Inscribed angles Inscribed angle theorem 1 case Ray BO coincides with the side of the angle ABC Inscribed angle theorem 1 case Ray BO coincides with the side of the angle ABC Given: Okr (O; R) ABC – inscribed angle Prove: ABC = ½ AC Proof: 1. AOB is isosceles, since OB = OA = R, which means B = A. 2. SOA – external corner, therefore, SOA = OVA + OAV SOA = 2 OVA, which means OVA = ½ SOA SBA = ½ AC.
°
Review game “Believe it or not” Do you believe that if the central angle is 90˚, then the inscribed angle subtended by this arc is 45˚? Do you believe that the tangent segments to a circle are equal and make equal angles with a line passing through the center of the circle? Do you believe that the angle passing through the center of a circle is called its central angle? Do you believe that an inscribed angle is measured by half the arc on which it subtends? Do you believe that the central angle is twice as large? greater than the value the arc on which it rests? Do you believe that the inscribed angle of a semicircle is 180˚? Do you believe that an angle whose sides intersect a circle is called an inscribed angle? Do you believe that inscribed angles subtending the same arc are equal? Do you believe that with further study of the material, not only angles, but also triangles and quadrilaterals will be associated with a circle? No, the tangent segments to a circle (drawn from one point) are equal and make equal angles with a straight line passing through (this point and) the center of the circle. YES, if the central angle is 90˚, then the inscribed angle subtended by this arc is 45˚. No, the angle passing through (out from) the center of the circle is called its central angle. Yes, an inscribed angle is measured by half the arc it subtends. No, the magnitude of the central angle is twice as large (equal) to the magnitude of the arc on which it rests. No, an inscribed angle subtended by a semicircle is equal to 180˚ (straight line). No, an angle whose sides intersect a circle (and whose vertex lies on the circle) is called an inscribed angle. Yes, inscribed angles subtending the same arc are equal. Yes, with further study of the material, not only angles, but also triangles and quadrilaterals will be associated with a circle.
Inscribed angles Work on the test with programmed control of the solution. Option Angle ACB is 38° less than angle AOB. Find the sum of angles AOB and ACB a) 96 °; b) 114 °; c) 104 °; d) 76 °; 2. MR – diameter, O – center of the circle. OM=OK=MK. Find the angle RKO. a) 60°; b)40°; c) 30°; d) 45°; 3. Angle ABC is inscribed, angle AOC is central. Find angle ABC if angle AOC = 126 ° a) 112 °; b) 123 °; c) 117°; d) 113 °; Option Angle MSK is 34° less than angle MOK. Find the sum of the angles MSC and MOC. a) 112°; b) 102°; c) 96°; d) 68°; 2. AC is the diameter of the circle, O is its center. AB=OB=OA. Find the angle OBC. a) 50°; b) 60°; c) 30°; d) 45°; 3. O – center of the circle, angle L = 136 °. Find angle B. a) 292 °; b) 224 °; c) 112 °; d) 146 °;
A chord that does not pass through the center is equal to the diameter. Let diameter AB be drawn in a circle. Through point B we draw some chord BC that does not pass through the center, then through the middle of this chord D and point A we draw new chord AE. Finally, connect points E and C with a straight line segment. Let's look at ABD and EDC. In these triangles: ВD = DC (by construction), A = C (as inscribed ones, based on the same arc). In addition, BDA = EDC (as vertical). If the side and two angles of one triangle are respectively equal to the side and two angles of another triangle, then such triangles are congruent. This means that BDA = EDC, and in equal triangles against equal angles lie equal sides. Therefore, AB=EC.
Let's find the error According to the theorem on the sign of equality of a triangle: If a side and two adjacent angles of one triangle are respectively equal to the side and two adjacent angles of another triangle, then such triangles are congruent. And in our case, angle A is not adjacent to side BD.
Inscribed Angles Test for optical illusion according to the pictures with an alternative answer. We observe optical illusion quite often and even use it in our practice, but we know very little about its essence. The illusion of vision is used by architects when constructing buildings, fashion designers when creating models, and artists when creating scenery. We know that a body painted in light colors appears larger than a body of the same size painted in a dark tone. There are reasons that cause optical illusions. Inscribed angles Test 2 Test 3 Test 2 Test 3 The following is inscribed in a circle: 1. a square 2. a figure close to a square Test 2, 3: Circles are dominant here. Angles inscribed in a circle form a square in the first case, and in the second regular triangle. These figures, due to the many circles, present themselves as figures close to a square and a triangle. The sides appear to be concave inward. So, we can use illusion in practice, in Everyday life. For example, it can be used to hide imperfections in the shape of the face and figure. Inscribed in a circle: 1. a triangle 2. a figure close to a triangle
Inscribed angles Having mastered the theorem about the size of an inscribed angle in a circle, we draw a conclusion, because from all points of the circle, except for the ends of the chord, this chord is visible at the same angle, we can plant rose bushes at any point on the circumference of the flowerbed, except for points M and N. This is one of practical applications theorems on the size of an inscribed angle in a circle.
Inscribed angles Homework. p. 71, learn the definition of an inscribed angle; learn the inscribed angle theorem (by writing down the proof of case 3) and two corollaries from it;
In this article I will tell you how to solve problems that use .
First, as usual, let us recall the definitions and theorems that you need to know in order to successfully solve problems in .
1.Inscribed angle is an angle whose vertex lies on a circle and whose sides intersect the circle:
2.Central angle is the angle whose vertex coincides with the center of the circle:
Degree value of a circular arc measured by the magnitude of the central angle that rests on it.
IN in this case the degree value of the arc AC is equal to the value of the angle AOS.
3. If the inscribed and central angles are based on the same arc, then the inscribed angle is half the size of the central angle:
4. All inscribed angles that rest on one arc are equal to each other:
5. The inscribed angle subtended by the diameter is 90°:
Let's solve several problems.
1 . Task B7 (No. 27887)
Let's find the value of the central angle that rests on the same arc:
Obviously, the angle AOC is equal to 90°, therefore, angle ABC is equal to 45°
Answer: 45°
2.Task B7 (No. 27888)
Find the size of angle ABC. Give your answer in degrees.
Obviously, angle AOC is 270°, then angle ABC is 135°.
Answer: 135°
3. Task B7 (No. 27890)
Find the degree value of the arc AC of the circle on which the angle ABC. Give your answer in degrees.
Let's find the value of the central angle that rests on the arc AC:
The angle AOC is 45°, therefore, degree measure arc AC is 45°.
Answer: 45°.
4 . Task B7 (No. 27885)
Find the angle ACB if the inscribed angles ADB and DAE rest on circular arcs whose degree values are equal to and respectively. Give your answer in degrees.
Angle ADB rests on arc AB, therefore, the value of the central angle AOB is equal to 118°, therefore, angle BDA is equal to 59°, and the adjacent angle ADC is equal to 180°-59° = 121° 
Similarly, the angle DOE is 38° and the corresponding inscribed angle DAE is 19°.
Consider triangle ADC:
The sum of the angles of a triangle is 180°.
The angle ACB is equal to 180°- (121°+19°)=40°
Answer: 40°
5 . Task B7 (No. 27872)
The sides of the quadrilateral ABCD AB, BC, CD and AD subtend circumscribed circle arcs whose degree values are equal to , , and , respectively. Find angle B of this quadrilateral. Give your answer in degrees.
Angle B rests on arc ADC, the value of which is equal to the sum of the values of arcs AD and CD, that is, 71°+145°=216°
Inscribed angle B is equal to half the magnitude of arc ADC, that is, 108°
Answer: 108°
6. Task B7 (No. 27873)
Points A, B, C, D, located on a circle, divide this circle into four arcs AB, BC, CD and AD, the degree values of which are in the ratio 4:2:3:6 respectively. Find angle A of quadrilateral ABCD. Give your answer in degrees.
(see drawing of the previous task)
Since we have given the ratio of the magnitudes of the arcs, we introduce the unit element x. Then the magnitude of each arc will be expressed by the following ratio:
AB=4x, BC=2x, CD=3x, AD=6x. All arcs form a circle, that is, their sum is 360°.
4x+2x+3x+6x=360°, hence x=24°.
Angle A is supported by arcs BC and CD, which together have a value of 5x=120°.
Therefore, angle A is 60°
Answer: 60°
7. Task B7 (No. 27874)
Quadrangle ABCD inscribed in a circle. Corner ABC equal to , angle CAD
Inscribed angle, theory of the problem. Friends! In this article we will talk about tasks for which you need to know the properties of an inscribed angle. This is a whole group of tasks, they are included in the Unified State Exam. Most of them can be solved very simply, in one action.
There are more difficult problems, but they won’t present much difficulty for you; you need to know the properties of an inscribed angle. Gradually we will analyze all the prototypes of tasks, I invite you to the blog!
Now necessary theory. Let us remember what a central and inscribed angle, a chord, an arc are, on which these angles rest:
The central angle in a circle is a plane angle withapex at its center.
The part of a circle located inside a plane anglecalled an arc of a circle.
The degree measure of an arc of a circle is called the degree measurethe corresponding central angle.
An angle is said to be inscribed in a circle if the vertex of the angle lieson a circle, and the sides of the angle intersect this circle.
A segment connecting two points on a circle is calledchord. The largest chord passes through the center of the circle and is calleddiameter.
To solve problems involving angles inscribed in a circle,you need to know the following properties:
1. The inscribed angle is equal to half the central angle, based on the same arc.
2. All inscribed angles subtending the same arc are equal.
3. All inscribed angles based on the same chord and whose vertices lie on the same side of this chord are equal.
4. Any pair of angles based on the same chord, the vertices of which lie along different sides chords add up to 180°.
Corollary: the opposite angles of a quadrilateral inscribed in a circle add up to 180 degrees.
5. All inscribed angles subtended by a diameter are right angles.
In general, this property is a consequence of property (1), this is its special case. Look - the central angle is equal to 180 degrees (and this unfolded angle is nothing more than a diameter), which means, according to the first property, the inscribed angle C is equal to half of it, that is, 90 degrees.
Knowledge of this property helps in solving many problems and often allows you to avoid unnecessary calculations. Having mastered it well, you will be able to solve more than half of the problems of this type orally. Two conclusions that can be drawn:
Corollary 1: if a triangle is inscribed in a circle and one of its sides coincides with the diameter of this circle, then the triangle is right-angled (vertex right angle lies on the circle).
Corollary 2: the center of the described about right triangle circle coincides with the middle of its hypotenuse.
Many prototypes of stereometric problems are also solved by using this property and these consequences. Remember the fact itself: if the diameter of a circle is a side of an inscribed triangle, then this triangle is right-angled (the angle opposite the diameter is 90 degrees). You can draw all other conclusions and consequences yourself; you don’t need to teach them.
As a rule, half of the problems on an inscribed angle are given with a sketch, but without symbols. To understand the reasoning process when solving problems (below in the article), notations for vertices (angles) are introduced. You don't have to do this on the Unified State Examination.Let's consider the tasks:
What is the acute inscribed angle subtended by the chord? equal to the radius circles? Give your answer in degrees.
Let's construct a central angle for a given inscribed angle and designate the vertices:
According to the property of an angle inscribed in a circle:
Angle AOB is equal to 60 0, since the triangle AOB is equilateral, and in equilateral triangle all angles are equal to 60 0. The sides of the triangle are equal, since the condition says that the chord is equal to the radius.
Thus, the inscribed angle ACB is equal to 30 0.
Answer: 30
Find the chord supported by an angle of 30 0 inscribed in a circle of radius 3.
This is essentially inverse problem(previous). Let's construct the central angle.
It is twice as large as the inscribed one, that is, angle AOB is equal to 60 0. From this we can conclude that triangle AOB is equilateral. Thus, the chord is equal to the radius, that is, three.
Answer: 3
The radius of the circle is 1. Find the magnitude of the obtuse inscribed angle subtended by the chord, equal to the root out of two. Give your answer in degrees.
Let's construct the central angle:
Knowing the radius and chord, we can find the central angle ASV. This can be done using the cosine theorem. Knowing the central angle, we can easily find the inscribed angle ACB.
Cosine theorem: square any side of the triangle equal to the sum squares of the other two sides, without doubling the product of these sides by the cosine of the angle between them.
Therefore, the second central angle is 360 0 – 90 0 = 270 0 .
Angle ACB, according to the property of an inscribed angle, is equal to half of it, that is, 135 degrees.
Answer: 135
Find the chord subtended by an angle of 120 degrees inscribed in a circle of radius root of three.
Let's connect points A and B to the center of the circle. Let's denote it as O:
We know the radius and inscribed angle ASV. We can find the central angle AOB (greater than 180 degrees), then find the angle AOB in triangle AOB. And then, using the cosine theorem, calculate AB.
According to the property of the inscribed angle, the central angle AOB (which is greater than 180 degrees) will be equal to twice the inscribed angle, that is, 240 degrees. This means that angle AOB in triangle AOB is equal to 360 0 – 240 0 = 120 0.
According to the cosine theorem:
Answer:3
Find the inscribed angle subtended by an arc that is 20% of the circle. Give your answer in degrees.
According to the property of an inscribed angle, it is half the size of the central angle based on the same arc, in this case we are talking about the arc AB.
It is said that arc AB is 20 percent of the circumference. This means that the central angle AOB is also 20 percent of 360 0.*A circle is an angle of 360 degrees. Means,
Thus, the inscribed angle ACB is 36 degrees.
Answer: 36
Arc of a circle A.C., not containing a point B, is 200 degrees. And the arc of a circle BC, not containing a point A, is 80 degrees. Find the inscribed angle ACB. Give your answer in degrees.
For clarity, let us denote the arcs whose angular measures are given. Arc corresponding to 200 degrees – Blue colour, the arc corresponding to 80 degrees is red, the remaining part of the circle is yellow.
Thus, the degree measure of the arc AB (yellow), and therefore the central angle AOB is: 360 0 – 200 0 – 80 0 = 80 0 .
The inscribed angle ACB is half the size of the central angle AOB, that is, equal to 40 degrees.
Answer: 40
What is the inscribed angle subtended by the diameter of the circle? Give your answer in degrees.
It is necessary to know the property of an inscribed angle; understand when and how to use the cosine theorem, learn more about it.
That's all! I wish you success!
Sincerely, Alexander Krutitskikh
Mathematics teacher at school in the third grade:
- Children, tell me, how much is 6*6?
The children answer in unison:
- Seventy six!
- Well, what are you saying, kids! Six by six will be thirty-six... well, maybe another 37, 38, 39... well, maximum 40... but not seventy-six!
P.S: I would be grateful if you tell me about the site on social networks.
Lesson objectives: formation of knowledge on the topic, organization of work on mastering concepts and scientific facts.
Educational objectives:
- introduce the concept of inscribed angle;
- teach to recognize inscribed angles in drawings;
- anticipate an additional construction containing an inscribed angle leading to the solution of the problem;
- consider the inscribed angle theorem and its consequences;
- show the application of the theorem in solving problems;
- introduce optical illusions
Educational tasks: activating the independence of cognitive activity of students. skills formation teamwork, development of a sense of responsibility for one’s knowledge, a culture of communication, familiarization with the knowledge of optical illusion and its application in practice, education of an aesthetic culture.
Developmental tasks: continue to develop the ability to analyze, compare, compare, highlight the main thing, establish cause-and-effect relationships; improve graphic culture.
Technology: problem-based learning using information technology.
Lesson type: lesson in the formation of new knowledge.
Lesson form: lesson - problem presentation.
Lesson equipment: presentation: presentation, self-analysis sheets.
Lesson steps
- Motivation to educational activities-1 minute.
- Statement of the problem and creation of a plan to solve it – 2 minutes.
- Updating knowledge - 4 minutes.
- Discovery of a new concept - 10 minutes.
- Research to identify the properties of a new concept - 4 minutes.
- Application of new knowledge - 11 minutes.
- The game “Believe it or not” in order to consolidate new theoretical material - 2 minutes.
- Individual work with dough - 5 minutes.
- Applying new knowledge in unfamiliar situations - 4 minutes.
- Reflection - 3 minutes.
During the classes
1. Motivation for educational activities
Hello guys. Sit down. I hope that the knowledge you gain in this lesson will be useful to you in life.
2. Statement of the problem and creation of a plan to solve it
Dana flower bed round shape, on one of the chords of which roses are planted. In what different places in the flower bed should three rose bushes be planted so that from these points all the roses are visible from the same angle? (Slide 2).Presentation
What versions do you have for solving this problem?
Arises problematic situation. Students lack knowledge.
To answer this question, we need to use the properties of an inscribed angle. Then let's make a lesson plan together. What are the goals of the lesson and how will we achieve them?” During the discussion, a lesson plan appears on the screen. (C lead 3)
3. Updating knowledge
Teacher: “Give the definition of an angle. What is the central angle called? (C lead 4)
Tasks (Slide 5
4. Discovery of a new concept
Now you see six drawings. What groups would you divide them into and why? (Slide 6)
Sharp, straight, blunt.
Angles 1, 3, 5 and 2, 4, 6 according to the location of the vertex of the angle? What are angles 1, 3, 5 called?
And angles 2, 4, 6 are called inscribed. These are what we will talk about today.
How are angles ABC and KRO similar and different? (Slide 7)
After answering this question, students try to define an inscribed angle, after which the teacher displays the statement on the screen, emphasizing the important points: (C lead 8)
- the vertex lies on the circle
- the sides intersect the circle.
Find pictures that show inscribed angles.
Exercise. Express the magnitude of the inscribed angle, knowing how the magnitude of the central angle is expressed in terms of the arc on which it rests. Working with slide 10
What additional construction needs to be done to complete this task? If students do not immediately guess, clarify: what central angle should be connected to this inscribed angle?
Next, students see that the resulting central angle is an external angle of an isosceles triangle and come to the conclusion that one of the angles (in particular, the inscribed one), equal to their half-sum, is equal to half of the central one, i.e. half of the arc on which it rests.
The exact formulation of the theorem is given and projected on the screen. (C lead 11).
Students transfer the drawing to their notebooks ( slide 12), then write down the condition in your notebook. One of the students comments on the posts. The next student writes down and comments on the proof of the theorem. The logic and completeness of the design is checked using slide 12). Thus, the proof of the theorem is formalized for the case when the side of the inscribed angle passes through the center of the circle.
The case when the center of a circle lies inside an angle is discussed orally using slide 13.
The teacher offers to justify the next case, when the center of the circle lies outside the angle, home preparation. (C lead 14). In the class according to the drawing slide 15 find out that a given inscribed angle can be considered as the difference of two angles, each of which has one side that is one side of the given angle, and the second side is common and passes through the center of the circle.
5. Research work to identify the properties of a new concept
Working with slide 15.
Exercise. How to quickly use a compass and ruler to construct several angles equal to this angle? They notice that their methods are irrational. A problematic situation arises: old knowledge does not provide a rational solution to the problem.
Think about how to use new material, this problem can be solved. You can draw a circle passing through the vertex of an angle without specifying the center and construct various inscribed angles based on the same arc. The problematic situation has been resolved. After which Corollary 1 is formulated: “Inscribed angles subtended by the same arc are equal.”
The work leading to the formulation of Corollary 2 is carried out in a similar way. (C lead 16)
How to quickly construct a right angle using a compass and ruler? It is clarified that “fast” should be understood as a “minimum number of steps”. We come to the irrationality of this construction. If the students have not guessed how to complete the construction, the teacher asks the question: on which arc should the right inscribed angle rest? After this, students outline the construction step by step:
- Draw a circle of arbitrary radius.
- Draw the diameter.
- Select any point on the circle except the ends of the diameter.
- Draw rays from the selected point through the ends of the diameter.
After this, the teacher says that this construction Corollary 2 from the inscribed angle theorem was used. Try to formulate it.
The clarified wording is projected on the screen. ( Slides 17-19)
6. Application of new knowledge
Solving problems to consolidate new material. Working with slides 20-26.
7. Game of repetition for the purpose of consolidation theoretical material.(C lead 27)
Game “Believe it or not”
- Do you believe that if the central angle is 90˚, then the inscribed angle subtended by this arc is 45˚?
- Do you believe that the tangent segments to a circle are equal and make equal angles with a line passing through the center of the circle? Do you believe that the angle passing through the center of the circle is called its central angle?
- Do you believe that an inscribed angle is measured by half the arc on which it subtends?
- Do you believe that the magnitude of the central angle is twice the magnitude of the arc on which it rests?
- Do you believe that the inscribed angle of a semicircle is 180˚?
- Do you believe that an angle whose sides intersect a circle called an inscribed angle?
- Do you believe that inscribed angles subtending the same arc are equal?
- Do you believe that with further study of the material, not only angles, but also triangles and quadrilaterals will be associated with a circle?
8. Individual work with the test. (C leads 28-30)
The answer sheets are given to the teacher. The teacher then comments on the solutions.
Option 1.
1. Angle ACB is 38° less than angle AOB. Find the sum of angles AOB and ACB
a) 96°; b) 114°; c) 104°; d) 76°;
2. MR – diameter, O – center of the circle. OM=OK=MK. Find the angle RKO.
a) 60°; b)40°; c) 30°; d) 45°;
3. Angle ABC is inscribed, angle AOC is central. Find angle ABC if angle AOC = 126°
a) 112 °; b) 123 °; c) 117°; d) 113 °;
Option 2.
1. The MSK angle is 34 ° less than the MOK angle. Find the sum of the angles MSC and MOC.
a) 112°; b) 102°; c) 96°; d) 68°;
2. AC is the diameter of the circle, O is its center. AB=OB=OA. Find the angle OBC.
a) 50°; b) 60°; c) 30°; d) 45°;
3. O – center of the circle, angle L = 136 °. Find angle B.
a) 292 °; b) 224 °; c) 112 °; d) 146 °;
Answers to assignments are checked after completing the test.
| Tasks | 1 | 2 | 3 |
| 1 Option | B | IN | IN |
| Option 2 | B | IN | IN |
9. Application of new knowledge in unfamiliar situations
a) Working with slides 31-33.
Teacher: “At home you solved the problem of calculating the angles of a five-pointed star inscribed in a circle. How did you solve it?
How to solve this problem using the inscribed angle theorem.
Method II: When the vertices of a pentagonal star divide the circle into equal arcs, the problem is solved very simply: 360°: 5:2 *5=180°.
b) Analysis of mathematical sophism on the application of the theorem on the magnitude of the inscribed angle.
A chord not passing through the center is equal to the diameter.(C laid 34-36) Find the error in the reasoning.
Solution. Let diameter AB be drawn in a circle. Through point B we draw some chord BC that does not pass through the center, then through the middle of this chord D and point A we draw a new chord AE. Finally, connect points E and C with a straight line segment. Let's consider ▲АВD and ▲ЭДС. In these triangles: BD = DC (by construction), Ð A = Ð C (as inscribed triangles based on the same arc). In addition, Ð BDA = Ð EDC (as vertical). If the side and two angles of one triangle are respectively equal to the side and two angles of another triangle, then such triangles are congruent. Means,
▲ BDA = ▲ EDC, and in equal triangles equal sides lie opposite equal angles.
Therefore, AB=EC.
Find the error in reasoning.
c) Optical illusion test using pictures with an alternative answer. ( Slides 37-39)
Show what illusory deformation is caused by sharp central angles and inscribed angles.
Test1. Here, sharp central angles cause illusory deformation. Although the angles AOB, BOC, COD are equal, due to the many sharp corners, on which two angles are divided, they pretend to be larger than the average angle.
Test 2-3. Circles are dominant here. Angles inscribed in a circle form a square in the first case, and a regular triangle in the second. These figures, due to the many circles, present themselves as figures close to a square and a triangle. The sides appear to be concave inward.
So, we can use illusion in practice, in everyday life. For example, it can be used to hide imperfections in the shape of the face and figure.
10. Reflection
Let's go back to the lesson plan and see if we answered all the questions?
You and I have not answered one question. So how should you plant three roses? (Slide 40-41)
Having mastered the theorem about the size of the inscribed angle in a circle, we conclude, because from all points on the circle, except the ends of the chord, this chord is visible at the same angle, we can plant rose bushes at any point on the circumference of the flowerbed, except for points M and N. This is one of the practical applications of the theorem on the size of the inscribed angle in a circle.
At the end of the lesson, students can be given a questionnaire to fill out, which allows them to carry out self-analysis, give a qualitative and quantitative assessment of the lesson, and, in addition, can be formulated a task to justify their answer:
1. During the lesson I worked...;
2. With my work in class I...;
3. The lesson seemed to me...;
4. For the lesson I...;
5. The lesson material was for me...;
6. Homework seems to me...
Homework. (C light 42)
- P. 71, learn the definition of an inscribed angle;
- learn the inscribed angle theorem (by writing down the proof of case 3) and two corollaries from it;
- № 654 № 656 № 657.
Bibliography:
- Geometry: Textbook. For 7–9 grades. general images. institutions / L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev and others - 12th ed., - M.: Education, 2002.
- Ziv B.G., Mailer V.M., Didactic materials in geometry for 8th grade. – 6th ed. – M.: Education, 2002.
- Smirnova I.M., Smirnov V.A. Oral exercises in geometry for grades 7–11. Book for teachers. M.; Enlightenment, 2003
- Rabinovich E.M. Tasks and exercises for finished drawings. Geometry grades 7–9. “Ilexa”, “Gymnasium”, Moscow-Kharkov, 2003
Centers and Internet sites:
- Workshop. Multimedia presentations for math lessons. http://www.intergu.ru/infoteka/
- Internet state teachers in the Infotech-Mathematics section. http://www.intergu.ru/infoteka/
- TsORs from the portal “Network of Creative Teachers”.