Presentation of proportional segments definition of similar triangles. Presentation on the topic "definition of similar triangles"

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proportional cuts. Definition similar triangles The ratio of the areas of similar triangles The first sign of the similarity of triangles (Proof) The second sign of the similarity of triangles (Proof) The third sign of the similarity of triangles (Proof) Practical application

Slide 9

Continuation

Basic information Measuring work on the ground Determining the height of an object Determining the distance to an inaccessible point Determining the distance by building similar triangles (1) (2) (5) (4) (3)

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Proportional segments

The ratio of the segments AB and CD is the ratio of their lengths, i.e. AB / CD. They say that the segments AB and CD are proportional to the segments A1 B1 and C1 D1, if AB / A1B1 \u003d CD / C1D1. The concept of proportionality is also introduced for a large number of segments

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Definition of similar triangles.

Two triangles are said to be similar if their angles are respectively congruent and the sides of one triangle are proportional to the similar sides of the other

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The ratio of the areas of similar triangles

Theorem The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient

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Proof.

Let the triangles ABC and A1B1C1 be similar and the similarity coefficient be equal to r. Let S and S1 denote the areas of these triangles. Since angle A=angleA1, then S/S1=AB*AC/A1B1*A1C1(according to the area ratio theorem, similarity ratios for triangles having equal angles). According to formulas (2), we have: AB / A1B1 \u003d R, AC / A1C1 \u003d R, therefore S / S \u003d R 2

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The first sign of the similarity of triangles

If two angles of one triangle are respectively equal to two angles of another, then such triangles are equal to A B C

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The second sign of the similarity of triangles

If two sides of another triangle are proportional to two sides of another triangle and the angles included between these sides are equal, then such triangles are similar.

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The third sign of the similarity of triangles

If three sides of one triangle are proportional to three sides of another, then the triangles are similar. A B C

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Proof.(1)

Given: ABC and A1B1C1 are two triangles, in which angle A \u003d angle A1, angle B \u003d angle B1 Let's prove that triangle ABC is triangle A!B1C1

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Proof.

According to the theorem on the sum of angles of a triangle, angle C \u003d 180 degrees-angle A-angle B, angle C \u003d 180 degrees-angle A - angle B, and, therefore, angle C \u003d corner C. Thus, the angles of triangle ABC are respectively equal to the angles of triangle A B C 1 1 1 1 1 1 1

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Let us prove that the sides of the triangle ABC are proportional to the similar sides of the triangle A B C. c \u003d CA * CB / C A * C B. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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From these equalities it follows that AB / A B \u003d BC / B C Similarly, using the equalities angle A \u003d corner A Angle B \u003d corner B, we get, BC / B C \u003d CA / C A. So the sides of the triangle ABC are proportional to the similar sides of the triangle A In C The theorem is proved. 1 1 1 1 1 1 1 1

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Proof (2)

Given: two triangles ABC and A B C, in which AB / A B = AC / A C, angle A \u003d angle A B = corner B 1 1 1 1 1 1 1 1 1 1

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Consider a triangle ABC, in which angle1 = angleA, angle2 = angle B. Triangles ABC A B C are similar in the first sign of similarity of triangles, therefore AB / A B \u003d AC / A C. On the other hand, by the condition AB / A B \u003d AC /А С. From these two equalities we get AC=AC. 2 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 2

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Triangles ABC and ABC are equal on the two sides between them (AB is the common side, AC=AC and angle A=angle 1 because angle A=angle A and angle 1=angle A). It follows that angle B = angle 2, and since angle 2 = angle B, then angle B = angle B. The theorem is proved. 2 2 1 1 1 1

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Proof (3)

Given: the sides of triangles ABC and ABC are proportional. Let us prove that triangle ABC is a triangle A B C 1 1 1

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Proof

To do this, taking into account the second sign of similarity of triangles, it is enough to prove that angle A \u003d angle A. Consider a triangle ABC, in which angle 1 \u003d angle A, angle 2 \u003d angle B. Triangles ABC and A BC are similar in the first sign of similarity of triangles, therefore AB / A B \u003d BC / B C \u003d C A / C A.

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Comparing these equalities with equalities (1) we get: ВС=ВС, СА= С А. Triangles ABC and ABC are equal on three sides. It follows that angle A = angle 1 and since angle1 = angle A, then angle A = angle A. The theorem is proved. 2 2 2 1 1

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Practical applications of similar triangles

When solving many problems on the construction of triangles, the so-called similarity method is used. It consists in the fact that, first, on the basis of some data, there is a triangle similar to the desired one, and then, using the rest of the data, the desired triangle is built.

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Task #1

Construct a triangle given two angles and the bisector at the vertex of the third angle

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Solution

First, let's construct some kind of triangle similar to the desired one. To do this, draw an arbitrary segment A B and construct a triangle A B C, in which the angles A and B, respectively, are equal to the given angles

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Continuation

Next, we construct the bisector of angle C and plot the segment CD on it, which are equal to this segment. Draw a line through point D parallel to A B. It intersects the sides of angle C at some points A and B. The triangle ABC is the desired

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In fact, since AB is parallel to AB, then angle A = angle A, angle B = angle B, and, therefore, two angles of the triangle ABC are respectively equal to the given angles. By construction, the bisector CD of the triangle ABC is equal to this segment. So, the triangle ABC satisfies all the conditions of the problem.

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Basic information(1)

1. Triangle ABC is similar to triangle A B C if and only if one of the following equivalent conditions is met. 1 1 1

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Terms

A) AB: BC: CA \u003d A B: B C: C A; C) AB: BC \u003d A B: B C and angle ABC \u003d corner A B C; C) angle ABC = angle A B C and angle BAC = angle B A C. 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Basic information(2)

2) if parallel lines cut off triangles ABC and ABC from an angle with vertex A, then these triangles are similar and AB:AB = AC:AC (points B and B lie on one side of the angle, C and C on the other). 1 1 2 2 1 2 1 2 1 2 1 2

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Basic Information(3)

3) the middle line of a triangle is a segment connecting the midpoints of the sides. This segment is parallel to the third side and equal to half of its length. The median line of a trapezoid is a segment that connects the midpoints of the sides of the trapezoid. This segment is parallel to the bases and equal to half the sum of their lengths

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Basic information (4)

4) the ratio of the areas of similar triangles is equal to the square of the similarity coefficient, i.e. the square of the ratio of the lengths of the corresponding sides. This follows, for example, from the formula Savs=0.5*AB*ACsinA.

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Keynote (5)

Polygons A A ... A and B B ... B are called similar if A A: A A: ...: A A \u003d B B: B B: ... B B and the angles at the vertices A ..., A. Are equal, respectively, to the angles at the vertices A, ….,A are equal The ratio of the corresponding diagonals of similar polygons is equal to the similarity coefficient; for similar polygons described, the ratio of the radii of the inscribed circles is also equal to the similarity coefficient 1 2 n 1 2 n 1 2 2 3 n 1 1 2 2 3 n 1 1 n 1 n

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Measuring work on the ground

The properties of such triangles can be used to carry out various measurements on the ground. We will consider two tasks: determining the height of an object on the ground and the distance to an inaccessible point.

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Task #1

Determining the height of an object

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Continuation

Suppose that we need to determine the height of some object, for example, the height of a telegraph pole A C, for this we put a pole AC with a rotating bar at a certain distance from the pole and direct the bar to the top point A of the pole. We mark point B on the surface of the earth, at which the straight line A A intersects with the surface of the earth. 1 1 1 1

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Right-angled triangles A C B and DIA are similar in the first sign of triangles (angle C \u003d angle C \u003d 90 degrees, angle B is common). From the similarity of triangles it follows A C /AC \u003d BC / BC, from which A C \u003d AC * BC / BC by measuring the distance BC and BC and knowing the length AC of the pole, using the obtained formula, we determine the height AC of the telegraph pole 1 1 1 1 1 1 1 1 1 one

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Task (2)

Determining the distance to an inaccessible point

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Continuation

Suppose we need to find the distance from point A to inaccessible point B. To do this, we select point C on the ground, hang the segment AC and measure it. Then, using an astrolabe, we measure angles A and C. On a piece of paper we build some triangle A B C, in which angle A \u003d angle A, angle C \u003d corner C, and measure the lengths of the sides A B and A C of this triangle. 1 1 1 1 1 1 1 1 1

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Since the triangle ABC and A B C are similar (according to the first sign of similarity of triangles), then AB / A B \u003d AC A C, from which we get AB \u003d AC * A B / A C. This formula allows for known distances AC, A C and A B, find the distance AB. 1 1 1 1 1 1 1 1 1 1 1 1 1

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To simplify the calculations, it is convenient to construct a triangle A B C in such a way that A C: AC \u003d 1: 1000. for example, if AC = 130m, then the distance A C is taken equal to 130mm. In this case, AB \u003d AC / A C * A B \u003d 1000 * A B, therefore, by measuring the distance A B in millimeters, we immediately get the distance AB in meters 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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Example

Let AC = 130m, angle A = 73 degrees, angle C = 58 degrees. We build a triangle A B C on paper so that angle A = 73 degrees, angle C = 58 degrees, A C = 130 mm, and measure the segment A B. It is 153 mm, therefore, the required distance is early 153m. 1 1 1 1 1

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Determining distance by building similar triangles

When determining the distance to distant or inaccessible objects, you can use the following technique. On an ordinary match, two-millimeter divisions must be applied with ink or a pencil. You also need to know the approximate height of the object to which the distance is determined. So a person's height is 1.7-1.8 m, a car wheel is 0.5 m, a rider is 2.2 m, a telegraph pole is 6 m, a one-story house without a roof is 2.5-4 m.

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Continuation

Suppose we need to determine the distance to the pole. We direct a match at it on an outstretched arm, the length of which is approximately 60 cm. 4 mm. Having such data, we will make up the proportion: 0.6 / x \u003d 0.004 / 6.0; x \u003d (0.6 * 6) / 0y004 \u003d 900. Thus, up to the pillar 900m.

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"Geometry "Similar Triangles"" - Basic trigonometric identity. The second sign of the similarity of triangles. Sine, cosine and tangent. Sine, cosine and tangent values ​​for 30°, 45°, 60° angles. Similar triangles. Similar to right triangles. Continuation of the sides. proportional cuts. Theorem on the ratio of the areas of similar triangles. Sine, cosine and tangent values. The two sides of the triangle are connected by a segment that is not parallel to the third.

"Finding the area of ​​a trapezoid" - Results. Properties right triangle. Find the area of ​​the trapezoid. Compare areas. Designate the bases. Tasks for self-control. The area of ​​the trapezoid. Repetition of the material covered. Trap. Write down the formulas. Develop the ability to apply the formula. Find the area. Cell area. The solution of the task. Let's summarize. Area.

"Quadangles, their signs and properties" - Rhombus. Quadrilaterals, their signs and properties. Learn about types of quadrilaterals. Rectangle. Parallelogram properties. A rectangle with all sides equal. A quadrilateral whose vertices are at the midpoints of the sides. Diagonals. Types of quadrilaterals. Tests. Of which two equal triangles you can make a square. Types of trapezoid. Rhombus corners. Square. Features of a parallelogram. Quadrangles.

"Inscribed angle theorem" - The radius of the circle is 4 cm. Answer. Sharp corner. Consolidation of the studied material. Updating students' knowledge. Knowledge update. Learning new material. Circle radius. What is the name of the angle with the vertex at the center of the circle? Find the angle between the chords. The concept of an inscribed angle. Triangle. Find the angle between them. Solution. Test yourself. Correct answer. The circles intersect. Inscribed angle theorem.

"Pythagorean theorem for a right triangle" - Right triangle. The name of Pythagoras. A combination of two contradictory principles. Herodotus. Statement of the theorem. ancient authors. Pythagoras of Samos. Coin with the image of Pythagoras. Pythagorean theorem. The teachings of Pythagoras.

"The concept of the area of ​​a polygon" - Adjacent sides of a parallelogram. Area of ​​a triangle. Mathematical dictation. Parallelogram. Rhombus area. The concept of the area of ​​a polygon. The area of ​​the rectangle. The area of ​​the trapezoid. Heights. The area of ​​polygons. Area of ​​a right triangle. Theorem. Sharp corner. The area of ​​a parallelogram. Calculate the area of ​​the rhombus. Find the area of ​​a right triangle. Triangles. Area units.

SIMILAR TRIANGLES

MBOU Gymnasium №14

Mathematics teacher: E.D. Lazareva

Proportional segments

attitude segments AB and CD is the ratio of their lengths, i.e.

Segments AB and CD proportional segments A 1 B 1 and C 1 D 1 if

Definition of Similar Triangles

The two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the corresponding sides of the other.

The number k, equal to the ratio of similar sides of triangles, is called similarity coefficient

B 1

A 1

C 1

The ratio of the areas of similar triangles

The ratio of the areas of two similar triangles is the square of the similarity coefficient

The bisector of a triangle divides the opposite side into segments proportional to the adjacent sides of the triangle.

B 1

A 1

C 1

I

If two angles of one triangle are respectively equal to two angles of another triangle, then such triangles are similar

 ABC,  A 1 B 1 C 1 ,

 A =  A 1 ,  B =  B 1

Prove:

 ABC  A 1 B 1 C 1

B 1

A 1

C 1

Signs of similarity of triangles

II triangle similarity sign

If two sides of one triangle are proportional to two sides of another triangle and the angles included between these sides are equal, then such triangles are similar

 ABC,  A 1 B 1 C 1 ,

Prove:

 ABC  A 1 B 1 C 1

B 1

A 1

C 1

Signs of similarity of triangles

III triangle similarity sign

If three sides of one triangle are proportional to three sides of another triangle, then such triangles are similar

 ABC,  A 1 B 1 C 1 ,

Prove:

 ABC  A 1 B 1 C 1

B 1

A 1

C 1

Middle line of the triangle

The midline of a triangle is the line segment that connects the midpoints of two sides.

Middle line of the triangle

parallel to one of its sides

and equal to half of this side

 ABC, MN - center line

Prove:

MN  AC, MN = AC

The medians of a triangle intersect at one point, which divides each median in a ratio of 2:1, counting from the top

A 1

C 1

B 1

Applying Similarity to Problem Solving

Height of a right triangle drawn from a vertex right angle, divides the triangle into two similar right triangles, each of which is similar to the given triangle.

 ABC  ACD,

Application of similarity to theorem proving

1. The height of a right triangle drawn from the vertex of the right angle is the average proportional between the segments into which the hypotenuse is divided by this height

Application of similarity to theorem proving

2. The leg of a right triangle is the mean proportional between the hypotenuse and the segment of the hypotenuse enclosed between the leg and the height drawn from the vertex of the right angle.

1.1. Proportional segments Definition of similar triangles 1.2. Definition of similar triangles 1.3. Ratio of areas of similar triangles Ratio of areas of similar triangles Properties of similarity.

1.1 Proportional segments. The ratio of the segments AB and CD is the ratio of their lengths, i.e. It is said that the segments AB and CD are proportional to the segments A 1 B 1 and C 1 D 1, if EXAMPLE 1. The segments AB and CD, whose lengths are 2 cm and 1 cm, proportional to the segments A 1 B 1 and C 1 D 1, the segments of which are 3 cm and 1.5 cm. Indeed,

1.2. Definition of similar triangles. IN Everyday life there are objects of the same shape but different sizes, such as soccer and tennis balls, a round plate and a large round dish. In geometry, figures of the same shape are called similar. So, any two squares, any two circles are similar. Let us introduce the concept of similar triangles.

1.2. Definition of similar triangles. SIMILARITY, a geometric concept that characterizes the presence of the same shape in geometric shapes, regardless of their size. Two figures F1 and F2 are called similar if a one-to-one correspondence can be established between their points, in which the ratio of the distances between any pairs of corresponding points of the figures F1 and F2 is equal to the same constant k, called the similarity coefficient. The angles between the corresponding lines of similar figures are equal. Similar figures are F1 and F2.

Definition. Two triangles are said to be similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other triangle. In other words, two triangles are similar if they can be denoted by the letters ABC and A 1 B 1 C 1 so that A= A 1, B= B 1, C= C 1. The number k equal to the ratio of the similar sides of the triangles is called the similarity coefficient .

1.3. The ratio of the areas of similar triangles. Theorem. The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient. Proof. Let triangles ABC and A1B1C1 be similar and the similarity coefficient be equal to k. Let S and S1 denote the areas of these triangles. Since A= A1, then

similarity properties. Problem 2. Prove that the bisector of a triangle divides the opposite side into segments proportional to the adjacent sides of the triangle Solution. Let AD be the bisector of triangle ABC. We prove that Triangles ABD and ACD have a common height AH, so 12 A H B D C

Proof: By the angle sum theorem: C \u003d A - B, and C 1 \u003d A 1 - B 1, then C \u003d C 1. Since A \u003d A 1 and C \u003d C 1, it follows from this: It turns out, that the corresponding sides are proportional. Given: ABC and A 1 B 1 C 1 A= A 1 B= B 1 Prove: ABC A 1 B 1 C 1 A C B A1A1 B1B1 C1C1

ABC 2 A 1 B 1 C 1 (according to the first sign), which means, on the other hand, from these equalities we get AC = = AC 2. ABC = ABC 2 - on two sides and the angle between them (AB-common side, AC = AC 2 and, because i). So and, then ABC A1B1C1 Given: ABC and A 1 B 1 C 1 D-th: Proof: Consider ABC 2, for which and

Proof: A 1 B 1 is the middle line, and A 1 B 1 // AB, and therefore AOB A 1 OB 1 (at two corners), then But AB \u003d A 1 B 1, therefore AO \u003d 2A 1 O and BO \u003d 2В 1 O. So the point O is the intersection of the medians AA 1 and BB 1 divides each of them in a ratio of 2: 1, counting from the top. Similarly, it is proved that the point O - the intersection of the medians BB 1 and CC 1 divides each of them in a ratio of 2: 1, counting from the top. So the point O - the intersection of the medians AA 1, BB 1 and SS 1 divides them in a ratio of 2: 1, counting from the top.

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Slides captions:

Similar triangles

Similar figures Figures are called similar if they have the same shape (similar in appearance).

Similarity in life (maps of the area)

Proportional Segments Definition: Segments are called proportional if their lengths are proportional. 12 6 8 4 A 1 B 1 AB C 1 K 1 SK They say that the segments A 1 B 1 and C 1 K 1 are proportional to the segments AB and SK. Are the segments AB and SK proportional to the segments EP and HT if: a) AB = 15 cm, SC = 2.5 cm, EP = 3 cm, HT = 0.5 cm? b) AB = 12 cm, SC = 2.5 cm, EP = 36 cm, HT = 5 cm? c) AB = 24 cm, SC = 2.5 cm, EP = 12 cm, HT = 5 cm? yes no no A B 6 cm C K 4 cm A 1 B 1 12 cm C 1 8 cm K 1

b Proportional segments Test 1. Indicate the correct statement: a) the segments AB and PH are proportional to the segments SK and ME; b) the segments ME and AB are proportional to the segments PH and SK; c) the segments AB and ME are proportional to the segments PH and SK. A B 3 cm C K 2cm M E 9 cm RN 6 cm Appendix: the equality ME AB RN SK can be written with three more equalities: RN SK ME AB; ME RN AB SK; AB SK ME RN.

Proportional segments 2 . Test F Y Z R L S N 1 c m 2 cm 4 cm 2 cm 3 cm a) RL ; b) RS; c) SN a) RL

Proportional segments (desired property) The bisector of a triangle divides the opposite side into segments proportional to the adjacent sides of the triangle. H Given: ABC, AK - bisector. Proof: 1 A B K C 2 Since AK is a bisector, then 1 \u003d 2, which means that ABK and ASK have an equal angle, therefore AVK and ASK have a common height AN, so S AVK S ASK VC K C AB A C BK K C VC AB KS AC Therefore, Let's draw AN VS.

Similar Triangles Definition: Triangles are said to be similar if the angles of one triangle are equal to the angles of another triangle and the sides of one triangle are proportional to the similar sides of the other. A 1 B 1 C 1 A B C Similar sides in similar triangles are sides that lie opposite equal angles. A 1 \u003d A, B 1 \u003d B, C 1 \u003d C A 1 B 1 B 1 C 1 A 1 C 1 AB BC AC k A 1 B 1 C 1 ABC K - similarity coefficient ~

Similar triangles A 1 B 1 C 1 A B C Desired property: A 1 \u003d A, B 1 \u003d B, C 1 \u003d C, AB BC AC A 1 B 1 B 1 C 1 A 1 C 1 1 k ABC ~ A 1 B 1 C 1 , – similarity coefficient 1 k A 1 B 1 C 1 ABC , K – similarity coefficient ~

Solve problems 3. According to the data in the drawing, find the sides AB and B 1 C 1 of similar triangles ABC and A 1 B 1 C 1: A B C A 1 C 1 B 1 6 3 4 2.5? ? Find sides A 1 B 1 C 1 similar to ABC if AB = 6, BC = 12. AC = 9 and k = 3. 2. Find sides A 1 B 1 C 1 similar to ABC if AB = 6, BC = 12. AC = 9 and k = 1/3.

Theorem 1. The ratio of the perimeters of similar triangles is equal to the similarity coefficient. M K E A B C Given: MKE ~ ABC, K is the similarity coefficient. Prove: P MKE: P ABC = k Proof: K , MK AB KE BC ME AC Hence, MK = k ∙ AB, KE = k ∙ BC, ME = k ∙ AC. Since according to the condition MKE ~ ABC, k is the similarity coefficient, then P MKE \u003d MK + KE + ME \u003d k ∙ AB + k ∙ BC + k ∙ AC = k ∙ (AB + BC + AC) \u003d k ∙ P ABC. Hence, R MKE: R ABC \u003d k.

Theorem 2. The ratio of the areas of similar triangles is equal to the square of the similarity coefficient a. M K E A B C Given: MKE ~ ABC, K is the similarity coefficient. Prove: S MKE: S АВС = k 2 AS. S MKE S ABC MK ∙ ME AB ∙ AC k ∙ AB ∙ k ∙ AC AB ∙ AC k 2

Solve the problems Two similar sides of similar triangles are 8 cm and 4 cm. The perimeter of the second triangle is 12 cm. What is the perimeter of the first triangle? 24 cm 2. Two similar sides of similar triangles are 9 cm and 3 cm. The area of ​​the second triangle is 9 cm 2. What is the area of ​​the first triangle? 81 cm 2 3. Two similar sides of similar triangles are 5 cm and 10 cm. The area of ​​the second triangle is 32 cm 2. What is the area of ​​the first triangle? 8 cm 2 4. The areas of two similar triangles are 12 cm 2 and 48 cm 2. One of the sides of the first triangle is 4 cm. What is the similar side of the second triangle? 8 cm

Solution of the problem The areas of two similar triangles are 50 dm 2 and 32 dm 2, the sum of their perimeters is 117 dm. Find the perimeter of each triangle. Find: R ABC, R REC Solution: Since by condition the triangles ABC and REC are similar, then: Given: ABC, REC are similar, S ABC = 50 dm 2, S REC = 32 dm 2, R ABC + R REC = 117dm. S ABC S REC 50 32 25 16 K 2 . Hence, k \u003d 5 4 K, R ABC R REK R ABC R REK 5 4 1.25 Hence, R ABC \u003d 1.25 R REK Let R REK \u003d x dm, then R ABC \u003d 1.25 x dm T. to .by condition R ABC + R REC = 117dm, then 1.25 x + x = 117, x = 52. Hence, R REC = 52 dm, R ABC = 117 - 52 = 65 (dm). Answer: 65 dm, 52 dm.

“Mathematics should be taught later, that it puts the mind in order” M. V. Lomonosov I wish you success in your studies! Mikhailova L.P. GOU TsO No. 173.