas of which are equal to - , and the area of ​​the hypotenuse face is - . Then

Alternative formulation: For a tetrahedral pyramid, in which at one of the vertices all plane angles are right, the sum of the squares of the areas of the lateral faces is equal to the square of the area of ​​the base.

Of course, if the usual Pythagorean theorem is formulated for the lengths of the sides of triangles, then our theorem is formulated for the areas of the sides of the pyramid. Proving this theorem in three dimensions is very easy if you know a little vector algebra.

Proof

Let's express the areas in terms of the lengths of the vectors.

Where .

Let's imagine the area as half the area of ​​a parallelogram built on the vectors and

As is known, the vector product of two vectors is a vector whose length is numerically equal to the area of ​​the parallelogram constructed on these vectors.
That's why

Thus,

Q.E.D!

Of course, as a person professionally engaged in research, this has already happened in my life, more than once. But this moment was the brightest and most memorable. I experienced the full range of feelings, emotions, and experiences of a discoverer. From the birth of a thought, the crystallization of an idea, the discovery of evidence - to the complete misunderstanding and even rejection that my ideas met with among my friends, acquaintances and, as it seemed to me then, the whole world. It was unique! I felt like I was in the shoes of Galileo, Copernicus, Newton, Schrödinger, Bohr, Einstein and many many other discoverers.

Afterword

In life, everything turned out to be much simpler and more prosaic. I was late... But by how much! Just 18 years old! Under terrible prolonged torture and not the first time, Google admitted to me that this theorem was published in 1996!

This article was published by Texas Tech University Press. The authors, professional mathematicians, introduced terminology (which, by the way, largely coincided with mine) and also proved a generalized theorem that is valid for a space of any dimension greater than one. What happens in dimensions higher than 3? Everything is very simple: instead of faces and areas there will be hypersurfaces and multidimensional volumes. And the statement, of course, will remain the same: the sum of the squares of the volumes of the side faces is equal to the square of the volume of the base - just the number of faces will be greater, and the volume of each of them will be equal to half the product of the generating vectors. It's almost impossible to imagine! One can only, as philosophers say, think!

Surprisingly, when I learned that such a theorem was already known, I was not at all upset. Somewhere in the depths of my soul, I suspected that it was quite possible that I was not the first, and I understood that I needed to always be prepared for this. But that emotional experience that I received lit a spark of researcher in me, which, I am sure, will never fade now!

P.S.

An erudite reader sent a link in the comments
De Gois' theorem

Excerpt from Wikipedia

In 1783, the theorem was presented to the Paris Academy of Sciences by the French mathematician J.-P. de Gois, but it was previously known to René Descartes and before him Johann Fulgaber, who was probably the first to discover it in 1622. In a more general form, the theorem was formulated by Charles Tinsault (French) in a report to the Paris Academy of Sciences in 1774

So I was not 18 years late, but at least a couple of centuries late!

Sources

Readers provided several useful links in the comments. Here are these and some other links:

Pythagorean theorem: Sum of areas of squares resting on legs ( a And b), equal to the area of ​​the square built on the hypotenuse ( c).

Geometric formulation:

The theorem was originally formulated as follows:

Algebraic formulation:

That is, denoting the length of the hypotenuse of the triangle by c, and the lengths of the legs through a And b :

a 2 + b 2 = c 2

Both formulations of the theorem are equivalent, but the second formulation is more elementary; it does not require 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:

Proof

At the moment, 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.

Of course, conceptually all of them can be divided into a small number of classes. The most famous of them: proofs by the method of areas, axiomatic and exotic proofs (for example, using differential equations).

Through 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 base by H. Triangle ACH similar to a triangle ABC at two corners. Likewise, triangle CBH similar ABC. By introducing the notation

we get

What is equivalent

Adding it up, we get

Proofs using the area method

The proofs below, despite their apparent simplicity, are not so simple at all. They all use properties of area, the proof of which is more complex than the proof of the Pythagorean theorem itself.

Proof via equicomplementation

1. Let's arrange four equal right triangles as shown in Figure 1.
2. Quadrangle with sides c is a square, since the sum of two acute angles is 90°, and the straight angle is 180°.
3. The area of ​​the entire figure is equal, on the one hand, to the area of ​​a square with side (a + b), and on the other hand, to the sum of the areas of four triangles and two internal squares.

Q.E.D.

Proofs through equivalence

Elegant proof using permutation

An example of one such proof is shown in the drawing on the right, where a square built on the hypotenuse is rearranged into two squares built on the legs.

Euclid's proof

Drawing for Euclid's proof

Illustration for Euclid's proof

The idea of ​​Euclid's proof is as follows: let's try to prove that half the area of ​​the square built on the hypotenuse is equal to the sum of the half areas of the squares built on the legs, and then the areas of the large and two small squares are equal.

Let's look at the drawing on the left. On it we constructed squares on the sides of a right triangle and drew a ray s from the vertex of the right angle C perpendicular to the hypotenuse AB, it cuts the square ABIK, built on the hypotenuse, into two rectangles - BHJI and HAKJ, respectively. It turns out that the areas of these rectangles are exactly equal to the areas of the squares built on the corresponding legs.

Let's try to prove that the area of ​​the square DECA is equal to the area of ​​the rectangle AHJK. To do this, we will use an auxiliary observation: The area of ​​a triangle with the same height and base as the given rectangle is equal to half the area of ​​the given rectangle. This is a consequence of defining the area of ​​a triangle as half the product of the base and the height. From this observation it follows that the area of ​​triangle ACK is equal to the area of ​​triangle AHK (not shown in the figure), which in turn is equal to half the area of ​​rectangle AHJK.

Let us now prove that the area of ​​triangle ACK is also equal to half the area of ​​square DECA. The only thing that needs to be done for this is to prove the equality of triangles ACK and BDA (since the area of ​​triangle BDA is equal to half the area of ​​the square according to the above property). This equality is obvious, the triangles are equal on both sides and the angle between them. Namely - AB=AK,AD=AC - the equality of the angles CAK and BAD is easy to prove by the method of motion: we rotate the triangle CAK 90° counterclockwise, then it is obvious that the corresponding sides of the two triangles in question will coincide (due to the fact that the angle at the vertex of the square is 90°).

The reasoning for the equality of the areas of the square BCFG and the rectangle BHJI is completely similar.

Thus, we have proven that the area of ​​a square built on the hypotenuse is composed of the areas of squares built on the legs. The idea behind this proof is further illustrated by the animation above.

Proof of Leonardo da Vinci

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and motion.

Let's consider the drawing, as can be seen from the symmetry, a segment CI cuts the square ABHJ into two identical parts (since triangles ABC And JHI equal in construction). Using a 90 degree counterclockwise rotation, we see the equality of the shaded figures CAJI And GDAB . Now it is clear that the area of ​​the figure we have shaded is equal to the sum of half the areas of the squares built on the legs and the area of ​​the original triangle. On the other hand, it is equal to half the area of ​​the square built on the hypotenuse, plus the area of ​​the original triangle. The last step in the proof is left to the reader.

Proof by the infinitesimal method

The following proof using differential equations is often attributed to the famous English mathematician Hardy, who lived in the first half of the 20th century.

Looking at the drawing shown in the figure and observing the change in side a, we can write the following relation for infinitesimal side increments With And a(using triangle similarity):

Proof by the infinitesimal method

Using the method of separation of variables, 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

c 2 = a 2 + b 2 + constant.

Thus we arrive at the desired answer

c 2 = a 2 + b 2 .

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 associated with 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 increment (in this case, the leg b). Then for the integration constant we obtain

Variations and generalizations

• If instead of squares we construct other similar figures on the sides, then the following generalization of the Pythagorean theorem is true: In a right triangle, the sum of the areas of similar figures built on the sides is equal to the area of ​​the figure built on the hypotenuse. In particular:
  • The sum of the areas of regular triangles built on the legs is equal to the area of ​​a regular triangle built on the hypotenuse.
  • The sum of the areas of semicircles built on the legs (as on the diameter) is equal to the area of ​​the semicircle built on the hypotenuse. This example is used to prove the properties of figures bounded by the arcs of two circles and called Hippocratic lunulae.

Story

Chu-pei 500–200 BC. On the left is the inscription: the sum of the squares of the lengths of the height and base is the square of the length of the hypotenuse.

The ancient Chinese book Chu-pei talks about a Pythagorean triangle with sides 3, 4 and 5: The same book offers a drawing that coincides with one of the drawings of the Hindu geometry of Bashara.

Cantor (the greatest German historian of mathematics) believes that the equality 3² + 4² = 5² was already known to the Egyptians around 2300 BC. e., during the time of King Amenemhat I (according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonaptes, or “rope pullers,” built right angles using right triangles with sides of 3, 4, and 5.

It is very easy to reproduce their method of construction. Let's take a rope 12 m long and tie a colored strip to it at a distance of 3 m. from one end and 4 meters from the other. The right angle will be enclosed between sides 3 and 4 meters long. It could be objected to the Harpedonaptians that their method of construction becomes superfluous if one uses, for example, a wooden square, which is used by all carpenters. Indeed, Egyptian drawings are known in which such a tool is found, for example, drawings depicting a carpenter's workshop.

Somewhat more is known about the Pythagorean theorem among the Babylonians. In one text dating back to the time of Hammurabi, that is, to 2000 BC. e., an approximate calculation of the hypotenuse of a right triangle is given. From this we can conclude that in Mesopotamia they were able to perform calculations with right triangles, at least in some cases. Based, on the one hand, on the current level of knowledge about Egyptian and Babylonian mathematics, and on the other hand, on a critical study of Greek sources, Van der Waerden (Dutch mathematician) came to the following conclusion:

Literature

In Russian

• Skopets Z. A. Geometric miniatures. M., 1990
• Elensky Shch. In the footsteps of Pythagoras. M., 1961
• Van der Waerden B. L. Awakening Science. Mathematics of Ancient Egypt, Babylon and Greece. M., 1959
• Glazer G.I. History of mathematics at school. M., 1982
• W. Litzman, “The Pythagorean Theorem” M., 1960.
  • A site about the Pythagorean theorem with a large number of proofs, material taken from the book by V. Litzmann, a large number of drawings are presented in the form of separate graphic files.
• The Pythagorean theorem and Pythagorean triples chapter from the book by D. V. Anosov “A look at mathematics and something from it”
• About the Pythagorean theorem and methods of proving it G. Glaser, academician of the Russian Academy of Education, Moscow

In English

• Pythagorean Theorem at WolframMathWorld
• Cut-The-Knot, section on the Pythagorean theorem, about 70 proofs and extensive additional information (English)

Wikimedia Foundation. 2010.