Complete lessons - Knowledge Hypermarket. Sum of angles of a triangle. Triangle Angle Sum Theorem

“Tell me and I will forget,
Show me and I will remember
Involve me and I will learn”
Eastern proverb

Goal: Prove the theorem about the sum of the angles of a triangle, practice solving problems using this theorem, develop students’ cognitive activity using additional material from different sources, and develop the ability to listen to others.

Equipment: Protractor, ruler, triangle models, mood strip.

DURING THE CLASSES

1. Organizing time.

Mark your mood at the beginning of the lesson on the mood tape.

2. Repetition.

Review the concepts that will be used in proving the theorem: properties of angles for parallel lines, definition of a straight angle, degree measure of a straight angle.

3. New material.

3.1. Practical work.

Each student has three models of a triangle: acute, rectangular and obtuse. It is proposed to measure the angles of a triangle and find their sum. Analyze the result. You can get values ​​of 177, 178, 179, 180, 181, 182, 183 degrees. Calculate the arithmetic mean (=180°). It is suggested to remember when angles have a degree measure of 180 degrees. Students remember that this is a straight angle and the sum of one-sided angles.

Let's try to get the sum of the angles of a triangle using origami.

Historical reference

Origami (Japanese, lit.: “folded paper”) is the ancient art of folding paper figures. The art of origami has its roots in ancient China, where paper was discovered.

3.2. Proof of the theorem from the textbook by Atanasyan L.S.

Theorem on the sum of the angles of a triangle.

Let us prove one of the most important theorems of geometry - the theorem on the sum of the angles of a triangle.

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

Proof. Consider an arbitrary triangle ABC and prove that A + B + C = 180°.

Let us draw a straight line a through vertex B, parallel to side AC. Angles 1 and 4 are cross-lying angles when parallel lines a and AC are intersected by secant AB, and angles 3 and 5 are cross-lying angles when the same parallel lines are intersected by secant BC. Therefore, angle 4 is equal to angle 1, angle 5 is equal to angle 3.

Obviously, the sum of angles 4, 2 and 5 is equal to the developed angle with vertex B, i.e. angle 4 + angle 2 + angle 5 = 180°. From here, taking into account the previous equalities, we obtain: angle 1 + angle 2+ angle 3 = 180°, or A + B+ C = 180°. The theorem has been proven.

3.3. Proof of the theorem from A. V. Pogorelov’s textbook.

Prove: A + B + C = 180°

Proof:

1. Draw a line BD // AC through vertex B

2. DBC=ACB, lying crosswise at AC//BD and secant BC.

3. ABD =ACB +CBD

Hence, A + B+C = ABD+BAC

4. ABD and BAC are one-sided with BD // AC and secant AB, which means their sum is equal to 180 °, i.e. A+B + C=180°, which is what needed to be proven.

3. 4. Proof of the theorem from the textbook by Kiselev A.N., Rybkina N.A.

Given: ABC

Prove: A+B +C=180°

Proof:

1. Let's continue the AC side. We will carry out SE//AV

2. A=ESD, as corresponding with AB//CE and AD - secant

3. B=ALL, lying crosswise at AB//CE and BC - the secant.

4. ESD + ALL + C = 180 °, which means A + B + C = 180 °, which was what needed to be proven.

3.5. Corollaries 1. In any triangle, all angles are acute, or two angles are acute, and the third is obtuse or straight.

Corollary 2.

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

3.6. The theorem allows us to classify triangles not only by sides, but also by angles.

Triangle view	Isosceles	Equilateral	Versatile
rectangular
obtuse
acute-angled

4. Consolidation.

4.1. Solving problems using ready-made drawings.

Find the unknown angles of the triangle.

4.2. Check of knowledge.

1. At the end of our lesson, answer the questions:

Are there triangles with angles:

a) 30, 60, 90 degrees,

b) 46, 4, 140 degrees,

c) 56, 46, 72 degrees?

2. Can a triangle have:

a) two obtuse angles,

b) obtuse and right angles,

c) two right angles?

3. Determine the type of triangle if one angle is 45 degrees, the other is 90 degrees.

4. In which triangle is the sum of the angles greater: acute, obtuse or rectangular?

5. Is it possible to measure the angles of any triangle?

This is a joke question, because... There is a Bermuda Triangle, located in the Atlantic Ocean between Bermuda, the state of Puerto Rico and the Florida Peninsula, whose angles cannot be measured. (Annex 1)

5. Lesson summary.

Mark your mood at the end of the lesson on the mood tape.

Homework.

P. 30–31; No. 223 a, b; No. 227 a; workbook № 116, 118.

. (Slide 1)

Lesson type: lesson of learning new material.

Lesson objectives:

• Educational:
  • consider the theorem on the sum of the angles of a triangle,
  • show the application of the theorem in solving problems.
• Educational:
  • fostering a positive attitude of students towards knowledge,
  • Instill self-confidence in students through lessons.
• Developmental:
  • development of analytical thinking,
  • development of “skills to learn”: use knowledge, skills and abilities in educational process,
  • development logical thinking, the ability to clearly formulate your thoughts.

Equipment: interactive whiteboard, presentation, cards.

DURING THE CLASSES

I. Organizational moment

– Today in class we will remember the definitions of right, isosceles, and equilateral triangles. Let us repeat the properties of the angles of triangles. Using the properties of internal one-sided and internal cross-lying angles, we will prove the theorem about the sum of the angles of a triangle and learn how to apply it when solving problems.

II. Orally(Slide 2)

1) Find rectangular, isosceles, equilateral triangles in the pictures.
2) Define these triangles.
3) Formulate the properties of the angles of an equilateral and isosceles triangle.

4) In the picture KE II NH. (slide 3)

– Specify secants for these lines
– Find interior one-sided angles, interior angles lying crosswise, name their properties

III. Explanation of new material

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

According to the formulation of the theorem, the guys build a drawing, write down the condition and conclusion. By answering questions, they independently prove the theorem.

Given: Prove:

Proof:

1. Through vertex B of the triangle we draw a straight line BD II AC.
2. Specify secants for parallel lines.
3. What can be said about the angles CBD and ACB? (make a note)
4. What do we know about angles CAB and ABD? (make a note)
5. Replace angle CBD with angle ACB
6. Draw a conclusion.

IV. Finish the sentence.(Slide 4)

1. The sum of the angles of a triangle is...
2. In a triangle, one of the angles is equal, the other, third angle of the triangle is equal to...
3. The sum of the acute angles of a right triangle is...
4. The angles of an isosceles right triangle are equal...
5. The angles of an equilateral triangle are equal...
6. If the angle between the lateral sides of an isosceles triangle is 1000, then the angles at the base are equal...

V. A little history.(Slides 5-7)

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

The following types of polygon with three vertices are distinguished:

• 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 equal sides there 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 CN (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, the acute angles add up to 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 ∟H = 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 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 such properties characteristic of an equilateral triangle:

• the median, bisector, height in such a geometric figure coincide, and their length is calculated as (a x √3): 2;
• if you describe it around given polygon circle, 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 theorem, that the sum of the angles obtuse triangle equal to 180 degrees. Again, this theorem does not need to be proven again.

Theorem on the sum of interior angles of a triangle

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

Proof:

• Given triangle ABC.
• Through vertex B we draw a straight line DK parallel to the base AC.
• \angle CBK= \angle C as internal crosswise lying with parallel DK and AC, and secant BC.
• \angle DBA = \angle A internal crosswise lying with DK \parallel AC and secant AB. Angle DBK is reversed and equal to
• \angle DBK = \angle DBA + \angle B + \angle CBK
• Since the unfolded angle is equal to 180 ^\circ , and \angle CBK = \angle C and \angle DBA = \angle A , we get 180 ^\circ = \angle A + \angle B + \angle C.

The theorem is proven

Corollaries from the theorem on the sum of angles of a triangle:

1. The sum of the acute angles of a right triangle is equal to 90°.
2. In an isosceles right triangle, each acute angle is equal to 45°.
3. In an equilateral triangle, each angle is equal 60°.
4. In any triangle, either all the angles are acute, or two angles are acute, and the third is obtuse or right.
5. An exterior angle of a triangle is equal to the sum of two interior angles that are not adjacent to it.

Triangle Exterior Angle Theorem

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

Proof:

• Given a triangle ABC, where BCD is the exterior angle.
• \angle BAC + \angle ABC +\angle BCA = 180^0
• From the equalities the angle \angle BCD + \angle BCA = 180^0
• We get \angle BCD = \angle BAC+\angle ABC.

Following on from yesterday:

Let's play with a mosaic based on a geometry fairy tale:

Once upon a time there were triangles. So similar that they are just copies of each other.
They somehow stood side by side in a straight line. And since they were all the same height -
then their tops were at the same level, under the ruler:

Triangles loved to tumble and stand on their heads. Climbed in top row and stood on the corner like acrobats.
And we already know - when they stand with their tops exactly in a line,
then their soles also follow a ruler - because if someone is the same height, then they are also the same height upside down!

They were the same in everything - the same height, and the same soles,
and the slides on the sides - one steeper, the other flatter - are the same in length
and they have the same slope. Well, just twins! (only in different clothes, each with their own piece of the puzzle).

- Where do the triangles have identical sides? Where are the corners the same?

The triangles stood on their heads, stood there, and then decided to slide off and lie down in bottom row.
They slid and slid down a hill; but their slides are the same!
So they fit exactly between the lower triangles, without gaps, and no one pushed anyone aside.

We looked around the triangles and noticed an interesting feature.
Wherever their angles come together, all three angles will certainly meet:
the largest is the “head angle”, the most acute angle and the third, medium largest angle.
They even tied colored ribbons so that it would be immediately obvious which was which.

And it turned out that the three angles of the triangle, if you combine them -
make up one large angle, an “open corner” - like the cover of an open book,

______________________O ___________________

it's called a turned angle.

Any triangle is like a passport: three angles together are equal to the unfolded angle.
Someone knocks on your door: - knock-knock, I'm a triangle, let me spend the night!
And you tell him - Show me the sum of the angles in expanded form!
And it’s immediately clear whether this is a real triangle or an impostor.
Failed verification - Turn around one hundred and eighty degrees and go home!

When they say "turn 180°" it means to turn around backwards and
go in the opposite direction.

The same thing in more familiar expressions, without “once upon a time”:

Let us perform a parallel translation of triangle ABC along the OX axis
to vector AB equal to length AB bases.
Line DF passing through vertices C and C 1 of triangles
parallel to the OX axis, due to the fact that perpendicular to the OX axis
segments h and h 1 (heights equal triangles) are equal.
Thus, the base of the triangle A 2 B 2 C 2 is parallel to the base AB
and equal to it in length (since the vertex C 1 is shifted relative to C by the amount AB).
Triangles A 2 B 2 C 2 and ABC are equal on three sides.
Therefore, the angles ∠A 1 ∠B ∠C 2 forming a straight angle are equal to the angles of triangle ABC.
=> The sum of the angles of a triangle is 180°

With movements - “translations”, the so-called proof is shorter and clearer,
even a child can understand the pieces of the mosaic.

But traditional school:

based on the equality of internal cross-lying angles cut off on parallel lines

valuable in that it gives an idea of ​​why this is so,
Why the sum of the angles of a triangle is equal to the reverse angle?

Because otherwise parallel lines would not have the properties familiar to our world.

The theorems work both ways. From the axiom of parallel lines it follows
equality of crosswise lying and vertical angles, and from them - the sum of the angles of a triangle.

But the opposite is also true: as long as the angles of a triangle are 180°, there are parallel lines
(such that through a point not lying on a line one can draw a unique line || of the given one).
If one day a triangle appears in the world whose sum of angles is not equal to the unfolded angle -
then the parallel ones will cease to be parallel, the whole world will be bent and skewed.

If stripes with triangle patterns are placed one above the other -
you can cover the entire field with a repeating pattern, like a floor with tiles:

you can trace different shapes on such a grid - hexagons, rhombuses,
star polygons and get a variety of parquets

Tiling a plane with parquet is not only an entertaining game, but also a relevant mathematical problem:

________________________________________ _______________________-------__________ ________________________________________ ______________
/\__||_/\__||_/\__||_/\__||_/\__|)0(|_/\__||_/\__||_/\__||_/\__||_/\=/\__||_/ \__||_/\__||_/\__||_/\__|)0(|_/\__||_/\__||_/\__||_/\__||_/\

Since every quadrilateral is a rectangle, square, rhombus, etc.,
can be composed of two triangles,
respectively, the sum of the angles of a quadrilateral: 180° + 180° = 360°

The same isosceles triangles fold into squares in different ways.
A small square of 2 parts. Average of 4. And the largest of the 8.
How many figures are there in the drawing, consisting of 6 triangles?