Constructing a triangle using 2 sides. Video tutorial “Constructing a triangle using three elements

We present to your attention a video tutorial on the topic “Constructing a triangle using three elements.” You will be able to solve several examples from the class of construction problems. The teacher will analyze in detail the problem of constructing a triangle using three elements, and will also recall the theorem on the equality of triangles.

This topic has wide practical application, so we will consider some types of problem solving. Let us remind you that any constructions are carried out exclusively with the help of a compass and a ruler.

Example 1:

Construct a triangle using two sides and the angle between them.

Given: Suppose the analyzed triangle looks like this

Rice. 1.1. Analyzed triangle example 1

Let the given segments be c and a, and the given angle be

Rice. 1.2. Given elements for example 1

Construction:

First you should set aside corner 1

Rice. 1.3. Deferred angle 1 for example 1

Then, on the sides of a given angle, we draw two given sides with a compass: measure the length of the side with a compass A and place the tip of the compass at the vertex of angle 1, and with the other part we make a notch on the side of angle 1. We do a similar procedure with the side With

Rice. 1.4. Set aside sides A And With for example 1

Then we connect the resulting notches, and we get the desired triangle ABC

Rice. 1.5. Constructed triangle ABC for example 1

Will this triangle be equal to the expected one? It will, because the elements of the resulting triangle (two sides and the angle between them) are respectively equal to the two sides and the angle between them given in the condition. Therefore, by the first property of equality of triangles - - the desired one.

The construction is completed.

Note:

Let us recall how to plot an angle equal to a given one.

Example 2

Subtract an angle from a given ray equal to a given one. Angle A and ray OM are given. Build.

Construction:

Rice. 2.1. Condition for example 2

1. Construct a circle Okr(A, r = AB). Points B and C are the points of intersection with the sides of angle A

Rice. 2.2. Solution for example 2

1. Construct a circle Okr(D, r = CB). Points E and M are the points of intersection with the sides of angle A

Rice. 2.3. Solution for example 2

1. Angle MOE is the desired one, since .

The construction is completed.

Example 3

Construct triangle ABC using a known side and two adjacent angles.

Let the analyzed triangle look like this:

Rice. 3.1. Condition for example 3

Then the given segments look like this

Rice. 3.2. Condition for example 3

Construction:

Let's plot the angle on the plane

Rice. 3.3. Solution for example 3

On the side of a given angle we plot the length of the side A

Rice. 3.4. Solution for example 3

Then we set aside the angle C from the vertex. Non-common sides of angles γ and α intersect at point A

Rice. 3.5. Solution for example 3

Is the constructed triangle the desired one? Is, since the side and two adjacent angles of the constructed triangle are respectively equal to the side and the angle between them given in the condition

Searched for by the second criterion for the equality of triangles

Construction completed

Example 4

Construct a triangle on 2 legs

Let the analyzed triangle look like this

Rice. 4.1. Condition for example 4

Known elements - legs

Rice. 4.2. Condition for example 4

This task differs from the previous ones in that the angle between the sides can be determined by default - 90 0

Construction:

Let's set aside an angle equal to 90 0. We will do this in exactly the same way as shown in example 2

Rice. 4.3. Solution for example 4

Then on the sides of this angle we plot the lengths of the sides A And b, given in the condition

Rice. 4.4. Solution for example 4

As a result, the resulting triangle is the desired one, because its two sides and the angle between them are respectively equal to the two sides and the angle between them given in the condition

Note that you can set aside an angle of 90 0 by constructing two perpendicular lines. Let's look at how to accomplish this task in an additional example.

Additional example

Restore the perpendicular to line p passing through point A,

Line p, and point A lying on this line

Rice. 5.1. Condition for additional example

Construction:

First, let's construct a circle of arbitrary radius with a center at point A

Rice. 5.2. Solution to additional example

This circle intersects a line r at points K and E. Then we construct two circles Okr(K, R = KE), Okr(E, R = KE). These circles intersect at points C and B. The segment NE is the required one,

Rice. 5.3. Answer to additional example

Unified collection of digital educational resources ().
Mathematics tutor ().

No. 285, 288. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I., edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
Construct an isosceles triangle using a side and an angle opposite the base.
Construct a right triangle using the hypotenuse and an acute angle
Construct a triangle using the angle, altitude, and bisector drawn from the vertex of the given angle.

The three theorems on the equality of triangles proved in paragraph 188 show that a triangle is completely defined if three of its sides, two sides and the angle enclosed between them, a side and two angles adjacent to it (or even two angles of any kind) are given.

The existence of a triangle, determined by specifying certain specific values of the sides or angles, is revealed when solving the problem of constructing a triangle using these elements: the uniqueness of the solution to the construction problem once again proves the signs of equality from paragraph 188. In accordance with the three signs of equality, three main problems arise on construction of triangles.

Problem 1. Given three segments a, b, c. Construct a triangle with these segments as its sides.

Solution. Let c be the largest of the three segments: in order for the problem to have a solution, it is necessary that the condition be satisfied. We will assume that this condition is satisfied. On an arbitrary straight line (Fig. 226), we plot the segment in an arbitrary place. Let's take its ends as the two vertices of the desired triangle. The third vertex must lie at a distance b from point A (or from point B) and at a distance a from B (or A). To construct the missing vertex, draw a circle of radius b with center A and a circle of radius a with center B.

These two circles will intersect, since, according to the condition, the distance between their centers is less than the sum of the radii and greater than their difference, since c is the largest segment among the data. We obtain two intersection points C and C, i.e., two possible positions of vertex C; the corresponding two triangles, however, are equal, as symmetrically located relative to AB. In Fig. 226 also shows how to obtain two more positions of the third vertex by swapping the radii of the circles.

Task 2. Construct a triangle using two sides and the angle between them.

Task 3. Construct a triangle using a side and adjacent angles whose sum is less than .

When analyzing the signs of equality of triangles, two circumstances attract attention:

1) There are no signs in which the equality of triangles would be ensured only by the equality of three angles. This is explained by the fact that two triangles having equal angles are not necessarily equal (similar triangles, see Chapter XVI for more details).

2) The sign of equality of triangles on two sides requires the equality of not arbitrary angles, but certainly those concluded between equal sides. To find out the reason for this, we pose the following problem.

Task 4. Construct a triangle using two sides and an angle opposite one of them.

Solution. Let, for example, be given sides a and b and an angle a lying opposite a (Fig. 227). To construct a triangle, let’s plot segment b on an arbitrary straight line AC and from one of its vertices, for example A, draw a ray AM at an angle a to segment AC. The unknown third side of the triangle must lie on this ray; its end is the missing vertex of the triangle. It is known, however, that this third vertex lies at a distance a from C and, therefore, is located on a circle with center C of radius a. Let's draw such a circle. The points of its intersection with the ray AM will give the possible positions of the third vertex. Since a circle and a ray may not have common points, or have one or two common points, then the problem may not have solutions, or have one or two solutions.

In Fig. 227 presents the case when the angle a is acute, and there are four options for the side for which the problem, accordingly, has no solutions, has one solution, two solutions, and again one solution. Both solutions are shown for A complete analysis of this problem is given in Section 223 in connection with problems solving triangles.

You can pose various other tasks to construct triangles using certain data. In all cases, to be able to construct a triangle, either three of its linear elements must be specified (i.e., three segments: sides, medians, altitudes, etc.), or two segments and one angle, or one segment and two corner.

Problem 5. Given two sides a, c of a triangle and the median . Construct a triangle.

Solution. Let's start solving the problem with analysis. This is the name of the solution stage, when we conditionally assume that the problem has already been solved and find out its features that will actually help us solve it. So, let's assume that triangle ABC (Fig. 228, a) is the desired one. Then in it

Note that the segment VM, by definition of the median, is half c, i.e., it can be considered known. But now all three sides of the IUD triangle are known! Here is the key to solving the problem, the rest is simple. We construct (Fig. 228, b) a triangle BMC along three sides and then extend the side VM to a distance equal to , thereby obtaining the third vertex A of the triangle. The correctness of the construction performed is clear.

The condition for the solvability of the problem is that it is possible to construct a “partial” triangle using side a, the median and half of the other side.

Lesson topic:Constructing a triangle using three elements

Objective of the lesson: learn to build triangles usingthree elements

Lesson objectives: constructing a triangle using a ruler and compass

Lesson progress:

Stage 1: organizational moment, greeting, checking homework

Stage 2: new topic

Constructing a triangle using two sides and the angle between them .

Given two segmentsaAndb, they are equal to the sides of the desired triangle, and the angle∡ 1 , equal to the angle of the triangle between the sides. It is necessary to construct a triangle with elements equal to the given segments and angle.

1. Draw a straight line.

Aa.

∡ 1 (vertex of angleA

4. On the other side of the angle, set aside a segment equal to this segmentb.

5. Connect the ends of the segments.

According to the criterion of equality of triangles along two sides and the angle between them, the constructed triangle is equal to all triangles that have these elements.

Constructing a triangle using a side and two adjacent angles .

Given a segmentaand two corners∡ 1 And∡ 2 , equal to the angles of the triangle adjacent to a given side. It is necessary to construct a triangle with elements equal to the given segment and angles.

1. Draw a straight line.

2. On a straight line from the selected pointAset aside a segment equal to a given segmentaB.

3. Construct an angle equal to the given one∡ 1 (vertex of angleA, one side of the angle lies on the straight line).

4. Construct an angle equal to the given one∡ 2 (vertex of angleB, one side of the angle lies on the straight line).

5. The point of intersection of the other sides of the angles is the third vertex of the desired triangle.

According to the criterion of equality of triangles along a side and two adjacent angles, the constructed triangle is equal to all triangles that have these elements.

Constructing a triangle using three sides .

Three segments are given:a, bAndc, equal to the sides of the desired triangle. It is necessary to construct a triangle with sides equal to these segments.

In this case, before starting construction, you need to make sure whether the triangle inequality is satisfied (the length of each segment is less than the sum of the lengths of the other two segments), and these segments can be sides of the triangle.

1. Draw a straight line.

2. On a straight line from the selected pointAset aside a segment equal to a given segmenta, and mark the other end of the segmentB.

3. Draw a circle with the centerAand a radius equal to the segmentb.

4. Draw a circle with the centerBand a radius equal to the segmentc.

5. The point of intersection of the circles is the third vertex of the desired triangle

According to the criterion of equality of triangles on three sides, the constructed triangle is equal to all triangles that have these sides.

Stage 3: problem solving

№ 239 page 74

construct a right triangle using two sides

Stage 4: summing up

Stage 5: homework No. 240 page 74

Example 1:

Construct a triangle using two sides and the angle between them.

Given: Suppose the analyzed triangle looks like this

Rice. 1.1. Analyzed triangle example 1

Let the given segments be c and a, and the given angle be

Rice. 1.2. Given elements for example 1

Construction:

First you should set aside corner 1

Rice. 1.3. Deferred angle 1 for example 1

Rice. 1.4. Set aside sides A And With for example 1

Then we connect the resulting notches, and we get the desired triangle ABC

Rice. 1.5. Constructed triangle ABC for example 1

The construction is completed.

Note:

Let us recall how to plot an angle equal to a given one.

Example 2

Subtract an angle from a given ray equal to a given one. Angle A and ray OM are given. Build.

Construction:

Rice. 2.1. Condition for example 2

1. Construct a circle Okr(A, r = AB). Points B and C are the points of intersection with the sides of angle A

Rice. 2.2. Solution for example 2

1. Construct a circle Okr(D, r = CB). Points E and M are the points of intersection with the sides of angle A

Rice. 2.3. Solution for example 2

1. Angle MOE is the desired one, since .

The construction is completed.

Example 3

Construct triangle ABC using a known side and two adjacent angles.

Let the analyzed triangle look like this:

Rice. 3.1. Condition for example 3

Then the given segments look like this

Rice. 3.2. Condition for example 3

Construction:

Let's plot the angle on the plane

Rice. 3.3. Solution for example 3

On the side of a given angle we plot the length of the side A

Rice. 3.4. Solution for example 3

Then we set aside the angle C from the vertex. Non-common sides of angles γ and α intersect at point A

Rice. 3.5. Solution for example 3

Searched for by the second criterion for the equality of triangles

Construction completed

Example 4

Construct a triangle on 2 legs

Let the analyzed triangle look like this

Rice. 4.1. Condition for example 4

Known elements - legs

Rice. 4.2. Condition for example 4

This task differs from the previous ones in that the angle between the sides can be determined by default - 90 0

Construction:

Let's set aside an angle equal to 90 0. We will do this in exactly the same way as shown in example 2

Rice. 4.3. Solution for example 4

Then on the sides of this angle we plot the lengths of the sides A And b, given in the condition

Rice. 4.4. Solution for example 4

As a result, the resulting triangle is the desired one, because its two sides and the angle between them are respectively equal to the two sides and the angle between them given in the condition

Note that you can set aside an angle of 90 0 by constructing two perpendicular lines. Let's look at how to accomplish this task in an additional example.

Additional example

Restore the perpendicular to line p passing through point A,

Line p, and point A lying on this line

Rice. 5.1. Condition for additional example

Construction:

First, let's construct a circle of arbitrary radius with a center at point A

Rice. 5.2. Solution to additional example

This circle intersects a line r at points K and E. Then we construct two circles Okr(K, R = KE), Okr(E, R = KE). These circles intersect at points C and B. The segment NE is the required one,

Rice. 5.3. Answer to additional example

Unified collection of digital educational resources ().
Mathematics tutor ().

No. 285, 288. Atanasyan L. S., Butuzov V. F., Kadomtsev S. B., Poznyak E. G., Yudina I. I., edited by Tikhonov A. N. Geometry grades 7-9. M.: Enlightenment. 2010
Construct an isosceles triangle using a side and an angle opposite the base.
Construct a right triangle using the hypotenuse and an acute angle
Construct a triangle using the angle, altitude, and bisector drawn from the vertex of the given angle.

Finally, consider a problem whose solution leads to the construction of a triangle using a side and two angles:

On the other side of the river (Fig. 72) a milestone is visible A. It is required, without crossing the river, to find out the distance to it from the milestone IN on this shore.

Let's do this. Let's measure from the point IN any distance in a straight line Sun and at the ends of it IN And WITH Let's measure angles 1 and 2 (Fig. 73). If we now measure the distance on a convenient area DE, equal Sun, and build angles at its ends A And b(Fig. 74), equal to angles 1 and 2, then at the point of intersection of their sides we get the third vertex F triangle DEF. It is easy to verify that the triangle DEF equal to a triangle ABC; indeed, if we imagine that the triangle DEF superimposed on ABC so that side DE coincided with its equal side Sun, then ug. A will coincide with angle 1, angle b – with angle 2, and side DF will go to the side VA, and the side E.F. on the side SA. Since two lines can intersect only at one point, then the vertex F should coincide with the top A. So the distance DF equal to the required distance VA.

The problem, as we see, has only one solution. In general, using a side and two angles adjacent to this side, only one triangle can be constructed; There cannot be other triangles with the same side and the same two angles adjacent to it in the same places. All triangles that have one identical side and two identical angles adjacent to it in the same places can be brought into complete coincidence by superposition. This means that this is a sign by which one can establish the complete equality of triangles.

Together with the previously established signs of equality of triangles, we now know the following three:

Triangles:

on three sides;

at the two sides and at the corner between them;

on the side and two sides.

For the sake of brevity, we will further denote these three cases of equality of triangles as follows:

on three sides: SSS;

on two sides and the angle between them: SUS;

along the side and two corners: USU.

Applications

14. To find out the distance to a point A on the other side of the river from the point IN on this bank (Fig. 5), measure some line in a straight line sun, then at point IN construct an angle equal to ABC, on the other side Sun, and at the point WITH- in the same way, an angle equal to DIA Point distance D intersection of the sides of both sides of the angles to the point IN equal to the required distance AB. Why?

Solution: Triangles ABC And BDC equal on one side ( Sun) and two angles (ang. DCB= ug. DIA; ug. DBC= ug. ABC.) Hence, AB= ВD, as sides lying in equal triangles against equal angles.