Teaching planimetry in a school course.
Lyceum No. 000
Lyceum No. 000.
"If the same task is entrusted
two equally ignorant of it
people and one of them is a mathematician,
then a mathematician will do it better,”
Introduction
Mastering almost any modern profession requires certain mathematical knowledge. Idea about the role of mathematics in modern world, mathematical knowledge has become a necessary component general culture. For life self-realization, the possibility of productive activity in information world A fairly strong mathematical background is required.
The role and place of mathematics in science and the life of society, the value of mathematical education, humanization and humanization of education, understanding of the subject of mathematics, and personality structure determine the goals of mathematical education. Three groups of goals are distinguished, correlating them with general educational, educational and practical functions.
Ø Mathematical education includes mastering a system of mathematical knowledge, abilities and skills that gives an idea of the subject of mathematics, its language and symbolism, periods of development, mathematical modeling, special mathematical techniques, and basic general scientific methods of cognition.
Ø Formation of students’ worldview, logical and heuristic components of thinking, education of morality, culture of communication, independence, activity, education of hard work, responsibility for decision-making, and the desire for self-realization.
Ø Specification of individual component goals is important for constructing a set of lesson goals and adequacy to the subject content of the educational material. Transformation of educational goals into actions will make it possible to diagnose and manage the process of acquiring knowledge, skills, development and education of a student.
At the level of the actual educational process, learning goals are formed taking into account the characteristics of students and the possibilities of differentiating their learning.
In the process of students’ mathematical activity, the arsenal of techniques and methods of thinking includes induction and deduction, generalization and specification, analysis and synthesis, classification and systematization, abstraction, and analogy. Objects of mathematical inferences and the rules for their construction reveal the mechanism of logical constructions, develop the ability to formulate, justify and prove judgments, thereby developing logical thinking. The leading role belongs to mathematics in the formation of algorithmic thinking, developing the ability to act according to a given algorithm and construct new ones in the course of solving problems, the basis of educational activities in mathematics lessons. The creative and applied sides of thinking develop.
- objections to the reduction of the school mathematics course;
· assessment of the course program as overloaded with unnecessary or too special information (for example, a lot of formulas to be memorized);
· conversations about the obvious insufficiency of the hours allocated for mathematics (as the main tool for development logical thinking schoolchildren, etc.);
- requirements of the school mathematics course and entrance examinations; qualifications of mathematics teachers, because any educational reform, any restructuring of the program is doomed to success only if teachers are prepared for this in advance and comprehensively.
Currently, teachers have quite a few textbooks for each parallel in their arsenal. When choosing a particular system, each teacher naturally proceeds from his own criteria and the specifics of the educational institution. However, it is necessary to take into account the possibility of implementing successive connections between courses, as well as analyze the possibility of organizing differentiated training. The teacher, depending on the specific working conditions and the level of preparation of the students, can organize a full-fledged educational process. The student gets a real opportunity, studying in the same class and according to the same program, to choose the level of learning that suits their needs, interests, and abilities. The mandatory minimum in mathematics determines the list of questions that must be presented in the program and textbooks in mathematics, regardless of their level and focus. In other words, specific programs and textbooks used in a particular institution may expand this level, but not reduce or cut it.
The choice of level of mathematical preparation should be determined by the needs of students, therefore educational institutions in the humanities, law and other fields, it is advisable to use an in-depth program in mathematics, since their graduates also go to technical universities; in addition, serious mathematics studies are necessary for the formation and development of logical thinking.
The essence of geometry is contradictory: “... it directly studies ideal geometric figures that do not exist in reality, but its conclusions are applicable to real things, to practical problems.” The task of any teacher is to bring students closer to their understanding, without obscuring geometry itself from students with numerous surveys, tests, tests, and allowing children to choose their own level of knowledge of geometry. Every choice is worthy of the goal facing the student, sometimes determined intuitively, but freely. Often persistent adherence to a set goal and persistence in achieving it are completely meaningless, especially if the teacher’s goal is not the student’s goal. It’s probably worth trying to learn how to organize students’ activities in geometry classes so that they are not constrained by our goals, our questions, so that they are open to all sorts of perceptions. A child goes to school with a lot of questions, but the school itself has prepared several times more questions for him. She answers her own questions, and even gets angry when her answers are poorly received.
One of the ways of knowledge consists of the following stages: a thought, a chain of thoughts, and finally, a strictly logically justified, desired result of the search. I would like to implement the second, more open way through a generous set of tasks proposed for each topic. When using this path, thought does not slow down fantasy, does not close the intuitive search, there is no pursuit of thoughts, there is no quick leap to the goal, but calm, unhurried perception and observation reign, sensitivity appears, it would seem to be extraneous, but sometimes it is this extraneousness that enriches the search , leads to the goal. How often in class do we rush children, urging them on like a whip with the words: “Think. Think." Or maybe it’s true that someone who searches excessively may not have time to find?
Our task is to search for paths leading to knowledge of geometry. We will think about how to help children discover truths for themselves, the truths of geometry. What should a teacher be guided by, what tactics and strategy should he choose? What should a teacher do in class? Should we stand up for knowledge, abilities, skills, arguing that knowledge is power, or try with all our might to organize the educational process so that knowledge does not overshadow knowledge, does not turn the child’s soul away from knowledge.
Perhaps the wisdom of a teacher lies in knowing the secrets of discovery, the secrets of cognition and, in particular, the secrets of geometry, in the ability to create an atmosphere in the classroom that facilitates the mastery of these methods of perception and cognition. The teacher’s logic and the student’s logic, in what relationship should they be in the lesson? What more? Perhaps when the teacher offers not a series of clearly thought-out questions, but a sequence of tasks, reflecting on which the student, his thought does all the work necessary for the moment preceding the discovery. Then the teacher’s logic is in the necessary relationship with the student’s logic. Or maybe the basis of the search should be to choose intuition, liberate it, stimulate it, rely on it? Or something else?
Perhaps, among all the textbooks and among all the lessons, the textbook for grade 7 is the most important and the first lesson is the most responsible, because they introduce a systematic course to the study. From the very first lessons, from reading the very first pages of the textbook, it depends whether the learning process will be successful and whether the students will be able to develop a sustainable interest in the subject. No student is prevented from studying a geometry course at any level. The only obstacle may not be the complexity of the material, not the difficulty of presentation, but the lack of interest in reading further pages of the textbook. However, having studied the theory even at the very first (visual) level, the student can solve any problem on this topic, since he will have enough knowledge to solve it.
Let's move on to characterizing the levels of mastery of educational material and tell the teacher how he can discover material related to each of them.
The first level is general education, humanitarian. It includes content that every student should master. In geometry, the study of such material occurs at a visual level, which is why we call the first level visual. It includes definitions of concepts, accompanied by a large number of illustrations, formulations of theorems, explanation of their meaning in drawings, and simple logical deductions.
At the second level, the material of the first level expands, applied problems are solved, it is shown how geometric knowledge is applied to knowledge of the world. We call this level the application level. At this level, students are expected to master the proofs of most theorems.
Finally, the third level is a significant deepening of the material of the first level, a fairly complete logical justification is given. This advanced level includes the most difficult proofs of theorems and theoretical problems. The third level is also problematic.
We have identified the first level of assimilation - visually - practical, at which schoolchildren, like physicists, obtain information through experience. The student must imagine an object, describe it, solve problems related to it simple task. And it doesn’t matter if at the same time he cannot pronounce the definition accurately. At this level, visual and operational knowledge of the subject is essential, containing visual representations and the ability to operate with them correctly.
When studying geometry, it is necessary to invite students to independently formulate a definition of a particular concept. This is not done so that the children will then memorize it, but so that by participating in this process they will delve deeper into the meaning of the concept, learn the structure of the definition itself and several formulations of the theorems. This will contribute to a deeper assimilation of the relevant educational material. Children's discoveries are a great incentive for learning.
It is generally accepted that a geometry course should teach logical thinking. However, many students often do not so much assimilate the logic of formulations and proofs as formally memorize them. One of the first means of overcoming this danger is to reduce the number of formulations and evidence that the student must know (learn, remember). If we want to teach logical thinking, then we need to teach this, and not mechanical memorization of ready-made reasoning. Therefore, formulations should be considered more like exercises in the development of logical thinking, and not as postulates that must be known by heart. It is useful for students to understand, and not mindlessly memorize, as many proofs as possible and solve as many proof problems as possible: it is much more pleasant and useful for the student if he figures it out on his own and does at least small conclusion, and will not memorize other people’s reasonings (not counting, of course, those that are especially instructive, witty and elegant).
The logic of geometry lies not only in individual formulations, but in their entire system as a whole. The meaning of each definition, each theorem, and proof is ultimately determined only by this system. Which makes geometry a holistic theory, and not a collection of individual definitions and statements. Therefore, we suggest that our colleagues try not to ask students to evaluate a single proof of theorems for a certain time, but to take this survey to the end of a rather extensive topic as a theoretical test, which is what we do at the Lyceum. The children must get used to the very concepts and terms “theorem”, “given”, “prove”, “proof”, and understand their meaning. Of course, theorems must be proven. It may be necessary to analyze their proofs more than once in class: frontally in pairs, on different drawings. It is quite acceptable that, from our point of view, before proving the theorem, immediately after analyzing its formulation, begin solving problems. And when students get used to the formulation and understand its meaning, they can begin to analyze the proof. By this point, students will have developed to some extent a taste for searching for truth. Respect for her.
Of course, if teaching is completely limited only to geometric knowledge itself, then the development of logical thinking skills and elements of a scientific worldview will be carried out within the framework of only this science. Therefore, the teacher must constantly draw students’ attention to the connection between geometry and other sciences and practice and show the universal (and not for geometry alone) importance of the requirement of evidence and accuracy in establishing the truth. This point is especially important for those students who have insufficient motivation to study geometry as a science, in contrast to motivated and interested children who do not need to be pushed and stimulated to solve complex, non-standard problems, consider various options solutions. Practice also shows that students love to listen to the teacher’s stories about the history of the subject. In the first lesson, stronger students who are interested can be asked to simply solve problems that are beautiful, interesting, and unusual in form and methods of solving. Problems that would allow students to discover something new. For unmotivated students, the process is important, they want to build and draw geometric shapes with their own hands, and it is necessary to meet their expectations especially in the very first lessons, offer to draw them ornaments that include various geometric shapes, and then the emotional start of these lessons will be ensured. The first lesson is important; it, like a tuning fork, sets the tone for the entire work.
I would like to emphasize one thing: now, when there is no compulsory exam in geometry, perhaps the pursuit of knowledge should not turn a child away from such a beautiful, incredibly useful science as geometry? Maybe work in peace once in your life. So that the sword of Damocles of marks and assessments does not hang over you. So that in the lesson the teacher and student are equal in knowledge and potential. To have GEOMETRY.
What problems in elementary mathematics are considered the most difficult? Probably most readers will answer: geometric. Why? Yes, because in algebra, trigonometry, and the beginnings of mathematical analysis, entire series of solution algorithms have been developed typical tasks. If there is an algorithm, then there is a program of action, and therefore difficulties, if they occur, are most often of a technical rather than fundamental nature.
Geometric problems are a different matter. As a rule, there are no algorithms for solving them, and choosing the most suitable one this case a theorem from an extensive list of theorems is not easy. Therefore, the main recipe is more philosophical than didactic in nature: if you want to learn how to solve geometric problems, solve them! However, there are some general principles that are useful to know when solving geometric problems. About these general provisions we would like to talk.
When solving geometric problems, three main methods are usually used: geometric- when the required statement is derived using logical reasoning from a number of well-known theorems; algebheavenly- when the desired geometric quantity is calculated based on various dependencies between elements of geometric shapes directly or using equations; combined- when at some stages the solution is carried out by the geometric method, and at others - by the algebraic method.
Whatever solution path is chosen, the success of its use depends, naturally, on knowledge of the theorems and the ability to apply them. Without considering all the theorems of planimetry here, let us pay attention to those that, on the one hand, are actively used in solving problems, but, on the other hand, as experience shows, are not always “at the first level of memory” among students. You need to love these theorems, make them your assistants, so that your students give preference to them.
Let's voice these theorems and show how they work using specific problems.
When solving problems, as a rule, individual stages of reasoning are recorded. This is done for convenience, to make it easier to follow the progress of reasoning. And I would also like to note: the tasks will be of varying difficulty, but those that are most useful to the teacher from a methodological point of view.
TRIANGLES AND QUADRAGONS.
When solving problems about triangles and quadrilaterals, let us pay attention to the following theorems:
THEOREM 1. Equality of angles with mutually perpendicular sides:
If both are sharp or both are obtuse and , then 
.
THEOREM 2. Properties of the midline of a trapezoid:
A) middle line the trapezoid is parallel to the bases of the trapezoid;
B) the middle line is equal to half the sum of the bases of the trapezoid;
C) the middle line (and only it) bisects any segment enclosed between the bases of the trapezoid.
These properties are also valid for the midline of a triangle, if we consider the triangle to be a “degenerate” trapezium, one of the bases of which has a length equal to zero.
THEOREM 3. On the intersection points of medians, bisectors, altitudes of a triangle:
A) three medians of a triangle intersect at one point (it is called the center of gravity of the triangle) and divide at this point in a ratio of 2: 1, counting from the vertex;
B) three bisectors of a triangle intersect at one point;
C) three altitudes intersect at one point (it is called the orthocenter of the triangle).
THEOREM 4. Property of the median in a right triangle:
in a right triangle, the median drawn to the hypotenuse is equal to half of it.
The converse theorem is also true: if in a triangle one of the medians is equal to half the side to which it is drawn, then this triangle is right angled
THEOREM 5. property of the bisector of the interior angle of a triangle:
The bisector of an internal angle of a triangle divides the side to which it is drawn into parts proportional to the opposite sides:
THEOREM 6. Metric relations in a right triangle:
Ifa andb – legs,c – hypotenuse,h is the height, and are the projections of the legs onto the hypotenuse, then: a) ; b) ; V) ; G) ; d)
THEOREM 7. Determination of the type of triangle based on its sides:
Leta,b,c are the sides of the triangle, with c being the largest side; Then:
A) if , then the triangle is acute;
B) if , then the triangle is right-angled;
C) if , then the triangle is obtuse.
THEOREM 8. Metric relations in a parallelogram:
The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides:
.
When solving geometric problems, you often have to establish the equality of two segments (or angles). Let's indicate three main ways of geometrically proving the equality of two segments:
1) consider the segments as sides of two triangles and prove that these triangles are equal;
2) represent the segments as sides of a triangle and prove that this triangle is isosceles;
3) replace the segment A an equal segment https://pandia.ru/text/78/456/images/image008_12.gif" width="17" height="19 src=">and prove the equality of the segments and .
Task 1.Two mutually perpendicular lines intersect the sidesAB,B.C.CD,AD squareABCD at pointsE,F,K,L accordingly. Prove thatEK =FL (see figure for task No. 1).
Solution:
1.
Using the first of the above paths for the equality of two segments, we draw the segments and - then the segments of interest to us E.K. And FL become sides of two right triangles EPK And FML(see figure for task No. 1) .
2
. We have: PK =FM(more details: PK =A.D.AD=AB,AB =FM meansPK =FM),(as angles with mutually perpendicular sides, Theorem 1). This means (along the leg and acute angle). From the equality of right triangles follows the equality of their hypotenuses, i.e., segments E.K. And FL. ■
Note that when solving geometric problems, you often have to make additional constructions, for example the following: drawing a straight line parallel or perpendicular to one of those in the figure (as we did in task 1); doubling the median of the triangle in order to complete the triangle to a parallelogram (we will do this in Problem 2), drawing an auxiliary bisector. There are useful additional constructions related to the circle.
Task 2.The sides are equala,b,c. Calculate the median drawn to side c (see figure for problem 2).
Solution:
Let us double the median, building it up to the parallelogram ACVR, and apply Theorem 8 to this parallelogram. We obtain: , i.e. 
, where we find: 
Task 3.Prove that in any triangle the sum of the medians is greater than ¾ of the perimeter, but less than the perimeter.
Solution:
1.
Consider https://pandia.ru/text/78/456/images/image036_6.gif" width="131" height="41">; 
. Because AM + MS > AC, That
https://pandia.ru/text/78/456/images/image039_4.gif" alt=" Signature:" align="left" width="148" height="32">Проведя аналогичные рассуждения для треугольников АМВ и ВМС, получим:!}
https://pandia.ru/text/78/456/images/image041_3.gif" width="111" height="41 src="> (3)
Adding inequalities (1), (2), (3), we obtain: 
,
i.e. we have proven that the sum of the medians is greater than ¾ of the perimeter.
2. Let's double the median BD, completing the triangle to a parallelogram (see figure for problem 3)..gif" width="80" height="24 src="> (4)
Similarly: https://pandia.ru/text/78/456/images/image039_4.gif" alt=" Caption: Fig. for task No. 3" align="left hspace=12" width="148" height="32"> (6)!}
Adding inequalities (4), (5), (6), we get: https://pandia.ru/text/78/456/images/image049_2.gif" align="left" width="159" height="93 "> Solution: Let DIA be a right triangle, https://pandia.ru/text/78/456/images/image051_2.gif" width="233" height="21"> (see figure for problem 4).
1. as angles with mutually perpendicular sides (https://pandia.ru/text/78/456/images/image054_2.gif" alt=" Signature:" align="left" width="148" height="33">!} 2. Since (see Theorem 4), then SM = MV, and then from this we conclude that So,
3. Since and (after all, CD is a bisector), that is what needed to be proved. ■
Task 5. In a parallelogram with sidesa Andb bisectors of internal angles are drawn (see figure for problem 5). Find the lengths of the diagonals of the quadrilateral formed at the intersection of the bisectors.
Solution:
1
..gif" width="27 height=17" height="17">(see figure). since in a parallelogram 
i.e. then This means that in triangle ABC the sum of angles A and B is equal to 900, then angle K is equal to 900, i.e. the bisectors AE and BP are mutually perpendicular.
Similarly, the mutual perpendicularity of the bisectors AE and DQ, BP and CF, CF and DQ is proved.
OUTPUT: KLMN is a quadrilateral with right angles, i.e. a rectangle. A rectangle has equal diagonals, so it is enough to find the length of one of them, for example KM.
2.
Let's consider He has AK - both the bisector and the height. This means, firstly, the triangle ABP is isosceles, i.e. AB = AP = b, and, secondly, that the segment AK is simultaneously the median of the triangle ABP, i.e. K is the midpoint of the bisector BP.
It is proved in a similar way that M is the midpoint of the bisector DQ.
3. Consider the segment KM. It bisects the segments BP and DQ. But the middle line of a parallelogram (note that a parallelogram is a special case of a trapezoid; if we can talk about the middle line of a trapezoid, then we can equally well talk about the middle line of a parallelogram, which has the same properties) passes through points K and M (see theorem 2). This means that KM is a segment on the midline, and therefore .
4. Since and , then KMDP is a parallelogram, and therefore
Answer: ■
In fact, in the process of solving the problem (at stages 1 and 2), we proved a rather important property: the bisectors of the angles adjacent to the side of the trapezoid intersect at right angles at a point lying on the midline of the trapezoid.
It should be noted that the main method of composing equations in geometric problems is method supporting element, which is as follows: the same element (side, angle, area, radius, etc.) is expressed through known and unknown quantities by two different ways and the resulting expressions are equated.
Quite often, an area is chosen as a reference element figures. Then we say that to construct the equation we use area method.
It is necessary to teach schoolchildren how to solve basic problems, i.e. technical problems. Which are included as components in many other tasks. These are, for example, problems of finding the basic elements of a triangle: median, height, bisector, inscribed and circumscribed circle radii, area.
Task 6. In triangle ABC, sides AB and BC are equal, and BH is the height. A point is taken on the BC sideD so that (see figure for problem 6). In what ratio is the segmentAD divides the height of VN?
Solution: 1. Let BD = a, then CD = 4 a, AB = 5a.
2.
Let's draw a segment (see figure for problem 6) Since NK is the middle line of the triangle ACD DK = KC = 2
a .
3. Consider the triangle VNK. We have: BD = a,
DK = 2a and https://pandia.ru/text/78/456/images/image080_2.gif" width="84" height="41"> but 
This means that ■
If the problem requires finding the ratio of some - or quantities, then, as a rule, the problem is solved using the auxiliary parameter method. This means that at the beginning of solving the problem we declare some linear quantity known, denoting it, for example, with the letter A, and then express it through A those quantities whose ratio is required to be found. When the required relation is compiled, the auxiliary parameter A is shrinking. This is exactly how we acted in the problem . Our advice: when solving problems in which it is necessary to find the ratio of quantities (in particular, in problems of determining an angle - after all, as a rule, when calculating an angle we are talking about finding its trigonometric function, i.e. about the relationship of the parties right triangle), students should be taught to highlight the introduction of an auxiliary parameter as the first stage of solving. The auxiliary parameter method is also used in problems where geometric figure determined up to similarity.
Task 7. A rectangle is inscribed in a triangle with sides equal to 10, 17 and 21 cm so that its two vertices are on one side of the triangle, and the other two vertices are on the other two sides of the triangle. Find the sides of the rectangle if it is known that its perimeter is 22.5 cm.
Solution.
1.
First of all, let's determine the type of triangle. We have: 102 = 100; 172 = 289; 212 = 441. Since 212 > 102 + 172, the triangle is obtuse (see Theorem 7), which means that a rectangle can be inscribed in it only in one way: by placing its two vertices on the larger side of the triangle ABC (see the figure for Problem 7 ), where AC = 21 cm, AB = 10 cm, BC = 17 cm.
2. Find the height ВН of triangle ABC. BH = 8 cm.
3. Let's put ED=x. Then EF = 11.25 –x(since the perimeter of the rectangle DEFK equal to 22.5 cm), BP = 8 – x. Triangles BEF and ABC are similar, which means (in similar triangles the ratio of the corresponding heights is equal to the similarity coefficient), i.e. 
from where we find x = 6.
Answer: 6 cm, 5.25 cm. ■
When solving the problem, we used the statement that in similar triangles not only the sides, but also the corresponding heights are proportional. A more general factor is the following, which is, so to speak, a generalized similarity theorem:
If two triangles are similar, then any line element (or sum of line elements) of one triangle is related to the corresponding line element (or sum of corresponding line elements) of the other triangle as corresponding sides.
In particular, the radii of circumscribed or inscribed circles, perimeters, corresponding altitudes, medians, and bisectors of two similar triangles are related as corresponding sides.
Task 8.In triangle ABC, angle A is 2 times larger than angle C, side BC is 2 cm larger than side AB, and AC = 5 cm. Find AB and BC.
Solution.
1.
Let's draw the bisector AD of angle A..gif" alt=" Signature:" align="left" width="148" height="33">!} 3.
Triangles ABC and ABC are similar, since the angle B of these triangles is common. From the similarity of triangles we conclude that 
i.e. 
4. To find X And at a system of two equations with two unknowns is obtained: 
where 
Subtracting the second equation from the first, we get 5y – 10 = 2y, i.e. y = . This means i.e.x=4.
Answer: AB = 4 cm; BC = 6 cm. ■
Very often, when composing the relations of corresponding sides in similar triangles in non-trivial cases (trivial cases of similarity were in problems 6 and 7 - the triangle was cut off from the latter by a straight line parallel to one of its sides), those who solve the problem. They make purely technical mistakes: either they confuse the order of the triangles (which one is first and which is second), or they unsuccessfully choose pairs of sides as corresponding ones. Our advice: if the similarity of triangles ABC and DEF is established, then we recommend proceeding as follows: “drive” the sides of one triangle into the numerators, for example like this: 
Considering that the corresponding sides in similar triangles are those that lie opposite equal angles, find the simplest pairs of corresponding sides; if these are AB and DE, BC and DF, then write: https://pandia.ru/text/78/456/images/image100_1.gif" align="left" width="121" height="96 src=" >b) so that you can fit approx.circumference, it is necessary and sufficient that the sums of the lengths of its opposite sides be equal.
THEOREM 5. Metric ratios in a circle:
https://pandia.ru/text/78/456/images/image103_2.gif" alt=" Signature: Fig. 2" align="left" width="76" height="29">!}
https://pandia.ru/text/78/456/images/image105_2.gif" alt=" Signature: Fig. 3" align="left" width="76" height="28">!}
https://pandia.ru/text/78/456/images/image107_1.gif" width="13 height=19" height="19">, hypotenuse - c (see figure). Calculate the radius r of the inscribed circle.
Solution. 1. From the center O of the inscribed circle, draw radii to the points of its tangency with the sides of the triangle; taking into account that they are perpendicular to the corresponding sides (see Theorem 1, a), and then using Theorem 1, b, we mark pairs of equal segments: CD= SE, AE= AF,BD =B.F.(see picture).
2. 
Because EODC- square (corners E,D, C - straight and EU= CD), then OE =O.D.= CD = CE= r. Then BD= A -r, AE =b –r And ,
respectively, BF=BD = a– r,AF=AE =b–r.
3. Since AB= AF+FB, That c = (b –r) + (a –r), from where .■
Note that if the problem concerns a circle inscribed in a triangle (or quadrilateral), then it is almost always advisable to draw the radii at the points of contact of the circle with the sides, taking into account that the radii will be perpendicular to the corresponding sides, and immediately mark pairs of equal segments in the drawing (for two tangents drawn to the circle from a given point). This is what we did when solving the above problem.
Let's pay attention to the formula https://pandia.ru/text/78/456/images/image110_1.gif" width="43" height="44">, where S is the area, R– semiperimeter of a triangle.
Regarding the radius R circle circumscribed about a triangle, then for a right triangle (the hypotenuse is the diameter of a circle circumscribed about a right triangle), for a non-right triangle, the formula is usually used https://pandia.ru/text/78/456/images/image114_1.gif" width="59 " height="41 src=">.
Problem 10. Given a rectangular circular sector.A circle of the same radius is drawn with the center at the end of the sector arc; it divides the sector into two curvilinear triangles. A circle is inscribed in the smaller of these triangles (see figure). Find the ratio of the radii of the inscribed circle and the sector.
Solution.
1. Let us carry out the necessary additional constructions, which are usually done when it comes to the internal or external tangency of circles or the tangency of a circle and a straight line: O2O3– line of centers; IN- point of contact; O1O3– line of centers; A– point of contact; O3C O1C; WITH– point of contact (see figure).
https://pandia.ru/text/78/456/images/image119_1.gif" width="43" height="41">. So, .
Answer: . ■
Let us give two more additions about useful additional constructions: 1) if two circles touch (internally or externally), then it is necessary to draw a line of centers, that is, a straight line passing through the centers of the tangent circles, and take into account that the point of contact lies on the line of centers (this is what we did when solving the above problem, which was the key to success); 2) sometimes it is useful (as additional constructions) to make a so-called “remote” drawing, i.e., take out a fragment of an existing rather complex drawing separately for special study (for example, when solving a problem, we took out a separate fragment containing ∆ O1O2O3– see fig.).
Problem 11. Circle radiusR passes through two adjacent vertices A andD square (see figure). The segment BM tangent to the circle drawn from the third vertex B of the square is twice the side of the latter. Find the side of the square.
Solution.
Let us introduce the notation VA= x, VM = 2x. Let's continue the segment VA until it intersects the circle at the point TO. Then VK ∙ VA = VM2(see Theorem 5, c), i.e. VK ∙ x= 4x2, where we find: VC= 4x- Means, AK= Zx. Further, KAD =
= 90°, which means KD– diameter of the circle. From a right triangle ADK we find: AD2+ AK2= KD2, i.e. x2+9x2= 4R 2, where from X= https://pandia.ru/text/78/456/images/image125_0.gif" width="45" height="45 src=">. ■
The orthocenter, i.e., the point of intersection of the altitudes of a triangle, has a number of interesting properties: the orthocenter of an acute-angled triangle coincides with the center of a circle inscribed in a triangle, the vertices of which are the bases of the altitudes of a given triangle; in a non-right triangle ABC, the distance from the orthocenter to vertex B is twice the distance from the center of the circumscribed circle around the triangle to side AC. We use the last property to introduce the concept of the Euler straight line. For visual reasons, we will limit ourselves to an acute-angled triangle.
So let N– orthocenter, O – circumcenter, O.D. AC,OD║BH,AD= DC(see picture).
Let's draw the median BD and segment HE. Triangles VNM And MOD similar, which means https://pandia.ru/text/78/456/images/image128_0.gif" width="56" height="41 src=">.gif" width="17" height="16 src =">C = 90°, then Euler’s straight line is a straight line passing through vertex C right angle and middle ABOUT hypotenuse AB, i.e. the median.
Let's continue the conversation about solving planimetric problems. Let's move on to solving problems related to the concept of area of a plane figure.
Let us begin, as in previous cases, by identifying “working” theorems. There are two such theorems on calculating areas.
THEOREM 1. The ratio of the areas of similar figures is equal to the square of the similarity coefficient.
THEOREM 2. A) If two triangles have equalbases, then their areas are related to their heights.
b) If two triangles have equal heights, then theirareas are treated as bases.
And, of course, it makes sense to give the basic formulas for calculating the areas of plane figures.
1. Formulas for the area of a triangle:
a) https://pandia.ru/text/78/456/images/image131_0.gif" width="84" height="41 src=">; c) ;
d) S = Rr, Where R=; R– radius of the circumscribed circle; r- radius of the inscribed circle;
e) S = https://pandia.ru/text/78/456/images/image136_0.gif" align="left hspace=12" width="159" height="139"> a) S= A.C.∙ BDsin;
THEOREM 1.Equality of angles with mutually perpendicular sides:If 
both sharp or both obtuse and 
, 
, That 
.
THEOREM 2. Properties of the midline of a trapezoid:A) the midline of the trapezoid is parallel to the bases of the trapezoid;B) the middle line is equal to half the sum of the bases of the trapezoid;C) the middle line (and only it) bisects any segment enclosed between the bases of the trapezoid. These properties are also valid for the midline of a triangle, if we consider the triangle to be a “degenerate” trapezium, one of the bases of which has a length equal to zero. THEOREM 3. On the intersection points of medians, bisectors, altitudes of a triangle:A) three medians of a triangle intersect at one point (it is called the center of gravity of the triangle) and divide at this point in a ratio of 2: 1, counting from the vertex;B) three bisectors of a triangle intersect at one point;C) three altitudes intersect at one point (it is called the orthocenter of the triangle).THEOREM 4. Property of the median in a right triangle:in a right triangle, the median drawn to the hypotenuse is equal to half of it. The converse theorem is also true: if in a triangle one of the medians is equal to half the side to which it is drawn, then this triangle is right angledTHEOREM 5. property of the bisector of the interior angle of a triangle:The bisector of an internal angle of a triangle divides the side to which it is drawn into parts proportional to the opposite sides: 
THEOREM 6. Metric relations in a right triangle:IfaAndb– legs,c– hypotenuse,h- height, 
And 
- projections of the legs onto the hypotenuse, then: a) 
; b) 
; V) 
; G) 
; d) 
THEOREM 7. Determination of the type of triangle based on its sides:Leta,
b,
c– sides of the triangle, with c being the largest side; Then:And if 
, then the triangle is acute;B) if 
, then the triangle is right-angled;B) if 
, then the triangle is obtuse.THEOREM 8. Metric relations in a parallelogram:The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of all its sides:
.
When solving geometric problems, you often have to establish the equality of two segments (or angles). Let's indicate three main ways of geometrically proving the equality of two segments:
1) consider the segments as sides of two triangles and prove that these triangles are equal; 2) represent the segments as sides of a triangle and prove that this triangle is isosceles; 3 
) replace the segment A an equal segment 
, and the segment b –
equal to it and prove the equality of the segments and . Task 1.Two mutually perpendicular lines intersect the sidesAB,
B.C.,
CD,
ADsquareABCDat pointsE,
F,
K,
Lrespectively. Prove thatE.K. =
FL(see figure for task No. 1).R 
Rice. to task No. 1
Solution: 1.
Using the first of the above paths for the equality of two segments, we draw the segments 
And 
- then the segments we are interested in E.K.
And
FL
become sides of two right triangles EPK And FML(see figure for task No. 1) . 2
Rice. to task No. 1
We have: PK =
FM(more details: PK =
AD,
AD =
AB,
AB =
FM, Means,PK =
FM),
(as angles with mutually perpendicular sides, Theorem 1). This means (along the leg and acute angle). From the equality of right triangles it follows that their hypotenuses are equal, i.e. segments E.K. And FL. ■ Note that when solving geometric problems, you often have to make additional constructions, for example the following: drawing a straight line parallel or perpendicular to one of those in the figure (as we did in task 1); doubling the median of the triangle in order to complete the triangle to a parallelogram (we will do this in Problem 2), drawing an auxiliary bisector. There are useful additional constructions related to the circle. Task 2.Parties 
equala,
b,
c. Calculate median 
, drawn to side c. (see figure for problem 2).R 
Rice. to problem No. 2
Solution: Double the median by completing 
to the parallelogram ACVR, and apply Theorem 8 to this parallelogram. We obtain: , i.e. 
, where we find: 
Task 3.Prove that in any triangle the sum of the medians is greater than ¾ of the perimeter, but less than the perimeter.R 
solution:
1.
Let's consider 
(see figure for problem 3) We have: 
; 
. Because AM + MS > AC, That 
(1) P
Rice. to problem No. 3
Carrying out similar reasoning for triangles AMB and BMC, we obtain: 
(2)
(3) Adding inequalities (1), (2), (3), we obtain: 
, T 
.e. we have proven that the sum of the medians is greater than ¾ of the perimeter. 2.
Let's double the median BD by building the triangle into a parallelogram (see the figure for Problem 3). Then from 
we get: B.K. <
B.C. +
CK,
those. 
(4)
Likewise: 
(5)
Rice. to problem No. 3
(6) Adding inequalities (4), (5), (6), we obtain: , i.e. the sum of the medians is less than the perimeter. ■ Task 4.Prove that in a non-isosceles right triangle, the bisector of a right angle bisects the angle between the median and the altitude drawn from the same vertex.R 
solution:
Let ACB be a right triangle, 
, CH – height, CD – bisector, SM – median. Let us introduce the following notation: (see figure for problem 4) . 1.
as angles with mutually perpendicular sides (). 2
Rice. to problem No. 4
Because 
(see Theorem 4), then SM = MV, and then from 
we conclude that 
So, 3.
Since and (after all, CD is a bisector), that is what needed to be proved. ■ Task 5.In a parallelogram with sidesa
Andbbisectors of internal angles are drawn (see figure for problem 5). Find the lengths of the diagonals of the quadrilateral formed at the intersection of the bisectors.Solution:
1
. AE – bisector 
, BP – bisector 
(see figure) . since in a parallelogram 
those. then This means that in triangle ABC the sum of angles A and B is equal to 90 0, then angle K is equal to 90 0, i.e., the bisectors AE and BP are mutually perpendicular. A 
The mutual perpendicularity of the bisectors AE and DQ, BP and CF, CF and DQ is proved logically. OUTPUT: KLMN is a quadrilateral with right angles, i.e. rectangle. A rectangle has equal diagonals, so it is enough to find the length of one of them, for example KM. 2
Rice. to problem No. 5
Let's consider 
He has AK - both bisector and height. This means, firstly, the triangle ABP is isosceles, i.e. AB = AP = b, and, secondly, that the segment AK is at the same time the median of the triangle ABP, i.e. K – the middle of the bisector BP. It is proved in a similar way that M is the midpoint of the bisector DQ. 3.
Consider the segment KM. It bisects the segments BP and DQ. But the middle line of a parallelogram (note that a parallelogram is a special case of a trapezoid; if we can talk about the middle line of a trapezoid, then we can equally well talk about the middle line of a parallelogram, which has the same properties) passes through points K and M (see theorem 2). This means that KM is a segment on the midline, and therefore 
.4.
Because 
And 
, then KMDP is a parallelogram, and therefore. Answer:
■ In fact, in the process of solving the problem (at stages 1 and 2), we proved a rather important property: the bisectors of angles adjacent to the side of a trapezoid intersect at right angles at a point lying on the midline of the trapezoid. It should be noted that the main method of composing equations in geometric problems is methodsupporting element, which is as follows: the same element (side, angle, area, radius, etc.) is expressed through known and unknown quantities in two different ways and the resulting expressions are equated. Quite often, an area is chosen as a reference elementfigures.
Then we say that to construct the equation we use area method. It is necessary to teach schoolchildren how to solve basic problems, i.e. those. Which are included as components in many other tasks. These are, for example, problems of finding the basic elements of a triangle: median, height, bisector, inscribed and circumscribed circle radii, area. Z 
problem 6.In triangle ABC, sides AB and BC are equal, and BH is the height. A point is taken on the BC sideDSo 
(see figure for problem 6). In what ratio is the segmentADdivides the height of the VN?Solution:
1.
Let BD = a,
then CD = 4
a, AB = 5a.2
Rice. to problem No. 6
Let's draw a segment 
(see figure for problem 6) Since NK is the middle line of the triangle ACD DK = KC = 2
a
.3.
Consider the triangle VNK. We have: BD = a,DK = 2
a And 
. According to Thales' theorem 
But 
That means 
■ If the problem requires finding the ratio of any number of quantities, then, as a rule, the problem is solved using the auxiliary parameter method. This means that at the beginning of solving the problem we declare some linear quantity known, denoting it, for example, with the letter A, and then express it through A those quantities whose ratio is required to be found. When the required relation is compiled, the auxiliary parameter A is shrinking. This is exactly how we acted in the problem . Our advice:
when solving problems in which it is necessary to find the ratio of quantities (in particular, in problems on determining an angle - after all, as a rule, when calculating an angle we are talking about finding its trigonometric function, i.e. about the ratio of the sides of a right triangle), students should be taught The first stage of the solution is the introduction of an auxiliary parameter. The auxiliary parameter method is also used in problems where a geometric figure is defined up to similarity. Task 7.A rectangle is inscribed in a triangle with sides equal to 10, 17 and 21 cm so that its two vertices are on one side of the triangle, and the other two vertices are on the other two sides of the triangle. Find the sides of the rectangle if it is known that its perimeter is 22.5 cm.R 
decision
.
1.
First of all, let's determine the type of triangle. We have: 10 2 = 100; 17 2 = 289; 21 2 = 441. Since 21 2 > 10 2 + 17 2, the triangle is obtuse-angled (see Theorem 7), which means that a rectangle can be inscribed in it only in one way: by placing its two vertices on the larger side of triangle ABC (see fig. . to problem 7), where AC = 21 cm, AB = 10 cm, BC = 17 cm. 2
53.Angles (internal angles) of a triangle three angles are called, each of which is formed by three rays emerging from the vertices of the triangle and passing through the other two vertices.
54. Triangle Angle Sum Theorem. The sum of the angles of a triangle is 180°.
55. External corner of a triangle is an angle adjacent to some angle of this triangle.
56. External corner triangle equal to the sum two angles of a triangle that are not adjacent to it.
57. If all three corners triangle spicy, then the triangle is called acute-angled. 
58. If one of the corners triangle blunt, then the triangle is called obtuse-angled. 
59. If one of the corners triangle straight, then the triangle is called rectangular. 
60. The side of a right triangle lying opposite the right angle is called hypotenuse(Greek word gyipotenusa - “contracting”), and two sides forming a right angle - legs(Latin word katetos - “plumb”) .
61. Theorem on the relationships between the sides and angles of a triangle. In a triangle the larger angle is opposite the larger side, and back, The larger side lies opposite the larger angle.
62. In a right triangle The hypotenuse is longer than the leg.
because The larger side always lies opposite the larger angle.
Signs of an isosceles triangle.
If in a triangle two angles are equal, then it is isosceles;
If in a triangle the bisector is the median or height,
then this triangle is isosceles;
If in a triangle the median is the bisector or height, That
this triangle is isosceles;
If in a triangle height is median or bisector,
then this triangle is isosceles.
64. Theorem. Triangle inequality. The length of each side of a triangle is greater than the difference and less than the sum of the lengths of the other two sides:
Properties of the angles of a right triangle.
Sum of two sharp corners of a right triangle is 90°.
A + B = 90°
66. Right Triangle Property.
A leg of a right triangle opposite an angle of 30° is equal to half the hypotenuse.
If/
A = 30°, then BC = ½ AB
67. Properties of a right triangle.
a) If a leg of a right triangle is equal to half the hypotenuse, then the angle opposite this leg is 30°.
If BC = ½ AB, then /
B = 30°
B) The median drawn to the hypotenuse is equal to half the hypotenuse.
median CF = ½ AB
Sign of equality of right triangles on two sides.
If the legs of one right triangle are respectively equal to the legs of another, then such triangles are congruent.
For angles with respectively parallel sides the following propositions are valid:
1. If sides a and b of one angle are respectively parallel to sides a and b of another angle and have the same directions as them, then the angles are equal.
2. If, under the same condition of parallelism, sides a and b are adjusted opposite to sides a and b, then the angles are also equal.
3. If, finally, the sides a and are parallel and in the same direction, and the sides are parallel and opposite in direction, then the angles complement each other until they are reversed.
Proof. Let us prove the first of these propositions. Let the sides of the angles be parallel and equally directed (Fig. 191). Let's connect the vertices of the corners with a straight line.
In this case, two cases are possible: the straight line passes inside the corners or outside these corners (Fig. 191, b). In both cases the proof is obvious: so, in the first case
but where do we get it from? In the second case we have
and the result again follows from the equalities
We leave the proofs of Propositions 2 and 3 to the reader. We can say that if the sides of the angles are respectively parallel, then the angles are either equal or add up to the opposite angle.
Obviously, they are equal if both are simultaneously acute or both are obtuse, and their sum is equal if one of them is acute and the other is obtuse.
Angles with correspondingly perpendicular sides are equal or complementary to each other up to a straight angle.
Proof. Let a be some angle (Fig. 192), and O be the vertex of the angle formed by straight lines; therefore, let there be any of the four angles formed by these two straight lines). Let us rotate the angle (i.e., both of its sides) around its vertex O at a right angle; we obtain an angle equal to it, but one whose sides are perpendicular to the sides of the rotated angle indicated in Fig. 192 through They are parallel to the straight lines forming a given angle a. Therefore, angles mean that angles are either equal or form a complete angle in total.
An angle is a part of a plane bounded by two rays emanating from one point. The rays that limit the angle are called the sides of the angle.
The point from which the rays emerge is called the vertex of the angle. Corner designation scheme
Let's look at the example of the angle shown in Figure 1.
The angle shown in Figure 1 can be designated in three ways:
Angles are called equal angles if they can be combined. If the intersection of two lines produces four equal angles , then such angles are called right angles (Fig. 2). Intersecting straight lines that form right angles are called.
perpendicular lines If through a point A, not lying on a line l, a line is drawn perpendicular to line l and intersecting line l to point B, then they say that from the point B perpendicular AB is dropped onto line l (Fig. 3). Point B is called.
Note. The length of segment AB is called distance from point A to straight line l.
Angle of 1° (one degree) called the angle that makes up one ninetieth part right angle.
An angle k times greater than an angle of 1° is called an angle of k° (k degrees).
Angles are also measured in radians. You can read about radians in the section of our reference book “Measuring angles. Degrees and radians".
Table 1 - Types of angles depending on the value in degrees
| Drawing | Types of angles | Properties of corners |
 | Right angle | A right angle is 90° |
 | Sharp corner | Acute angle less than 90° |
 | Obtuse angle | Obtuse angle greater than 90° but less than 180° |
 | Straight angle | The rotated angle is 180° |
 | This angle is greater than 180° but less than 360° | |
 | Full Angle | Full angle is 360° |
 | Angle equal to zero | This angle is 0° |
| Right angle |
Property: A right angle is 90° |
| Sharp corner |
Property: Acute angle less than 90° |
| Obtuse angle |
Property: Obtuse angle greater than 90° but less than 180° |
| Straight angle |
Property: The rotated angle is 180° |
| Angle greater than straight |
Property: This angle is greater than 180° but less than 360° |
| Full Angle |
Property: Full angle is 360° |
| Angle equal to zero |
Property: This angle is 0° |
Table 2 - Types of angles depending on the location of the sides
| Drawing | Types of angles | Properties of corners |
 | Vertical angles | Vertical angles are equal |
 | Adjacent angles | The sum of adjacent angles is 180° |
 | Angles with respectively parallel sides are equal if both are acute or both are obtuse | |
 | The sum of angles with correspondingly parallel sides is equal to 180°, if one of them is acute and the other is obtuse | |
 | Angles with respectively perpendicular sides are equal if both are acute or both are obtuse | |
 | The sum of angles with correspondingly perpendicular sides is equal to 180°, if one of them is acute and the other is obtuse |
| Vertical angles |
Property of vertical angles: Vertical angles are equal |
| Adjacent angles |
Property of adjacent angles: The sum of adjacent angles is 180° |
| Angles with correspondingly parallel sides |
Angles with respectively parallel sides are equal if both are acute or both are obtuse |
Property of angles with correspondingly parallel sides: The sum of angles with correspondingly parallel sides is equal to 180°, if one of them is acute and the other is obtuse |
| Angles with correspondingly perpendicular sides |
Angles with respectively perpendicular sides are equal if both are acute or both are obtuse |
Property of angles with correspondingly perpendicular sides: The sum of angles with correspondingly perpendicular sides is equal to 180°, if one of them is acute and the other is obtuse |
Definition . The bisector of an angle is the ray that bisects the angle.
Task . Prove that the bisectors of adjacent angles are perpendicular.
Solution . Consider Figure 4.
In this figure, angles AOB and BOC are adjacent, and rays OE and OD are bisectors of these angles. Because the
2α + 2β = 180°.
Q.E.D.
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