What is the sum of the triangle? Triangle Angle Sum Theorem

home Theorem.

The sum of the interior angles 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 has been proven.

2. Property of the external angle of a triangle. Theorem. External angle of a triangle equal to the sum

two internal angles not adjacent to it.

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.

2. Property of the external angle of a triangle. 3. Property of a right triangle with an angle of 30°. Leg right triangle

, lying opposite an angle of 30°, is equal to half the hypotenuse.

Let angle B in the right triangle ACB be equal to 30° (Fig. 210). Then its other acute angle will be equal to 60°.

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

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's mark point D on it so that points A and D lie on opposite sides of straight 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 has been proven.

The idea of ​​this proof is to carry out parallel line and designation of equality of the desired angles. Let us reconstruct the idea of ​​such an additional construction by 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“constructed” proofs of some school geometric theorems were carried out 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 basic methods obtaining and discovering geometric theorems. It is necessary to develop a methodology for transferring the method to the student. Remains open question 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 theoretical thinking in schoolchildren, is its basis and, therefore, the ability for mental experimentation needs to be developed.

The materials located on this page are copyrighted. Copying for posting on other sites is permitted only with the express consent of the author and site administration.

Sum of angles of a triangle.

Smirnova I. N., mathematics teacher.
Information prospectus for an open lesson.

The purpose of the methodological lesson: introduce teachers to modern methods and techniques for using ICT tools in various types educational activities.
Lesson topic: Sum of angles of a triangle.
Lesson name:“Knowledge is only knowledge when it is acquired through the efforts of one’s thoughts, and not through memory.” L. N. Tolstoy.
Methodological innovations that will form the basis of the lesson.
The lesson will show methods scientific research using ICT (use of mathematical experiments as one of the forms of obtaining new knowledge; experimental testing of hypotheses).
Overview of the lesson model.

1. Motivation for studying the theorem.
2. Disclosure of the content of the theorem during a mathematical experiment using the educational and methodological set “Living Mathematics”.
3. Motivation for the need to prove the theorem.
4. Work on the structure of the theorem.
5. Finding a proof of the theorem.
6. Proof of the theorem.
7. Consolidating the formulation of the theorem and its proof.
8. Application of the theorem.

Geometry lesson in 7th grade
according to the textbook "Geometry 7-9"
on the topic: “Sum of the angles of a triangle.”

Lesson type: lesson of learning new material.
Lesson objectives:
Educational: prove the theorem on the sum of the angles of a triangle; gain skills in working with the “Living Mathematics” program, developing interdisciplinary connections.
Educational: improving the ability to consciously carry out such thinking techniques as comparison, generalization and systematization.
Educational: fostering independence and the ability to work in accordance with the planned plan.
Equipment: multimedia cabinet, interactive board, cards with a plan practical work, “Living Mathematics” program.

Lesson structure.

1. Updating knowledge.
   1. Mobilizing start to the lesson.
   2. Statement of a problematic problem in order to motivate the study of new material.
   3. Setting a learning task.
2. 1. Practical work “Sum of angles of a triangle.”
   2. Proof of the theorem on the sum of the angles of a triangle.
3. 1. Solving a problematic problem.
   2. Solving problems using ready-made drawings.
   3. Summing up the lesson.
   4. Setting homework.

During the classes.

1. Updating knowledge.

   Lesson plan:

   1. Establish and put forward a hypothesis experimentally about the sum of the angles of any triangle.
   2. Prove this assumption.
   3. Reinforce the established fact.
2. Formation of new knowledge and methods of action.
   1. Practical work “Sum of angles of a triangle.”

      Students sit down at their computers and are given cards with a plan for practical work.

      Practical work on the topic “Sum of angles of a triangle” (sample card)

      Print the card
      

      Students hand over the results of practical work and sit at their desks.
      After discussing the results of practical work, a hypothesis is put forward that the sum of the angles of a triangle is 180°.
      Teacher: Why can’t we yet say that the sum of the angles of absolutely any triangle is equal to 180°?
      Student: It is impossible to make absolutely accurate constructions, nor to make absolutely accurate measurements, even on a computer.
      The statement that the sum of the angles of a triangle is 180° applies only to the triangles we have considered. We cannot say anything about other triangles, since we did not measure their angles.
      Teacher: It would be more correct to say: the triangles we have considered have a sum of angles approximately equal to 180°. To make sure that the sum of the angles of a triangle is exactly equal to 180°, and for any triangles, we still need to carry out the appropriate reasoning, that is, prove the validity of the statement suggested to us by experience.
      

   2. Proof of the theorem on the sum of the angles of a triangle.

      Students open their notebooks and write down the topic of the lesson “Sum of the angles of a triangle.”

      Work on the structure of the theorem.

      To formulate the theorem, answer the following questions:
      • What triangles were used in the measurement process?
      • What is included in the conditions of the theorem (what is given)?
      • What did we find during the measurements?
      • What is the conclusion of the theorem (what needs to be proven)?
      • Try to formulate the theorem on the sum of the angles of a triangle.

      Construction of the drawing and brief recording of the theorem

      At this stage, students are asked to make a drawing and write down what is given and what needs to be proven.

      Construction of the drawing and brief recording of the theorem.

      Given: Triangle ABC.
      Prove:
      டA + டB + டC = 180°.

      Finding a proof of the theorem

      When searching for a proof, you should try to expand the condition or conclusion of the theorem. In the theorem on the sum of the angles of a triangle, attempts to expand the condition are hopeless, so it is reasonable to work with students on developing the conclusion.
      Teacher: Which statements talk about angles whose sum is equal to 180°?
      Student: If two parallel lines are intersected by a transversal, then the sum of the interior one-sided angles is 180°.
      The sum of adjacent angles is 180°.
      Teacher: Let's try to use the first statement to prove it. In this regard, it is necessary to construct two parallel lines and a secant, but this must be done in such a way that greatest number the corners of the triangle became internal or included in them. How can this be achieved?
      

      Finding a proof of the theorem.

      

      Student: Draw a straight line parallel to the other side through one of the vertices of the triangle, then the side will be a secant. For example, through vertex B.
      Teacher: Name the internal one-sided angles formed by these lines and the transversal.
      Student: Angles DBA and BAC.
      Teacher: Which angles add up to 180°?
      Student:டDBA and டBAC.
      Teacher: What can be said about the magnitude of angle ABD?
      Student: Its value is equal to the sum of the angles ABC and SVK.
      Teacher: What statement do we need to prove the theorem?
      Student:டDBC = டACB.
      Teacher: What are these angles?
      Student: Internal ones lying crosswise.
      Teacher: On what basis can we say that they are equal?
      Student: According to the property of internal crosswise angles for parallel lines and transversals.

      As a result of searching for a proof, a plan for proving the theorem is drawn up:
      

      Plan of proof of the theorem.

      1. Draw a straight line through one of the vertices of the triangle parallel to the opposite side.
      2. Prove the equality of internal crosswise angles.
      3. Write down the sum of interior one-sided angles and express them in terms of the angles of the triangle.

      Proof and its recording.

      
      1. Let's do BD || AC (parallel lines axiom).
      2. ட3 = ட4 (since these are crosswise angles with BD || AC and secant BC).
      3. டA + டАВD = 180° (since these are one-sided angles with BD || AC and secant AB).
      4. டA + டАВD = ட1 + (ட2 + ட4) = ட1 + ட2 + ட3 = 180°, which is what needed to be proven.

      Consolidating the formulation of the theorem and its proof.

      To master the formulation of the theorem, students are asked to complete the following tasks:

      1. State the theorem we just proved.
      2. Highlight the condition and conclusion of the theorem.
      3. What shapes does the theorem apply to?
      4. Formulate a theorem with the words “if... then...”.
3. Application of knowledge, development of skills and abilities.

A triangle is a polygon that has three sides (three angles). Most often, the sides are indicated by small letters corresponding to the capital letters that represent the opposite vertices. In this article we will get acquainted with the types of these geometric figures, the theorem that determines what the sum of the angles of a triangle equals.

Types by angle size

Distinguish the following types polygon with three vertices:

• acute-angled, in which all the corners are acute;
• rectangular, having one right angle, its generators are called legs, and the side that is located opposite right angle, is called the hypotenuse;
• obtuse when one ;
• isosceles, in which two sides are equal, and they are called lateral, and the third is the base of the triangle;
• equilateral, having all three equal sides.

Properties

There are basic properties that are characteristic of each type of triangle:

• Opposite the larger side there is always a larger angle, and vice versa;
• opposite sides of equal size are equal angles, and vice versa;
• any triangle has two acute angles;
• an external angle is larger than any internal angle not adjacent to it;
• the sum of any two angles is always less than 180 degrees;
• the external angle is equal to the sum of the other two angles that do not intersect with it.

Triangle Angle Sum Theorem

The theorem states that if you add up all the angles of a given geometric figure, which is located on the Euclidean plane, then their sum will be 180 degrees. Let's try to prove this theorem.

Let us have an arbitrary triangle with vertices KMN.

Through vertex M we draw KN (this line is also called the Euclidean straight line). We mark point A on it so that points K and A are located on different sides of the straight line MH. We obtain equal angles AMN and KNM, which, like the internal ones, lie crosswise and are formed by the secant MN together with the straight lines KH and MA, which are parallel. It follows from this that the sum of the angles of the triangle located at the vertices M and H is equal to the size of the angle KMA. All three angles make up a sum that is equal to the sum of the angles KMA and MKN. Since these angles are internal one-sided relative to the parallel straight lines KN and MA with a secant KM, their sum is 180 degrees. The theorem has been proven.

Consequence

The following corollary follows from the theorem proved above: any triangle has two acute angles. To prove this, let us assume that this geometric figure has only one acute angle. It can also be assumed that none of the corners are acute. In this case, there must be at least two angles whose magnitude is equal to or greater than 90 degrees. But then the sum of the angles will be greater than 180 degrees. But this cannot happen, since according to the theorem, the sum of the angles of a triangle is equal to 180° - no more and no less. This is what needed to be proven.

Property of external angles

What is the sum of the exterior angles of a triangle? The answer to this question can be obtained using one of two methods. The first is that it is necessary to find the sum of the angles, which are taken one at each vertex, that is, three angles. The second implies that you need to find the sum of all six vertex angles. First, let's look at the first option. So, the triangle contains six external angles - two at each vertex.

Each pair has equal angles because they are vertical:

∟1 = ∟4, ∟2 = ∟5, ∟3 = ∟6.

In addition, it is known that the external angle of a triangle is equal to the sum of two internal ones that do not intersect with it. Hence,

∟1 = ∟A + ∟C, ∟2 = ∟A + ∟B, ∟3 = ∟B + ∟C.

From this it turns out that the sum of the external angles, which are taken one at each vertex, will be equal to:

∟1 + ∟2 + ∟3 = ∟A + ∟C + ∟A + ∟B + ∟B + ∟C = 2 x (∟A + ∟B + ∟C).

Taking into account the fact that the sum of the angles is equal to 180 degrees, we can say that ∟A + ∟B + ∟C = 180°. This means that ∟1 + ∟2 + ∟3 = 2 x 180° = 360°. If the second option is used, then the sum of the six angles will be, accordingly, twice as large. That is, the sum of the external angles of the triangle will be:

∟1 + ∟2 + ∟3 + ∟4 + ∟5 + ∟6 = 2 x (∟1 + ∟2 + ∟2) = 720°.

Right triangle

What is the sum of the acute angles of a right triangle? The answer to this question, again, follows from the theorem, which states that the angles in a triangle add up to 180 degrees. And our statement (property) sounds like this: in a right triangle sharp corners the total is 90 degrees. Let's prove its veracity.

Let us be given a triangle KMN, in which ∟Н = 90°. It is necessary to prove that ∟К + ∟М = 90°.

So, according to the theorem on the sum of angles ∟К + ∟М + ∟Н = 180°. Our condition says that ∟Н = 90°. So it turns out, ∟К + ∟М + 90° = 180°. That is, ∟К + ∟М = 180° - 90° = 90°. This is exactly what we needed to prove.

In addition to the properties of a right triangle described above, you can add the following:

• angles that lie opposite the legs are acute;
• the hypotenuse is triangular larger than any of the legs;
• the sum of the legs is greater than the hypotenuse;
• The leg of the triangle, which lies opposite the angle of 30 degrees, is half the size of the hypotenuse, that is, equal to half of it.

As another property of this geometric figure, we can highlight the Pythagorean theorem. She states that in a triangle with an angle of 90 degrees (rectangular), the sum of the squares of the legs is equal to the square of the hypotenuse.

Sum of angles of an isosceles triangle

Earlier we said that an isosceles polygon with three vertices and containing two equal sides is called. This property of this geometric figure is known: the angles at its base are equal. Let's prove it.

Let's take the triangle KMN, which is isosceles, KN ​​is its base.

We are required to prove that ∟К = ∟Н. So, let's say that MA is the bisector of our triangle KMN. The triangle MKA, taking into account the first sign of equality, is equal to the triangle MNA. Namely, by condition it is given that KM = NM, MA is the common side, ∟1 = ∟2, since MA is a bisector. Using the fact that these two triangles are equal, we can state that ∟К = ∟Н. This means the theorem is proven.

But we are interested in what is the sum of the angles of a triangle (isosceles). Since in this respect it does not have its own peculiarities, we will build on the theorem discussed earlier. That is, we can say that ∟К + ∟М + ∟Н = 180°, or 2 x ∟К + ∟М = 180° (since ∟К = ∟Н). This property We will not prove it, since the theorem on the sum of the angles of a triangle itself was proven earlier.

In addition to the properties discussed about the angles of a triangle, the following important statements also apply:

• at which it was lowered onto the base, is at the same time the median, the bisector of the angle that is between equal sides, as well as its base;
• the medians (bisectors, heights) that are drawn to the lateral sides of such a geometric figure are equal.

Equilateral triangle

It is also called regular, this is the triangle in which all sides are equal. And therefore the angles are also equal. Each one is 60 degrees. Let's prove this property.

Let's say that we have a triangle KMN. We know that KM = NM = KN. This means that, according to the property of the angles located at the base in an isosceles triangle, ∟К = ∟М = ∟Н. Since, according to the theorem, the sum of the angles of a triangle is ∟К + ∟М + ∟Н = 180°, then 3 x ∟К = 180° or ∟К = 60°, ∟М = 60°, ∟Н = 60°. Thus, the statement is proven.

As can be seen from the above proof based on the theorem, the sum of the angles, like the sum of the angles of any other triangle, is 180 degrees. There is no need to prove this theorem again.

There are also properties characteristic of equilateral triangle:

• the median, bisector, height in such a geometric figure coincide, and their length is calculated as (a x √3): 2;
• if we describe a circle around a given polygon, then its radius will be equal to (a x √3): 3;
• if you inscribe a circle in an equilateral triangle, then its radius will be (a x √3): 6;
• The area of ​​this geometric figure is calculated by the formula: (a2 x √3) : 4.

Obtuse triangle

By definition, one of its angles is between 90 and 180 degrees. But given that the other two angles of this geometric figure are acute, we can conclude that they do not exceed 90 degrees. Therefore, the triangle angle sum theorem works in calculating the sum of angles in an obtuse triangle. It turns out that we can safely say, based on the above-mentioned theorem, that the sum of the angles of an obtuse triangle is equal to 180 degrees. Again, this theorem does not need to be proven again.

Preliminary information

First, let's look directly at the concept of a triangle.

Definition 1

We will call it a triangle geometric figure, which is made up of three points connected to each other by segments (Fig. 1).

Definition 2

Within the framework of Definition 1, we will call the points the vertices of the triangle.

Definition 3

Within the framework of Definition 1, the segments will be called sides of the triangle.

Obviously, any triangle will have 3 vertices, as well as three sides.

Theorem on the sum of angles in a triangle

Let us introduce and prove one of the main theorems related to triangles, namely the theorem on the sum of angles in a triangle.

Theorem 1

The sum of the angles in any arbitrary triangle is $180^\circ$.

Proof.

Consider the triangle $EGF$. Let us prove that the sum of the angles in this triangle is equal to $180^\circ$. Let's make an additional construction: draw the straight line $XY||EG$ (Fig. 2)

Since the lines $XY$ and $EG$ are parallel, then $∠E=∠XFE$ lie crosswise at the secant $FE$, and $∠G=∠YFG$ lie crosswise at the secant $FG$

Angle $XFY$ will be reversed and therefore equals $180^\circ$.

$∠XFY=∠XFE+∠F+∠YFG=180^\circ$

Hence

$∠E+∠F+∠G=180^\circ$

The theorem has been proven.

Triangle Exterior Angle Theorem

Another theorem on the sum of angles for a triangle can be considered the theorem on the external angle. First, let's introduce this concept.

Definition 4

An external angle of a triangle will be called an angle that will be adjacent to any angle of the triangle (Fig. 3).

Let us now consider the theorem directly.

Theorem 2

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

Proof.

Consider an arbitrary triangle $EFG$. Let it have an external angle of the triangle $FGQ$ (Fig. 3).

By Theorem 1, we will have that $∠E+∠F+∠G=180^\circ$, therefore,

$∠G=180^\circ-(∠E+∠F)$

Since the angle $FGQ$ is external, it is adjacent to the angle $∠G$, then

$∠FGQ=180^\circ-∠G=180^\circ-180^\circ+(∠E+∠F)=∠E+∠F$

The theorem has been proven.

Sample tasks

Example 1

Find all angles of a triangle if it is equilateral.

Since all the sides of an equilateral triangle are equal, we will have that all the angles in it are also equal to each other. Let us denote their degree measures by $α$.

Then, by Theorem 1 we get

$α+α+α=180^\circ$

Answer: all angles equal $60^\circ$.

Example 2

Find all angles of an isosceles triangle if one of its angles is equal to $100^\circ$.

Let us introduce the following notation for angles in an isosceles triangle:

Since we are not given in the condition exactly what angle $100^\circ$ is equal to, then two cases are possible:

An angle equal to $100^\circ$ is the angle at the base of the triangle.

Using the theorem on angles at the base of an isosceles triangle, we obtain

$∠2=∠3=100^\circ$

But then only their sum will be greater than $180^\circ$, which contradicts the conditions of Theorem 1. This means that this case does not occur.

An angle equal to $100^\circ$ is the angle between equal sides, that is