How to make the construction of an angle equal to a given one. How to construct an angle equal to a given

Often it is necessary to draw (“build”) an angle that would be equal to a given angle, and the construction must be done without the help of a protractor, but using only a compass and a ruler. Knowing how to build a triangle on three sides, we can solve this problem. Let on a straight line MN(dev. 60 and 61) is required to be built at the point K angle equal to angle B. This means that it is necessary from the point K draw a straight line constituting MN angle equal to B.

To do this, mark a point on each side of a given angle, for example BUT and FROM, and connect BUT and FROM straight line. Get a triangle ABC. Let's build now on a straight line MN this triangle so that its apex AT was at the point To: then this point will have an angle equal to the angle AT. Build a triangle on three sides Sun, VA and AU we can: postpone (dev. 62) from the point To line segment kl, equal sun; get a point L; around K, as near the center, we describe a circle with a radius VA, and around L- radius SA. Point R connect the intersections of the circles with To and Z, - we get a triangle KPL, triangular ABC; it has a corner To= ang. AT.

This construction is faster and more convenient if from the top AT set aside equal segments (with one dissolution of the compass) and, without moving its legs, describe with the same radius a circle around the point TO, like near the center.

How to cut a corner in half

Let it be required to divide the angle BUT(Fig. 63) into two equal parts using a compass and ruler, without using a protractor. We'll show you how to do it.

From the top BUT draw equal segments on the sides of the angle AB and AU(Fig. 64; this is done with one dissolution of the compass). Then we put the tip of the compass at the points AT and FROM and describe with equal radii the arcs intersecting at the point D. straight line connecting BUT and D divides the angle BUT in half.

Let's explain why. If the point D connect with AT and C (Fig. 65), then you get two triangles ADC and adb, u which have a common side AD; side AB equal to the side AU, a BD is equal to CD. Triangles are equal on three sides, so the angles are equal. bad and DAC, lying opposite equal sides BD and CD. Therefore, a straight line AD divides the angle YOU in half.

Applications

12. Construct an angle of 45° without a protractor. At 22°30'. At 67°30'.

Solution. Dividing the right angle in half, we get an angle of 45 °. Dividing the angle of 45° in half, we get an angle of 22°30'. By constructing the sum of the angles 45° + 22°30', we get an angle of 67°30'.

How to draw a triangle given two sides and an angle between them

Let it be required on the ground to find out the distance between two milestones BUT and AT(device 66), separated by an impenetrable swamp.

How to do it?

We can do this: aside from the swamp, we choose such a point FROM, from where both milestones are visible and it is possible to measure distances AU and Sun. Corner FROM we measure with the help of a special goniometric device (called an astrolabe). According to these data, i.e., according to the measured sides AC and sun and corner FROM between them, build a triangle ABC somewhere in a convenient location as follows. Having measured one known side in a straight line (Fig. 67), for example AU, build with it at the point FROM corner FROM; on the other side of this angle, a known side is measured Sun. Ends of known sides, i.e. points BUT and AT connected by a straight line. It turns out a triangle in which two sides and the angle between them have pre-specified dimensions.

It is clear from the method of construction that only one triangle can be constructed given two sides and the angle between them. therefore, if two sides of one triangle are equal to two sides of another and the angles between these sides are the same, then such triangles can be superimposed on each other by all points, that is, they must also have third sides and other angles equal. This means that the equality of the two sides of the triangles and the angle between them can serve as a sign of the complete equality of these triangles. Shortly speaking:

Triangles are equal under two sides and angles between them.

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

With a ruler, you can:

arbitrary line;

an arbitrary 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.

A compass can be used to draw a segment on a given line from a given point.

Consider the main tasks for the construction.

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

Solution. With the help of a ruler, draw an arbitrary straight line and take an arbitrary point B on it. With 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 the center B, and with a compass opening equal to b - a circle from the 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 line segments to serve as sides of a triangle, it is necessary that the larger of them be less than the sum of the other two (and< b + с).

Task 2.

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

Draw an arbitrary circle centered at the 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). Let's draw a circle with radius AB with the center at the point O - the starting point of this ray (Fig. 3, b). The point of intersection of this circle with the given ray will be denoted as С 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 \u003d Δ OB 1 C 1 (the third criterion for the equality of triangles).

Task 3. Construct the bisector of the given 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 with the same radius we describe circles. Let D be their intersection point, different from A. Ray AD divides angle A in half. This follows from the equality ΔABD = ΔACD (the third criterion for the equality of triangles).

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

Solution. With an arbitrary but identical compass opening (large 1/2 AB), we describe two arcs with centers at points A and B, which intersect each other at some points C and D. The straight line CD will be the required 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 section 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. Two cases are possible:

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

From point O we draw a circle with an arbitrary radius intersecting the straight line a at points A and B. From points A and B we draw circles with the same radius. Let О 1 be their intersection point different from О. We get ОО 1 ⊥ AB. In fact, the 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: angle A. A Constructed angle O. B C O D E Prove: A \u003d O Proof: consider triangles ABC and ODE. 1.AC=OE, as radii of one circle. 2.AB=OD, as the radii of one circle. 3.BC=DE, as radii of one circle. ABC \u003d ODE (3 prizes) A ​​\u003d O

Let us prove that the ray AB is a bisector A P L A N 1. Additional construction. 2. Let's prove the equality of triangles ACB and ADB. 3. Conclusions A B C D 1.AC=AD, as radii of one circle. 2.CB=DB, as radii of one circle. 3.AB - common side. ASV \u003d ADB, according to the III sign of equality of triangles Beam AB - bisector Construction of the bisector of the angle.

A N B A C 1 = 2 12 In the r/b triangle AMB, the segment MC is a bisector, and hence the height. Then, a 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 is the common side. MBN= MAN, on three sides Construction of perpendicular lines. M a

Q P VA APQ \u003d BPQ, on three sides \u003d 2 Triangle ARV r / b. The segment RO is a bisector, and therefore a median. Then point O is the midpoint of AB. О Let us prove that О is the midpoint of the segment AB. Construction of the middle of the segment

D С Construction of a triangle given two sides and an angle between them. Angle hk h 1. Let's build a beam a. 2. Set aside the segment AB, equal to P 1 Q 1. 3. Construct an angle equal to this one. 4. Set aside the segment AC, equal to P 2 Q 2. B A The triangle ABC is the desired one. Justify using the I sign. Given: Segments P 1 Q 1 and P 2 Q 2 Q1Q1 P1P1 P2P2 Q2Q2 a k

D С Construction of a triangle by a side and two angles adjacent to it. Angle h 1 k 1 h2h2 1. Let's build a beam a. 2. Set aside the 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. B 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. 2. Set aside the segment AB, equal to P 1 Q 1. 3. Construct an arc centered at point A and radius P 2 Q 2. 4. Construct an arc centered at point B and radius P 3 Q 3. B A Triangle ABC desired. Justify using the III sign. Given: segments P 1 Q 1, P 2 Q 2, P 3 Q 3. Q1Q1 P1P1 P3P3 Q2Q2 and P2P2 Q3Q3 Construction of a triangle on three sides.

The purpose of the lesson: Formation of the ability to build an angle equal to a given one. Task: Create conditions for mastering the construction algorithm using a compass and a ruler of an angle equal to a given one; create conditions for mastering the sequence of actions when solving a construction problem (analysis, construction, proof); improve the skill of using the properties of a circle, signs of equality of triangles to solve the problem of proof; provide the opportunity to apply new skills in solving problems

In geometry, construction tasks are distinguished that can be solved only with the help of two tools: a compass and a ruler without scale divisions. The ruler allows you to draw an arbitrary straight line, as well as build a straight line passing through two given points; using a compass, you can draw a circle of arbitrary radius, as well as a circle with a center at a given point and a radius equal to a given segment. I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I IIII I

Given: angle A. A Constructed: angle O. B C O D E Prove: A = O Proof: consider triangles ABC and ODE. 1.AC=OE, as radii of one circle. 2.AB=OD, as the radii of one circle. 3.BC=DE, as radii of one circle. ABC \u003d ODE (3 prizes) A ​​\u003d O Task 2. Set aside an angle equal to this one from a given beam

Let us prove that the ray AB is the bisector of A 3. Proof: Additional construction (let's connect the point B with the points D and C). Consider ASV and ADB: A B C D 1.AC=AD as radii of one circle. 2.CB=DB, as radii of one circle. 3. AB - common side. DIA \u003d ADB, according to the III sign of equality of triangles Ray AB is a bisector 4. Research: The problem always has a unique solution.

Scheme for solving construction problems: Analysis (drawing the desired figure, establishing links between the given and desired elements, construction plan). Building according to plan. Proof that the figure satisfies the conditions of the problem. Research (when and how many solutions does the problem have?).