If a straight line cuts off equal segments. Thales' theorem. Middle line of the triangle

About parallel and secant.

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:

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))).)

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

Variations and generalizations

Converse theorem

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

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

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 ).

If the secants are parallel, then it is necessary to require that the segments on both secants be equal to each other, otherwise this statement becomes false (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).

This theorem is used in navigation: a collision between 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 on a line l (\displaystyle l) and straight m (\displaystyle m). Then the set of lines will be the set of tangents to some conic section (possibly degenerate).

In the case of Thales's theorem, the conic will be the 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)- projective transformation of a conic. Then the envelope of the set of straight lines X f (X) (\displaystyle Xf(X)) will be a conic (possibly degenerate).

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

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. 1).

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 A 3, then B 1 B 2 = B 2 B 3.

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 has been proven.

Using Thales' theorem, the following theorem is established.

Theorem 2. middle line of a 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 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.

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 to create from the Egyptian art of measurement 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 modern world, and back in 585, even BC.

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 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 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 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. 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 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 one important roles in mathematics. She was famous 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 Thales’ theorem.

Thales' theorem in culture

Thales' theorem became famous not only in mathematics, but it was also introduced to culture. Once upon a time Argentine musical group Les Luthiers (Spanish) presented a song to the audience, which it dedicated to the 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”
Subjects > Mathematics > Mathematics 8th grade

Plan:

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

Introduction

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

Thales's theorem- one of the theorems of planimetry.

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


Proof in the case of secants

Proof of Thales' Theorem

Let's consider the option 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 straight line BC. Angles ABC and BCD are equal as internal crosswise lying with parallel lines AB and CD and secant BC, and angles ACB and CBD are equal as internal crosswise lying with parallel lines AC and BD and secant BC. Then, by the first criterion for the equality of triangles, triangles ABC and DCB are congruent. It follows that AC = BD and AB = CD. ■

There is also generalized Thales' theorem:

Parallel lines cut off proportional segments at secants:

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


1. 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:

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

Thus (see figure) from what follows that straight lines .

If the secants are parallel, then it is necessary to require that the segments on both secants be equal to each other, otherwise this statement becomes false (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).


2. Thales' theorem in culture

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

3. Interesting facts

  • 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.

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 abstract is based on an article from Russian Wikipedia. Synchronization completed 07/16/11 23:06:34
Similar abstracts:

Planimetry theorem 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:

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))).)

Notes [ | ]

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

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

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 ).

If the secants are parallel, then it is necessary to require that the segments on both secants be equal to each other, otherwise this statement becomes false (a counterexample is a trapezoid intersected by a line passing through the midpoints of the bases).

This theorem is used in navigation: a collision between 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 on a line l (\displaystyle l) and straight m (\displaystyle m). Then the set of lines will be the set of tangents to some conic section (possibly degenerate).

In the case of Thales's theorem, the conic will be the 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)- projective transformation of a conic. Then the envelope of the set of straight lines X f (X) (\displaystyle Xf(X)) will be a conic (possibly degenerate).

| ]

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