Thales's theorem and its proof. Generalized Thales' theorem; Formulation

Thus, Thales' theorem is proven.

Using this theorem will greatly facilitate and speed up the solution of geometric problems. Good luck in mastering this entertaining science of mathematics!

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If the sides of an angle are intersected by straight parallel lines that divide one of the sides into several segments, then the second side, straight lines, will also be divided into segments equivalent to the other side.

Thales' theorem proves the following: C 1, C 2, C 3 are the places where parallel lines intersect on any side of the angle. C 2 is in the middle relative to C 1 and C 3.. Points D 1, D 2, D 3 are the places where the lines intersect, which correspond to the lines on the other side of the angle. We prove that when C 1 C 2 = C 2 C h, then D 1 D 2 = D 2 D 3.
We draw in place D 2 a straight segment KR, parallel to section C 1 C 3. In the properties of a parallelogram, C 1 C 2 = KD 2, C 2 C 3 = D 2 P. If C 1 C 2 = C 2 C 3, then KD 2 = D 2 P.

The resulting triangular figures D 2 D 1 K and D 2 D 3 P are equal. And D 2 K=D 2 P by proof. The angles with the upper point D 2 are equal as vertical, and the angles D 2 KD 1 and D 2 PD 3 are equal as internal crosswise lying with parallel C 1 D 1 and C 3 D 3 and the dividing KP.
Since D 1 D 2 =D 2 D 3 the theorem is proven by the equality of the sides of the triangle

The note: If we take not the sides of the angle, but two straight segments, the proof will be the same.
Any straight segments parallel to each other, which intersect the two lines we are considering and divide one of them into equal sections, do the same with the second.

Let's look at a few examples

First example

The condition of the task is to split the straight line CD into P identical segments.
From point C we draw a semi-line c, which does not lie on the line CD. Let's mark parts of the same size. SS 1, C 1 C 2, C 2 C 3 .....C p-1 C p. Connect C p with D. Draw straight lines from points C 1, C 2,...., C p-1 which will be parallel with respect to C p D. The straight lines will intersect CD in places D 1 D 2 D p-1 and divide the straight line CD into n equal segments.

Second example

Point CK is marked on side AB of triangle ABC. The segment SC intersects the median AM of the triangle at point P, while AK = AP. It is required to find the ratio of VC to RM.
We draw a straight segment through point M, parallel to SC, which intersects AB at point D

By Thales' theoremВD=КD
Using the proportional segment theorem, we find that
РМ = КD = ВК/2, therefore, ВК: РМ = 2:1
Answer: VK: RM = 2:1

Third example

In triangle ABC, side BC = 8 cm. Line DE intersects sides AB and BC parallel to AC. And cuts off the segment EC = 4 cm on the side BC. Prove that AD = DB.

Since BC = 8 cm and EC = 4 cm, then
BE = BC-EC, therefore BE = 8-4 = 4(cm)
By Thales' theorem, since AC is parallel to DE and EC = BE, therefore, AD = DB. Q.E.D.

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Lesson topic

Lesson Objectives

• Get acquainted with new definitions and remember some already studied.
• Formulate and prove the properties of a square, prove its properties.
• Learn to apply the properties of shapes when solving problems.
• Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
• Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.

Lesson Objectives

• Test students' problem-solving skills.

Lesson Plan

1. Historical reference.
2. Thales as a mathematician and his works.
3. It's useful to remember.

Historical reference

• Thales's theorem is still used in maritime navigation as a rule that a collision between ships moving at a constant speed is inevitable if the ships maintain a heading towards each other.

• Outside the Russian-language literature, Thales's theorem is sometimes called another theorem of planimetry, namely, the statement that the inscribed angle based on the diameter of a circle is right. The discovery of this theorem is indeed attributed to Thales, as evidenced by Proclus.

• Thales learned the basics of geometry in Egypt.

Discoveries and merits of its author

Did you know that Thales of Miletus was one of the seven most famous at that time, the sage of Greece. He founded the Ionian school. The idea that Thales promoted in this school was the unity of all things. The sage believed that there is a single beginning from which all things originated.

The great merit of Thales of Miletus is the creation of scientific geometry. This great teaching was able, from the Egyptian art of measurement, to create a deductive geometry, the basis of which is common grounds.

In addition to his enormous knowledge of geometry, Thales was also well versed in astronomy. He was the first to predict a total eclipse of the Sun. But this did not happen in the modern world, but back in 585, even before our era.

Thales of Miletus was the man who realized that north could be accurately determined by the constellation Ursa Minor. But this was not his last discovery, since he was able to accurately determine the length of the year, divide it into three hundred and sixty-five days, and also established the time of the equinoxes.

Thales was in fact a comprehensively developed and wise man. In addition to the fact that he was famous as an excellent mathematician, physicist, and astronomer, he was also a real meteorologist and was able to quite accurately predict the olive harvest.

But the most remarkable thing is that Thales never limited his knowledge only to the scientific and theoretical field, but always tried to consolidate the evidence of his theories in practice. And the most interesting thing is that the great sage did not focus on any one area of ​​his knowledge, his interest had various directions.

The name Thales became a household name for the sage even then. His importance and significance for Greece was as great as the name of Lomonosov for Russia. Of course, his wisdom can be interpreted in different ways. But we can say for sure that he was characterized by ingenuity, practical ingenuity, and, to some extent, detachment.

Thales of Miletus was an excellent mathematician, philosopher, astronomer, loved to travel, was a merchant and entrepreneur, was engaged in trade, and was also a good engineer, diplomat, seer and actively participated in political life.

He even managed to determine the height of the pyramid using a staff and a shadow. And it was like that. One fine sunny day, Thales placed his staff on the border where the shadow of the pyramid ended. Next, he waited until the length of the shadow of his staff was equal to its height, and measured the length of the shadow of the pyramid. So, it would seem that Thales simply determined the height of the pyramid and proved that the length of one shadow is related to the length of another shadow, just as the height of the pyramid is related to the height of the staff. This is what struck Pharaoh Amasis himself.

Thanks to Thales, all knowledge known at that time was transferred to the field of scientific interest. He was able to convey the results to a level suitable for scientific consumption, highlighting a certain set of concepts. And perhaps with the help of Thales the subsequent development of ancient philosophy began.

Thales' theorem plays an important role in mathematics. It was known not only in Ancient Egypt and Babylon, but also in other countries and was the basis for the development of mathematics. And in everyday life, during the construction of buildings, structures, roads, etc., one cannot do without Thales’ theorem.

Thales' theorem in culture

Thales' theorem became famous not only in mathematics, but it was also introduced to culture. One day, the Argentine musical group Les Luthiers (Spanish) presented a song to the audience, which they dedicated to a famous theorem. Members of Les Luthiers, in their video clip specifically for this song, provided proofs for the direct theorem for proportional segments.

Questions

1. Which lines are called parallel?
2. Where is Thales's theorem practically applied?
3. What does Thales' theorem say?

List of sources used

1. Encyclopedia for children. T.11. Mathematics/Editor-in-chief M.D.Aksenova.-m.: Avanta+, 2001.
2. “Unified State Exam 2006. Mathematics. Educational and training materials for preparing students / Rosobrnadzor, ISOP - M.: Intellect-Center, 2006"
3. L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina “Geometry, 7 – 9: textbook for educational institutions”

About parallels and secants.

Outside the Russian-language literature, Thales' theorem is sometimes called another theorem of planimetry, namely, the statement that the inscribed angle subtended by the diameter of a circle is a right angle. The discovery of this theorem is indeed attributed to Thales, as evidenced by Proclus.

Formulations

If several equal segments are laid out in succession on one of two lines and parallel lines are drawn through their ends that intersect the second line, then they will cut off equal segments on the second line.

A more general formulation, also called proportional segment theorem

Parallel lines cut off proportional segments at secants:

(\displaystyle (\frac (A_(1)A_(2))(B_(1)B_(2)))=(\frac (A_(2)A_(3))(B_(2)B_(3) ))=(\frac (A_(1)A_(3))(B_(1)B_(3))).)

• Notes

• The theorem has no restrictions on the relative position of secants (it is true for both intersecting and parallel lines). It also does not matter where the segments on the secants are located.

Thales's theorem is a special case of the proportional segments theorem, since equal segments can be considered proportional segments with a proportionality coefficient equal to 1.

Proof in the case of secants Let's consider the option with unconnected pairs of segments: let the angle be intersected by straight lines A A 1 | |.

1. B B 1 | | C C 1 | C (\displaystyle C) straight lines parallel to the other side of the angle. A B 2 B 1 A 1 (\displaystyle AB_(2)B_(1)A_(1)) C C 1 | C D 2 D 1 C 1 (\displaystyle CD_(2)D_(1)C_(1)). According to the property of a parallelogram: A B 2 = A 1 B 1 (\displaystyle AB_(2)=A_(1)B_(1)) C C 1 | C D 2 = C 1 D 1 (\displaystyle CD_(2)=C_(1)D_(1)).
2. Triangles △ A B B 2 (\displaystyle \bigtriangleup ABB_(2)) C C 1 | △ C D D 2 (\displaystyle \bigtriangleup CDD_(2)) are equal based on the second sign of equality of triangles

Proof in the case of parallel lines

Let's make a direct B.C.. Angles ABC C C 1 | BCD equal as internal crosswise lying with parallel lines AB C C 1 | CD and secant B.C., and the angles ACB C C 1 | CBD equal as internal crosswise lying with parallel lines A.C. C C 1 | BD and secant B.C.. Then, by the second criterion for the equality of triangles, triangles ABC C C 1 | DCB are equal. It follows that A.C. = BD C C 1 | AB = CD. ■

Variations and generalizations

Converse theorem

If in Thales’s theorem equal segments start from the vertex (this formulation is often used in school literature), then the converse theorem will also be true. For intersecting secants it is formulated as follows:

Thus (see figure) from the fact that C B 1 C A 1 = B 1 B 2 A 1 A 2 = … (\displaystyle (\frac (CB_(1))(CA_(1)))=(\frac (B_(1)B_(2))(A_ (1)A_(2)))=\ldots ), follows that A 1 B 1 |.

|

A 2 B 2 |

|

… (\displaystyle A_(1)B_(1)||A_(2)B_(2)||\ldots )

m (\displaystyle m)

. Then the set of lines

X f (X) (\displaystyle Xf(X))

Let us draw a straight line EF through point B 2, parallel to straight line A 1 A 3. By the property of a parallelogram A 1 A 2 = FB 2, A 2 A 3 = B 2 E.

And since A 1 A 2 = A 2 A 3, then FB 2 = B 2 E.

Triangles B 2 B 1 F and B 2 B 3 E are equal according to the second criterion. They have B 2 F = B 2 E according to what has been proven. The angles at the vertex B 2 are equal as vertical, and the angles B 2 FB 1 and B 2 EB 3 are equal as internal crosswise lying with parallel A 1 B 1 and A 3 B 3 and the secant EF. From the equality of triangles follows the equality of sides: B 1 B 2 = B 2 B 3. The theorem is proven.

Using Thales' theorem, the following theorem is established.

Theorem 2. The middle line of the triangle is parallel to the third side and equal to half of it.

The midline of a triangle is the segment connecting the midpoints of its two sides. In Figure 2, segment ED is the middle line of triangle ABC.

ED - midline of triangle ABC

Example 1. Divide this segment into four equal parts.

Solution. Let AB be a given segment (Fig. 3), which must be divided into 4 equal parts.

Dividing a segment into four equal parts

To do this, draw an arbitrary half-line a through point A and plot on it sequentially four equal segments AC, CD, DE, EK.

Let's connect points B and K with a segment. Let us draw straight lines parallel to line BK through the remaining points C, D, E, so that they intersect the segment AB.

According to Thales' theorem, the segment AB will be divided into four equal parts.

Example 2. The diagonal of a rectangle is a. What is the perimeter of a quadrilateral whose vertices are the midpoints of the sides of the rectangle?

Solution. Let Figure 4 meet the conditions of the problem.

Then EF is the midline of triangle ABC and, therefore, by Theorem 2. $$ EF = \frac(1)(2)AC = \frac(a)(2) $$

Similarly $$ HG = \frac(1)(2)AC = \frac(a)(2) , EH = \frac(1)(2)BD = \frac(a)(2) , FG = \frac( 1)(2)BD = \frac(a)(2) $$ and therefore the perimeter of the quadrilateral EFGH is 2a.

Example 3. The sides of a triangle are 2 cm, 3 cm and 4 cm, and its vertices are the midpoints of the sides of another triangle. Find the perimeter of the large triangle.

Solution. Let Figure 5 meet the conditions of the problem.

Segments AB, BC, AC are the middle lines of triangle DEF. Therefore, according to Theorem 2 $$ AB = \frac(1)(2)EF\ \ ,\ \ BC = \frac(1)(2)DE\ \ ,\ \ AC = \frac(1)(2)DF $$ or $$ 2 = \frac(1)(2)EF\ \ ,\ \ 3 = \frac(1)(2)DE\ \ ,\ \ 4 = \frac(1)(2)DF $$ whence $$ EF = 4\ \ ,\ \ DE = 6\ \ ,\ \ DF = 8 $$ and, therefore, the perimeter of triangle DEF is 18 cm.

Example 4. In a right triangle, through the middle of its hypotenuse there are straight lines parallel to its legs. Find the perimeter of the resulting rectangle if the sides of the triangle are 10 cm and 8 cm.

Solution. In triangle ABC (Fig. 6)

∠ A is a straight line, AB = 10 cm, AC = 8 cm, KD and MD are the midlines of triangle ABC, whence $$ KD = \frac(1)(2)AC = 4 cm. \\ MD = \frac(1) (2)AB = 5 cm. $$ The perimeter of rectangle K DMA is 18 cm.