There are no restrictions on the mutual arrangement of secants in the theorem (it is true both for intersecting lines and for parallel ones). It also doesn't matter where the line segments are on the secants.

Proof in the case of parallel lines

Let's draw a line BC. Angles ABC and BCD are equal as interior crosses lying under parallel lines AB and CD and secant BC, and angles ACB and CBD are equal as interior crosses lying under parallel lines AC and BD and secant BC. Then, according to the second criterion for the equality of triangles, triangles ABC and DCB are congruent. This implies that AC = BD and AB = CD. ■

Also exists proportional segment theorem:

Parallel lines cut proportional segments at secants: $$

\frac(A_1A_2)(B_1B_2)=\frac(A_2A_3)(B_2B_3)=\frac(A_1A_3)(B_1B_3).

The Thales 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.

Inverse theorem

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

Thus (see Fig.) from the fact that $\backslash frac(CB\_1)(CA\_1)=\backslash frac(B\_1B\_2)(A\_1A\_2)=\backslash ldots\; =\; (\backslash rm\; idem)$ it follows that the direct $A\_1B\_1||A\_2B\_2||\backslash ldots$.

If the secants are parallel, then it is necessary to require the equality of the segments on both secants between themselves, otherwise this statement becomes incorrect (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).

Variations and Generalizations

The following statement is dual to Sollertinsky's lemma:

Thales' theorem is still used today in maritime navigation as the rule that a collision between ships moving at a constant speed is unavoidable if the ships keep heading towards each other. Outside of the Russian-language literature, the Thales theorem is sometimes called another theorem of planimetry, namely, the statement that an inscribed angle based on the diameter of a circle is a right one. The discovery of this theorem is indeed attributed to Thales, as evidenced by Proclus. Write a review on the article "Theorem of Thales" Literature Atanasyan L. S. and others. Geometry 7-9. - Ed. 3rd. - M .: Enlightenment, 1992. Notes see also Thales' theorem on an angle based on a diameter of a circle An excerpt characterizing the Thales Theorem “I don’t think anything, I just don’t understand it ... - Wait, Sonya, you will understand everything. See what kind of person he is. Don't think bad things about me or him. “I don’t think bad things about anyone: I love everyone and feel sorry for everyone. But what am I to do? Sonya did not give up on the gentle tone with which Natasha addressed her. The softer and more searching Natasha's expression was, the more serious and stern was Sonya's face. “Natasha,” she said, “you asked me not to talk to you, I didn’t, now you yourself started. Natasha, I don't believe him. Why this secret? - Again, again! Natasha interrupted. - Natasha, I'm afraid for you. - What to be afraid of? “I am afraid that you will ruin yourself,” Sonya said decisively, herself frightened by what she said. Natasha's face again expressed anger. “And I will destroy, I will destroy, I will destroy myself as soon as possible. None of your business. Not to you, but to me it will be bad. Leave, leave me. I hate you. - Natasha! Sonya called out in fear. - I hate it, I hate it! And you are my enemy forever! Natasha ran out of the room. Natasha did not speak to Sonya anymore and avoided her. With the same expression of agitated surprise and criminality, she paced the rooms, taking up first this and then another occupation and immediately abandoning them. No matter how hard it was for Sonya, she kept her eyes on her friend. On the eve of the day on which the count was supposed to return, Sonya noticed that Natasha had been sitting all morning at the living room window, as if waiting for something and that she had made some kind of sign to the passing military man, whom Sonya mistook for Anatole. Sonya began to observe her friend even more attentively and noticed that Natasha was in a strange and unnatural state all the time of dinner and evening (she answered inappropriately to questions put to her, began and did not finish phrases, laughed at everything). After tea, Sonya saw a timid maid waiting for her at Natasha's door. She let it through, and, eavesdropping at the door, learned that the letter had again been handed over. And suddenly it became clear to Sonya that Natasha had some kind of terrible plan for this evening. Sonya knocked on her door. Natasha didn't let her in. “She will run away with him! Sonya thought. She is capable of anything. To-day there was something particularly pathetic and resolute in her face. She burst into tears, saying goodbye to her uncle, Sonya recalled. Yes, that's right, she runs with him - but what should I do? thought Sonya, now recalling those signs that clearly proved why Natasha had some kind of terrible intention. "There is no count. What should I do, write to Kuragin, demanding an explanation from him? But who tells him to answer? Write to Pierre, as Prince Andrei asked in case of an accident? ... But maybe, in fact, she had already refused Bolkonsky (she sent a letter to Princess Marya yesterday). There are no uncles!” It seemed terrible to Sonya to tell Marya Dmitrievna, who believed so much in Natasha. But one way or another, Sonya thought, standing in a dark corridor: now or never the time has come to prove that I remember the good deeds of their family and love Nicolas. No, I won’t sleep for at least three nights, but I won’t leave this corridor and won’t let her in by force, and won’t let shame fall on their family, ”she thought. Anatole recently moved to Dolokhov. The plan for the abduction of Rostova had already been thought out and prepared by Dolokhov for several days, and on the day when Sonya, having overheard Natasha at the door, decided to protect her, this plan was to be carried out. Natasha promised to go out to Kuragin on the back porch at ten o'clock in the evening. Kuragin was supposed to put her in a prepared troika and take her 60 miles from Moscow to the village of Kamenka, where a trimmed priest was prepared, who was supposed to marry them. In Kamenka, a set-up was ready, which was supposed to take them to the Varshavskaya road, and there they were supposed to ride abroad on postage. Anatole had a passport, and a traveler's, and ten thousand money taken from his sister, and ten thousand borrowed through Dolokhov. Two witnesses—Khvostikov, the former clerk whom Dolokhov and Makarin used to play games, a retired hussar, a good-natured and weak man who had boundless love for Kuragin—were sitting in the first room at tea. In Dolokhov's large office, decorated from wall to ceiling with Persian carpets, bearskins and weapons, Dolokhov sat in a traveling beshmet and boots in front of an open bureau, on which lay bills and wads of money. Anatole, in his unbuttoned uniform, walked from the room where the witnesses were sitting, through the study to the back room, where his French footman and others were packing the last things. Dolokhov counted money and wrote it down. “Well,” he said, “Khvostikov should be given two thousand. - Well, let me, - said Anatole. - Makarka (that's what they called Makarina), this one disinterestedly for you through fire and into water. Well, the scores are over, - said Dolokhov, showing him a note. - So? “Yes, of course, that’s how it is,” said Anatole, apparently not listening to Dolokhov and with a smile that did not leave his face, looking ahead of himself.

About parallel and secant.

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

Wording

If on one of the two straight lines several equal segments are sequentially laid aside and parallel lines are drawn through their ends, intersecting the second straight line, then they will cut off equal segments on the second straight line.

A more general formulation, also called proportional segment theorem

Parallel lines cut proportional segments at secants:

A 1 A 2 B 1 B 2 = A 2 A 3 B 2 B 3 = A 1 A 3 B 1 B 3 . (\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))).)

Remarks

• There are no restrictions on the mutual arrangement of secants in the theorem (it is true both for intersecting lines and for parallel ones). It also doesn't matter where the line segments are on the secants.

• The Thales 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

Consider a variant with unconnected pairs of segments: let the angle be intersected by straight lines A A 1 | | B B 1 | | C C 1 | | D D 1 (\displaystyle AA_(1)||BB_(1)||CC_(1)||DD_(1)) and wherein A B = C D (\displaystyle AB=CD).

1. Pass through the dots A (\displaystyle A) and 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)) and C D 2 D 1 C 1 (\displaystyle CD_(2)D_(1)C_(1)). According to the parallelogram property: A B 2 = A 1 B 1 (\displaystyle AB_(2)=A_(1)B_(1)) and C D 2 = C 1 D 1 (\displaystyle CD_(2)=C_(1)D_(1)).
2. triangles △ A B B 2 (\displaystyle \bigtriangleup ABB_(2)) and △ C D D 2 (\displaystyle \bigtriangleup CDD_(2)) are equal on the basis of the second criterion for the equality of triangles

Proof in the case of parallel lines

Let's draw a straight line BC. corners ABC and BCD are equal as internal crosses lying at parallel lines AB and CD and secant BC, and the angles ACB and CBD are equal as internal crosses lying at parallel lines AC and BD and secant BC. Then, according to the second criterion for the equality of triangles, the triangles ABC and DCB are equal. Hence it follows that AC = BD and AB = CD. ■

Variations and Generalizations

Inverse theorem

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

Thus (see Fig.) 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 ).

If the secants are parallel, then it is necessary to require the equality of the segments on both secants between themselves, otherwise this statement becomes incorrect (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).

This theorem is used in navigation: a collision of ships moving at a constant speed is inevitable if the direction from one ship to another is maintained.

Lemma of Sollertinsky

The following statement is dual to Sollertinsky's lemma:

Let f (\displaystyle f)- projective correspondence between points of the line l (\displaystyle l) and direct m (\displaystyle m). Then the set of lines X f (X) (\displaystyle Xf(X)) will be the set of tangents to some

Plan:

Introduction

• 1 Inverse theorem
• 2 Thales' theorem in culture
• 3 Interesting Facts
Notes

Introduction

This is the parallel lines theorem. For an angle based on a diameter, see another theorem.

Thales' theorem- one of the theorems of planimetry.

There are no restrictions on the mutual arrangement of secants in the theorem (it is true both for intersecting lines and for parallel ones). It also doesn't matter where the line segments are on the secants.

Proof in the case of secants

Proof of Thales' theorem

Consider a variant with unconnected pairs of segments: let the angle be intersected by straight lines AA 1 | | BB 1 | | CC 1 | | DD 1 and wherein AB = CD .

Proof in the case of parallel lines

Let's draw a line BC. Angles ABC and BCD are equal as interior crosses lying under parallel lines AB and CD and secant BC, and angles ACB and CBD are equal as interior crosses lying under parallel lines AC and BD and secant BC. Then, according to the first criterion for the equality of triangles, triangles ABC and DCB are congruent. This implies that AC = BD and AB = CD. ■

Also exists generalized Thales theorem:

Parallel lines cut proportional segments at secants:

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

1. Inverse theorem

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

In the inverse Thales theorem, it is important that equal segments start from the vertex

Thus (see Fig.) from what follows that the lines .

If the secants are parallel, then it is necessary to require the equality of the segments on both secants between themselves, otherwise this statement becomes incorrect (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).

2. Thales' theorem in culture

Argentine musical group Les Luthiers ( Spanish) presented a song dedicated to the theorem. The video clip for this song provides a proof for the direct theorem for proportional intervals.

3. Interesting facts

• Thales' theorem is still used today in maritime navigation as the rule that a collision between ships moving at a constant speed is unavoidable if the ships keep heading towards each other.
• Outside of Russian-language literature, the Thales theorem is sometimes called another theorem of planimetry, namely, the statement that an inscribed angle based on the diameter of a circle is a right one. The discovery of this theorem is indeed attributed to Thales, as evidenced by Proclus.
• Thales comprehended the basics of geometry in Egypt.

Notes

1. El Teorema de Thales por Les Luthiers en You Tube - www.youtube.com/watch?v=czzj2C4wdxY
2. 3. Travel to Egypt / Home / Ancient Literature and Philosophy. Thales from Miletus - www.fales-iz-mileta.narod.ru/3_puteshestvie_v_egipet

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This tomb is small, but the glory over it is immense.
In it, before you, the many-minded Thales is hidden.

Inscription on the tomb of Thales of Miletus

Imagine such a picture. 600 BC Egypt. Before you is a huge Egyptian pyramid. To surprise the pharaoh and remain among his favorites, you need to measure the height of this pyramid. You have… nothing at your disposal. You can fall into despair, or you can do what Thales of Miletus: use the triangle similarity theorem. Yes, it turns out that everything is quite simple. Thales of Miletus waited until the length of his shadow and his height coincided, and then, using the triangle similarity theorem, found the length of the shadow of the pyramid, which, accordingly, was equal to the shadow cast by the pyramid.

Who is this Thales of Miletus? A man who gained fame as one of the "seven wise men" of antiquity? Thales of Miletus is an ancient Greek philosopher who excelled in astronomy, as well as mathematics and physics. The years of his life have been established only approximately: 625-645 BC

Among the proofs of Thales's knowledge of astronomy is the following example. May 28, 585 BC the prediction of a solar eclipse by Miletus helped to end the war between Lydia and Media that had already lasted for 6 years. This phenomenon so frightened the Medes that they agreed to unfavorable conditions for making peace with the Lydians.

The legend that characterizes Thales as a resourceful person is quite widely known. Thales often heard unflattering comments about his poverty. Once he decided to prove that philosophers can, if they wish, live in abundance. Even in winter, Thales, by observing the stars, determined that there would be a good harvest of olives in the summer. Then he hired oil presses in Miletus and Chios. It cost him quite cheaply, since in winter there is practically no demand for them. When the olives gave a rich harvest, Thales began to rent out his oil presses. A large amount of money collected by this method was regarded as proof that philosophers can earn with their minds, but their vocation is higher than such earthly problems. This legend, by the way, was repeated by Aristotle himself.

As for geometry, many of his "discoveries" were borrowed from the Egyptians. And yet this transfer of knowledge to Greece is considered one of the main merits of Thales of Miletus.

The achievements of Thales are the formulation and proof of the following theorems:

• vertical angles are equal;
• equal triangles are those in which the side and two adjacent angles are respectively equal;
• the angles at the base of an isosceles triangle are equal;
• the diameter bisects the circle;
• An inscribed angle based on a diameter is a right angle.

Another theorem is named after Thales, which is useful in solving geometric problems. There is its generalized and particular form, the inverse theorem, the formulations may also differ slightly depending on the source, but the meaning of all of them remains the same. Let's consider this theorem.

If parallel lines intersect the sides of an angle and cut off equal segments on one of its sides, then they cut off equal segments on its other side.

Let's say points A 1, A 2, A 3 are the points of intersection of parallel lines on one side of the angle, and B 1, B 2, B 3 are the points of intersection of parallel lines with the other side of the angle. It is necessary to prove that if A 1 A 2 \u003d A 2 A 3, then B 1 B 2 \u003d B 2 B 3.

Draw a line through point B 2 parallel to line A 1 A 2 . Let's designate a new straight line С 1 С 2 . Consider the parallelograms A 1 C 1 B 2 A 2 and A 2 B 2 C 2 A 3 .

The parallelogram properties allow us to assert that A1A2 = C 1 B 2 and A 2 A 3 = B 2 C 2 . And since according to our condition A 1 A 2 \u003d A 2 A 3, then C 1 B 2 \u003d B 2 C 2.

And finally, consider the triangles ∆ C 1 B 2 B 1 and ∆ C 2 B 2 B 3 .

C 1 B 2 = B 2 C 2 (proved above).

And this means that Δ C 1 B 2 B 1 and Δ C 2 B 2 B 3 will be equal according to the second sign of equality of triangles (along the side and adjacent angles).

Thus, the Thales theorem is proved.

The use of 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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Theorem 6.6 (Thales' theorem).If parallel lines intersecting the sides of an angle cut off equal segments on one side of it, then they cut off equal segments on the other side.(Fig. 131).

Proof. Let A 1, A 2, A 3 be the intersection points of parallel lines with one of the sides of the angle and A 2 lies between A 1 and A 3 (Fig. 131). Let B 1 , B 2 , B 3 be the corresponding points of intersection of these lines with the other side of the angle. Let us prove that if A 1 A 2 = A 2 Az, then B 1 B 2 = B 2 B 3.

Let us draw a line EF through the point B 2 parallel to the line A 1 A 3 . By the property of a parallelogram A 1 A 2 \u003d FB 2, A 2 A 3 \u003d B 2 E. And since A 1 A 2 \u003d A 2 A 3, then FB 2 \u003d B 2 E.

Triangles B 2 B 1 F and B 2 B 3 E are equal in the second criterion. They have B 2 F=B 2 E by the 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 a secant EF.

From the equality of triangles follows the equality of the sides: B 1 B 2 \u003d B 2 B 3. The theorem has been proven.

Comment. In the condition of the Thales theorem, instead of the sides of the angle, you can take any two straight lines, while the conclusion of the theorem will be the same:

parallel lines intersecting two given lines and cutting off equal segments on one line, cut off equal segments on the other line.

Sometimes Thales' theorem will be applied in this form as well.

Problem (48). Divide the given segment AB into n equal parts.

Solution. Let us draw from the point A a half-line a not lying on the line AB (Fig. 132). Set aside equal segments on the half-line a: AA 1, A 1 A 2, A 2 A 3, .... A n - 1 A n. Connect the points A n and B. Draw through the points A 1, A 2, .... A n -1 straight lines parallel to the line A n B. They intersect the segment AB at points B 1, B 2, B n-1, which divide the segment AB into n equal segments (according to the Thales theorem).

A. V. Pogorelov, Geometry for grades 7-11, Textbook for educational institutions