The Pythagorean theorem: background, evidence, examples of practical application. Different ways to prove the Pythagorean theorem: examples, descriptions and reviews Pythagorean theorem you know

Various ways to prove the Pythagorean theorem

student of 9 "A" class

MOU secondary school №8

Scientific adviser:

mathematic teacher,

MOU secondary school №8

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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 the theoretical and practical course of geometry in the future. The theorem is surrounded by the richest historical material related to its appearance and methods of proof. The study of the history of the development of geometry instills a love for this subject, contributes to 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 is to replenish and generalize knowledge on the proof of the Pythagorean theorem. It was possible to find and consider various ways of proof and deepen knowledge on the topic, going beyond the pages of a school textbook.

The collected material convinces even more that the Pythagorean theorem is the great theorem of geometry and is of great theoretical and practical importance.

Introduction. Historical background 5 Main body 8

3. Conclusion 19

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 HISTORICAL REFERENCE.

The essence of 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 indisputable, impeccable.

To celebrate, the gods were given a vow by Pythagoras:

For touching infinite wisdom,

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

He offered prayers and praises to the victim after.

Since then, bulls, when they smell, pushing,

What leads people to the new truth again,

They roar furiously, so there is no urine to listen,

Such Pythagoras instilled terror in them forever.

Bulls, powerless to resist the new truth,

What remains? - Just close your eyes, roar, tremble.

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, while he himself studied with Egyptian priests for more than 20 years. There is a legend that says that, having proved 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, one can read that he "forbidden even killing animals, and even more so feeding them, because animals have a soul, 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 of the famous English writer Huxley "Young Archimedes". The same Proof, but for the particular case of an isosceles right triangle, is given in Plato's dialogue Meno.

Fairy tale house.

“Far, far away, where even planes do not 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 get a room, but wherever she applied, she was refused everywhere. At last she approached the rickety house and knocked. She was opened by a man who called himself the Right Angle, and he invited the Hypotenuse to live with him. The hypotenuse remained in the house where Right Angle and his two little sons, named Katet, lived. Since then, life in the Right Angle House has changed in a new way. The hypotenuse planted flowers in the window, and spread red roses in the front garden. The house took the form of a right triangle. Both legs liked Hypotenuse very much 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 it. Once during a game, Right Angle noticed an interesting property: if he manages to find the legs, then finding the Hypotenuse is not difficult. So 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 playful 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 build the legs in a square,

We find the sum of degrees -

And in such a simple way

We will come to the result.

Studying algebra and the beginnings of analysis and geometry in the 10th grade, I was convinced that in addition to the method of proving the Pythagorean theorem considered in the 8th grade, there are other ways of proving it. I present them for your consideration.
2. MAIN PART.

Theorem. Square in a right triangle

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

1 WAY.

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

Proof.

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

Let's prove that c²=a²+b².

Proof.

We complete the triangle to a square with a side a + b as shown in fig. 1b. 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, the area of each of which is ½ aw, and a square with a side With, so S = 4 * ½ av + s² = 2av + s².

Thus,

(a + b)² = 2 av + s²,

c²=a²+b².

The theorem has been proven.
2 WAY.

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 for the hypotenuse and the segment of the hypotenuse enclosed between the leg and the height drawn from the vertex of the right angle.

Consider a right-angled triangle with a right angle C, CD is the height (Fig. 2). Let's prove that AC² + SW² = AB² .

Proof.

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

AC = , CB = .

We 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 WAY.

The definition of the cosine of an acute angle of a right triangle can be applied to the proof of the Pythagorean theorem. Consider Fig. 3.

Proof:

Let ABC be a given right triangle with a right angle C. Draw a height CD from the vertex of the right angle C.

By definition of the cosine of an angle:

cos A \u003d AD / AC \u003d AC / AB. Hence AB * AD = AC²

Likewise,

cos B \u003d BD / BC \u003d BC / AB.

Hence AB * BD \u003d BC².

Adding the resulting equalities term by term and noticing that AD + DВ = AB, we get:

AC² + Sun² \u003d AB (AD + DB) \u003d AB²

The proof is complete.
4 WAY.

Having studied the topic "Ratios between the sides and angles of a right triangle", I think that the Pythagorean theorem can be proved in another way.

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

Let's prove that c²=a²+b².

Proof.

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

sin² B= in²/s²; cos² IN\u003d a² / s².

Adding them up, we get:

sin² IN+ cos² B= v² / s² + a² / s², where sin² IN+ cos² B=1,

1 \u003d (v² + a²) / s², therefore,

c² = a² + b².

The proof is complete.

5 WAY.

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

6 WAY.

For proof on the cathete sun 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 WAY.

Given(Fig. 7):

ABS,= 90° , sun= a, AC=b, AB = c.

Prove:c2 = a2 +b2.

Proof.

Let the leg b A. Let's continue the segment SW per point IN and build a triangle bmd so that the points M And A lay on one side of a straight line CD and besides, B.D.=b, BDM= 90°, DM= a, then bmd= ABC on two sides and the angle between them. Points A and M connect by segments AM. We have MD CD And AC CD, means straight AC parallel to a straight line MD. Because MD< АС, then straight CD And AM are 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 divided into three non-overlapping right triangles, then by the area axioms

(a+b)(a+b)

Dividing all the terms of the inequality by , we obtain

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

c2 = a2 + b2.

The proof is complete.

8 WAY.

This method is based on the hypotenuse and legs of a right triangle ABC. He builds 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- overall height.

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

5) Similarly, one can prove that

6) Adding term by term, we get:

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

9 WAY.

Proof.

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

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

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

3) Let EL DC, AM EL. It can be easily proved that ABC = BDK = DEL = EAM (with legs A And b). Then KS= CM= ML= LK= A -b.

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

The proof is complete.

10 WAY.

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

ABC shift, as shown by the arrow, and it takes the position KDN. The rest of the figure AKDCB equal to the area of a square AKDC- it's a parallelogram AKNB.

Made a parallelogram model AKNB. We shift the parallelogram as sketched in the content of the work. To show the transformation of a parallelogram into an equal triangle, in front of the students, we cut off a triangle on the model and shift it down. So the area of the square AKDC is equal to the area of the rectangle. Similarly, we convert the area of a square to 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-sized parallelogram (Fig. 11.6):

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

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

Proof:

PCL- straight (Fig. 14);

KLOA= ACPF= ACED= a2;

LGBO= CVMR =CBNQ= b 2;

AKGB= AKLO +LGBO= c2;

c2 = a2 + b2.

Proof over .

12 WAY.

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

Here: triangle ABC with right angle C; line segment bf perpendicular SW and equal to it, the segment BE perpendicular AB and equal to it, the segment AD perpendicular AC and equal to him; points F, C,D belong to one straight line; quadrangles ADFB And ACBE are equal because ABF = ECB; triangles ADF And ACE are equal; we subtract from both equal quadrangles a common triangle for them abc, we get

, c2 = a2 + b2.

The proof is complete.

13 WAY.

The area of this right triangle, on the one hand, is equal to , with another, ,

3. CONCLUSION

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

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

4. LITERATURE USED.

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

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

3. Geometry 7-9. and etc.

4. Geometry 7-9. and etc.

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's draw a height from the vertex $C$ to the hypotenuse $A B$, denote the base of the height as $H$ .

Right triangle $A C H$ is similar to triangle $A B C$ in two angles ($\angle A C B=\angle C H A=90^(\circ)$, $\angle A$ is common). Similarly, triangle $C B H$ is similar to $A B C$ .

Introducing the notation

$$B C=a, A C=b, A B=c$$

from the similarity of triangles we get that

$$\frac(a)(c)=\frac(H B)(a), \frac(b)(c)=\frac(A H)(b)$$

Hence we have that

$$a^(2)=c \cdot H B, b^(2)=c \cdot A H$$

Adding the obtained equalities, we obtain

$$a^(2)+b^(2)=c \cdot H B+c \cdot A H$$

$$a^(2)+b^(2)=c \cdot(H B+A H)$$

$$a^(2)+b^(2)=c \cdot A B$$

$$a^(2)+b^(2)=c \cdot c$$

$$a^(2)+b^(2)=c^(2)$$

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. You are given a right triangle $A B C$ whose legs are 6 cm and 8 cm. Find the hypotenuse of this triangle.

Solution. According to the condition of the leg $a=6$ cm, $b=8$ cm. Then, according to the Pythagorean theorem, the square of the hypotenuse

$c^(2)=a^(2)+b^(2)=6^(2)+8^(2)=36+64=100$

Hence we get that the required 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 longer than the other, and the hypotenuse is 25 cm.

Solution. Let $x$ cm be the length of the smaller leg, then $(x+5)$ cm is 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 get that

$x_(1)=15$ (cm) , $x_(2)=-20$ (cm)

The value of $x_(2)$ does not satisfy the condition of the problem, which means that the smaller leg is 15 cm, and the larger one is 20 cm.

The area of a right triangle is 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 suan jing" speaks of a Pythagorean triangle with sides 3, 4 and 5. The largest German historian of mathematics Moritz Kantor (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, the builders then built right angles using right-angled 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.

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 a variety can only be explained by the fundamental significance of the theorem for geometry.

Pythagorean theorem: The sum of the areas of the squares supported by the legs ( a And b), is 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 through 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.

Inverse Pythagorean theorem:

Proof

Of course, conceptually, all of them can be divided into a small number of classes. The most famous of them: proofs by the area method, 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 built directly from the axioms. In particular, it does not use the concept of figure area.

Let ABC there is a right angled triangle C. Let's draw a height from C and denote its base by H. Triangle ACH similar to a triangle ABC at two corners. Likewise, the triangle CBH similar ABC. Introducing the notation

we get

What is equivalent

Adding, we get

Area proofs

The following proofs, despite their apparent simplicity, are not so simple at all. All of them use the properties of the area, the proof of which is more complicated than the proof of the Pythagorean theorem itself.

Proof via Equivalence

Arrange four equal right triangles as shown in Figure 1.
Quadrilateral with sides c is a square because the sum of two acute angles is 90° and the straight angle is 180°.
The area of the whole figure is equal, on the one hand, to the area of a square with a side (a + b), and on the other hand, the sum of the areas of four triangles and two inner squares.

Q.E.D.

Evidence through Equivalence

An elegant permutation proof

An example of one of these proofs is shown in the drawing on the right, where the square built on the hypotenuse is converted by permutation 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.

Consider the drawing on the left. We built squares on the sides of a right-angled triangle on it and drew a ray s from the vertex of 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 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), 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 by the above property). This equality is obvious, the triangles are equal in two sides and the angle between them. Namely - AB=AK,AD=AC - the equality of angles CAK and BAD is easy to prove by the motion method: let's rotate the triangle CAK 90 ° counterclockwise, then it is obvious that the corresponding sides of the two triangles under consideration will coincide (due to the fact that the angle at the vertex of the square is 90°).

The argument about the equality of the areas of the square BCFG and the rectangle BHJI is completely analogous.

Thus, we have proved that the area of the square built on the hypotenuse is the sum of the areas of the squares built on the legs. The idea behind this proof is further illustrated with the animation above.

Proof of Leonardo da Vinci

The main elements of the proof are symmetry and movement.

Consider the drawing, as can be seen from the symmetry, the segment CI dissects the square ABHJ into two identical parts (since triangles ABC And JHI are 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 shaded by us 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.

Considering 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 similar triangles):

Proof by the infinitesimal method

Using the method of separation of variables, we find

A more general expression for changing the hypotenuse in the case of increments of both legs

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 due 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 increment (in this case, the leg b). Then for the integration constant we get

Variations and Generalizations

If, instead of squares, other similar figures are constructed on the legs, 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 legs 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 the 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 arcs of two circles and bearing the name hippocratic lunula.

Story

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

The ancient Chinese book Chu-pei speaks of a Pythagorean triangle with sides 3, 4 and 5: In the same book, a drawing is proposed that coincides with one of the drawings of the Hindu geometry of Baskhara.

Kantor (the largest German historian of mathematics) believes that the equality 3 ² + 4 ² = 5² was already known to the Egyptians around 2300 BC. e., during the time of King Amenemhet I (according to papyrus 6619 of the Berlin Museum). According to Cantor, the harpedonapts, or "stringers", built right angles using right triangles with sides 3, 4 and 5.

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

Somewhat more is known about the Pythagorean theorem among the Babylonians. In one text dating back to the time of Hammurabi, i.e., 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-angled triangles, at least in some cases. Based, on the one hand, on the current level of knowledge of Egyptian and Babylonian mathematics, and on the other, on a critical study of Greek sources, Van der Waerden (a Dutch mathematician) concluded the following:

Literature

In Russian

Skopets Z. A. Geometric miniatures. M., 1990
Yelensky Sh. Following 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, the material is taken from the book by W. Litzman, a large number of drawings are presented as 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”
On the Pythagorean theorem and methods of its proof G. Glaser, Academician of the Russian Academy of Education, Moscow

In English

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

Wikimedia Foundation. 2010 .

The Pythagorean theorem says:

In a right triangle, the sum of the squares of the legs is equal to the square of the hypotenuse:

a 2 + b 2 = c 2,

a And b- legs forming a right angle.
With is the hypotenuse of the triangle.

Formulas of the Pythagorean theorem

a = \sqrt(c^(2) - b^(2))
b = \sqrt (c^(2) - a^(2))
c = \sqrt (a^(2) + b^(2))

Proof of the Pythagorean Theorem

The area of a right triangle is calculated by the formula:

S = \frac(1)(2)ab

To calculate the area of an arbitrary triangle, the area formula is:

p- semiperimeter. p=\frac(1)(2)(a+b+c) ,
r is the radius of the inscribed circle. For a rectangle r=\frac(1)(2)(a+b-c).

Then we equate the right sides of both formulas for the area of a triangle:

\frac(1)(2) ab = \frac(1)(2)(a+b+c) \frac(1)(2)(a+b-c)

2 ab = (a+b+c) (a+b-c)

2 ab = \left((a+b)^(2) -c^(2) \right)

2ab = a^(2)+2ab+b^(2)-c^(2)

0=a^(2)+b^(2)-c^(2)

c^(2) = a^(2)+b^(2)

Inverse 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 the triangle is a right triangle. That is, for any triple of positive numbers a, b And c, such that

a 2 + b 2 = c 2,

there is a right triangle with legs a And b and hypotenuse c.

Pythagorean theorem- one of the fundamental theorems of Euclidean geometry, establishing the relationship between the sides of a right triangle. It was proved by the scientist mathematician and philosopher Pythagoras.

The meaning of the theorem in that it can be used to prove other theorems and solve problems.

Additional material:

However, the name is received in honor of the scientist only for the reason that he is the first and even the only person who was able to prove the theorem.

The German historian of mathematics Kantor claimed that the theorem was already known to the Egyptians around 2300 BC. e. He believed that right angles used to be built thanks to right triangles with sides 3, 4 and 5.

The famous scientist Kepler said that geometry has an irreplaceable treasure - this is the Pythagorean theorem, thanks to which it is possible to derive most of the theorems in geometry.

Previously, the Pythagorean theorem was called the “bride theorem” or the “nymph theorem”. And the thing is that her drawing was very similar to a butterfly or a nymph. The Arabs, when they translated the text of the theorem, decided that the nymph means the bride. This is how the interesting name of the theorem appeared.

Pythagorean theorem, formula

Theorem

- in a right triangle, the sum of the squares of the legs () is equal to the square of the hypotenuse (). This is one of the fundamental theorems of Euclidean geometry.

Formula:

As already mentioned, there are many different proofs of the theorem with versatile mathematical approaches. However, area theorems are more commonly used.

Construct squares on the triangle ( blue, green, red)

That is, the sum of the areas of the squares built on the legs is equal to the area of the square built on the hypotenuse. Accordingly, the areas of these squares are equal -. This is the geometric explanation of Pythagoras.

Proof of the theorem by the area method: 1 way

Let's prove that .

Consider the same triangle with legs a, b and hypotenuse c.

We complete the right triangle to a square. From leg “a” we continue the line up to the distance of leg “b” (red line).
Next, we draw the line of the new leg “a” to the right (green line).
We connect two legs with the hypotenuse “c”.

It turns out the same triangle, only inverted.

Similarly, we build on the other side: from the leg “a” we draw the line of the leg “b” and down “a” and “b” And from the bottom of the leg “b” we draw the line of the leg “a”. In the center of each leg, a hypotenuse “c” was drawn. Thus the hypotenuses formed a square in the center.

This square consists of 4 identical triangles. And the area of each right triangle = half the product of its legs. Respectively, . And the area of the square in the center = , since all 4 hypotenuses have sides. The sides of a quadrilateral are equal and the angles are right. How can we prove that the angles are right? Very simple. Let's take the same square:

We know that the two angles shown in the figure are 90 degrees. Since the triangles are equal, then the next leg angle “b” is equal to the previous leg “b”:

The sum of these two angles = 90 degrees. Accordingly, the previous angle is also 90 degrees. Of course, the same is true on the other side. Accordingly, we really have a square with right angles.

Since the acute angles of a right triangle are 90 degrees in total, the angle of the quadrilateral will also be 90 degrees, because 3 angles in total = 180 degrees.

Accordingly, the area of a square consists of four areas of identical right-angled triangles and the area of the square, which is formed by the hypotenuses.

Thus, we got a square with side . We know that the area of a square with a side is the square of its side. That is . This square consists of four identical triangles.

And this means that we have proved the Pythagorean theorem.

IMPORTANT!!! If we find the hypotenuse, then we add two legs, and then we derive the answer from the root. When finding one of the legs: from the square of the length of the second leg, subtract the square of the length of the hypotenuse and find the square root.