CHAPTER VIII.

PROPORTIONALITY OF LINES. SIMILARITY OF FIGURES.

§ 93. CONSTRUCTION OF SIMILAR FIGURES.

1. Construction of similar triangles.

We already know that to construct a triangle similar to the given one, it is enough to draw a line parallel to the side of the triangle from some point taken on the side of the triangle. We get a triangle similar to this one (Fig. 382):

/\ DIA /\ A"C"B"

2. Construction of similar polygons.

To construct a polygon similar to the given one, we can proceed as follows: we divide the given polygon into triangles by diagonals drawn from any of its vertices (Fig. 383). On some side of the given polygon ABCDE, for example, on side AE, we take some point E" and draw a line parallel to side ED until it intersects diagonal AD, for example, at point D".

From point D" draw a line parallel to side DC until it intersects diagonal AC at point C". From point C" draw a line parallel to side CB until it intersects with side AB at point B". The resulting polygon AB"C"D"E" is similar to the given polygon ABCDE.

The validity of this statement is proved independently.

If it is required to build a polygon similar to the given one with the specified similarity coefficient, then the starting point E" is taken on the side AE ​​or its continuation, respectively, according to the given similarity coefficient.

3. Shooting a plan of the land plot.

a) The shooting of the plan is carried out using a special device called beaker(dev. 384).

The menzula is a square board placed on a tripod. When drawing a plan, the board is brought to a horizontal position, which is checked using a level. To draw straight lines in the desired direction, an alidade equipped with diopters is used. Each diopter has a slot in which the hair is stretched, which allows you to accurately direct the alidade in the right direction. A sheet of white paper is fastened to the scale with buttons, on which the plan is drawn.

In order to take a plan from the land plot ABCDE, choose some point O inside the plot so that all the tops of the land plot are visible from it (Fig. 385).

With the help of a fork with a plumb line (Fig. 386), the scale is set so that the point O, marked on a sheet of paper, falls against the point O chosen on the site.

Then, from point O on a sheet of paper attached to the beaker, rays are drawn with an alidade in directions to points A, B, C, D and E; measure distances
OA, OB, OS, OD and OE and lay on these rays in the accepted scale segments
OA", OB", OS, OD" and OE".

Points A, B, C, D, and E are connected. It turns out the polygon A "B" C "D" E, which is a plan of the given land plot in the accepted scale.

The method of scale shooting described by us is called polar.

There are other ways to shoot a plane with a scale, which you can read about in special guides for scale shooting.

On each plan, a scale is usually given by which the true dimensions of the removed area can be established, as well as its area.

The plan also indicates the direction of the cardinal points.

Practical work.

a) Make the simplest scale model in the school workshop and use it to make a plan of some small plot of land.

b) Surveying the plan of the land plot can be done with the help of an astrolabe.

Suppose it is necessary to remove the plan of the land plot ABCDE. Let's take one of the vertices of the section, for example A, as the initial one and use the astrolabe to measure the angles at the vertex A, i.e.
/ 1, / 2, / 3 (dev. 387).

Then, using a measuring chain, we measure the distances AE, AD, AC and AB. Depending on the size of the plot and the size of the sheet of paper on which the plan is applied, the scale for drawing the plan is selected.

At point A, which is taken as the vertex of the polygon, we build three angles, respectively equal to / 1, / 2 and / 3; then, on the selected scale on the sides of these corners from the point A "put off the segments A" E ", A" D", A "C" and A "B". Connecting the points A "and E", E "and D", D "and C, C" and B", B" and A", we get a polygon A"B"C"D"E", similar to the polygon ABCDE. This will be a plan of this land plot, drawn in the chosen scale.

When solving many construction problems, the similarity method is used, the essence of which is as follows: first, a figure similar to the given one is constructed, then this figure increases (decreases) in the required ratio (i.e., a similar figure is built) that satisfies the condition of the problem.

The process of learning how to apply similarity to solving construction problems should be divided into four stages: preparatory, introductory, skill-forming, skill-improving. Each stage has its own didactic goal, which is achieved when students complete specially designed tasks.

The didactic goal of the preparatory stage is to form students' skills: to highlight the data that determine the shape of the figure, a lot of pairs of figures similar to each other; build a figure according to the data that defines the shape; move from the constructed figure to the desired one.

After studying the first sign of similarity of triangles, we can propose the following set assignments:

Construct a triangle with two corners. How many solutions does the problem have? What elements determine the shape of the constructed triangles?

Name similar triangles in Figure 35.

The following elements of a triangle are known: a) angles of 75 and 25; b) height 1.5 cm; c) angles of 75 and 25, height 1.5 cm. Which of these data determine the only figure in Fig. 35?

What angles determine the shape of the triangles in Figure 35?

Will it be possible to determine the dimensions of one of the triangles in Fig. 35 if the following data become known: a) the angles at the base of the triangle; b) the height of the triangle; c) side and corners at the base?

Are triangles ABC and ABC similar in Figure 36 if ACAC? if they are similar, what is their coefficient of similarity?

The set of tasks presented to students after studying the 2nd and 3rd signs of the similarity of triangles are compiled in a similar way. However, when moving from this feature to the next, the questions become somewhat more complicated, namely: the location of the triangles in the figures changes, moving away from the standard one, the set of the element that defines the only figure varies. Tasks, for example, could be:

1. Are triangles ABC and ABC similar if:

a) AB=5cm, BC=7cm, B=30º, AB=10cm, BC=14cm, B=60º;

b) AB=5cm, BC=7cm, B=30º, AB=10cm, BC=14cm, H=30º;

c) AB=3cm, BC=5cm, CA=7cm, AB=4.5cm, BC=7.5cm, CA=10.5cm;

d) AB=1.7cm, BC=3cm, SA=4.2cm, AB=34cm, BC=60cm, SA=84cm.

2. In a triangle ABC with an acute angle C, heights AE and BD are drawn (Fig. 37). Prove that ABC is similar to EDC.

3. Prove that the perimeters of similar triangles are related as the corresponding sides.

The didactic purpose of the introductory stage is to explain to students the structure of the construction process by the similarity method.

The explanation starts with the problem.

Task. Construct a triangle given two given angles and and a bisector of length d drawn from the vertex of the third angle.

Analyzing the task with the students, the teacher offers tasks - questions, the answers to which are briefly recorded on the board. Questions might be:

1. What data determines the shape of the required triangle?

2. What data determine the dimensions of the desired triangle?

3. How many triangles can be built with two corners? What will be the construction form of all constructed triangles?

4. What segment should be drawn in a triangle similar to the desired one?

5. How to build the desired triangle?

Answers to questions are accompanied by a freehand drawing on the board (Fig. 38).

a) ABC: A=, B=;

b) construct the bisector of angle C in triangle ABC,

c) construct СN=d, NCD;

d) draw a straight line through point N, AB;

e) AC=A, BC=B;

f) ABC - the desired one: A=, B= (since ABC ABC by 1 feature) and CN=d by construction. The didactic purpose of the stage, which forms the ability to solve problems of the type under consideration, is already clear from its name. The main form of activity at this stage is individual search. It ends with a summarizing conversation.

Here are some examples of tasks that can be proposed at this stage.

Task. A point F is given inside the angle AOB. Construct a point M on the side OA, equally distant from F and from the side OB

Decision.

1. Analysis. Let's turn to Figure 39. Let the point M be built, then MF=MP. This means that the desired point M is the center of a circle of radius MF with center M, touching the side OB at point P.

If we take an arbitrary point M on OA and drop MP on CB and find F the intersection of the circle with center M of radius MP with the line OF, then MFP will be similar to MFP. From this follows the required construction.

2. Construction. We draw OF, take an arbitrary point M on CA and lower MP to CB. We draw a circle of radius MP centered at point M. Let F be the intersection point of this circle with OF. We draw FM and then we draw a straight line through the point FFM. The point M of the intersection of this line with OA is the required one.

3. Proof. It is obvious from the analysis carried out.

4. Research. The problem has 2 solutions. This follows from the fact that the circle intersects with OF at 2 points.

Task. Construct a triangle with 2 corners and a perimeter.

Decision.

1. Analysis. Let and be given angles and P be the perimeter of the desired triangle (Fig. 40). Let us assume that the desired triangle is built, then if we consider any ABC similar to the desired one, the ratio of the perimeter P ABC to the perimeter P ABC is equal to the ratio of the sides AC and AC.

2. Construction. Let's construct an ABC similar to the desired one. On the ray AB, set aside the segments AD=P and AD=P, then connect the point D and C, and draw a line DC through the point D. Let C be the point of intersection of the line with the ray AC. Draw a line CB through point C and denote the point of intersection of this line with AD, then ABC is the required one.

3. Proof. Obviously, ACD is similar to ACD, therefore. The aspect ratio is equal to the ratio of the perimeters of similar ABC and ABC, therefore the perimeter ABC \u003d P, therefore, ABC is the desired one.

4. Research. Since the sum of any two angles of a triangle<180, то условие +<180 является необходимым условием для данного построения оно и достаточно. Затем указанным выше способом строится искомый АВС. Такой треугольник единственный, ибо любой другой с такими же данными будет иметь периметр Р и следовательно, будет подобен построенному с коэффициентом подобия равным 1, а два подобных треугольника с одним коэффициентом равны.

Task. Given AOB and point M, located in the inner region of this corner. Construct a circle passing through point A and touching the sides of angle AOB.

Decision.

1. Analysis. Let AOB be given and point M, located in the inner region of the corner (Fig. 41).

Let's draw another circle touching the sides of AOB. Let M be the point of intersection of the circle with the straight line OM and consider OMN and OMN (N and N centers of the circle and).

These triangles are similar in two angles, so the construction of the desired circle can be done as follows:

2. Construction. Since the center of the desired circle lies on the bisector AOB, we draw the bisector of the angle. Further, we take the point N here and construct a circle with center N touching AOB. Then we draw the line SM and denote by M - the point of intersection of the line with the circle (there are two such points - M and M - we take one of them). We draw the line MN and its line through the point M. Then N is the intersection of the line with the bisector of the angle and is the center of the desired circle, and its radius is equal to MN. Let's get her through.

3. Proof. By construction, the circle is similar, O is the center of similarity. This follows from the similarity of triangles OMN and OMN, therefore, since the circle touches the sides of the angle, then the circle will also touch the sides of the angle.

4. Research. The problem has two solutions, because OM intersects with the circle at two points M and M, each of which will correspond to its own circle passing through the point M and touching the sides of AOB.

The didactic goal of the stage that improves the ability to solve problems of the type considered above is the transfer of the formed skill to more complex problems, in particular to the following situations: the desired figure occupies a certain position in relation to given points or lines, while the elimination of one of the conditions of the problem leads to a system of similar or homothetic figures. Let us give an example of such a task.

Task. Inscribe a square in a given triangle so that two of its vertices lie on one side of the triangle, and the other two lie on the other two sides.

Tasks corresponding to the objectives of this stage are excluded from the mandatory level tasks. Therefore, they are offered only to well-performing students. At this stage, the main attention is paid to the individual search activity of students.

As a rule, two triangles are considered similar if they have the same shape, even if they are different sizes, rotated or even upside down.

The mathematical representation of two similar triangles A 1 B 1 C 1 and A 2 B 2 C 2 shown in the figure is written as follows:

∆A 1 B 1 C 1 ~ ∆A 2 B 2 C 2

Two triangles are similar if:

1. Each angle of one triangle is equal to the corresponding angle of another triangle:
∠A 1 = ∠A 2 , ∠B 1 = ∠B 2 and ∠C1 = ∠C2

2. The ratios of the sides of one triangle to the corresponding sides of another triangle are equal to each other:
$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)$

3. Relationships two sides of one triangle to the corresponding sides of another triangle are equal to each other and at the same time
the angles between these sides are equal:
$\frac(B_1A_1)(B_2A_2)=\frac(A_1C_1)(A_2C_2)$ and $\angle A_1 = \angle A_2$
or
$\frac(A_1B_1)(A_2B_2)=\frac(B_1C_1)(B_2C_2)$ and $\angle B_1 = \angle B_2$
or
$\frac(B_1C_1)(B_2C_2)=\frac(C_1A_1)(C_2A_2)$ and $\angle C_1 = \angle C_2$

Similar triangles should not be confused with equal triangles. Congruent triangles have corresponding side lengths. So for equal triangles:

$\frac(A_1B_1)(A_2B_2)=\frac(A_1C_1)(A_2C_2)=\frac(B_1C_1)(B_2C_2)=1$

It follows from this that all equal triangles are similar. However, not all similar triangles are equal.

Although the above notation shows that in order to find out whether two triangles are similar or not, we need to know the values ​​of the three angles or the lengths of the three sides of each triangle, to solve problems with similar triangles, it is enough to know any three values ​​from the above for each triangle. These values ​​can be in various combinations:

1) three angles of each triangle (the lengths of the sides of the triangles do not need to be known).

Or at least 2 angles of one triangle must be equal to 2 angles of another triangle.
Since if 2 angles are equal, then the third angle will also be equal. (The value of the third angle is 180 - angle1 - angle2)

2) the lengths of the sides of each triangle (no need to know the angles);

3) the lengths of the two sides and the angle between them.

Next, we consider the solution of some problems with similar triangles. First, we will look at problems that can be solved by using the above rules directly, and then we will discuss some practical problems that can be solved using the similar triangles method.

Practical problems with similar triangles

Example #1: Show that the two triangles in the figure below are similar.

Decision:
Since the lengths of the sides of both triangles are known, the second rule can be applied here:

$\frac(PQ)(AB)=\frac(6)(2)=3$ $\frac(QR)(CB)=\frac(12)(4)=3$ $\frac(PR)(AC )=\frac(15)(5)=3$

Example #2: Show that two given triangles are similar and find the lengths of the sides PQ and PR.

Decision:
∠A = ∠P and ∠B = ∠Q, ∠C = ∠R(because ∠C = 180 - ∠A - ∠B and ∠R = 180 - ∠P - ∠Q)

It follows from this that the triangles ∆ABC and ∆PQR are similar. Hence:
$\frac(AB)(PQ)=\frac(BC)(QR)=\frac(AC)(PR)$

$\frac(BC)(QR)=\frac(6)(12)=\frac(AB)(PQ)=\frac(4)(PQ) \Rightarrow PQ=\frac(4\times12)(6) = 8$ and
$\frac(BC)(QR)=\frac(6)(12)=\frac(AC)(PR)=\frac(7)(PR) \Rightarrow PR=\frac(7\times12)(6) = 14$

Example #3: Determine the length AB in this triangle.

Decision:

∠ABC = ∠ADE, ∠ACB = ∠AED and ∠A common => triangles ΔABC and ΔADE are similar.

$\frac(BC)(DE) = \frac(3)(6) = \frac(AB)(AD) = \frac(AB)(AB + BD) = \frac(AB)(AB + 4) = \frac(1)(2) \Rightarrow 2\times AB = AB + 4 \Rightarrow AB = 4$

Example #4: Determine length AD(x) geometric figure in the figure.

Triangles ∆ABC and ∆CDE are similar because AB || DE and they have a common top corner C.
We see that one triangle is a scaled version of the other. However, we need to prove it mathematically.

AB || DE, CD || AC and BC || EU
∠BAC = ∠EDC and ∠ABC = ∠DEC

Based on the foregoing and taking into account the presence of a common angle C, we can state that triangles ∆ABC and ∆CDE are similar.

Hence:
$\frac(DE)(AB) = \frac(7)(11) = \frac(CD)(CA) = \frac(15)(CA) \Rightarrow CA = \frac(15 \times 11)(7 ) = $23.57
x = AC - DC = 23.57 - 15 = 8.57

Practical examples

Example #5: The factory uses an inclined conveyor belt to transport products from level 1 to level 2, which is 3 meters above level 1, as shown in the figure. The inclined conveyor is serviced from one end to level 1 and from the other end to a workstation located at a distance of 8 meters from the level 1 operating point.

The factory wants to upgrade the conveyor to access the new level, which is 9 meters above level 1, while maintaining the conveyor angle.

Determine the distance at which you need to set up a new work station to ensure that the conveyor works at its new end at level 2. Also calculate the additional distance that the product will travel when moving to a new level.

Decision:

First, let's label each intersection point with a specific letter, as shown in the figure.

Based on the reasoning given above in the previous examples, we can conclude that the triangles ∆ABC and ∆ADE are similar. Hence,

$\frac(DE)(BC) = \frac(3)(9) = \frac(AD)(AB) = \frac(8)(AB) \Rightarrow AB = \frac(8 \times 9)(3 ) = 24 m$
x = AB - 8 = 24 - 8 = 16 m

Thus, the new point must be installed at a distance of 16 meters from the existing point.

And since the structure is made up of right triangles, we can calculate the product travel distance as follows:

$AE = \sqrt(AD^2 + DE^2) = \sqrt(8^2 + 3^2) = 8.54 m$

Similarly, $AC = \sqrt(AB^2 + BC^2) = \sqrt(24^2 + 9^2) = 25.63 m$
which is the distance that the product travels at the moment when it hits the existing level.

y = AC - AE = 25.63 - 8.54 = 17.09 m
This is the extra distance that a product must travel to reach a new level.

Example #6: Steve wants to visit his friend who recently moved into a new house. The road map to get to Steve and his friend's house, along with the distances known to Steve, is shown in the figure. Help Steve get to his friend's house in the shortest way.

Decision:

The roadmap can be represented geometrically in the following form, as shown in the figure.

We see that triangles ∆ABC and ∆CDE are similar, therefore:
$\frac(AB)(DE) = \frac(BC)(CD) = \frac(AC)(CE)$

The task statement states that:

AB = 15 km, AC = 13.13 km, CD = 4.41 km and DE = 5 km

Using this information, we can calculate the following distances:

$BC = \frac(AB \times CD)(DE) = \frac(15 \times 4.41)(5) = 13.23 km$
$CE = \frac(AC \times CD)(BC) = \frac(13.13 \times 4.41)(13.23) = 4.38 km$

Steve can get to his friend's house using the following routes:

A -> B -> C -> E -> G, the total distance is 7.5+13.23+4.38+2.5=27.61 km

F -> B -> C -> D -> G, the total distance is 7.5+13.23+4.41+2.5=27.64 km

F -> A -> C -> E -> G, the total distance is 7.5+13.13+4.38+2.5=27.51 km

F -> A -> C -> D -> G, the total distance is 7.5+13.13+4.41+2.5=27.54 km

Therefore, route #3 is the shortest and can be offered to Steve.

Example 7:
Trisha wants to measure the height of the house, but she doesn't have the right tools. She noticed that a tree was growing in front of the house and decided to use her resourcefulness and knowledge of geometry received at school to determine the height of the building. She measured the distance from the tree to the house, the result was 30 m. Then she stood in front of the tree and began to back away until the top edge of the building was visible above the top of the tree. Trisha marked the spot and measured the distance from it to the tree. This distance was 5 m.

The height of the tree is 2.8 m and the height of Trisha's eyes is 1.6 m. Help Trisha determine the height of the building.

Decision:

The geometric representation of the problem is shown in the figure.

First we use the similarity of triangles ∆ABC and ∆ADE.

$\frac(BC)(DE) = \frac(1.6)(2.8) = \frac(AC)(AE) = \frac(AC)(5 + AC) \Rightarrow 2.8 \times AC = 1.6 \times (5 + AC) = 8 + 1.6 \times AC$

$(2.8 - 1.6) \times AC = 8 \Rightarrow AC = \frac(8)(1.2) = 6.67$

We can then use the similarity of triangles ∆ACB and ∆AFG or ∆ADE and ∆AFG. Let's choose the first option.

$\frac(BC)(FG) = \frac(1.6)(H) = \frac(AC)(AG) = \frac(6.67)(6.67 + 5 + 30) = 0.16 \Rightarrow H = \frac(1.6 )(0.16) = 10 m$

Task 1. Construct a triangle, knowing its two angles and perimeter.

Decision. Knowing the angles of a triangle already determines it up to a similarity transformation. Therefore, to solve the problem, we build any triangle LS, with the given angles (Fig. 277). It remains to similarly transform the triangle so that its perimeter becomes equal to the given value.

To do this, set aside its sides on the extensions of the side, the segment will be equal to the perimeter of the triangle. Take any segment KL parallel to the segment but equal to the given perimeter. We connect the ends of both parallel segments and take the point O of the intersection of the lines as the center of similarity. The construction of vertices A and C of the desired triangle can be seen from Fig. 277, its sides AB and CB are parallel to the corresponding sides of the triangle.

In the case of a triangle - already desired.

Task 2. Given the angle formed by the rays OA and OB, and the point N inside this angle. Construct a circle tangent to the sides of the corner and passing through the given point N (Fig. 278).

Decision. A circle tangent to the sides of an angle must be centered on the bisector of that angle. Let's take an arbitrary point on this bisector and construct a circle centered on the sides of the corner (its radius is simply equal to the distance of the point from the sides of the corner). If we now transform this circle similarly with the center of similarity at the vertex of angle O, then again we get a circle centered on the bisector; such a circle will again touch the sides of the corner, since its radius leading to the point of contact will, by virtue of conservation of angles, go over into a radius perpendicular to the side of the corner. It remains to ensure the fulfillment of the second condition: the transformed circle must pass through the point N. This implies the solution of the problem. Draw the ray ON to the intersection with the circle at points and construct its radii leading to these points. Through a given point N we draw lines NC and NC parallel to these radii; the points of their intersection C, C with the bisector and give the possible positions of the center of the desired circle. The problem has two solutions. How will the solution change if the point N lies on the bisector of the angle?

Exercises

1. The perimeter of a triangle is 10 cm, and its area What is the perimeter of a similar triangle if its area is?

2. Prove that isosceles triangles with equal vertex angles are similar.

3. Construct a triangle similar to the given one and inscribed in a circle of given radius.

4. Inscribe a square in a given triangle ABC so that one of its sides lies on the side BC of the triangle, and two vertices are on the other two sides of the triangle.