How to construct an angle equal to a given one. Constructing an angle equal to a given one

Lesson objectives:

• Formation of the ability to analyze the studied material and the skills of applying it to solve problems;
• Show the significance of the concepts being studied;
• Development of cognitive activity and independence in acquiring knowledge;
• Cultivating interest in the subject and a sense of beauty.

Lesson objectives:

• Develop skills in constructing an angle equal to a given one using a scale ruler, compass, protractor and drawing triangle.
• Test students' problem-solving skills.

Lesson plan:

1. Repetition.
2. Constructing an angle equal to a given one.
3. Analysis.
4. Construction example first.
5. Construction example two.

Repetition.

Corner.

Flat angle- an unlimited geometric figure formed by two rays (sides of an angle) emerging from one point (vertex of the angle).

An angle is also called a figure formed by all points of the plane enclosed between these rays (Generally speaking, two such rays correspond to two angles, since they divide the plane into two parts. One of these angles is conventionally called internal, and the other - external.
Sometimes, for brevity, the angle is called the angular measure.

There is a generally accepted symbol to denote an angle: , proposed in 1634 by the French mathematician Pierre Erigon.

Corner is a geometric figure (Fig. 1), formed by two rays OA and OB (sides of the angle), emanating from one point O (vertex of the angle).

An angle is denoted by a symbol and three letters indicating the ends of the rays and the vertex of the angle: AOB (and the letter of the vertex is the middle one). Angles are measured by the amount of rotation of ray OA around vertex O until ray OA moves to position OB. There are two widely used units for measuring angles: radians and degrees. For radian measurement of angles, see below in the paragraph “Arc Length”, as well as in the chapter “Trigonometry”.

Degree system for measuring angles.

Here the unit of measurement is a degree (its designation is °) - this is a rotation of the beam by 1/360 of a full revolution. Thus, a full rotation of the beam is 360 o. One degree is divided into 60 minutes (symbol ‘); one minute – respectively for 60 seconds (designation “). An angle of 90° (Fig. 2) is called right; an angle less than 90° (Fig. 3) is called acute; an angle greater than 90° (Fig. 4) is called obtuse.

Straight lines forming a right angle are called mutually perpendicular. If the lines AB and MK are perpendicular, then this is denoted: AB MK.

Constructing an angle equal to a given one.

Before starting construction or solving any problem, regardless of the subject, you need to carry out analysis. Understand what the assignment says, read it thoughtfully and slowly. If after the first time you have doubts or something was not clear or clear but not completely, it is recommended to read it again. If you are doing an assignment in class, you can ask the teacher. Otherwise, your task, which you misunderstood, may not be solved correctly, or you may find something that is not what was required of you, and it will be considered incorrect and you will have to redo it. As for me - It’s better to spend a little more time studying the task than to redo the task all over again.

Analysis.

Let a be the given ray with vertex A, and the angle (ab) be the desired one. Let's choose points B and C on rays a and b, respectively. By connecting points B and C, we get triangle ABC. In congruent triangles, the corresponding angles are equal, and this is where the method of construction follows. If on the sides of a given angle we select points C and B in some convenient way, and from a given ray into a given half-plane we construct a triangle AB 1 C 1 equal to ABC (and this can be done if we know all the sides of the triangle), then the problem will be solved.

When carrying out any constructions Be extremely careful and try to carry out all constructions carefully. Since any inconsistencies can result in some kind of errors, deviations, which can lead to an incorrect answer. And if a task of this type is performed for the first time, the error will be very difficult to find and fix.

Construction example first.

Let's draw a circle with its center at the vertex of this angle. Let B and C be the points of intersection of the circle with the sides of the angle. With radius AB we draw a circle with the center at point A 1 – the starting point of this ray. Let us denote the point of intersection of this circle with this ray as B 1 . Let us describe a circle with center at B 1 and radius BC. The intersection point C 1 of the constructed circles in the indicated half-plane lies on the side of the desired angle.

Triangles ABC and A 1 B 1 C 1 are equal on three sides. Angles A and A 1 are the corresponding angles of these triangles. Therefore, ∠CAB = ∠C 1 A 1 B 1

For greater clarity, you can consider the same constructions in more detail.

Construction example two.

The task remains to also set aside an angle equal to a given angle from a given half-line into a given half-plane.

Construction.

Step 1. Let us draw a circle with an arbitrary radius and centers at vertex A of a given angle. Let B and C be the points of intersection of the circle with the sides of the angle. And let's draw segment BC.

Step 2. Let's draw a circle of radius AB with the center at point O - the starting point of this half-line. Let us denote the point of intersection of the circle with the ray as B 1 .

Step 3. Now we describe a circle with center B 1 and radius BC. Let point C 1 be the intersection of the constructed circles in the indicated half-plane.

Step 4. Let's draw a ray from point O through point C 1. Angle C 1 OB 1 will be the desired one.

Proof.

Triangles ABC and OB 1 C 1 are congruent triangles with corresponding sides. And therefore angles CAB and C 1 OB 1 are equal.

Interesting fact:

In numbers.

In objects of the surrounding world, you first of all notice their individual properties that distinguish one object from another.

The abundance of particular, individual properties obscures the general properties inherent in absolutely all objects, and it is therefore always more difficult to detect such properties.

One of the most important general properties of objects is that all objects can be counted and measured. We reflect this general property of objects in the concept of number.

People mastered the process of counting, that is, the concept of number, very slowly, over centuries, in a persistent struggle for their existence.

In order to count, one must not only have objects that can be counted, but also already have the ability to abstract when considering these objects from all their other properties except number, and this ability is the result of a long historical development based on experience.

Every person now learns to count with the help of numbers imperceptibly in childhood, almost simultaneously with the time he begins to speak, but this counting, which is familiar to us, has gone through a long path of development and has taken different forms.

There was a time when only two numerals were used to count objects: one and two. In the process of further expansion of the number system, parts of the human body were involved, primarily fingers, and if this kind of “numbers” was not enough, then also sticks, pebbles and other things.

N. N. Miklouho-Maclay in his book "Trips" talks about a funny method of counting used by the natives of New Guinea:

Questions:

1. Define angle?
2. What types of angles are there?
3. What is the difference between diameter and radius?

List of sources used:

1. Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”
2. Mathematical savvy. B.A. Kordemsky. Moscow.
3. L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina “Geometry, 7 – 9: textbook for educational institutions”

Worked on the lesson:

Levchenko V.S.

Poturnak S.A.

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In construction tasks we will consider the construction of a geometric figure, which can be done using a ruler and compass.

Using a ruler you can:

arbitrary straight line;

an arbitrary straight line passing through a given point;

a straight line passing through two given points.

Using a compass, you can describe a circle of a given radius from a given center.

Using a compass you can plot a segment on a given line from a given point.

Let's consider the main construction tasks.

Task 1. Construct a triangle with given sides a, b, c (Fig. 1).

Solution. Using a ruler, draw an arbitrary straight line and take an arbitrary point B on it. Using a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a compass opening equal to c, we describe a circle from center B, and with a compass opening equal to b, we describe a circle from center C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.

Comment. In order for three straight segments to serve as sides of a triangle, it is necessary that the largest of them be less than the sum of the other two (and< b + с).

Task 2.

Solution. This angle with vertex A and the ray OM are shown in Figure 2.

Let us draw an arbitrary circle with its center at vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). With radius AB we draw a circle with the center at point O - the starting point of this ray (Fig. 3, b). Let us denote the point of intersection of this circle with this ray as C 1 . Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC = Δ OB 1 C 1 (the third sign of equality of triangles).

Task 3. Construct the bisector of this angle (Fig. 4).

Solution. From vertex A of a given angle, as from the center, we draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C we describe circles with the same radius. Let D be their intersection point, different from A. Ray AD bisects angle A. This follows from the equality Δ ABD = Δ ACD (the third criterion for the equality of triangles).

Task 4. Draw a perpendicular bisector to this segment (Fig. 5).

Solution. Using an arbitrary but identical compass opening (larger than 1/2 AB), we describe two arcs with centers at points A and B, which will intersect each other at some points C and D. The straight line CD will be the desired perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the perpendicular bisector to segment AB.

Task 5. Divide this segment in half. It is solved in the same way as problem 4 (see Fig. 5).

Task 6. Through a given point draw a line perpendicular to the given line.

Solution. There are two possible cases:

1) a given point O lies on a given straight line a (Fig. 6).

From point O we draw a circle with an arbitrary radius intersecting line a at points A and B. From points A and B we draw circles with the same radius. Let O 1 be the point of their intersection, different from O. We obtain OO 1 ⊥ AB. In fact, points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the perpendicular bisector to this segment.

Constructing an angle equal to a given one. Given: half-line, angle. Construction. V.A.S. 7. For proof, it is enough to note that triangles ABC and OB1C1 are congruent as triangles with respectively equal sides. Angles A and O are the corresponding angles of these triangles. It is necessary: ​​to put off from a given half-line into a given half-plane an angle equal to a given angle. C1. IN 1. A. 1. Let's draw an arbitrary circle with its center at vertex A of the given angle. 2. Let B and C be the points of intersection of the circle with the sides of the angle. 3. Using radius AB we draw a circle with the center at point O - the starting point of this half-line. 4. Let us denote the point of intersection of this circle with this half-line as B1. 5. Let us describe a circle with center B1 and radius BC. 6. Point C1 of intersection of the constructed circles in the indicated half-plane lies on the side of the desired angle.

Geometry 7th grade

“Isosceles triangle” - Theorem. A triangle is the simplest closed rectilinear figure. Problem solving. Find angle KBA. Equality of triangles. Guess the rebus. ABC - isosceles. List the congruent elements of triangles. Classification of triangles by sides. In an isosceles triangle AMK AM = AK. Classification of triangles according to the size of their angles. Sides. A triangle with all sides equal. Isosceles triangle.

“Measuring segments and angles” - Comparison of segments. http://www.physicsdepartment.ru/blog/images/0166.jpg. Ф3 = Ф4. MN > CD. 1m =. The middle of the segment. 1km. What is the greatest number of parts that 4 different straight lines can divide a plane into? Other units of measurement. Comparing shapes using overlay. Comparison of angles. The sides of the VM and the EU came together. How many parts can a plane be divided into by 3 different straight lines? http://www.robertagor.it/calibro.jpg.

“Right triangle, its properties” - One of the corners of a right triangle. Solution. Which triangle is called a right triangle? Right triangle. Properties of a right triangle. Warm up. Development of logical thinking. Bisector. Leg of a right triangle. Let's create an equation. Let's look at the drawing carefully. Property of a right triangle. Residents of three houses. Triangle.

“Definition of an angle” - Concepts of angles. Draw the rays. Preparatory stage of the lesson. Corner. Explanation of new material. An angle divides a plane. Concepts of internal and external areas of an angle. Get interested in the subject. The ray in the figure divides the angle. Definition of a straight angle. Development of logical thinking. Obtuse angle. Sharp corner. Opening words. Paint the inside area of ​​the corner. Angles. Ray BM divides angle ABC into two angles.

“The second and third signs of equality of triangles” - Sides. Median in an isosceles triangle. The second and third signs of equality of triangles. Solution. Three sides of one triangle. Base. Prove. Properties of an isosceles triangle. Signs of equality of triangles. Problem solving. Mathematical dictation. Angles. Task. Perimeter of an isosceles triangle.

“Cartesian coordinate system on a plane” - The plane on which the Cartesian coordinate system is specified. Coordinates in people's lives. Geographic coordinate system. Cartesian coordinate system on a plane. Algebra project. Scientists who are the authors of the coordinates. Ancient Greek astronomer Claudius. A cell on the playing field. The point of intersection of the axes. Introduction of simpler notation to algebra. A place in a cinema. The meaning of the Cartesian coordinate system.

Constructing an angle equal to a given one. Given: angle A. A Constructed angle O. B C O D E Prove: A = O Proof: consider triangles ABC and ODE. 1.AC = OE, like the radii of one circle. 2.AB=OD, as the radii of one circle. 3.ВС=DE, as the radii of one circle. ABC = ODE (3rd prize) A = O

Let us prove that the ray AB is a bisector A P L A N 1.Additional construction. 2. Let us prove the equality of triangles ACB and ADB. 3. Conclusions A B C D 1.AC = AD, as the radii of one circle. 2.CB=DB, as the radii of one circle. 3.AB – common side. ACB = ADB, according to the III criterion of equality of triangles Ray AB - bisector Construction of the bisector of an angle.

A N B A C 1 = 2 12 In the r/b triangle AMB, the segment MC is a bisector, and therefore a height. Then, and MN. M Let's prove that a MN Let's look at the location of the compasses. AM=AN=MB=BN, as equal radii. MN-common side. MВN= MAN, on three sides Construction of perpendicular lines. M a

Q P BA ARQ = BPQ, on three sides = 2 Triangle ARV r/b. The segment PO is a bisector, and therefore a median. Then, point O is the middle of AB. О Let us prove that O is the midpoint of the segment AB. Constructing the midpoint of a segment

D C Constructing a triangle using two sides and the angle between them. Angle hk h 1. Let's construct ray a. 2. Set aside a segment AB equal to P 1 Q 1. 3. Construct an angle equal to this one. 4. Let us set aside the segment AC equal to P 2 Q 2. VA Triangle ABC is the desired one. Justify using the first sign. Given: Segments P 1 Q 1 and P 2 Q 2 Q1Q1 P1P1 P2P2 Q2Q2 a k

D C Constructing a triangle using a side and two adjacent angles. Angle h 1 k 1 h2h2 1. Construct ray a. 2. Set aside a segment AB equal to P 1 Q 1. 3. Construct an angle equal to the given h 1 k 1. 4. Construct an angle equal to h 2 k 2. BA A Triangle ABC is the desired one. Justify using the second sign. Given: Segment P 1 Q 1 Q1Q1 P1P1 a k2k2 h1h1 k1k1 N

C 1. Let's build a ray a. 2. Set aside a segment AB equal to P 1 Q 1. 3. Construct an arc with a center at point A and radius P 2 Q 2. 4. Construct an arc with a center at point B and radius P 3 Q 3. BA A Triangle ABC sought after Justify using the third sign. Given: segments P 1 Q 1, P 2 Q 2, P 3 Q 3. Q1Q1 P1P1 P3P3 Q2Q2 a P2P2 Q3Q3 Construction of a triangle using three sides.