Theorem parallel lines cut off equal segments. Thales' theorem. Middle line of the triangle


Lesson topic

Lesson Objectives

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

Lesson objectives

  • Check students' ability to solve problems.

Lesson plan

  1. History reference.
  2. Thales as a mathematician and his works.
  3. Good to remember.

History reference

  • Thales' theorem is still used in maritime navigation as a rule that the collision of ships moving from constant speed, is inevitable if the course of the ships towards each other is maintained.


  • 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.

Discoveries and merits of its author

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

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

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

Thales of Miletus was the man who realized that the north can be accurately determined by the constellation Ursa Minor. But it wasn't him either. latest discovery, since he was able to accurately determine the length of the year, break it into three hundred and sixty-five days, and also set the time of the equinoxes.

Thales was in fact comprehensively developed and wise man. In addition to being famous as an excellent mathematician, physicist, and astronomer, he was also, as a real meteorologist, 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 different directions.

The name of 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 definitely say that he was characterized by both ingenuity, and 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 with the help of 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. Then he waited until the length of the shadow of his staff equaled his 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 the other shadow, just as the height of the pyramid is related to the height of the staff. This struck the pharaoh Amasis himself.

Thanks to Thales, all knowledge known at that time was transferred to the field of scientific interest. He was able to bring 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 one important roles in mathematics. She was known not only in Ancient Egypt and Babylon, but also in other countries and was the basis for the development of mathematics. Yes and in Everyday life, during the construction of buildings, structures, roads, etc., one cannot do without the Thales theorem.

Thales' theorem in culture

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

Questions

  1. What lines are called parallel?
  2. Where is the Thales theorem applied in practice?
  3. What is the Thales theorem about?

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 the preparation of 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: a textbook for educational institutions"
Subjects > Mathematics > Mathematics Grade 8

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 \frac(CB_1)(CA_1)=\frac(B_1B_2)(A_1A_2)=\ldots = (\rm idem) it follows that the direct A_1B_1||A_2B_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).

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.

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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 recent times 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, a former clerk, whom Dolokhov and Makarin used for playing, a retired hussar, good-natured and weak person, who had boundless love for Kuragin - sat in the first room for 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 him.


          1. wording;

          2. Proof;

  1. Theorem on proportional segments;

  2. Ceva's theorem;

          1. wording;

          2. Proof;

  1. Theorem of Menelaus;

          1. wording;

          2. Proof;

  1. Tasks and their solutions;

  2. Conclusion;

  3. List of used sources and literature.

Introduction.

All the little things are needed

To be significant...

I. Severyanin
This abstract is devoted to the application of the method of parallel lines to the proof of theorems and problem solving. Why are we using this method? In that academic year At the school Olympiad in mathematics, a geometric problem was proposed, which seemed to us very difficult. It was this task that gave impetus to the beginning of work on the study and development of the method of parallel lines in solving problems on finding the ratio of the lengths of segments.

The idea of ​​the method itself is based on the use of the generalized Thales theorem. The Thales theorem is studied in the eighth grade, its generalization and the topic “Similarities of Figures” in the ninth grade and only in the tenth grade, in an introductory plan, two important theorems of Ceva and Menelaus are studied, with the help of which a number of problems are relatively easily solved for finding the ratio of the lengths of segments. Therefore, at the level of basic education, we can decide quite narrow circle assignments for this study material. Although at the final certification for the course of the main school and at the USE in mathematics, tasks on this topic (Thales' theorem. Similarity of triangles, similarity coefficient. Signs of similarity of triangles) are offered in the second part of the examination paper and are of a high level of complexity.

In the process of working on the abstract, it became possible to deepen our knowledge on this topic. The proof of the theorem on proportional segments in a triangle (the theorem is not included in the school curriculum) is based on the method of parallel lines. In turn, this theorem allowed us to propose another way to prove the theorems of Ceva and Menelaus. And as a result, we were able to learn how to solve a wider range of problems for comparing the lengths of segments. This is the relevance of our work.

Generalized Thales theorem.

Formulation:

Parallel lines intersecting two given lines cut proportional segments on these lines.
Given:

Straight a cut by parallel lines ( BUT 1 AT 1 , BUT 2 AT 2 , BUT 3 AT 3 ,…, BUT n B n) into segments BUT 1 BUT 2 , BUT 2 BUT 3 , …, A n -1 A n, and the straight line b- into segments AT 1 AT 2 , AT 2 AT 3 , …, AT n -1 AT n .


Prove:

Proof:

Let us prove, for example, that

Consider two cases:

1 case (Fig. b)

Direct a and b are parallel. Then the quadrilaterals

BUT 1 BUT 2 AT 2 AT 1 and BUT 2 BUT 3 AT 3 AT 2 - parallelograms. That's why

BUT 1 BUT 2 =AT 1 AT 2 and BUT 2 BUT 3 =AT 2 AT 3 , whence it follows that


2 case (fig. c)

Lines a and b are not parallel. Through the dot BUT 1 let's draw a straight line With, parallel to the line b. She will cross the lines BUT 2 AT 2 and BUT 3 AT 3 at some points FROM 2 and FROM 3 . triangles BUT 1 BUT 2 FROM 2 and BUT 1 BUT 3 FROM 3 are similar in two angles (angle BUT 1 – general, angles BUT 1 BUT 2 FROM 2 and BUT 1 BUT 3 FROM 3 equal as corresponding under parallel lines BUT 2 AT 2 and BUT 3 AT 3 secant BUT 2 BUT 3 ), that's why

1+

Or according to the property of proportions

On the other hand, by what was proved in the first case, we have BUT 1 FROM 2 =AT 1 AT 2 , FROM 2 FROM 3 =AT 2 AT 3 . Replacing in proportion (1) BUT 1 FROM 2 on the AT 1 AT 2 and FROM 2 FROM 3 on the AT 2 AT 3 , we arrive at the equality

Q.E.D.
Theorem on proportional segments in a triangle.

On the sides AC and sun triangle ABC points are marked To and M so AC:CS=m: n, BM: MC= p: q. Segments AM and VC intersect at a point O(Fig. 124b).


Prove:

Proof:
Through the dot M let's draw a straight line MD(Fig. 124a), parallel VC. She crosses the side AC at the point D, and according to the generalization of the Thales theorem

Let AK=mx. Then, in accordance with the condition of the problem KS=nx, and since KD: DC= p: q, then again we use the generalization of the Thales theorem:

Similarly, it is proved that .

Ceva's theorem.
The theorem is named after the Italian mathematician Giovanni Ceva, who proved it in 1678.

Formulation:

If on the sides AB, BC and CA of the triangle ABC points C are taken respectively 1 , BUT 1 and B 1 , then segments AA 1 , BB 1 and SS 1 intersect at one point if and only if


Given:

Triangle ABC and on its sides AB, sun and AC points are marked FROM 1 ,BUT 1 and AT 1 .


Prove:

2.cuts A A 1 , BB 1 and SS 1 intersect at one point.


Proof:
1. Let the segments AA 1 , BB 1 and SS 1 intersect at one point O. Let us prove that equality (3) holds. According to the theorem on proportional segments in triangle 1 we have:

The left parts of these equalities are the same, so the right parts are also equal. Equating them, we get


Dividing both parts into right side, we arrive at equality (3).

2. Let us prove the converse assertion. Let the points FROM 1 ,BUT 1 and AT 1 taken on the sides AB, sun and SA so that equality (3) holds. Let us prove that the segments AA 1 , BB 1 and SS 1 intersect at one point. Denote by letter O the point of intersection of the segments A A 1 and BB 1 and draw a straight line SO. She crosses the side AB at some point, which we denote FROM 2 . Since the segments AA 1 , BB 1 and SS 1 intersect at one point, then by what was proved in the first paragraph

Thus, equalities (3) and (4) hold.

Comparing them, we arrive at the equality = , which shows that the points C 1 and C 2 share a side AB C 1 and C 2 coincide, and hence the segments AA 1 , BB 1 and SS 1 intersect at a point O.

Q.E.D.
Theorem of Menelaus.

Formulation:

If on the sides AB and BC and the extension of the side AC (or on the extensions of the sides AB, BC and AC) points C are taken respectively 1 , BUT 1 , AT 1 , then these points lie on the same line if and only if

Given:

Triangle ABC and on its sides AB, sun and AC points are marked FROM 1 ,BUT 1 and AT 1 .


Prove:


2. points BUT 1 ,FROM 1 and AT 1 lie on the same line
Proof:
1. Let the points BUT 1 ,FROM 1 and AT 1 lie on the same line. Let us prove that equality (5) holds. Let's spend AD,BE and CF parallel to a straight line AT 1 BUT 1 (dot D lies on a straight line sun). According to the generalized Thales theorem, we have:


Multiplying the left and right parts of these equalities, we obtain


those. equality (5) holds.
2. Let us prove the converse assertion. Let the point AT 1 taken on the continuation side AC, and the points FROM 1 and BUT 1 - on the sides AB and sun, and in such a way that equality (5) holds. Let us prove that the points BUT 1 ,FROM 1 and AT 1 lie on the same line. Let the straight line A 1 C 1 intersect the continuation of the side AC at the point B 2, then, by what was proved in the first paragraph

Comparing (5) and (6), we arrive at the equality = , which shows that the points AT 1 and AT 2 share a side AC in the same respect. Therefore, the points AT 1 and AT 2 coincide, and hence the points BUT 1 ,FROM 1 and AT 1 lie on the same line. The converse assertion is proved similarly in the case when all three points BUT 1 ,FROM 1 and AT 1 lie on the extensions of the corresponding sides.

Q.E.D.

Problem solving.

It is proposed to consider a number of problems on the proportional division of segments in a triangle. As noted above, there are several methods for determining the location of the points needed in the problem. In our work, we settled on the method of parallel lines. The theoretical basis of this method is the generalized Thales theorem, which allows using parallel lines to transfer famous relationships proportions from one side of the angle to its second side, thus, you only need to draw these parallel lines in a convenient way for solving the problem.
Consider specific tasks:
Task №1 Point M is taken in the triangle ABC on the side BC so that VM:MC=3:2. Point P divides segment AM in a ratio of 2:1. Line BP intersects side AC at point B 1 . In what respect is point B 1 divides side AC?

Solution: It is necessary to find the ratio AB 1: B 1 C, AC is the desired segment on which the point B 1 lies.

The parallel method is as follows:


  1. cut the desired segment with parallel lines. One BB 1 is already there, and the second MN will be drawn through the point M, parallel to BB 1.

  2. Transfer the known ratio from one side of the angle to its other side, i.e. consider the angles of the side, which are cut by these straight lines.
The sides of the angle C are cut by straight lines BB 1 and MN and, according to the generalized Thales theorem, we conclude AT 1 N=3p, NC=2r. The sides of the MAC angle intersect the lines PB 1 and MN and divide its sides in a ratio of 2: 1, therefore AB 1: B 1 N \u003d 2: 1 and therefore AB 1 \u003d 2n, AT 1 N= n. Because AT 1 N=3p, and AT 1 N= n, then 3p=n.

Let's move on to the ratio of interest to us AB 1: B 1 C \u003d AB 1: (B 1 N + NC) \u003d 2n: (3p + 2p) \u003d (2 * 3p): (5p) \u003d 6: 5.

Answer: AB 1:B 1 C = 6:5.

Comment: This problem could be solved using the Menelaus theorem. Applying it to the triangle AMC. Then the line BB 1 intersects two sides of the triangle at points B 1 and P, and the continuation of the third at point B. So the equality applies: , Consequently
Task number 2 In the triangle ABC AN is the median. On the AC side, point M is taken so that AM: MC \u003d 1: 3. The segments AN and BM intersect at point O, and the ray CO intersects AB at point K. In what ratio does point K divide segment AB.

Solution: We need to find the ratio of AK to KV.

1) Draw a line NN 1 parallel to the line SK and a line NN 2 parallel to the line VM.

2) The sides of the angle ABC are intersected by straight lines SC and NN 1 and, according to the generalized Thales theorem, we conclude BN 1:N 1 K=1:1 or BN 1 = N 1 K= y.

3) The sides of the angle BCM are intersected by the lines BM and NN 2 and, according to the generalized Thales theorem, we conclude CN 2:N 2 M=1:1 or CN 2 = N 2 M=3:2=1.5.

4) The sides of the angle NAC are intersected by lines BM and NN 2 and according to the generalized Thales theorem we conclude AO: ON=1:1.5 or AO=m ON=1.5m.

5) The sides of the angle BAN are intersected by straight lines SK and NN 1 and, according to the generalized Thales theorem, we conclude AK: KN 1 \u003d 1: 1.5 or AK \u003d n KN 1 =1,5 n.

6) KN 1 \u003d y \u003d 1.5n.

Answer: AK:KV=1:3.

Comment: This problem could be solved using Ceva's theorem, applying it to the triangle ABC. By condition, the points N, M, K lie on the sides of the triangle ABC and the segments AN, CK and VM intersect at one point, which means that the equality is true: , we substitute the known relations, we have , AK:KV=1:3.

Task No. 3 On the side BC of triangle ABC, a point D is taken such that BD: DC \u003d 2: 5, and on the side AC, point E is such that . In what ratio are the segments BE and AD divided by the point K of their intersection?
Solution: Need to find 1) AK:KD=? 2) VK:KE=?

1) Draw the line DD 1 parallel to the line BE.

2) The sides of the angle ALL are intersected by lines BE and DD 1 and, according to the generalized Thales theorem, we conclude CD 1:D 1 E=5:2 or CD 1 = 5z, D 1 E=2z.

3) According to the condition AE:EC=1:2, i.e. AE \u003d x, EC \u003d 2x, but EC \u003d CD 1 + D 1 E, then 2y=5z+2 z=7 z, z=

4) The sides of the angle DCA are intersected by the lines BE and DD 1 and, according to the generalized Thales theorem, we conclude

5) To determine the ratio VK:KE, we draw a straight line EE 1 and, arguing in a similar way, we obtain


Answer: AK:KD=7:4; VK:KE=6:5.
Comment: This problem could be solved using the Menelaus theorem. Applying it to the triangle WEIGHT. Then the line DA intersects two sides of the triangle at points D and K, and the continuation of the third one at point A. So the equality applies: , therefore VK:KE=6:5. Arguing similarly with respect to the triangle ADC, we obtain , AK:KD=7:4.
Problem #4 In ∆ ABC, the bisector AD divides the side BC in a 2:1 ratio. In what ratio does the median CE divide this bisector?

Solution: Let O point the intersection of the bisector AD and the median CE. We need to find the ratio AO:OD.

1) Draw a line DD 1 parallel to line CE.

2) The sides of the angle ABC are intersected by the lines CE and DD 1 and, according to the generalized Thales theorem, we conclude BD 1:D 1 E=2:1 or BD 1 = 2p, D 1 E=p.

3) According to the condition AE:EB=1:1, i.e. AE=y, EB=y, but EB= BD 1 + D 1 E, so y=2p+ p=3 p, p =
4) The sides of the angle BAD are intersected by the lines OE and DD 1 and, according to the generalized Thales theorem, we conclude .

Answer: AO:OD=3:1.


Task #5 On the sides AB and AC ∆ABC, points M and N are given, respectively, such that the following equalities AM:MB=C are satisfiedN: NA=1:2. In what ratio does the point S of the intersection of the segments BN and CM divide each of these segments.

Problem №6 Point K is taken on the median AM of triangle ABC, and AK:KM=1:3. Find the ratio in which a line passing through point K parallel to side AC divides side BC.


Solution: Let M be 1 point intersection of a line passing through point K parallel to side AC and side BC. It is necessary to find the ratio of BM 1:M 1 C.

1) The sides of the angle AMC are intersected by straight lines KM 1 and AC and, according to the generalized Thales theorem, we conclude MM 1: M 1 C=3:1 or MM 1 \u003d 3z, M 1 C \u003d z

2) By condition VM:MS=1:1, i.e. VM=y, MC=y, but MC=MM 1 + M 1 C, so y=3z+ z=4 z,

3) .

Answer: VM 1:M 1 C = 7:1.


Problem №7 Triangle ABC is given. On the extension of side AC, a point is taken for point CN, and CN=AC; point K is the midpoint of side AB. In what respect is the line KNdivides side BC.

Comment: This problem could be solved using the Menelaus theorem. Applying it to the triangle ABC. Then the straight line KN intersects two sides of the triangle at points K and K 1, and the continuation of the third at point N. So the equality applies: , therefore VK 1:K 1 C=2:1.

Task #8

Sites:

http://www.problems.ru

http://interneturok.ru/

Unified State Examination 2011 Mathematics Task C4 R.K. Gordin M .: MTSNMO, 2011, - 148 s

Conclusion:

The solution of problems and theorems for finding the ratio of the lengths of segments is based on the generalized Thales theorem. We have formulated a method that allows, without applying the Thales theorem, to use parallel lines, transfer known proportions from one side of the angle to the other side and, thus, find the location of the points we need and compare the lengths. Working on the abstract helped us learn how to solve geometric problems high level difficulties. We realized the veracity of the words of the famous Russian poet Igor Severyanin: “Everything insignificant is needed to be significant ...” and we are sure that at the Unified State Examination we will be able to find a solution to the proposed tasks using the method of parallel lines.


1 The theorem on proportional segments in a triangle is the theorem described above.

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

Thales' theorem proves the following: С 1 , С 2 , С 3 - these are the places where parallel lines intersect on any side of the angle. C 2 is in the middle with respect 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 with the other side of the angle. We prove that when C 1 C 2 \u003d C 2 C z, then D 1 D 2 \u003d D 2 D 3 .
We draw a straight segment KR in place D 2, parallel to the section C 1 C 3. In the properties of a parallelogram C 1 C 2 \u003d KD 2, C 2 C 3 \u003d D 2 P. If C 1 C 2 \u003d C 2 C 3, then KD 2 \u003d 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 the proof. The angles with the top point D 2 are equal as vertical, and the angles D 2 KD 1 and D 2 PD 3 are equal as internal crosses lying with parallel C 1 D 1 and C 3 D 3 and separating KP.
Since D 1 D 2 =D 2 D 3 the theorem is proved 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 line segments parallel to each other, which intersect the two lines we are considering and divide one of them into identical 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 line CD into P identical segments.
We draw from the point C a semi-line c, which does not lie on the line CD. Let's mark the parts of the same size on it. SS 1, C 1 C 2, C 2 C 3 ..... C p-1 C p. We connect C p with D. We draw straight lines from points C 1, C 2, ...., C p-1 which will be parallel with respect to C p D. The lines will intersect CD at places D 1 D 2 D p-1 and divide the line CD into n identical segments.

Second example

Point CK is marked on side AB of triangle ABC. Segment SK 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 line through point M, parallel to SC, which intersects AB at point D

By Thales theoremВD=КD
By the theorem of proportional segments, we get that
PM \u003d KD \u003d VK / 2, therefore, VK: PM \u003d 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 on the BC side the segment EU = 4 cm. Prove that AD = DB.

Since BC = 8 cm and EU = 4 cm, then
BE = BC-EU, therefore BE = 8-4 = 4(cm)
By Thales theorem, since AC is parallel to DE and EC \u003d BE, therefore, AD \u003d DB. Q.E.D.

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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).

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:

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

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 will be the set of tangents to some (possibly degenerate) conic section.

In the case of the Thales theorem, the conic will be a point at infinity corresponding to the direction of parallel lines.

This statement, in turn, is a limiting case of the following statement:

Let f (\displaystyle f) is a projective transformation of a conic. Then the envelope of the set of lines X f (X) (\displaystyle Xf(X)) there will be a conic (possibly degenerate).

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