The theorem is the converse of Thales's theorem. Thales's theorem. Complete lessons – Knowledge Hypermarket

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 on the second line equal segments.

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’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: the collision of ships moving with constant speed, is inevitable if the direction from one vessel to another is maintained.

Sollertinsky's lemma

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).

About parallels and secants.

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 draw through the points 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 property of a parallelogram: 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)).
Triangles △ A B B 2 (\displaystyle \bigtriangleup ABB_(2)) And △ 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

Variations and generalizations

Converse theorem

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.

Sollertinsky's lemma

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 X f (X) (\displaystyle Xf(X)) will be a set of tangents to some

Formulation;
Proof;

Theorem on proportional segments;
Ceva's theorem;

Formulation;
Proof;

Menelaus' theorem;

Formulation;
Proof;

Problems and their solutions;
Conclusion;
List of used sources and literature.

Introduction.

Everything little is needed

To be significant...

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

The idea of the method itself is based on the use of the generalized Thales theorem. Thales' theorem is studied in the eighth grade, its generalization and the topic “Similarity of Figures” in the ninth and only in the tenth grade, in an introductory plan, two important theorems of Cheva and Menelaus are studied, with the help of which a number of problems on finding the ratio of the lengths of segments are relatively easily solved. Therefore, at the level of basic education we can decide quite narrow circle tasks for this educational material. Although at the final certification for a basic school course and at the Unified State Examination in mathematics, problems on this topic (Thales’ Theorem. Similarity of triangles, coefficient of similarity. Signs of similarity of triangles) are offered in the second part of the examination paper and relate to 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 about 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 made it possible 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 involving comparing the lengths of segments. This is the relevance of our work.

Generalized Thales' theorem.

Formulation:

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

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

Prove:

Proof:

Let us prove, for example, that

Let's consider two cases:

1 case (Fig. b)

Direct a And b parallel. Then quadrilaterals

A 1 A 2 IN 2 IN 1 And A 2 A 3 IN 3 IN 2 - parallelograms. That's why

A 1 A 2 =IN 1 IN 2 And A 2 A 3 =IN 2 IN 3 , from which it follows that

Case 2 (Fig. c)

Lines a and b are not parallel. Through the point A 1 let's make a direct With, parallel to the line b. She will cross the lines A 2 IN 2 And A 3 IN 3 at some points WITH 2 And WITH 3 . Triangles A 1 A 2 WITH 2 And A 1 A 3 WITH 3 similar at two angles (angle A 1 – general, angles A 1 A 2 WITH 2 And A 1 A 3 WITH 3 equal as corresponding when parallel lines A 2 IN 2 And A 3 IN 3 secant A 2 A 3 ), That's why

Or by the property of proportions

On the other hand, according to what was proved in the first case, we have A 1 WITH 2 =IN 1 IN 2 , WITH 2 WITH 3 =IN 2 IN 3 . Replacing in proportion (1) A 1 WITH 2 on IN 1 IN 2 And WITH 2 WITH 3 on IN 2 IN 3 , we come to equality

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

On the sides AC And Sun triangle ABC points marked TO And M So AK:KS=m: n, B.M.: M.C.= p: q. Segments AM And VC intersect at a point ABOUT(Fig. 124b).

Prove:

Proof:
Through the point M let's make a direct M.D.(Fig. 124a), parallel VC. She crosses the side AC at the point D, and according to a generalization of 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 a generalization of 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 points C are taken on sides AB, BC and CA of triangle ABC, respectively 1 , A 1 and B 1 , then the 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 marked WITH 1 ,A 1 And IN 1 .

Prove:

2.segments 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 ABOUT. Let us prove that equality (3) is satisfied. By the theorem on proportional segments in triangle 1 we have:

The left sides of these equalities are equal, which means the right sides are also equal. Equating them, we get

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

2. Let us prove the converse statement. Let the points WITH 1 ,A 1 And IN 1 taken on sides AB, Sun And SA so that equality (3) is satisfied. Let us prove that the segments AA 1 , BB 1 And SS 1 intersect at one point. Let us denote by the letter ABOUT point of intersection of segments A A 1 And BB 1 and let's make a direct CO. She crosses the side AB at some point, which we denote WITH 2 . Since the segments AA 1 , BB 1 And SS 1 intersect at one point, then by what was proven in the first point

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

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

Q.E.D.
Menelaus's theorem.

Formulation:

If on sides AB and BC and the continuation of side AC (or on the continuation of sides AB, BC and AC) points C are taken, respectively 1 , A 1 , IN 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 marked WITH 1 ,A 1 And IN 1 .

Prove:

2. points A 1 ,WITH 1 And IN 1 lie on the same straight line
Proof:
1. Let the points A 1 ,WITH 1 And IN 1 lie on the same straight line. Let us prove that equality (5) is satisfied. Let's carry out AD,BE And CF parallel to the line IN 1 A 1 (dot D lies on a straight line Sun). According to the generalized Thales theorem we have:

Multiplying the left and right sides of these equalities, we get

those. equality (5) is satisfied.
2. Let us prove the converse statement. Let the point IN 1 taken on the continuation side AC, and the points WITH 1 And A 1 – on the sides AB And Sun, and in such a way that equality (5) is satisfied. Let us prove that the points A 1 ,WITH 1 And IN 1 lie on the same straight line. Let straight line A 1 C 1 intersect the continuation of side AC at point B 2, then, by what was proven in the first point

Comparing (5) and (6), we arrive at the equality = , which shows that the points IN 1 And IN 2 share sides AC in the same respect. Therefore, the points IN 1 And IN 2 coincide, and, therefore, the points A 1 ,WITH 1 And IN 1 lie on the same straight line. The converse statement is proved similarly in the case when all three points A 1 ,WITH 1 And IN 1 lie on the continuations 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 you to transfer famous relationships proportions from one side of the angle to its second side, thus, you only need to draw these parallel straight lines in a way convenient for solving the problem.
Let's consider specific tasks:
Problem No. 1 In triangle ABC, point M is taken on side BC so that BM:MC = 3:2. Point P divides segment AM in a ratio of 2:1. Straight line BP intersects side AC at point B 1 . In what respect is point B 1 divides the AC side?

Solution: We need to find the ratio AB 1:B 1 C, AC the desired segment on which point B 1 lies.

The parallel method is as follows:

cut the required segment with parallel lines. One BB 1 already exists, and we will draw the second MN through point M, parallel to BB 1.
Transfer a known ratio from one side of the angle to its other side, i.e. consider the angles of the sides, which are cut by these straight lines.

The sides of angle C are dissected by straight lines BB 1 and MN and, using the generalized Thales theorem, we conclude IN 1 N=3r, NC=2р. The sides of the angle MAC intersect the lines PB 1 and MN and divide its sides in a ratio of 2:1, therefore AB 1:B 1 N=2:1 and therefore AB 1 =2n, IN 1 N= n. Because IN 1 N=3r, And IN 1 N= n, That 3р=n.

Let's move on to the relationship that interests us AB 1:B 1 C= AB 1:(B 1 N+ NC)= 2n:(3p+2p)=(2*3p):(5p)=6:5.

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

Comment: This problem could be solved using Menelaus' theorem. Applying it to triangle AMC. Then straight line BB 1 intersects two sides of the triangle at points B 1 and P, and the continuation of the third at point B. This means that the equality applies: , hence
Problem No. 2 In triangle ABC AN is the median. On the side AC, a point M is taken so that AM: MC = 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 the segment AB.

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

1) Let's draw a straight line NN 1 parallel to the straight line SK and a straight line NN 2 parallel to the straight line VM.

2) The sides of 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 ВСМ are intersected by straight 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 straight 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 = 1: 1.5 or AK = n KN 1 =1,5 n.

6) KN 1 =y=1.5n.

Answer: AK:KV=1:3.

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

Problem No. 3 On side BC of triangle ABC, point D is taken such that ВD: DC = 2:5, and on 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: We need to find 1) AK:KD=? 2) VK:KE=?

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

2) The sides of the angle ALL are intersected by straight lines BE and DD 1 and using 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=x, EC=2x, but EC= CD 1 + D 1 E, which means 2у=5z+2 z=7 z, z=

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

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

Answer: AK:KD=7:4; VK:KE=6:5.
Comment: This problem could be solved using Menelaus' theorem. Applying it to the triangle WEIGHT. Then straight line DA intersects two sides of the triangle at points D and K, and the continuation of the third at point A. This means that the equality applies:

, therefore VK:KE=6:5. Arguing similarly for triangle ADC, we obtain

, AK:KD=7:4.
Problem No. 4 In ∆ ABC, the bisector AD divides side BC in the ratio 2: 1. In what ratio does the median CE divide this bisector?

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

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

2) The sides of angle ABC are intersected by straight lines CE and DD 1 and, using the generalized Thales theorem, we conclude ВD 1:D 1 E=2:1 or ВD 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, which means y=2p+ p=3 p, p =
4) The sides of the angle BAD are intersected by straight lines OE and DD 1 and, using the generalized Thales theorem, we conclude .

Answer: AO:OD=3:1.

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

Problem No. 6 On the median AM of triangle ABC, point K is taken, 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 straight line passing through point K parallel to side AC and side BC. We need to find the ratio VM 1:M 1 C.

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

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

3) .

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

Problem No. 7 Given a triangle ABC. On the continuation of side AC, point C is taken as pointN, and CN=AC; point K is the middle of side AB. In what ratio is the straight line KNdivides the side of the sun.

Comment: This problem could be solved using Menelaus' theorem. Applying it to 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. This means that the equality applies: , therefore VK 1:K 1 C=2:1.

Problem No. 8

Websites:

http://www.problems.ru

http://interneturok.ru/

Unified State Exam 2011 Mathematics Problem C4 R.K. Gordin M.: MCNMO, 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 Thales' theorem, to use parallel straight 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 to solve geometric problems high level difficulties. We realized the truthfulness of the words of the famous Russian poet Igor Severyanin: “Everything insignificant is needed in order to be significant...” and we are confident that at the Unified State Exam we will be able to find a solution to the proposed problems using the method of parallel lines.

1 Theorem on proportional segments in a triangle - the theorem described above.

This tomb is small, but the glory over it is immense.
The multi-intelligent Thales is hidden in it before you.

Inscription on the tomb of Thales of Miletus

Imagine this picture. 600 BC Egypt. In front of 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 act like 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 theorem on the similarity of triangles, he found the length of the shadow of the pyramid, which, accordingly, was equal to the shadow cast by the pyramid.

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

Among the evidence of Thales's knowledge of astronomy, the following example can be given. May 28, 585 BC prediction by Miletus solar eclipse helped to end the war between Lydia and Media that had lasted for 6 years. This phenomenon frightened the Medes so much that they agreed to unfavorable terms for concluding peace with the Lydians.

There is a fairly widely known legend that characterizes Thales as a resourceful person. Thales often heard unflattering comments about his poverty. One day he decided to prove that philosophers, if they wish, can live in abundance. Even in winter, Thales, by observing the stars, determined that in summer it would be good harvest olives At the same time he hired oil presses in Miletus and Chios. This cost him quite little, since in winter there is practically no demand for them. When the olives produced a rich harvest, Thales began to rent out his oil presses. Collected a large number of making money using this method was regarded as proof that philosophers can make money with their minds, but their calling 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 considered to be the formulation and proof of the following theorems:

vertical angles are equal;
Equal triangles are those whose side and two adjacent angles are respectively equal;
the angles at the base of an isosceles triangle are equal;
diameter divides the circle in half;
the inscribed angle subtended by the 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 them all remains the same. Let's consider this theorem.

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

Let's say points A 1, A 2, A 3 are the points of intersection of parallel lines with 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 = A 2 A 3, then B 1 B 2 = B 2 B 3.

Through point B 2 we draw a line parallel to line A 1 A 2. Let's denote the new line C 1 C 2. Consider parallelograms A 1 C 1 B 2 A 2 and A 2 B 2 C 2 A 3 .

The properties of a parallelogram allow us to state 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 = A 2 A 3, then C 1 B 2 = B 2 C 2.

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

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

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 (by side and adjacent angles).

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!

website, when copying material in full or in part, a link to the source is required.

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's theorem is a special case of the generalized Thales's theorem, 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

El Teorema de Thales por Les Luthiers en You Tube - www.youtube.com/watch?v=czzj2C4wdxY

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
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