Each of the segments av and cd. Comparison of segments
7. Many points and lines are placed on a plane. Accept that you can construct points and straight lines on a plane; In practice, a ruler is used to construct a straight line.
The straight line stretches endlessly in both directions. For the hell of it. 4 straight line AB is constructed; with your imagination you can continue it endlessly in both directions. If you construct any point, for example, point O, on the straight line CD (Figure 4), then the straight line will be divided into 2 parts: one part stretches from point O to the right without end, and the other from point O to the left without end. Each of these parts is called a ray. Here we have 2 beams: the OD beam and the OC beam.
We can construct countless rays through each point.
If we take 2 points on a straight line, for example, on the straight line KL (Figure 4) points E and F, then the part of the straight line between these points is called a segment. In the drawing we have segment EF.
8. Compare 2 segment data AB and CD (draft 5).
Let's move the segment CD so that point C hits A, and rotate it around point A until the segment CD goes along the segment AB. When we achieve this, we note where point D falls: if it falls into B, then our segments are equal; if D falls somewhere between points A and B (for example, in M), then the segment CD is considered less than the segment AB, and if the point D falls behind the point B (for example, in N), then the segment CD is greater than the segment AB.
We understand “comparing” two segments in the sense of determining whether they are equal or one is greater than the other.
9. Find the sum of two given segments.
Two segments AB and CD are taken (Fig. 6); you need to add these segments.
To do this, we move the segment CD so that point C hits B, and then rotate it around B until it follows the continuation of segment AB. Note where point D falls; if it hits K, then the segment BK = CD and AK = AB + BK or AK = AB + CD.
Any segment can be divided by intermediate points into the sum of several terms; eg:
AB = AC + CD + DE + EF + FB (drawing 7)
It is clear to us that the sum of the segments does not change depending on the rearrangement of the terms .
10. Find the difference between two segments.
Given two segments AB and CD (Fig. 8); you need to subtract the smaller segment CD from the larger segment AB.
We move the segment CD so that point D hits point B, and we begin to rotate it around B until it goes in the direction of BA; Let us note, when we achieve this, where point C will fall. If C falls into K, then KB = CD and AK = AB – KB or AK = AB – CD.
You can multiply this segment by 2, 3, 4, etc., i.e., repeat it as a term 2, 3, etc. times.
From paragraphs. 8-10, it is important for us to understand that 1) the following concepts are applicable to segments, as well as to numbers: “equal,” “greater than,” and “less than”; 2) the concepts of “the sum and difference of two segments” have a very definite meaning.
In practice, to construct a segment equal to a given one, a compass is used.
11. Exercises. 1. Name the summand segments and their sum in each of the following images; write down (drawing A).
2. On the same drawings, indicate which segment can be considered the difference of two other segments; write down.
3. Divide this segment into 2, 3, and 4 terms; write down.
4. Present this segment as the difference of two other segments.
12. We can build a figure consisting of two rays emanating from one point, – such a figure is called an angle. For the hell of it. Figure 9 shows an angle consisting of rays OA and OB emanating from point O. This point is called the vertex of the angle, and each ray is called its side. The word “angle” is replaced with the sign ∠. An angle is called by three letters, one of which is placed at the vertex, and the other two somewhere on the sides of the angle - the letter at the vertex is placed in the middle of the name of the angle. For the hell of it. 9 we have ∠AOB or ∠BOA; sometimes an angle is called one letter placed at its vertex, saying ∠O. The sides of the angle (rays) must be considered to go without end.
A special case of an angle will arise when its sides form one straight line; such a special angle is called straightened or turned angle(Figure 12 shows right angles AOB and A 1 O 1 B 1).
Each angle divides the plane into 2 parts, into two regions. One of these parts is called internal area corner and say that it lies inside the corner, and the other is called external area corner and say that it lies outside the corner. Which of these two parts is called the external region and which internal is a matter of condition. Each time you should mark something internal, for example, an area. We will mark the inside area of the corner with curved lines drawn on the inside area between the sides of the corner; on black 10 marks the internal regions of the angles ABC, DEF and straightened ∠KLM.
It is useful to cut out corners from a sheet of thin cardboard: a piece of cardboard is a rough representation of part of the plane; by drawing two rays on it emanating from one point, and cutting this piece along the sides of the drawn angle, we will divide the piece of cardboard into 2 parts; Let's take one of these parts, about which we want to assume that it lies inside the angle, and remove the other - then we will have a model of the angle along with its internal region. To correctly interpret this model, one must keep in mind that a piece of cardboard is an image of only part of a plane, and the plane itself stretches without end.
13. Compare two given angles∠ABC and ∠DEF (drawing 11).
To “compare” two angles means to determine whether the angles are equal or one is larger than the other. To do this, we will begin to superimpose one angle on another so that their internal areas go along each other: if in this case it turns out that it is possible to achieve that the vertices and sides of our angles are aligned, then we say that these angles are equal; if the vertices on one side of our angles coincide, but the other sides do not coincide, then the angles are not equal, and we read the smaller one as the one whose interior area fits on the interior area of the other.
Exercise. Cut out models of corners from paper along with their internal areas and, by superimposing these models on top of each other, establish the possibility of the cases described above; Having cut out a model of one angle, then cut out a model of an angle equal to it and models of angles not equal to it (more or less).
Let's look at the angles ABC and DEF (Figure 11); the internal area of each of them is marked in the drawing. We move ∠DEF so that its vertex E hits point B and its side EF goes along side BC - then the internal areas of the corners will be located one after the other. If side ED goes along side BA, then ∠DEF = ∠ABC; if the ED side goes inside ∠ABC, for example, along the ray BM, then ∠DEF< ∠ABC (здесь внутренняя область угла DEF уляжется на внутренней области ∠ABC, и еще останется незанятой область ABM); если сторона ED пойдет вне ∠ABC, напр., по лучу BN, то ∠DEF >∠ABC.
It is useful to repeat the same reasoning for the angles ABC and DEF (with interior regions marked) given in Fig. 11 bis.
Let us apply the described method of comparing two angles to two straightened angles. Let us have 2 straightened angles ∠AOB and ∠A1O1B1 (drawing 12), the internal areas of which are marked in the drawing. By superimposing one of these angles on the other so that the vertex O 1 of one falls into the vertex O of the other and so that the side O 1 A 1 of one goes along the side OA of the other, we come to the conclusion that the other sides of these angles O 1 B 1 and OB coincide, since the lines A 1 O 1 B 1 and AOB are straight lines, the position of which is determined by two points. (Sometimes they say: “OB is a continuation of OA” instead of saying that the line AOB is a straight line). Therefore we come to the conclusion:
All right angles are equal to each other.
14. Straightened ∠AOB (drawing 12) divides the plane into 2 regions, internal and external. If you bend the plane along the straight line AOB, then both of these parts will coincide. Therefore, we can assume that the internal and external areas of a straightened angle are equal to each other.
If we have any non-rectified angle, for example, ∠DEF (drawing 11 or drawing 11 bis), then by continuing one of its sides, for example, side DE (no continuations are drawn on the drawings), we will see that about our angle, it can be established that it is either less than straightened (drawing 11), or greater than it (drawing 11 bis); It depends on which of the two parts of the plane is taken as the inner region of the corner. Usually the inner area of the angle is chosen so that this angle is smaller than the straightened one, and in this case we agree not to mark the inner area of the angle. Sometimes the origin of the angle will indicate that the internal region should be considered that part of the plane that the angle will be greater than the straightened one. These cases will sometimes occur in the future, and then we must mark the inner area of the corner.
15. Find the sum of two angles: ∠AOB and ∠PNM (drawing 13), or add ∠AOB and ∠PNM.
Here in the drawing the internal areas of the corners are not marked; according to the remark of the previous paragraph, this means that they must be chosen so that each angle is less than straightened, and we clearly see these areas.
Let us move ∠PNM so that its vertex N coincides with the vertex O of the angle AOB, and by rotating around the point O we will ensure that the side NP goes along the side OB; then the internal regions of our angles will be adjacent to each other - this circumstance is essential for the addition of angles. Let us then note how the NM side will go: let, for example, it go along the ray OC. Then we get a new ∠AOC, which is taken as the sum of the two given angles. We can write:
1) ∠BOC = ∠PNM, 2) ∠AOC = ∠AOB + ∠BOC
and 3) (based on 1) ∠AOC = ∠AOB + ∠PNM.
You can also fold several corners; You can break this angle into several terms. For the hell of it. 14 we have:
∠AOE = ∠AOB + ∠BOC + ∠COD + ∠DOE.
It is easy to construct two or more angles applied to each other so that their sum is equal to the straightened angle. It is possible that the sum of several angles will be greater than the straightened angle (Fig. 15), the inner region of this sum should be noted.
Another special case of adding angles is possible, when the internal regions of the added angles cover the entire plane when they are applied to each other. For the hell of it. 16 we have the following angles: ∠AOB, ∠BOC, ∠COD, ∠DOE, ∠EOF and ∠FOA. In this case, having constructed the ray OM, which is a continuation of the ray OA, we see that the sum of our angles consists of two straightened angles: 1) straightened ∠AOM, the inner region of which is marked by one curved line, and 2) straightened ∠AOM, the inner region of which is marked double curved line. Here we have:
∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOF + ∠FOA = 2 straightened corners.
They say: The sum of all successive angles surrounding a point is equal to two right angles.
If there are additional angles other than those constructed in drawing. 16, then they will have to be applied to the previous ones again along the first straightened angle, and then the sum turns out to be more than two straightened angles, equal to three straightened angles, more than three straightened angles, etc.
16. Find the difference of two angles: ∠AOB and ∠MNP (Dev. 17), or subtract ∠MNP from ∠AOB, assuming that ∠MNP< ∠AOB.
Let us move ∠MNP so that its vertex N falls into the vertex O of angle AOB; By rotating around point O, we will then achieve that side NM goes along side OB, and the inner areas of these angles are located one on top of the other. Let the NP side follow the OC beam; then we get a new ∠AOC, about which we know that ∠AOC + ∠COB = ∠AOB, from which, according to the definition of subtraction as the inverse action of addition, we get:
∠AOC = ∠AOB – ∠COB,
but ∠COB = ∠MNP; That's why
∠AOC = ∠AOB – ∠MNP.
From paragraphs. 13-16 we must grasp the idea that the following concepts are applicable to angles, as well as to segments: more, less, equal, and that the concepts of the sum and difference of two angles have a certain meaning.
17. Exercises. 1. Construct two angles attached to each other, name them with letters, indicate their sum and write down the addition of these angles.
2. In the same drawing, indicate that one of the angles is the difference between the other two; write it down.
3. In the following drawings (see drawing B), ∠AOB is expressed by the difference of the other two angles.
4. Divide this angle into 2, 3, and 4 terms; write it down every time; do the same with the straightened corner.
5. Present this angle as the difference between the straightened one and some other angle. What kind of structure is needed for this?
6. Add and subtract angles using angle models cut out of paper.
18. In the future, we will often number the angles in order to shorten the letter by calling them numbers. We will write the angle numbers inside each angle near the vertex.
Let's construct ∠AOB (drawing 18) and call it ∠1. Let's add this angle to a straight one. The problem has two solutions: construct a ray OC, which serves as a continuation of the ray OA; then we obtain ∠BOC or ∠2, which satisfies the requirement, since we see that
∠1 + ∠2 = straightened angle.
Here we have an example of adding two angles when the sum is equal to the straightened angle - such angles are called adjacent: ∠1 and ∠2 are adjacent angles. In order for 2 angles to be called “adjacent”, it is necessary that 1) they are attached to each other and 2) that their sum is equal to the straightened angle, or, what is the same, that these angles have a common vertex (at angles 1 and 2 common vertex O), one common side (our corners have a common side OB) and that the other two sides are a continuation of each other (OC is a continuation of OA).
The second solution to our problem will be obtained if we continue the side OB - let OD be a continuation of OB; then we get another ∠AOD or ∠4 adjacent to ∠1. Let us also call the resulting angle COD by ∠3.
Let us examine the 2 obtained solutions to our problem, i.e. ∠2 and ∠4. We see the peculiarity of the location of ∠2 and ∠4: they have a common vertex O, the sides of one of them are continuations of the sides of the other, namely OC is a continuation of OA and vice versa, and OB is a continuation of OD and vice versa - such two angles are called vertical.
Then we know that both ∠2 and ∠4 each complement ∠1 until rectified; from here we conclude that
Here's a more detailed summary of the latter consideration. According to the construction, we have:
1) ∠1 + ∠2 = straightened angle;
2) ∠1 + ∠4 = straightened angle.
We see that both additions lead to the same sum (all right angles are equal to each other), and, in addition, one term (namely ∠1) in both additions is the same; from here we conclude that the other terms must be equal to each other, i.e. ∠2 = ∠4.
If we construct two intersecting straight lines, we get two pairs of vertical angles. For the hell of it. 18 we have lines AC and BD, one pair of vertical angles is ∠2 and ∠4, and the other is ∠1 and ∠3. Everything above applies to every pair of vertical angles; for example, for the pair ∠1 and ∠3 we have that each of them complements ∠2 to the rectified one, therefore, ∠1 = ∠3. Therefore we have the theorem:
Vertical angles are equal to each other.
Exercise. Construct three straight lines through the point and indicate the resulting vertical angles; write down their equality.
Line segment. Length of the segment. Triangle.
1. In this paragraph you will be introduced to some concepts of geometry. Geometry- the science of "measuring the earth." This word comes from the Latin words: geo - earth and metr - measure, to measure. In geometry, various geometric objects, their properties, their connections with the outside world. The simplest geometric objects are a point, a line, a surface. More complex geometric objects, for example, geometric figures and bodies, are formed from the simplest.
If we apply a ruler to two points A and B and draw a line along it connecting these points, we get line segment, which is called AB or VA (we read: “a-be”, “be-a”). Points A and B are called ends of the segment(picture 1). The distance between the ends of a segment, measured in units of length, is called lengthcutka.
Units of length: m - meter, cm - centimeter, dm - decimeter, mm - millimeter, km - kilometer, etc. (1 km = 1000 m; 1 m = 10 dm; 1 dm = 10 cm; 1 cm = 10 mm). To measure the length of segments, use a ruler or tape measure. To measure the length of a segment means to find out how many times a particular length measure fits into it.
Equal are called two segments that can be combined by superimposing one on the other (Figure 2). For example, you can actually or mentally cut out one of the segments and attach it to another so that their ends coincide. If the segments AB and SK are equal, then we write AB = SK. Equal segments have equal lengths. The opposite is true: two segments of equal length are equal. If two segments have different lengths, then they are not equal. Of two unequal segments, the smaller is the one that forms part of the other segment. You can compare overlapping segments using a compass.
If we mentally extend the segment AB in both directions to infinity, then we will get an idea of straight AB (Figure 3). Any point lying on a line splits it into two beam(Figure 4). Point C splits line AB into two beam SA and SV. Tosca C is called the beginning of the ray.
2. If three points that do not lie on the same line are connected by segments, then we get a figure called triangle. These points are called peaks triangle, and the segments connecting them are parties triangle (Figure 5). FNM - triangle, segments FN, NM, FM - sides of the triangle, points F, N, M - vertices of the triangle. The sides of all triangles have the following property: d The length of any side of a triangle is always less than the sum of the lengths of its other two sides.
If you mentally extend, for example, the surface of a table top in all directions, you will get an idea of plane. Points, segments, straight lines, rays are located on a plane (Figure 6).
Block 1. Additional
The world in which we live, everything that surrounds us, the ancients called nature or space. The space in which we live is considered three-dimensional, i.e. has three dimensions. They are often called: length, width and height (for example, the length of a room is 4 m, the width of a room is 2 m and the height is 3 m).
The idea of a geometric (mathematical) point is given to us by a star in the night sky, a dot at the end of this sentence, a mark from a needle, etc. However, all of the listed objects have dimensions; in contrast, the dimensions of a geometric point are considered equal to zero (its dimensions are equal to zero). Therefore, a real mathematical point can only be imagined mentally. You can also tell where it is located. By placing a dot in a notebook with a fountain pen, we will not depict a geometric point, but we will assume that the constructed object is a geometric point (Figure 6). Points are designated in capital letters of the Latin alphabet: A, B, C, D, (read " point a, point be, point tse, point de") (Figure 7).
Wires hanging on poles, a visible horizon line (the boundary between sky and earth or water), a riverbed depicted on a map, a gymnastics hoop, a stream of water gushing from a fountain give us an idea of lines.
There are closed and open lines, smooth and non-smooth lines, lines with and without self-intersection (Figures 8 and 9).
A sheet of paper, laser disc, soccer ball shell, packaging box cardboard, Christmas plastic mask, etc. give us an idea of surfaces(Figure 10). When painting the floor of a room or a car, the surface of the floor or car is covered with paint.
Human body, stone, brick, cheese, ball, ice icicle, etc. give us an idea of geometric bodies (Figure 11).
The simplest of all lines is it's straight. Place a ruler on a sheet of paper and draw a straight line along it with a pencil. Mentally extending this line to infinity in both directions, we will get the idea of a straight line. It is believed that a straight line has one dimension - length, and its other two dimensions are equal to zero (Figure 12).
When solving problems, a straight line is depicted as a line that is drawn along a ruler with a pencil or chalk. Direct lines are designated by lowercase Latin letters: a, b, n, m (Figure 13). You can also denote a straight line by two letters corresponding to the points lying on it. For example, straight n in Figure 13 we can denote: AB or VA, ADorDA,DB or BD.
Points can lie on a line (belong to a line) or not lie on a line (not belong to a line). Figure 13 shows points A, D, B lying on line AB (belonging to line AB). At the same time they write. Read: point A belongs to line AB, point B belongs to AB, point D belongs to AB. Point D also belongs to line m, it is called general dot. At point D the lines AB and m intersect. Points P and R do not belong to straight lines AB and m:
Through any two points always you can draw a straight line and only one .
Of all types of lines connecting any two points, the segment whose ends are these points has the shortest length (Figure 14).
A figure that consists of points and segments connecting them is called a broken line (Figure 15). The segments that form a broken line are called links broken line, and their ends - peaks broken line A broken line is named (designated) by listing all its vertices in order, for example, the broken line ABCDEFG. The length of a broken line is the sum of the lengths of its links. This means that the length of the broken line ABCDEFG is equal to the sum: AB + BC + CD + DE + EF + FG.
A closed broken line is called polygon, its vertices are called vertices of the polygon, and its links parties polygon (Figure 16). A polygon is named (designated) by listing in order all its vertices, starting from any one, for example, polygon (heptagon) ABCDEFG, polygon (pentagon) RTPKL:
The sum of the lengths of all sides of a polygon is called perimeter polygon and is denoted by the Latin letterp(read: pe). Perimeters of polygons in Figure 13:
P ABCDEFG = AB + BC + CD + DE + EF + FG + GA.
P RTPKL = RT + TP + PK + KL + LR.
Mentally extending the surface of a table top or window glass to infinity in all directions, we get an idea of the surface, which is called plane (Figure 17). The planes are designated in small letters of the Greek alphabet: α, β, γ, δ, ... (we read: plane alpha, beta, gamma, delta, etc.).
Block 2. Vocabulary.
Make a dictionary of new terms and definitions from §2. To do this, enter words from the list of terms below in the empty rows of the table. In Table 2, indicate the term numbers in accordance with the line numbers. It is recommended that you carefully review §2 and block 2.1 before filling out the dictionary.
Block 3. Establish correspondence (CS).
Geometric figures.
Block 4. Self-test.
Measuring a segment using a ruler.
Let us recall that to measure a segment AB in centimeters means to compare it with a segment 1 cm long and find out how many such 1 cm segments fit in the segment AB. To measure a segment in other units of length, proceed in the same way.
To complete the tasks, work according to the plan given in the left column of the table. In this case, we recommend covering the right column with a sheet of paper. You can then compare your findings with the solutions in the table to the right.
Block 5. Establishing a sequence of actions (SE).
Constructing a segment of a given length.
Option 1. The table contains a mixed up algorithm (a mixed up order of actions) for constructing a segment of a given length (for example, let’s build a segment BC = 7 cm). In the left column is an indication of the action, in the right column is the result of performing this action. Rearrange the rows of the table so that you get the correct algorithm for constructing a segment of a given length. Write down the correct sequence of actions.
Option 2. The following table shows the algorithm for constructing the segment KM = n cm, where instead of n You can substitute any number. In this option there is no correspondence between action and result. Therefore, it is necessary to establish a sequence of actions, then for each action, select its result. Write the answer in the form: 2a, 1c, 4b, etc.
Option 3. Using the algorithm of option 2, construct segments in your notebook at n = 3 cm, n = 10 cm, n = 12 cm.
Block 6. Facet test.
Segment, ray, straight line, plane.
In the tasks of the facet test, pictures and records numbered 1 - 12, given in Table 1, are used. Task data is formed from them. Then the requirements of the tasks are added to them, which are placed in the test after the connecting word “TO”. Answers to the problems are placed after the word “EQUAL”. The set of tasks is given in Table 2. For example, task 6.15.19 is composed as follows: “IF the problem uses Figure 6 , s Then condition number 15 is added to it, the task requirement is number 19.”
13) construct four points so that every three of them do not lie on the same straight line;
14) draw a straight line through every two points;
15) mentally extend each of the surfaces of the box in all directions to infinity;
16) the number of different segments in the figure;
17) the number of different rays in the figure;
18) the number of different straight lines in the figure;
19) the number of different planes obtained;
20) length of segment AC in centimeters;
21) length of segment AB in kilometers;
22) length of segment DC in meters;
23) perimeter of triangle PRQ;
24) length of the broken line QPRMN;
25) quotient of the perimeters of triangles RMN and PRQ;
26) length of segment ED;
27) length of segment BE;
28) the number of resulting points of intersection of lines;
29) the number of resulting triangles;
30) the number of parts into which the plane was divided;
31) the perimeter of the polygon, expressed in meters;
32) the perimeter of the polygon, expressed in decimeters;
33) the perimeter of the polygon, expressed in centimeters;
34) the perimeter of the polygon, expressed in millimeters;
35) perimeter of the polygon, expressed in kilometers;
EQUALS (equal, has the form):
a) 70; b) 4; c) 217; d) 8; e) 20; e) 10; g) 8∙b; h) 800∙b; i) 8000∙b; j) 80∙b; k) 63000; m) 63; m) 63000000; o) 3; n) 6; p) 630000; c) 6300000; t) 7; y) 5; t) 22; x) 28
Block 7. Let's play.
7.1. Math labyrinth.
The labyrinth consists of ten rooms with three doors each. In each of the rooms there is one geometric object (it is drawn on the wall of the room). Information about this object is in the “guide” to the labyrinth. While reading it, you need to go to the room that is written about in the guidebook. As you walk through the rooms of the labyrinth, draw your route. The last two rooms have exits.
Guide to the Labyrinth
- You must enter the labyrinth through a room where there is a geometric object that has no beginning, but has two ends.
- The geometric object of this room has no dimensions, it is like a distant star in the night sky.
- The geometric object of this room is composed of four segments that have three common points.
- This geometric object consists of four segments with four common points.
- This room contains geometric objects, each of which has a beginning but no end.
- Here are two geometric objects that have neither beginning nor end, but with one common point.
- An idea of this geometric object is given by the flight of artillery shells
(trajectory of movement).
- This room contains a geometric object with three peaks, but they are not mountainous.
- The flight of a boomerang gives an idea of this geometric object (hunting
weapons of the indigenous people of Australia). In physics this line is called a trajectory
body movements.
- An idea of this geometric object is given by the surface of the lake in
calm weather.
Now you can exit the maze.
The maze contains geometric objects: plane, open line, straight line, triangle, point, closed line, broken line, segment, ray, quadrilateral.
7.2. Perimeter of geometric shapes.
In the drawings, highlight geometric shapes: triangles, quadrangles, pentagons and hexagons. Using a ruler (in millimeters), determine the perimeters of some of them.
7.3. Relay race of geometric objects.
Relay tasks have empty frames. Write down the missing word in them. Then move this word to another frame where the arrow points. In this case, you can change the case of this word. As you go through the stages of the relay, complete the required formations. If you complete the relay correctly, you will receive the following word at the end: perimeter.
7.4. Strength of geometric objects.
Read § 2, write down the names of geometric objects from its text. Then write these words in the empty cells of the “fortress”.
Segments are called equal if they can be superimposed on one another so that their ends coincide.Let us be given two segments AB and CD (Fig.). Let's superimpose segment AB onto segment CD so that point A coincides with point C, and direct segment AB along segment CD. If point B coincides with point D, then segments AB and CD are equal; AB = CD.
Let's compare two segments KO and EM (Fig.). 
Let's superimpose the segment KO onto the segment EM so that points K and E coincide. Let's direct the segment KO along the segment EM. If point O is somewhere between points E and M, then they say that the segment EM is greater than the segment KO; segment KO is less than segment EM.
It is written like this: EAT > KO, KO
Constructing a segment equal to a given one using a compass.
The construction of a segment equal to a given segment AB (Fig.) is performed using a compass in this way:
one leg of the compass is placed on one end of the segment AB, and the other - on its other end and, without changing the angle of the compass, transfer it to a certain straight line so that the end of one leg marks some point N, then the end of the other leg of the compass marks some point R on the same straight line. The segment NP will be equal to the segment AB.
Addition and subtraction of segments.
To find the sum of two segments, for example AB and CD (Fig.), you need to take a straight line and some point on it, for example point N (Fig., b), then, using a compass, first plot the segment NP on this straight line from point N, equal to the segment AB, and then from its end in the same direction lay off a segment PM equal to the segment CD. The segment NM will be called the sum of segments AB and CD.
It is written like this:
NM = AB + CD.
In the same way, the sum of several segments is found (Fig.)
MN = AB + CD + EF.
When adding segments, as in arithmetic when adding numbers, the laws are followed: commutative and associative.
AB + CD = CD + AB;
(AB + CD) + EF =AB + (CD + EF).
To find the difference between two segments AB and CD (Fig.),
It is necessary to set aside a smaller segment (CD) on a larger segment (AB) from its end, for example, point A. The remaining part (KB) of the larger segment will be the difference of these segments:
AB - CD = KV.
Multiplying and dividing a segment by an integer.
a) Multiply the segment AB by an integer, for example by 5, this means that the segment AB must be taken as a term 5 times (Fig.):
The segment MN is the product of the segment AB and the number 5.
b) In the figure, the segment MN is composed of five equal segments, i.e. the segment MN is divided into five equal parts. Each of them makes up 1/5 of the segment MN.
c) To divide a segment into equal parts using a compass, do this. For example, if you need to divide a segment into two equal parts, then the compass is moved apart by eye so that the opening of the compass is approximately half of the segment. Then, on a given segment from its end, two segments are laid out sequentially, one after the other, with this compass solution. If the resulting sum of segments is less than this segment, then the compass solution is increased; if the amount turns out to be more than this segment, then the compass solution is reduced. So, gradually correcting the error, you can find quite accurately half of the segment (Fig.).
In the same way, an approximate division of a segment into 3, 4, 5, etc. equal parts is performed. Only in this case should you take 1/3 by eye; 14 ; 1/5... of a segment and set aside the taken segment 3, 4, 5... times, depending on how many equal parts the given segment needs to be divided into.
Property of segments cut off by parallel lines on the sides of an angle
Theorem. If equal segments are laid out on one side of an angle and parallel lines are drawn through their ends, intersecting the other side of the angle, then equal segments will be laid out on this side of the angle.Let equal segments BM = MK = KS (Fig.) be laid out on the side AB of the angle ABN and parallel lines intersecting the side BN of the same angle are drawn through the division points M, K and C.
On this side three segments were formed: VM', M'K' and K'S'. It is required to prove that VM' = M'K' = K'C'.
To prove this, we draw straight lines parallel to AB through points M’ and K’. We get triangles ВММ', М'ЭК' and К'РС'. Let's compare these triangles.
First, compare the triangles MVM' and M'EK'. In these triangles we have:
∠1 = ∠2, as the corresponding angles for parallel BA and M'E and secant BN;
∠3 = ∠4, as acute angles 1 with correspondingly parallel sides (AB || M'E and MM' || KK').
VM = MK by construction;
MK = M'E, like opposite sides of a parallelogram.
Angles 1 and 4 may turn out to be both obtuse, but in this case they will remain equal, and therefore the proof of the theorem will not change.
Therefore, BM = M'E. Thus, ΔВММ’ = ΔМ’ЭК’ (on the side and two adjacent angles). It follows that VM' = M'K'.
It can also be proven that VM’ = K’C’, i.e. VM’ = M’K’ = K’C’. When proving the theorem, we started laying out segments from the vertex of the angle, but the theorem is also valid for the case when laying out segments starts not from the vertex of the angle, but from any point on its side.
In this case, the vertex of the corner need not be marked in the drawing (Fig.).
The theorem is also valid for the case when the lines KO and MR are parallel.
Proportional segments
From arithmetic we know that the equality of two ratios is called proportion. For example: 16 / 4 = 20 / 5 ; 2 / 3 = 4 / 6 We have the same thing in geometry: if two pairs of segments are given whose ratios are equal, then a proportion can be made.
If a / b= 4 / 3 and c / d= 4 / 3 (Drawn 351), then we get the proportion a / b = c / d ;
segments a, b, c, d are called proportional.
Attitude a / b is called, as in arithmetic, the first relation, c / d- second relation; A And d are called extreme terms of the proportion, b And With- middle members.
In a proportion, the ratios can be reversed; you can rearrange the extreme members, middle members; you can rearrange both at the same time.
Because in proportion a / b = c / d By letters we mean numbers expressing the lengths of segments, then the product of its extreme members is equal to the product of the middle members. From here, knowing the three terms of the proportion, you can find its unknown fourth term. Yes, in proportion a / x = c / d x = a d / c
Let us note some more properties of proportions, which will have to be used in the future when proving some theorems and solving problems.
a) If three terms of one proportion are respectively equal to three terms of another proportion, then the fourth terms of these proportions are also equal.
If a / b = c / x And a / b = c / y,That x = y. Indeed, x = b c / a , at = b c / a, i.e. and X And at equal to the same number b c / a .
b) If the previous terms are equal in proportion, then the subsequent ones are equal, i.e. if a / x = a / y, That x = y.
To verify this, let us rearrange the middle terms in this proportion.
We get: a / a = x / y. But a / a= 1. Therefore, and x / y = 1.
And this is only possible if the numerator and denominator of the fraction are equal, i.e.
x = y.
c) If subsequent terms are equal in proportion, then the previous ones are equal, i.e. if x / a = y / a, That x = y.
You are invited to verify the validity of this property for yourself. To do this, carry out reasoning similar to the previous one.
Construction of proportional segments
Theorem. If two lines are intersected by three parallel lines, then the ratio of the two segments obtained on one line is equal to the ratio of the two corresponding segments of the other line.Let two lines EF and OP be intersected by three parallel lines AB, CD and MN (Fig.).
It is required to prove that the segments AC, CM, BD and DN, enclosed between parallel secants, are proportional, i.e.
AC/CM = BD/DN
Let the length of segment AC be R, and the length of the segment CM is equal to q.
For example, R= 4 cm and q= 5 cm.
Let us divide AC and CM into segments equal to 1 cm, and from the division points we draw straight lines parallel to straight lines AB, CD and MN, as shown in the figure.
Then, equal segments will be deposited on the straight line OR, with 4 segments on the segment BD, and 5 segments on the segment DN.
The ratio of AC to CM is 4/5, and similarly the ratio of BD to DN is 4/5.
Hence AC/CM = BD/DN.
This means that the segments AC, CM, BD and DN are proportional. The segments AC, AM, BD and BN (overlapping each other) are also proportional, i.e. AC / AM = BD / BN,
since AC/AM = 4/9 and BD/BN = 4/9
The theorem will be valid for any other integer values R And q.
If the lengths of the segments AC and CM are not expressed in integers for a given unit of measurement (for example, a centimeter), then it is necessary to take a smaller unit (for example, a millimeter or micron), in which the lengths of the segments AC and CM are practically expressed in integers.
The proven theorem is also valid in the case when one of the parallel secants passes through the point of intersection of these lines. It is also valid in the case when the segments are not plotted directly one after another, but after a certain interval.
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