Proportional segments in a right triangle. Proportional segments in a right triangle Average proportional segments in a right triangle proof

Lesson objectives:

1. introduce the concept of the proportional mean (geometric mean) of two segments;
2. consider the problem of proportional segments in a right triangle: the property of the altitude of a right triangle drawn from the vertex of a right angle;
3. to develop students’ skills in using the studied topic in the process of solving problems.

Lesson type: lesson of learning new material.

Plan:

1. Org moment.
2. Updating knowledge.
3. Studying the property of the altitude of a right triangle drawn from the vertex of a right angle:
   - preparatory stage;
   – introduction;
   – assimilation.
4. Introduction of the concept of an average proportional to two segments.
5. Mastering the concept of the average proportional of two segments.
6. Proof of the consequences:
   – 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;
   – 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.
7. Problem solving.
8. Summarizing.
9. Setting homework.

During the classes

I. ORGANIZATIONAL MOMENT

- Hello guys, have a seat. Is everyone ready for class?

Let's start work.

II. KNOWLEDGE UPDATED

– What important mathematical concept did you learn in previous lessons? ( with the concept of similarity of triangles)

- Let's remember which two triangles are called similar? (two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to similar sides of the other triangle)

– What do we use to prove the similarity of two triangles? (

– Formulate these signs (formulate three signs of similarity of triangles)

III. STUDYING THE PROPERTIES OF THE HEIGHT OF A RECTANGULAR TRIANGLE, CONDUCTED FROM THE TOP OF A RIGHT ANGLE

a) preparatory stage

– Guys, please look at the first slide. ( Application) Shown here are two right triangles – and . and are the heights and respectively. .

Task 1. a) Determine whether and are similar.

– What do we use to prove the similarity of triangles? ( signs of similarity of triangles)

(the first sign, because in the problem nothing is known about the sides of the triangles)

. (Two pairs: 1. ∟B= ∟B1 (straight), 2. ∟A= ∟A 1)

– Draw a conclusion.( by the first criterion of similarity of triangles ~)

Task 1. b) Determine whether and are similar.

– What sign of similarity will we use and why? (the first sign, because in the problem nothing is known about the sides of the triangles)

– How many pairs of equal angles do we need to find? Find these pairs (since the triangles are right-angled, then one pair of equal angles is enough: ∟A= ∟A 1)

- Draw a conclusion. (based on the first criterion of similarity of triangles, we conclude that these triangles are similar).

As a result of the conversation, slide 1 looks like this:

b) discovery of the theorem

Task 2.

– Determine whether and are similar. As a result of the conversation, answers are built that are reflected on the slide.

– The picture indicated that . Did we use this degree measure when answering the assignment questions? ( No, we didn't use it)

– Guys, draw a conclusion: what triangles does a right triangle divide into by the altitude drawn from the vertex of the right angle? (conclude)

– The question arises: will these two right triangles, into which the height divides the right triangle, be similar to each other? Let's try to find pairs of equal angles.

As a result of the conversation, a record is built:

– Now let’s draw a full conclusion.( CONCLUSION: the altitude of a right triangle drawn from the vertex of the right angle divides the triangle into two similar

- That. We formulated and proved a theorem about the property of the height of a right triangle.

Let's establish the structure of the theorem and make a drawing. What is given in the theorem and what needs to be proven? Students write in their notebook:

– Let’s prove the first point of the theorem for the new drawing. What similarity feature will we use and why? (The first, because in the theorem nothing is known about the sides of triangles)

– How many pairs of equal angles do we need to find? Find these pairs. (In this case, one pair is sufficient: ∟A-general)

- Draw a conclusion. The triangles are similar. As a result, a sample of the theorem is shown

– Write out the second and third points at home yourself.

c) mastering the theorem

- So, formulate the theorem again (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 this one)

– How many pairs of similar triangles in the construction “in a right triangle the altitude is drawn from the vertex of a right angle” does this theorem allow you to find? ( Three pairs)

Students are given the following assignment:

IV. INTRODUCTION OF THE CONCEPT OF AVERAGE PROPORTIONAL OF TWO SEGMENTS

– And now we will study a new concept with you.

Attention!

Definition. Line segment XY called average proportional (geometric mean) between segments AB And CD, If

(write it down in a notebook).

V. UNDERSTANDING THE CONCEPT OF THE AVERAGE PROPORTIONAL OF TWO SEGMENTS

– Now let's turn to the next slide.

Exercise 1. Find the length of the average proportional segments MN and KP, if MN = 9 cm, KP = 16 cm.

– What is given in the problem? ( Two segments and their lengths: MN = 9 cm, KP = 16 cm)

– What do you need to find? ( The length of the average proportional to these segments)

– What formula expresses the proportional mean and how do we find it?

(Substitute the data into the formula and find the length of the average prop.)

Task No. 2. Find the length of segment AB if the proportional mean of segments AB and CD is 90 cm and CD = 100 cm

– What is given in the problem? (the length of the segment CD = 100 cm and the proportional average of the segments AB and CD is 90 cm)

– What should be found in the problem? ( Length of segment AB)

– How will we solve the problem? (Let’s write down the formula for the average proportional segments AB and CD, express the length AB from it and substitute the data in the problem.)

VI. CONCLUSION OF IMPLICATIONS

- Well done boys. Now let's return to the similarity of triangles, which we proved in the theorem. State the theorem again. ( 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 one)

– Let’s first use the similarity of triangles and . What follows from this? ( By definition, similarity sides are proportional to similar sides)

– What equality will result when using the basic property of proportion? ()

– Express CD and draw a conclusion (;.

Conclusion: 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)

– Now prove on your own that 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. We will find from -... the segments into which the hypotenuse is divided by this altitude )

A leg of a right triangle is the mean proportional between...(-...the hypotenuse and the segment of the hypotenuse enclosed between this leg and the height )

– Where do we apply the statements we have learned? ( When solving problems)

IX. SETTING HOMEWORK

d/z: No. 571, No. 572 (a, d), independent work in a notebook, theory.

Similarity test for right triangles

Let us first introduce the similarity criterion for right triangles.

Theorem 1

Similarity test for right triangles: two right triangles are similar when they each have one equal acute angle (Fig. 1).

Figure 1. Similar right triangles

Proof.

Let us be given that $\angle B=\angle B_1$. Since the triangles are right-angled, then $\angle A=\angle A_1=(90)^0$. Therefore, they are similar according to the first criterion of similarity of triangles.

The theorem has been proven.

Height theorem in right triangle

Theorem 2

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.

Proof.

Let us be given a right triangle $ABC$ with right angle $C$. Let's draw the height $CD$ (Fig. 2).

Figure 2. Illustration of Theorem 2

Let us prove that triangles $ACD$ and $BCD$ are similar to triangle $ABC$ and that triangles $ACD$ and $BCD$ are similar to each other.

Since $\angle ADC=(90)^0$, then the triangle $ACD$ is right-angled. Triangles $ACD$ and $ABC$ have a common angle $A$, therefore, by Theorem 1, triangles $ACD$ and $ABC$ are similar.

Since $\angle BDC=(90)^0$, then the triangle $BCD$ is right-angled. Triangles $BCD$ and $ABC$ have a common angle $B$, therefore, by Theorem 1, triangles $BCD$ and $ABC$ are similar.

Let us now consider the triangles $ACD$ and $BCD$

\[\angle A=(90)^0-\angle ACD\] \[\angle BCD=(90)^0-\angle ACD=\angle A\]

Therefore, by Theorem 1, the triangles $ACD$ and $BCD$ are similar.

The theorem has been proven.

Average proportional

Theorem 3

The altitude of a right triangle drawn from the vertex of a right angle is the average proportional to the segments into which the altitude divides the hypotenuse of the given triangle.

Proof.

By Theorem 2, we have that the triangles $ACD$ and $BCD$ are similar, therefore

The theorem has been proven.

Theorem 4

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 angle.

Proof.

In the proof of the theorem we will use the notation from Figure 2.

By Theorem 2, we have that triangles $ACD$ and $ABC$ are similar, therefore

The theorem has been proven.

Lesson 40. Proportional segments in a right triangle. C. b. a. h. S. bc. N. ac. A. B. The altitude of a right triangle drawn from the vertex of a right angle divides the triangle into 2 similar right triangles, each of which is similar to the given triangle. Similarity test for right triangles. Two right triangles are similar if they each have an equal acute angle. The segment XY is called the proportional mean (geometric mean) for the segments AB and CD if Property 1. The altitude of a right triangle drawn from the vertex of a right angle is the proportional mean between the projections of the legs onto the hypotenuse. Property 2. A leg of a right triangle is the mean proportional between the hypotenuse and the projection of this leg onto the hypotenuse.

Geometry 8th grade

“Solving problems on the Pythagorean theorem” - Triangle ABC is isosceles. Practical application of the Pythagorean theorem. ABCD is a quadrilateral. Area of ​​a square. Find the sun. Proof. Bases of an isosceles trapezoid. Consider the Pythagorean theorem. Area of ​​a quadrilateral. Right triangles. Pythagorean theorem. The square of the hypotenuse is equal to the sum of the squares of the legs.

“Finding the area of ​​a parallelogram” - Base. Height. Determining the height of a parallelogram. Signs of equality of right triangles. Area of ​​a parallelogram. Find the area of ​​the triangle. Properties of areas. Oral exercises. Find the area of ​​the parallelogram. Heights of a parallelogram. Find the perimeter of the square. Area of ​​a triangle. Find the area of ​​the square. Find the area of ​​the rectangle. Area of ​​a square.

""Square" 8th grade" - Black Square. Tasks for oral work around the perimeter of the square. Area of ​​a square. Signs of a square. The square is among us. A square is a rectangle with all sides equal. Square. Bag with a square base. Oral tasks. How many squares are shown in the picture? Properties of a square. Rich merchant. Assignments for oral work on the area of ​​a square. Perimeter of a square.

“Definition of axial symmetry” - Points lying on the same perpendicular. Draw two straight lines. Construction. Plot the points. Clue. Figures that do not have axial symmetry. Line segment. Missing coordinates. Figure. Figures that have more than two axes of symmetry. Symmetry. Symmetry in poetry. Construct triangles. Axes of symmetry. Construction of a segment. Construction of a point. Figures with two axes of symmetry. Peoples. Triangles. Proportionality.

“Definition of similar triangles” - Polygons. Proportional segments. Ratio of areas of similar triangles. Two triangles are called similar. Conditions. Construct a triangle using the given two angles and the bisector at the vertex. Let's say we need to determine the distance to the pillar. The third sign of similarity of triangles. Let's build some kind of triangle. ABC. Triangles ABC and ABC are equal on three sides. Determining the height of an object.

“Solution of the Pythagorean Theorem” - Parts of windows. The simplest proof. Hammurabi. Diagonal. Complete proof. Proof by subtraction method. Pythagoreans. Proof by decomposition method. History of the theorem. Diameter. Proof by addition method. Epstein's proof. Cantor. Triangles. Followers. Applications of the Pythagorean theorem. Pythagorean theorem. Statement of the theorem. Perigal's proof. Application of the theorem.

Today we bring to your attention another presentation on an amazing and mysterious subject - geometry. In this presentation we will introduce you to a new property of geometric shapes, in particular, the concept of proportional segments in right triangles.

First, we should remember what a triangle is? This is the simplest polygon, consisting of three vertices connected by three segments. A triangle in which one of the angles is equal to 90 degrees is called a right triangle. You have already become acquainted with them in more detail in our previous educational materials presented to your attention.

So, returning to our topic today, let’s denote in order that the altitude of a right triangle drawn from a 90 degree angle divides it into two triangles that are similar both to each other and to the original one. All drawings and graphs that interest you are given in the proposed presentation; we recommend that you refer to them, accompanied by the described explanation.

A graphic example of the above thesis can be seen on the second slide. Based on the first sign of similarity of triangles, the triangles are similar because they have two identical angles. If we specify in more detail, then the height lowered to the hypotenuse forms a right angle with it, that is, there are already identical angles, and each of the formed angles also has one common angle as the original one. The result is two angles equal to each other. That is, the triangles are similar.

Let us also designate what does the concept of “proportional mean” or “geometric mean” mean? This is a certain XY segment for segments AB and CD, when it is equal to the square root of the product of their lengths.

From which it also follows that the leg of a right triangle is the geometric mean between the hypotenuse and the projection of this leg onto the hypotenuse, that is, another leg.

Another property of a right triangle is that its height, drawn from an angle of 90°, is the average proportional between the projections of the legs onto the hypotenuse. If you turn to the presentation and other materials offered to your attention, you will see that there is evidence of this thesis in a very simple and accessible form. Previously, we have already proven that the resulting triangles are similar to each other and to the original triangle. Then, using the ratio of the legs of these geometric figures, we come to the conclusion that the height of a right triangle is directly proportional to the square root of the product of the segments that were formed as a result of lowering the height from the right angle of the original triangle.

The last thing in the presentation is that the leg of a right triangle is the geometric mean for the hypotenuse and its segment located between the leg and the altitude drawn from an angle equal to 90 degrees. This case should be considered from the point of view that the indicated triangles are similar to each other, and the leg of one of them turns out to be the hypotenuse of the other. But you will become more familiar with this by studying the proposed materials.

Similarity test for right triangles

Let us first introduce the similarity criterion for right triangles.

Theorem 1

Similarity test for right triangles: two right triangles are similar when they each have one equal acute angle (Fig. 1).

Figure 1. Similar right triangles

Proof.

Let us be given that $\angle B=\angle B_1$. Since the triangles are right-angled, then $\angle A=\angle A_1=(90)^0$. Therefore, they are similar according to the first criterion of similarity of triangles.

The theorem has been proven.

Height theorem in right triangle

Theorem 2

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.

Proof.

Let us be given a right triangle $ABC$ with right angle $C$. Let's draw the height $CD$ (Fig. 2).

Figure 2. Illustration of Theorem 2

Let us prove that triangles $ACD$ and $BCD$ are similar to triangle $ABC$ and that triangles $ACD$ and $BCD$ are similar to each other.

Since $\angle ADC=(90)^0$, then the triangle $ACD$ is right-angled. Triangles $ACD$ and $ABC$ have a common angle $A$, therefore, by Theorem 1, triangles $ACD$ and $ABC$ are similar.

Since $\angle BDC=(90)^0$, then the triangle $BCD$ is right-angled. Triangles $BCD$ and $ABC$ have a common angle $B$, therefore, by Theorem 1, triangles $BCD$ and $ABC$ are similar.

Let us now consider the triangles $ACD$ and $BCD$

\[\angle A=(90)^0-\angle ACD\] \[\angle BCD=(90)^0-\angle ACD=\angle A\]

Therefore, by Theorem 1, the triangles $ACD$ and $BCD$ are similar.

The theorem has been proven.

Average proportional

Theorem 3

The altitude of a right triangle drawn from the vertex of a right angle is the average proportional to the segments into which the altitude divides the hypotenuse of the given triangle.

Proof.

By Theorem 2, we have that the triangles $ACD$ and $BCD$ are similar, therefore

The theorem has been proven.

Theorem 4

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 angle.

Proof.

In the proof of the theorem we will use the notation from Figure 2.

By Theorem 2, we have that triangles $ACD$ and $ABC$ are similar, therefore

The theorem has been proven.