How to prove Thales's theorem 8. Thales's theorem

If the sides of an angle are intersected by straight parallel lines that divide one of the sides 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. The 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.

IN women's magazine- online, you will find a lot interesting information for myself. There is also a section dedicated to poems written by Sergei Yesenin. Come in, you won't regret it!

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. 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 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.

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

  • 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.
  • 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:

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



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

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