How to find the sum of the angles of a triangle. Reinforcing the material learned

This theorem is also formulated in the textbook by L.S. Atanasyan. , and in the textbook by Pogorelov A.V. . The proofs of this theorem in these textbooks do not differ significantly, and therefore we present its proof, for example, from the textbook by A.V. Pogorelov.

Theorem: The sum of the angles of a triangle is 180°

Proof. Let ABC - given triangle. Let us draw a line through vertex B parallel to line AC. Let us mark point D on it so that points A and D lie along different sides from the direct line BC (Fig. 6).

Angles DBC and ACB are equal as internal cross-lying ones, formed by the secant BC with parallel straight lines AC and BD. Therefore, the sum of the angles of a triangle at vertices B and C is equal to angle ABD. And the sum of all three angles of a triangle is equal to the sum of angles ABD and BAC. Since these are one-sided interior angles for parallel AC and BD and secant AB, their sum is 180°. The theorem is proven.

The idea of ​​this proof is to carry out parallel line and designation of equality of the desired angles. Let's reconstruct the idea of ​​such additional construction, proving this theorem using the concept of a thought experiment. Proof of the theorem using a thought experiment. So, the subject of our thought experiment is the angles of a triangle. Let us place him mentally in conditions in which his essence can be revealed with particular certainty (stage 1).

Such conditions will be such an arrangement of the corners of the triangle in which all three of their vertices will be combined at one point. Such a combination is possible if we allow the possibility of “moving” the corners by moving the sides of the triangle without changing the angle of inclination (Fig. 1). Such movements are essentially subsequent mental transformations (stage 2).

By designating the angles and sides of a triangle (Fig. 2), the angles obtained by “moving,” we thereby mentally form the environment, the system of connections in which we place our subject of thought (stage 3).

Line AB, “moving” along line BC and without changing the angle of inclination to it, transfers angle 1 to angle 5, and “moving” along line AC, transfers angle 2 to angle 4. Since with such a “movement” line AB does not change the angle of inclination to lines AC and BC, then the conclusion is obvious: rays a and a1 are parallel to AB and transform into each other, and rays b and b1 are a continuation of sides BC and AC, respectively. Since angle 3 and the angle between rays b and b1 are vertical, they are equal. The sum of these angles is equal to the rotated angle aa1 - which means 180°.

CONCLUSION

IN diploma work conducted “constructed” proofs of some school geometric theorems, using the structure of a thought experiment, which confirmed the formulated hypothesis.

The presented evidence was based on such visual and sensory idealizations: “compression”, “stretching”, “sliding”, which made it possible to transform the original geometric object in a special way and highlight its essential characteristics, which is typical for a thought experiment. In this case, a thought experiment acts as a certain “creative tool” that contributes to the emergence of geometric knowledge (for example, about midline trapezoid or about the angles of a triangle). Such idealizations make it possible to grasp the whole idea of ​​proof, the idea of ​​carrying out “additional construction,” which allows us to talk about the possibility of a more conscious understanding by schoolchildren of the process of formal deductive proof of geometric theorems.

A thought experiment is one of the basic methods for obtaining and discovering geometric theorems. It is necessary to develop a methodology for transferring the method to the student. The question remains open about the age of a student acceptable for “accepting” the method, about “ side effects» the evidence presented in this way.

These issues require further study. But in any case, one thing is certain: a thought experiment develops in schoolchildren theoretical thinking, is its basis and, therefore, the ability for mental experimentation needs to be developed.

>>Geometry: Sum of angles of a triangle. Complete lessons

LESSON TOPIC: Sum of angles of a triangle.

Lesson objectives:

• Consolidating and testing students’ knowledge on the topic: “Sum of angles of a triangle”;
• Proof of the properties of the angles of a triangle;
• Application of this property in solving simple problems;
• Usage historical material for development cognitive activity students;
• Instilling the skill of accuracy when constructing drawings.

Lesson objectives:

• Test students' problem-solving skills.

Lesson plan:

1. Triangle;
2. Theorem on the sum of the angles of a triangle;
3. Example tasks.

Triangle.

File:O.gif Triangle- the simplest polygon having 3 vertices (angles) and 3 sides; part of the plane bounded by three points and three segments connecting these points in pairs.
Three points in space that do not lie on the same straight line correspond to one and only one plane.
Any polygon can be divided into triangles - this process is called triangulation.
There is a section of mathematics entirely devoted to the study of the laws of triangles - Trigonometry.

Theorem on the sum of the angles of a triangle.

File:T.gif The triangle angle sum theorem is a classic theorem of Euclidean geometry that states that the sum of the angles of a triangle is 180°.

Proof" :

Let Δ ABC be given. Let us draw a line parallel to (AC) through vertex B and mark point D on it so that points A and D lie on opposite sides of line BC. Then the angle (DBC) and the angle (ACB) are equal as internal crosswise lying with parallel lines BD and AC and the secant (BC). Then the sum of the angles of the triangle at vertices B and C is equal to angle (ABD). But the angle (ABD) and the angle (BAC) at vertex A of triangle ABC are internal one-sided with parallel lines BD and AC and the secant (AB), and their sum is 180°. Therefore, the sum of the angles of a triangle is 180°. The theorem is proven.

Consequences.

External angle of a triangle equal to the sum two angles of a triangle that are not adjacent to it.

Proof:

Let Δ ABC be given. Point D lies on line AC so that A lies between C and D. Then BAD is external to the angle of the triangle at vertex A and A + BAD = 180°. But A + B + C = 180°, and therefore B + C = 180° – A. Hence BAD = B + C. The corollary is proven.

Consequences.

An exterior angle of a triangle is greater than any angle of the triangle that is not adjacent to it.

Task.

An exterior angle of a triangle is an angle adjacent to any angle of this triangle. Prove that outside corner of a triangle is equal to the sum of two angles of a triangle that are not adjacent to it.
(Fig.1)

Solution:

Let in Δ ABC ∠DAС be external (Fig. 1). Then ∠DAC=180°-∠BAC (by property adjacent corners), according to the theorem on the sum of the angles of a triangle ∠B+∠C = 180°-∠BAC. From these equalities we obtain ∠DAС=∠В+∠С

Interesting fact:

Sum of the angles of a triangle" :

In Lobachevsky geometry, the sum of the angles of a triangle is always less than 180. In Euclidean geometry it is always equal to 180. In Riemann geometry, the sum of the angles of a triangle is always greater than 180.

From the history of mathematics:

Euclid (3rd century BC) in his work “Elements” gives the following definition: “Parallel lines are lines that are in the same plane and, being extended in both directions indefinitely, do not meet each other on either side.” .
Posidonius (1st century BC) “Two straight lines lying in the same plane, equally spaced from each other”
The ancient Greek scientist Pappus (III century BC) introduced the symbol of parallel straight-sign=. Subsequently English economist Ricardo (1720-1823) used this symbol as an equals sign.
Only in the 18th century did they begin to use the symbol for parallel lines - the sign ||.
The living connection between generations is not interrupted for a moment; every day we learn the experience accumulated by our ancestors. Ancient Greeks based on observations and from practical experience they drew conclusions, expressed hypotheses, and then, at meetings of scientists - symposiums (literally “feast”) - they tried to substantiate and prove these hypotheses. At that time, the statement arose: “Truth is born in dispute.”

Questions:

1. What is a triangle?
2. What does the theorem about the sum of the angles of a triangle say?
3. What is the external angle of the triangle?

Goals and objectives:

Educational:

• repeat and generalize knowledge about the triangle;
• prove the theorem on the sum of the angles of a triangle;
• practically verify the correctness of the formulation of the theorem;
• learn to apply acquired knowledge when solving problems.

Educational:

• develop geometric thinking, interest in the subject, cognitive and creative activity students, math speech, the ability to independently acquire knowledge.

Educational:

• develop personal qualities students, such as determination, perseverance, accuracy, ability to work in a team.

Equipment: multimedia projector, triangles made of colored paper, teaching materials " Living mathematics", computer, screen.

Preparatory stage: The teacher gives the student the task to prepare historical information about the theorem “Sum of the angles of a triangle.”

Lesson type: learning new material.

During the classes

I. Organizational moment

Greetings. Psychological attitude students to work.

II. Warm-up

WITH geometric figure“triangle” we met in previous lessons. Let's repeat what we know about the triangle?

Students work in groups. They are given the opportunity to communicate with each other, each to independently build the process of cognition.

What happened? Each group makes their proposals, the teacher writes them on the board. The results are discussed:

Picture 1

III. Formulating the lesson objective

So, we already know quite a lot about the triangle. But not all. Each of you has triangles and protractors on your desk. What kind of problem do you think we can formulate?

Students formulate the task of the lesson - to find the sum of the angles of a triangle.

IV. Explanation of new material

Practical part(promotes updating knowledge and self-knowledge skills). Measure the angles using a protractor and find their sum. Write down the results in your notebook (listen to the answers received). We find out that the sum of the angles is different for everyone (this can happen because the protractor was not applied accurately, the calculation was carried out carelessly, etc.).

Fold along the dotted lines and find out what else the sum of the angles of a triangle is equal to:

A)
Figure 2

b)
Figure 3

V)
Figure 4

G)
Figure 5

d)
Figure 6

After completing the practical work, students formulate the answer: The sum of the angles of a triangle is equal to degree measure unfolded angle, i.e. 180°.

Teacher: In mathematics practical work It only makes it possible to make some kind of statement, but it needs to be proven. A statement whose validity is established by proof is called a theorem. What theorem can we formulate and prove?

Students: The sum of the angles of a triangle is 180 degrees.

Historical reference: The property of the sum of the angles of a triangle was established in Ancient Egypt. The proof presented in modern textbooks, contained in Proclus's comments to Euclid's Elements. Proclus claims that this proof (Fig. 8) was discovered by the Pythagoreans (5th century BC). In the first book of the Elements, Euclid sets out another proof of the theorem on the sum of the angles of a triangle, which can be easily understood with the help of a drawing (Fig. 7):

Figure 7

Figure 8

The drawings are displayed on the screen through a projector.

The teacher offers to prove the theorem using drawings.

Then the proof is carried out using the teaching and learning complex “Living Mathematics”. The teacher projects the proof of the theorem on the computer.

Theorem on the sum of angles of a triangle: “The sum of the angles of a triangle is 180°”

Figure 9

Proof:

A)

Figure 10

b)

Figure 11

V)

Figure 12

Students do in notebooks short note proof of the theorem:

Theorem: The sum of the angles of a triangle is 180°.

Figure 13

Given:Δ ABC

Prove: A + B + C = 180°.

Proof:

What needed to be proven.

V. Phys. just a minute.

VI. Explanation of new material (continued)

The corollary from the theorem about the sum of the angles of a triangle is deduced by students independently, this contributes to the development of the ability to formulate own point point of view, express and argue for it:

In any triangle, either all angles are acute, or two are acute and the third is obtuse or right..

If a triangle has all acute angles, then it is called acute-angled.

If one of the angles of a triangle is obtuse, then it is called obtuse-angled.

If one of the angles of a triangle is right, then it is called rectangular.

The theorem on the sum of angles of a triangle allows us to classify triangles not only by sides, but also by angles. (As students introduce types of triangles, students fill out the table)

Table 1

Triangle view	Isosceles	Equilateral	Versatile
Rectangular
Obtuse
Acute-angled

VII. Consolidation of the studied material.

1. Solve problems orally:

(Drawings are displayed on the screen through a projector)

Task 1. Find angle C.

Figure 14

Problem 2. Find the angle F.

Figure 15

Task 3. Find the angles K and N.

Figure 16

Problem 4. Find the angles P and T.

Figure 17

1. Solve problem No. 223 (b, d) yourself.
2. Solve the problem on the board and in notebooks, student No. 224.
3. Questions: Can a triangle have: a) two right angles; b) two obtuse angles; c) one right and one obtuse angle.
4. (done orally) The cards on each table show various triangles. Determine by eye the type of each triangle.

Figure 18

1. Find the sum of angles 1, 2 and 3.

Figure 19

VIII. Lesson summary.

Teacher: What have we learned? Is the theorem applicable to any triangle?

IX. Reflection.

Tell me your mood, guys! WITH reverse side use a triangle to depict your facial expressions.

Figure 20

Homework: paragraph 30 (part 1), question 1 ch. IV page 89 of the textbook; No. 223 (a, c), No. 225.

Theorem. Sum internal corners of a triangle is equal to two right angles.

Let's take some triangle ABC (Fig. 208). Let us denote its interior angles by numbers 1, 2 and 3. Let us prove that

∠1 + ∠2 + ∠3 = 180°.

Let us draw through some vertex of the triangle, for example B, a straight line MN parallel to AC.

At vertex B we got three angles: ∠4, ∠2 and ∠5. Their sum is a straight angle, therefore it is equal to 180°:

∠4 + ∠2 + ∠5 = 180°.

But ∠4 = ∠1 are internal crosswise angles with parallel lines MN and AC and secant AB.

∠5 = ∠3 - these are internal crosswise angles with parallel lines MN and AC and secant BC.

This means that ∠4 and ∠5 can be replaced by their equals ∠1 and ∠3.

Therefore, ∠1 + ∠2 + ∠3 = 180°. The theorem is proven.

2. Property of the external angle of a triangle.

In fact, in triangle ABC(Fig. 209) ∠1 + ∠2 = 180° - ∠3, but also ∠ВСD, the external angle of this triangle, not adjacent to ∠1 and ∠2, is also equal to 180° - ∠3.

Thus:

∠1 + ∠2 = 180° - ∠3;

∠BCD = 180° - ∠3.

Therefore, ∠1 + ∠2= ∠BCD.

The derived property of the exterior angle of a triangle clarifies the content of the previously proven theorem on the exterior angle of a triangle, which stated only that the exterior angle of a triangle is greater than each interior angle of a triangle not adjacent to it; now it is established that the external angle is equal to the sum of both internal angles not adjacent to it.

3. Property of a right triangle with an angle of 30°.

Let in right triangle ASV angle B is 30° (Fig. 210). Then the other one is his sharp corner will be equal to 60°.

Let us prove that leg AC is equal to half the hypotenuse AB. Let's continue leg AC beyond the top right angle C and set aside the segment CM, equal to the segment AC. Connect point M to point B. The resulting triangle ВСМ equal to a triangle DIA We see that each angle of triangle ABM is equal to 60°, therefore this triangle is an equilateral triangle.

Leg AC is equal to half AM, and since AM is equal to AB, leg AC will be equal to half the hypotenuse AB.