The presentation “Definition of similar triangles” covers the stage of introducing a new concept in a geometry lesson in the 8th grade - similarity of triangles. After clarifying the concept of proportionality of segments, on the basis of which the concept of similarity is built, students proceed to consider material that is quite complex for them - similarity. With the help of a presentation, the teacher, during the explanation, forms a clear understanding of the students about the subject being studied - the similarity of triangles, continues to develop skills in using mathematical speech, and develops skills in applying the studied concept to solve practical problems.
slides 1-2 (Presentation topic “Definition of similar triangles”, examples)
To explain the similarity property of triangles, the presentation uses the following tools:
- highlighting the main concepts in red;
- animated construction of the graphic part to clarify the definition and clarity when explaining the material;
- framing basic algebraic expressions on the topic;
- using pictures to understand the practical meaning of the concept being studied.
Such a demonstration allows you to deepen your understanding of the material and facilitate its memorization.
The presentation begins with a demonstration of objects on the outlines of which similar geometric figures are built. Examples include soccer and handball balls, patterned plates of different sizes. To the right of the objects are depicted the outlines of figures that are similar to each other - a large and small square, a large and small circle.
slides 3-4 (definition of similar triangles)
Such a demonstration, which introduces the student to the study of a given concept through practical application, is very effective and helps to solve one of the important goals of the lesson - to consolidate the student’s understanding of the subject being studied.
On the next slide, the concept of similarity is decomposed into its components using two constructed triangles ABC and A1B1C1. Using animation, gradually the corresponding angles are marked as equal. The corresponding angles are designated the same way - A and A1 by one semicircle, B and B1 by two, C and C1 by three. Given that these triangles have equal angles, their corresponding sides are called similar. This expression must be used in the future when solving geometric problems, so the expression is highlighted in green, indicating the need to remember it and use it in the future.
slide 5 (website)
Now we can formulate a definition of the similarity of triangles with the corresponding equality of angles and proportionality of similar sides. Next, an algebraic representation of the conditions for the similarity of triangles is demonstrated - equality of angles and proportionality of all three sides. The condition of proportionality of the sides is enclosed in a frame for memorization. The result of the ratio of each pair is the same number. It is denoted by k and is defined as the similarity coefficient of triangles.
Based on the studied concept, the following topics in the geometry course should be studied - ratios of areas of similar triangles, signs of similarity of triangles.
This presentation “Definition of similar triangles” can be recommended not only as demonstration material in a geometry lesson, accompanying the teacher’s explanation. It can help the student to independently study the material, and will also help explain the concept of similarity in a lesson during distance learning.
SIMILAR TRIANGLES
MBOU Gymnasium No. 14
Math teacher: E.D. Lazareva
Proportional segments
Attitude segments AB and CD is called 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
Two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the similar 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
Ratio of areas of similar triangles
The ratio of the areas of two similar triangles is squared 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 test
If two sides of one triangle are proportional to two sides of another triangle and the angles 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 test
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 segment connecting 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 2:1 ratio, counting from the vertex
A 1
C 1
B 1
Applying similarity to problem solving
The altitude of a right triangle drawn from the vertex of a right angle divides the triangle into two similar right triangles, each of which is similar to the given triangle.
ABC ACD,
Applying similarity to theorem proving
1. The height of a right triangle drawn from the vertex of a right angle is the average proportional between the segments into which the hypotenuse is divided by this height
Applying 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 altitude 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 segments AB and CD is the ratio of their lengths, i.e. It is said that segments AB and CD are proportional to segments A 1 B 1 and C 1 D 1 if EXAMPLE 1. Segments AB and CD, the lengths of which are 2 cm and 1 cm, are proportional to the segments A 1 B 1 and C 1 D 1, the segments of which are equal to 3 cm and 1.5 cm. Indeed,
1.2. Definition of similar triangles. In everyday life, there are objects of the same shape, but of different sizes, for example, football and tennis balls, a round plate and a large round dish. In geometry, figures of the same shape are usually 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 figures, 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 corresponding lines of similar figures are equal. Similar figures F1 and F2.
Definition. Two triangles are called 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 similar sides of the triangles, is called the similarity coefficient .
1.3. Ratio of 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 equal to k. Let us denote the areas of these triangles by the letters S and S1. Since A= A1, then
Properties of similarity. 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. Let us prove that Triangles ABD and ACD have a common height AH, therefore 12 A H B D C
Proof: By the theorem on the sum of angles: C = A - B, and C 1 = A 1 - B 1, which means C = C 1. Since A = A 1 and C = C 1, then it follows: It turns out, that similar 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 is the common side, AC = AC 2 and, since 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, therefore and So AOB A 1 OB 1 (at two angles), then But AB = A 1 B 1, therefore AO = 2A 1 O and VO = 2B 1 O. This means that point O is the intersection of the medians AA 1 and BB 1, dividing each of them in a ratio of 2:1, counting from the vertex. It is similarly proven that 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 vertex. This means that point O - the intersection of the medians AA 1, BB 1 and CC 1 divides them in a ratio of 2:1, counting from the top.
Slide 2
A little bit about yourself
Hello everyone, my name is Alesya, I am 15 years old and I study at school No. 11 in the 8th grade. I study at an amateur song club. My club is called KSP “Inspiration”. I love doing projects. One of which you see now.
Slide 3
Project goals
Do everything possible for the children so that they understand where similar triangles were used in ancient times and why they were needed
Slide 4
Motivational material
I think such triangles are needed to determine the distance to a point inaccessible to us and the height of an object
Slide 5
Uses in life.
Well, I think that such triangles would be useful for determining the distance to an inaccessible point and in the construction of a building.
Slide 6
Subject
Similar triangles
Slide 7
Definition of similar triangles
Slide 8
Proportional segments. Definition of similar triangles Ratio of areas of similar triangles First sign of similarity of triangles (Proof) Second sign of similarity of triangles (Proof) Third sign of 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 constructing similar triangles (1) (2) (5) (4) (3)
Slide 10
Proportional segments
The ratio of segments AB and CD is the ratio of their lengths, i.e. AB/CD. They say that segments AB and CD are proportional to segments A1 B1 and C1 D1, if AB/A1B1=CD/C1D1. The concept of proportionality is also introduced for large number segments
Slide 11
Definition of similar triangles.
Two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other
Slide 12
Ratio of areas of similar triangles
Theorem The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient
Slide 13
Proof.
Let triangles ABC and A1B1C1 be similar and let the similarity coefficient be equal to r. Let us denote the areas of these triangles by the letters S and S1. Since angle A = angle A1, then S/S1 = AB*AC/A1B1*A1C1 (according to the theorem on the ratio of areas, the similarity relations of triangles having equal angles). According to formulas (2) we have: AB/A1B1=R, AC/A1C1=R, therefore S/S=R 2
Slide 14
The first sign of similarity of triangles
If two angles of one triangle are respectively equal to two angles of another, then such triangles are equal A B C
Slide 15
The second sign of similarity of triangles
If two sides of another triangle are proportional to two sides of another triangle and the angles between these sides are equal, then the triangles are similar.
Slide 16
The third sign of similarity of triangles
If three sides of one triangle are proportional to three sides of another, then the triangles are similar. A B C
Slide 17
Proof.(1)
Given: ABC and A1B1C1 are two triangles in which angle A = angle A1, angle B = angle B1. Let us prove that triangle ABC is triangle A! B1C1
Slide 18
Proof.
According to the theorem on the sum of triangle angles, angle C = 180 degrees - angle A - angle B, angle C = 180 degrees - angle A - angle B, and, therefore, angle C = angle 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
Slide 19
Let us prove that the sides of triangle ABC are proportional to similar sides of triangle A B C. Since angle A = angle A and angle C = angle C, then S abs / Sa in c = AB * AC / A B * A C S abs / Sa in c = CA*SV/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
Slide 20
From these equalities it follows that AB/A B = BC/B C Similarly, using the equalities angle A = angle A Angle B = angle B, we obtain BC/B C = CA/C A. So the sides of triangle ABC are proportional to similar sides of triangle A In C The theorem is proven. 1 1 1 1 1 1 1 1
Slide 21
Proof (2)
Given: two triangles ABC and A B C, for which AB/A B = AC/A C, angle A = angle A. Prove that triangle ABC is triangle A B C. For this, taking into account the first sign of similarity of triangles, it is enough to prove that angle B = corner B 1 1 1 1 1 1 1 1 1 1
Slide 22
Let's consider triangle ABC, for which angle 1 = angle A, angle 2 = angle B. Triangles ABC A B C are similar according to the first criterion of similarity of triangles, therefore AB/A B = AC /A C. On the other hand, by condition AB/A B = AC /A C. From these two equalities we obtain AC = AC. 2 1 1 2 1 1 1 1 1 2 1 1 1 1 1 1 2
Slide 23
Triangles ABC and ABC are equal on two sides between them (AB is the common side, AC = AC and angle A = angle 1, since 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 proven. 2 2 1 1 1 1
Slide 24
Proof (3)
Given: the sides of triangles ABC and A B C are proportional. Let us prove that triangle ABC is triangle A B C 1 1 1
Slide 25
Proof
To do this, taking into account the second sign of similarity of triangles, it is enough to prove that angle A = angle A. Consider triangle ABC, in which angle 1 = angle A, angle 2 = angle B. Triangles ABC and A B C are similar according to the first sign of similarity of triangles, therefore AB /A B = BC / B C = C A/C A.
Slide 26
Comparing these equalities with equalities (1) we obtain: BC = BC, CA = C A. Triangles ABC and ABC are equal on three sides. It follows that angle A = angle 1 and since angle 1 = angle A, then angle A = angle A. The theorem is proven. 2 2 2 1 1
Slide 27
Practical applications of triangle similarity
When solving many problems involving the construction of triangles, the so-called similarity method is used. It consists of first constructing a triangle similar to the desired one based on some data, and then using the remaining data to construct the desired triangle.
Slide 28
Task No. 1
Construct a triangle given two angles and the bisector at the vertex of the third angle
Slide 29
Solution
First, let's construct some triangle similar to the one we are looking for. To do this, draw an arbitrary segment A B and construct a triangle A B C, whose angles A and B are respectively equal to the given angles
Slide 30
Continuation
Next, we will construct the bisector of angle C and plot the segment CD on it, equal to this segment. Through point D we draw a line parallel to A B. It intersects the sides of angle C at some points A and B. triangle ABC is the desired one
Slide 31
In fact, since AB is parallel to A B, then angle A = angle A, angle B = angle B, and, therefore, the two angles of triangle ABC are respectively equal to these angles. By construction, the bisector CD of triangle ABC is equal to the given segment. So, triangle ABC satisfies all the conditions of the problem.
Slide 32
Basics(1)
1. Triangle ABC is similar to triangle A B C if and only if one of the following equivalent conditions is satisfied. 1 1 1
Slide 33
Conditions
A)AB:BC:CA = A B: B C: C A; B)AB:BC=A B:B C and angle ABC=angle A B C; B) 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
Slide 34
Basics(2)
2) if parallel lines cut off triangles AB C and AB C 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
Slide 35
Basics(3)
3) the midline of a triangle is the segment connecting the midpoints of the lateral sides. This segment is parallel to the third side and equal to half its length. The midline of a trapezoid is the segment connecting the midpoints of the sides of the trapezoid. This segment is parallel to the bases and equal to half the sum of their lengths
Slide 36
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 Sabc = 0.5*AB*ACsinA.
Slide 37
Basic information (5)
Polygons A A...A and B B...B are called similar if A A:A A:...:A A =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 described similar polygons, 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
Slide 38
On-site measurement work
The properties of such triangles can be used to carry out various field measurements. We will consider two tasks: determining the height of an object on the ground and the distance to an inaccessible point.
Slide 39
Task No. 1
Determining the height of an object
Slide 40
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 place 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. Mark on the surface of the earth point B, at which the straight line And A intersects with the surface of the earth. 1 1 1 1
Slide 41
Right Triangles A C B and ACB are similar according to the first characteristic of triangles (angle C = angle C = 90 degrees, angle B is common). From the similarity of the triangles it follows A C /AC = BC /BC, from where A C = AC*BC /BC, measuring the distance BC and BC and knowing the length AC of the pole, using the resulting formula, we determine the height A C of the telegraph pole 1 1 1 1 1 1 1 1 1 1
Slide 42
Problem (2)
Determining the distance to an inaccessible point
Slide 43
Continuation
Let's assume that we need to find the distance from point A to an inaccessible point B. To do this, select point C on the ground, draw a 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 = angle A, angle C = corner C, and measure the lengths of sides A B and A C of this triangle. 1 1 1 1 1 1 1 1 1
Slide 44
Since triangle ABC and A B C are similar (based on the first sign of similarity of triangles), then AB/A B = AC A C, from which we obtain AB = AC*A B /A C. This formula allows, based on the 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
Slide 45
To simplify calculations, it is convenient to construct a triangle A B C in such a way that A C: AC = 1:1000. for example, if AC = 130m, then take the distance A C equal to 130mm. In this case, AB = AC/A C * A B =1000*A B, therefore, by measuring the distance A B in millimeters, we immediately obtain the distance AB in meters 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
Slide 46
Example
Let AC = 130m, angle A = 73 degrees, angle C = 58 degrees. On paper we build a triangle A B C so that angle A = 73 degrees, angle C = 58 degrees, A C = 130 mm, and measure the segment A B. It is equal to 153 mm, Therefore, the required distance is early 153m. 1 1 1 1 1
Slide 47
Determining distance by constructing similar triangles
When determining the distance to distant or inaccessible objects, you can use the following technique. On a regular match you need to apply two-millimeter divisions with ink or pencil. You also need to know the approximate height of the object to which the distance is being 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.
Slide 48
Continuation
Let's say we need to determine the distance to the pillar. We point a match at it at arm's length, the length of which is approximately 60 cm. Let's assume that the height of the pillar looks equal to two divisions of the match, i.e. 4 mm. Having such data, we will make a proportion: 0.6/x=0.004/6.0;x=(0.6*6)/01004=900. Thus, the distance to the pillar is 900m.
View all slides
