Report on Thales' theorem. Thales' theorem

The theorem has no restrictions on mutual arrangement secants (this is true for both intersecting and parallel lines). It also does not matter where the segments on the secants are located.

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, according to the second criterion for the equality of triangles, triangles ABC and DCB are equal. It follows that AC = BD and AB = CD. ■

There is also proportional segment theorem:

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

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.

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:

Thus (see figure) 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 straight $A\_1B\_1||A\_2B\_2||\backslash ldots$.

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).

Variations and generalizations

The following statement is dual to Sollertinsky's lemma:

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 is proven.

Using Thales' theorem, the following theorem is established.

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

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

ED - midline of triangle ABC

Example 1. Divide this segment into four equal parts.

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

Dividing a segment into four equal parts

To do this, draw an arbitrary half-line a through point A and plot on it successively 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.

1. 
   Formulation;
2. 
   Proof;

1. 
   Theorem on proportional segments;
2. 
   Ceva's theorem;

1. 
   Formulation;
2. 
   Proof;

1. 
   Menelaus' theorem;

1. 
   Formulation;
2. 
   Proof;

1. 
   Problems and their solutions;
2. 
   Conclusion;
3. 
   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 on 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 solve a rather narrow range of problems in this area. 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, similarity coefficient. Signs of similarity of triangles) are offered in the second part exam paper and belong to a high level of complexity.

In the process of working on the abstract, it became possible to deepen our knowledge on this topic. Proof of the theorem on proportional segments in a triangle (the theorem is not included in 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 in the end we were able to learn to solve more wide circle tasks to compare the lengths of segments. This is the relevance of our work.

Generalized Thales' theorem.

Wording:

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 .

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

1+

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 arrive at 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).

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.

Wording:

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 .

2.segments A A 1 , BB 1 And SS 1 intersect at one point.

Dividing both sides by the 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.

Wording:

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 .

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.

Solution: You 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:

1. 
   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.
2. 
   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.

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

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 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.

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.

If the sides of an angle intersect straight lines parallel lines which one of the sides is divided into several segments, then the second side, straight lines will also be divided into segments equivalent to the other side.

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

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

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

Let's look at a few examples

First example

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

Second example

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

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

Third example

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

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

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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:

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).

1. 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)).
2. 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

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:

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.

Sollertinsky's lemma

The following statement is dual to Sollertinsky's lemma: