How to draw a triangle with a compass. How to draw a triangle using a compass

The geometric construction of figures refers to the basic knowledge of the school geometry course. In addition to utilitarian use, the formation of spatial logic matters here. It is the corollary construction triangle, as a primitive polygonal figure, with help circle is considered in detail. The compass is a tool not only for constructing a circle. It also allows you to set aside equal segments of a given length. This will help us with help build a triangle.

You will need

Sheet of paper, compasses

Instruction

1. Take any piece of paper. Put a dot in the center of the sheet. This will be the first vertex A of the generated triangle .

2. Open the compass to a distance that correctly corresponds to the desired side of the created triangle. Firmly fix the legs of the compass in this location.

3. Place the compass needle at the marked point. Draw a circle with a stylus with a measured radius.

4. Place a dot anywhere along the circumference of the drawn arc. This will be the second vertex B of the created triangle .

5. Place the leg on the second vertex in the same way. Draw another circle so that it intersects with the first.

6. At the intersection point of both drawn arcs and is the third vertex C of the created triangle. Mark it in the picture.

7. Having obtained all three vertices, unite them with straight lines with help any flat surface (better than a ruler). Triangle ABC is built.

How to draw a triangle with a compass A compass is not only a tool for constructing a circle. It also allows you to set aside equal segments of a given length. This will help us build a triangle with it.

You will need: a sheet of paper, a compass, a ruler. Instruction. 1. Take any piece of paper. Put a dot in the center of the sheet. This will be the first vertex A of the triangle being created. A

Instruction 2. Open the compass to a distance corresponding to the desired side of the triangle being created. Firmly fix the legs of the compass in this position.

Instruction 3. Place the compass needle at the marked point. Draw a circle with a stylus with a measured radius.

Instruction 4. Put a dot anywhere along the circumference of the drawn arc. This will be the second vertex B of the triangle being created.

Instruction 5. In the same way, put the leg on the second peak. Draw another circle so that it intersects with the first.

Instruction 6. The third vertex C of the created triangle is located at the intersection point of both drawn arcs. Mark it on the picture.

Help! Granddaughter asked. Draw a regular triangle using a compass. and got the best answer

Answer from KINOholic[guru]
First, construct a segment with a length equal to the length of the future triangle.
Then dissolve the compass to the length of this segment and, placing the end of the compass at the beginning of the segment, draw a circle.
Place the compass at the other end of the segment and draw another circle.
The circles intersect at two points - above and below the segment. By connecting the ends of the segment to one of these points, you will get a regular (equilateral triangle).

Answer from Grisha Kolosov[newbie]
THX

Answer from Alexander Zhidaikin[newbie]
Divide the circle into 4 equal parts. Place the leg of the compass at the lowest point and draw a second circle of the same radius. We got two points of intersection - these are two points of the triangle. The third point is at the highest point of the first circle. We connect, we get)
figure 61 for help

Answer from grandfather07[guru]
Draw a circle. Mark a point on the circle (let A). From this point on a circle in both directions, measure 2 radii. Connect the received 3 points

Answer from *ORANGE*[guru]
en.wikibooks.org/wiki/.../Construction_of_a_regular_triangle

Answer from Elena Yakovleva[guru]
Draw a circle and divide it into 6 parts with the same radius (put 6 points), then connect three points (through one) with straight lines.

Answer from Antip[guru]
1) On a straight line, mark a segment of arbitrary length with a compass
2) from one end of the segment with a compass, open to the length of the marked segment, draw an arc (long enough)
3) do the same from the other end of the segment
4) the arcs will intersect
5) connect the intersection point with the ends of the segment
6) here we get an equilateral triangle - correct

Answer from Vega[guru]
draw a circle, then put the needle on the circle and make two serifs on the lines, then rearrange the compass so that you put a pencil on the serif, and move the needle further and make the next serif ... so connect all three serifs ... you get the right triangle. .

Answer from Yatyana Egorova[guru]
On a straight line, set aside a segment with a certain compass solution and draw arcs from both ends with the same solution. These arcs will intersect. This is the third vertex of your triangle.

Answer from 3 answers[guru]

Hey! Here is a selection of topics with answers to your question: Help! Granddaughter asked. Draw a regular triangle using a compass.

In construction problems, a compass and a ruler are considered ideal tools, in particular, a ruler has no divisions and has only one side of infinite length, and a compass can have an arbitrarily large or arbitrarily small opening.

Permissible constructions. The following operations are allowed in construction tasks:

1. Mark a point:

arbitrary point of the plane;

an arbitrary point on a given line;

an arbitrary point on a given circle;

the point of intersection of two given lines;

points of intersection/tangency of a given straight line and a given circle;

points of intersection/tangency of two given circles.

2. Using a ruler, you can build a straight line:

arbitrary straight line on the plane;

an arbitrary line passing through a given point;

a straight line passing through two given points.

3. Using a compass, you can build a circle:

arbitrary circle on the plane;

an arbitrary circle centered at a given point;

an arbitrary circle with a radius equal to the distance between two given points;

a circle centered at a given point and with a radius equal to the distance between two given points.

Solving building problems. The solution of the construction problem contains three essential parts:

Description of the method of constructing the desired object.

Proof that the object constructed in the described way is really the desired one.

Analysis of the described construction method for its applicability to different variants of initial conditions, as well as for the uniqueness or non-uniqueness of the solution obtained by the described method.

Construction of a segment equal to a given one. Let there be given a ray with origin at the point $O$ and a segment $AB$. To construct a segment $OP = AB$ on a ray, one needs to construct a circle centered at the point $O$ of radius $AB$. The intersection point of the ray with the circle will be the desired point $P$.

Constructing an angle equal to a given one. Let there be given a ray with the origin at the point $O$ and an angle $ABC$. With the center at the point $B$ we construct a circle with an arbitrary radius $r$. Denote the points of intersection of the circle with the rays $BA$ and $BC$ $A"$ and $C"$ respectively.

Let's construct a circle centered at the point $O$ of radius $r$. Denote the intersection point of the circle with the ray by $P$. Let's construct a circle centered at the point $P$ of radius $A"B"$. Denote the intersection point of the circles by $Q$. Let's draw a ray $OQ$.

We get the angle $POQ$ equal to the angle $ABC$, since the triangles $POQ$ and $ABC$ are equal on three sides.

Construction of a perpendicular bisector to a segment. We construct two intersecting circles of arbitrary radius with centers at the ends of the segment. Connecting the two points of their intersection, we get the perpendicular bisector.

Construction of the bisector of an angle. Let's draw a circle of arbitrary radius with the center at the corner vertex. Let's construct two intersecting circles of arbitrary radius with centers at the points of intersection of the first circle with the sides of the angle. By connecting the vertex of the angle to any of the intersection points of these two circles, we obtain the bisector of the angle.

Construction of the sum of two segments. To construct a segment on a given ray equal to the sum of two given segments, it is necessary to apply the method of constructing a segment equal to a given one twice.

Construction of the sum of two angles. In order to postpone from a given ray an angle equal to the sum of two given angles, it is necessary to apply the method of constructing an angle equal to a given one twice.

Finding the midpoint of a segment. In order to mark the midpoint of a given segment, you need to construct a median perpendicular to the segment and mark the point of intersection of the perpendicular with the segment itself.

Construction of a perpendicular line through a given point. Let it be required to construct a line perpendicular to the given one and passing through the given point. We draw a circle of arbitrary radius with a center at a given point (regardless of whether it lies on a straight line or not), intersecting a straight line at two points. We build a perpendicular bisector to the segment with ends at the points of intersection of the circle with the line. This will be the desired perpendicular line.

Constructing a parallel line through a given point. Let it be required to construct a line parallel to a given one and passing through a given point outside the line. We construct a line passing through a given point and perpendicular to a given line. Then we build a straight line passing through this point, perpendicular to the constructed perpendicular. The straight line thus obtained will be the required one.

In this lesson, we will consider tasks for constructing geometric objects using a compass and ruler.

To solve various practical problems, people have come up with many tools.

To measure the length of a segment or draw a segment of a given length, we use a ruler. To solve a similar problem for angles, there is a protractor.

Proving theorems and solving problems, we have not yet paid attention to such things as: “we will draw (construct) the median of a triangle ...”.

Median is a line segment that connects a vertex to the midpoint of the opposite side. Where is the top? Where is the middle of the opposite side? If we have a ruler at hand, then it will definitely not be difficult to solve this problem: we measured the length of the side, divided by 2, found the middle. With a protractor in the same way, it is not difficult to construct an angle bisector.
But what if there are no tools at hand? Let's say there is only a rope. What can we do with it? Draw a line (if stretched, then a straight line) and measure with it a segment equal to the given one, we can even draw a circle (see Fig. 1). Instead of a rope, we could do these operations with a ruler (without divisions) and a compass.

Rice. 1. Using a rope, you can draw a circle

In geometry, they talk about construction problems with the help of a compass and straightedge. There are tasks that can be solved with these two tools, and there are those that cannot. We will talk about this in today's lesson.

But first, let's try to answer the question: why a compass and a ruler without divisions? Why was it impossible to choose a ruler with divisions, a protractor, or some other tools? And why do you need to be able to solve such problems at all (we can reveal a terrible secret: even students of mathematical faculties and professional mathematicians do not study and do not solve such problems after graduation).

One consideration we have already voiced is that everything that can be done with a compass and a ruler (by default in this lesson we will mean that we mean a ruler without divisions) can also be done with a regular rope. And in some situations (for example, marking a site), these skills can come in handy.

But a more important argument is an example of tasks that are solved using the minimum possible resource. In life, we often face such tasks: to build an engine in order to travel the maximum distance for 100 liters of gasoline, or to spend the least possible time on homework, but get at least 4 for it, etc. That is. we often solve optimization problems with limited resources. In construction tasks, the tools we can use are limited.

Why learn to solve building problems?

Some may find the arguments presented unconvincing. Indeed, there are serious doubts about the need to study this topic. Nevertheless, we will give some more considerations that can help answer the formulated questions.

Mathematics works with absolutely accurate models (an ideal circle does not exist in life, but mathematics studies the properties of just such a circle so that it can be applied to describe real-life circles that are close to ideal).

Any measurement (using a ruler, protractor and other instrument) will contain an inaccuracy (we round to an accuracy that is determined by the purpose of the measurement). Therefore, from the point of view of mathematics, the solution to the problem - to divide the segment into two parts, measuring it with a ruler, is not correct.

In mathematics, a segment of length 1 must be divided into two segments of length 0.5. But if we start measuring the length of this segment with a ruler, it cannot be exactly equal to 1. And the lengths of the halves will differ from 0.5. Therefore, in order to work with ideal abstract objects, you need to use abstract ideal tools, which are a ruler without divisions and compasses.

But this is an explanation of why construction problems are studied in mathematics. Why do students need them? It seems that the most honest answer is for training. By and large, all such problems have an equivalent formulation: there are two operations; How to use them to get the required object from a given object?

For some people, solving such problems is fascinating (Gauss was so proud of being able to build a regular 17-gon using a compass and straightedge that he bequeathed to engrave it on his monument, although this is perhaps his least useful mathematical discovery from a practical point of view) . But this is not quite mathematics, but rather an intellectual game. The same as making words from sets of letters, solving crossword puzzles, etc.

Therefore, this lesson will be useful for those who enjoy solving mathematical problems, and the rest should just get acquainted with the idea and principle of solving building problems in order to have a general idea of such a mathematical tool.

So, in geometry, compasses and a ruler are considered classical tools for construction. The ruler has an infinite length. This means that if we do not have enough ruler length to solve a certain problem, we have a longer ruler, which will be enough. That is, the length of the ruler will never be a problem for us.

In the same way, the distance between the legs of the compass will not be a problem - we can move them apart to any distance (not enough - we take a larger compass). The same is paper. You yourself can explain what an endless sheet of paper, an endless plane means.

Compass functions

We can measure any given segment with it and put the same one from a point on a straight line in any direction, the resulting segment will be equal to the first one (see Fig. 2).
We can draw a circle with a center at any given point and a radius equal to any given segment (see Fig. 3).

Rice. 2. Using a compass, you can measure any given segment and set aside the same from a point on a straight line in any direction

Rice. 3. Using a compass, you can draw a circle with a center at any given point and a radius equal to any given segment

ruler function: We apply a ruler to two given points and draw a line that passes through them. We can also draw a segment or a ray. Recall that in this case we are talking about a ruler without marks (see Fig. 4).

Rice. 4. Using a ruler, you can draw a straight line passing through two given points

Basic constructions, which do not cause difficulties, but are constantly needed:

Draw a line through two given points.
Draw a circle of given radius centered at a given point.
Draw a line segment from a given point equal to the given one.

Let's move on to more interesting constructions. The problem already mentioned today is finding the middle of the segment. Or, what is the same, dividing a line in half.

So, let a segment be given. We need to get a point, which is its middle (see Fig. 5). In construction problems, we usually get a point as the intersection of lines, circles, or a line with a circle.

Rice. 5. A point that is the midpoint of a segment

Task 1. Construct the median (find the midpoint of the segment).

Decision

Suppose we want to find a point (middle) as the intersection of two lines and (see Fig. 6).

Rice. 6. Illustration for problem 1

We know that when two lines intersect, two pairs of angles are formed. But we do not have any additional conditions - only a segment, in which we are looking for a middle. Therefore, it would be strange to expect that the straight line will be inclined to the left or to the right (see Fig. 7).

Rice. 7. Illustration for problem 1

Consider the limiting case when the line is perpendicular to the segment (see Fig. 8).

Rice. 8. Illustration for problem 1

Then we know what is midperpendicular to the cut. And it has an important property: all its points are equidistant from the ends of the segment(see fig. 9). We will use this fact in the construction.

Rice. 9. Illustration for problem 1

To build a straight line, you need to find two of its points (more is possible, less is impossible). And any point of the perpendicular bisector, as we just found out, is equidistant from and. Let's construct two such equidistant points (see Fig. 10).

Rice. 10. Illustration for problem 1

Let's draw two circles of the same radius with centers at the points and . The radii must be taken large enough so that the circles intersect (see Fig. 11) (it is easy to obtain that the radius must be greater than half the length of the segment; in order to fulfill this condition exactly, you can draw circles with a radius that is equal to the length of the segment).

Rice. 11. Illustration for problem 1

The points of intersection belong to both circles, that is, they are removed from and at distances equal to the radii of the circles. But their radii are equal.

Hence, the points and are equidistant from and (see Fig. 12). So they belong to the perpendicular bisector. It remains to connect them and find the intersection point and . This is the desired point (see Fig. 13).

Rice. 12. Illustration for problem 1

Rice. 13. Illustration for problem 1

Problem solved.

Task 2. Draw a perpendicular to a line at a given point

Decision

Let a point be marked on the line (see Fig. 14). You need to draw a perpendicular at this point to this line. Or, as they say, "restore" the perpendicular to the line at a given point.

Rice. 14. Illustration for problem 2

Let's reduce the problem to the previous one - we already know how to build a perpendicular to the middle of the segment. So, you need to build a segment on this straight line, for which the point will be the midpoint.

Draw a circle of arbitrary radius centered at . We get two points of intersection of the circle and the straight line - and (see Fig. 15).

Rice. 15. Illustration for problem 2

Now the task has been reduced to an equivalent one - to construct a midperpendicular to the segment . We already know how to solve this problem, which means that the original problem is solved.

Problem solved.

So, we can build a median (find the midpoint of a segment) and restore a perpendicular to a straight line at a given point. And how to build a height or, what is the same, to drop a perpendicular to a straight line from a point that does not belong to it?

Task 3. Construct height (lower the perpendicular to the line from a point that does not belong to it).

Decision

Again, we use the tool known to us - the construction of the perpendicular bisector. So, let there be a line and a point not lying on it (see Fig. 16). It is necessary to draw a perpendicular from a point to a line.

Rice. 16. Illustration for problem 3

Let's draw a circle with a center at a point and a radius sufficient for this circle to intersect the line. The whole circle is usually not drawn in such cases, but only part of it, an arc, in order to obtain intersection points. We also got points on the straight line (see Fig. 17).

Rice. 17. Illustration for problem 3

Why do we need them? Obviously, it is equidistant from both of these points (the distance is equal to the radius of the circle) (see Fig. 18).

Rice. 18. Illustration for problem 3

But it means that it lies on the perpendicular bisector of the segment . And again we got an equivalent formulation of the problem: build a perpendicular bisector to the segment (it will pass through the point, and since only one perpendicular to the line can be drawn from the point, then it will be the desired one). And we know how to build it.

You can use the fact that the point lies on the perpendicular bisector and build circles with the same radius (see Fig. 19). And you can build two circles of a different radius, it doesn’t matter. The main thing is that we can build this perpendicular bisector, and it will be the desired one (see Fig. 20).

Rice. 19. Illustration for problem 3

Rice. 20. Illustration for problem 3

Problem solved.

These three tasks were very similar. In the first one, we built a median perpendicular to an existing segment. In the other two, we built a segment so that the given point lay on the perpendicular bisector, and then again built the perpendicular itself. Please note that we have learned how to build the perpendicular bisector, height and median. We will talk about the construction of the fourth remarkable line in a triangle, the bisector, later.

We have learned to build a line perpendicular to a given one. Is it possible to draw a straight line parallel to a given one using a compass and straightedge?

Task 4. Construct a line parallel to the given one.

Decision

Let there be a line and a point not lying on it (see Fig. 21). It is necessary to draw a line parallel to the line through the point. Again, we reduce the problem to the previous ones, using a sign of parallel lines: if two lines are perpendicular to a third, then they are parallel.

Rice. 21. Illustration for problem 4

Let's drop the perpendicular from the point to the line (we can do this) (see Fig. 22), and then draw another perpendicular through the point to the line just constructed (we can also) (see Fig. 23). As a result, we get the desired line (passes through and is parallel).

Rice. 22. Illustration for problem 4

Rice. 23. Illustration for problem 4

The fact that there can be only one such line guarantees us Euclid's fifth postulate: Through a point not on a line, only one line can be drawn parallel to the given line..

Problem solved.

Now we can return to the problem with segment division. We already know how to divide a segment into two equal parts. How about more parts? It is clear that into four parts - this is in half, and then each part in half again. And if on 3 or 7?

We have already considered this problem when we studied Thales theorem. Recall her wording: if parallel lines cut off equal segments on one side of an angle, then they cut off equal segments on the other side. This theorem can be used to divide a segment into any number of equal parts.

Task 5. Divide the segment into 7 equal parts.

Decision

Suppose you need to divide the segment into 7 equal parts. To do this, we draw a ray from the point that does not coincide with (see Fig. 24).

Rice. 24. Illustration for problem 5

We mark points on it at equal distances (see Fig. 25).

Rice. 25. Illustration for problem 5

Connect and (see Fig. 26).

Rice. 26. Illustration for problem 5

Through the remaining 6 points we draw straight lines that are parallel (we just learned how to do this). Since the segments are equal on one side of the angle, then, according to the Thales theorem, they are equal on the other side (see Fig. 27).

Rice. 27. Illustration for problem 5

Problem solved.

So, we already know how:

build a perpendicular bisector to a segment;
divide the segment in half using the perpendicular bisector;
divide the segment into an arbitrary number of equal parts using the Thales theorem;
construct a perpendicular to the line passing through the given point (moreover, the point can lie both on the line and outside it);
construct a parallel line through a point not on the given line.

The main elements of polygons are segments and angles. With segments, we have already learned a lot. Let's talk about corners.

The first task that arises for us is the construction of an angle equal to a given one. For segments, a similar problem was solved directly using a compass. The corners are a little more difficult.

Task 6. Set aside an angle from the beam equal to the given one.

Decision

Usually we need an equal angle not in an arbitrary place, but in a specific one, that is, one of its sides is already known. In this case, the problem is formulated as follows: to postpone the angle from the beam equal to the given one.

So, here is the corner with the top (see Fig. 28). The rays are its sides.

Rice. 28. Illustration for problem 6

There is a ray with a vertex (see Fig. 29). It is necessary to postpone from this beam an angle equal to the first angle.

Rice. 29. Illustration for problem 6

We usually met equal angles when proving the equality of triangles. We use this idea “on the contrary” - we construct equal triangles with corners at the vertices and from their equality we prove the equality of the angles.

Draw a circle of arbitrary radius from a point. We get points on the sides of the corner and a triangle (see Fig. 30).

Rice. 30. Illustration for problem 6

Let's construct a triangle equal to . Draw a circle from with the same radius. Let's get a point (see Fig. 31).

Rice. 31. Illustration for problem 6

In the first triangle, we “measure” the segment with a compass and draw a circle from the point with this radius. We get the intersection point of two circles - (see Fig. 32).

Rice. 32. Illustration for problem 6

Let's compare the two resulting triangles (see Fig. 33).

Rice. 33. Illustration for problem 6

(these are all equal radii of two circles)

(the point lies on a circle with a radius equal to )

It turns out that the triangles are equal on three sides (the third sign of the equality of triangles). So, the angles we need are also equal.

Problem solved.

Why are there two dots??

If two circles intersect, then at two points (see Fig. 34). We chose only one to build the corner. Why didn't we like the second one?

Rice. 34. Two circles intersect at points and

The fact is that the condition did not say in which direction from this beam an equal angle should be set aside (this can be done clockwise and counterclockwise). Accordingly, it is possible to construct two angles that satisfy this condition (see Fig. 35). We randomly chose one of them. But the second one is no worse, you could choose it (it depends on additional conditions).

Rice. 35. Two equal angles clockwise and counterclockwise from a given ray

In order to determine how many solutions a construction problem has, a research stage is usually carried out. We will talk more about it at the end of the lesson.

The task of constructing the median was reduced to dividing the segment in half. To build a bisector, you need to learn how to bisect an angle.

Task 7. Construct a bisector (divide the angle in half).

Decision

Consider an angle with a vertex at a point (see Fig. 36). Let's build two equal triangles again to get equal angles.

Rice. 36. Illustration for problem 7

Draw a circle with an arbitrary radius centered at the point . We get points on the sides of the corner and , where (see Fig. 37).

Rice. 37. Illustration for problem 7

From the points and draw another circle of equal radius (it can be the same, it can be different). The intersection of the circles will give a point (see Fig. 38). There will be two points, but you can choose any; if you drew circles of the same radius as in the first step, then the second point will coincide with - there will be no choice.

Rice. 38. Illustration for problem 7

We get that . Connect the dots and (see Fig. 39).

Rice. 39. Illustration for problem 7

The two resulting triangles are equal. Why, answer yourself. Well, since they are equal, then the angles are equal , is the bisector.

Problem solved.

By analogy with dividing a segment, I want to immediately proceed to dividing the angle into an arbitrary equal number of parts. Again, it is clear how to divide the angle into , etc. parts. Is it possible to divide an angle into three equal parts using a compass and straightedge? More on this below.

Dividing an angle into three parts

It turns out that in the general case, the division of an angle into three equal parts cannot be performed using only a compass and a straightedge. What does "generally" mean? For some special cases, for example, for a right angle, the problem is solved: you can simply construct an angle equal to (using the property of a right triangle - a leg that lies opposite the angle 2 times less than the hypotenuse).

But we are talking about an arbitrary angle (the degree measure of which we do not know in advance). In this case, the problem is not solved. This task is called angle trisection problem. And it is not the only one of the construction problems that cannot be solved with the help of a compass and a ruler (note: dividing an angle into three parts is generally and in principle not difficult - just take a protractor).

An example of another such well-known unsolvable problem is circle squaring problem. It requires constructing a square whose area would be equal to the area of the given circle. If we take a circle of radius 1, then the problem is reduced to constructing a square with a side equal to . It turns out that it also cannot be solved with a compass and straightedge.

Please note that this is not about the fact that at the moment they have not figured out how to do this, but they have proved that this cannot be done. That is, they proved that, no matter how they tried to use a compass and a ruler, it would not be possible to solve these problems.

Now practice on your own. Build a triangle on three sides. You are given three segments (see Fig. 40).

Rice. 40. Data segments

Construct a triangle whose sides are equal to these three segments. The decision can be found below.

Building a Triangle with Three Sides

Task. Construct a triangle on three sides (see Fig. 41).

Rice. 41. Illustration for the problem

Decision

To start somewhere, let's draw an arbitrary straight line and plot the first side of the triangle on it (see Fig. 42). Which side to take first does not matter, let it be the side.

Rice. 42. Illustration for the problem

From the ends of the postponed segment we draw two circles with radii and . The intersection of the circles will give us the third point (see Fig. 43).

Rice. 43. Illustration for the problem

There are two intersection points - you can choose any; build both versions of the triangles and make sure that they are the same triangles, symmetrical to each other with respect to the line (see Fig. 44).

Rice. 44. Illustration for the problem

Top opposite side a denoted as standard. Connect with the ends of the segment on a straight line. Obviously, the sides of the resulting triangle are equal to the given three segments. It remains to designate the two remaining vertices. Opposite the side is the top, opposite the side is the top (see Fig. 45).