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Wonderful triangle points. Four wonderful points of a triangle 4 wonderful points of a triangle definitions

The geometry lesson in the 8th grade is developed on the basis of the positional learning model.

Lesson Objectives:

  • The study of theoretical material on the topic "Four wonderful points of the triangle";
  • Development of thinking, logic, speech, imagination of students, the ability to analyze and evaluate work;
  • Development of group work skills;
  • Raising a sense of responsibility for the quality and result of the work performed.

Equipment:

  • cards with group names;
  • task cards for each group;
  • A-4 paper for recording the results of the work of the groups;
  • epigraph written on the board.

During the classes

1. Organizational moment.

2. Determining the goals and topic of the lesson.

Historically, geometry began with a triangle, so for two and a half millennia the triangle has been a symbol of geometry. School geometry can only become interesting and meaningful, only then can it become geometry proper when a deep and comprehensive study of the triangle appears in it. Surprisingly, the triangle, despite its apparent simplicity, is an inexhaustible object of study - no one, even in our time, dares to say that he has studied and knows all the properties of a triangle.

Who hasn't heard of the Bermuda Triangle, where ships and planes disappear without a trace? But the triangle itself is fraught with a lot of interesting and mysterious things.

The central place of the triangle is occupied by the so-called remarkable points.

I think that at the end of the lesson you will be able to say why the points are called wonderful and whether they are so.

What is the topic of our lesson? "Four Remarkable Points of the Triangle". The words of K. Weierstrass can serve as an epigraph to the lesson: “A mathematician who is not partly a poet will never achieve perfection in mathematics” (the epigraph is written on the blackboard).

Look at the wording of the topic of the lesson, at the epigraph and try to determine the goals of your work in the lesson. At the end of the lesson, we will check how you completed them.

3. Independent work of students.

Preparation for independent work

To work in the lesson, you must choose one of six groups: "Theorists", "Creativity", "Logic-constructors", "Practitioners", "Historians", "Experts".

briefing

Each group receives task cards. If the task is not clear, the teacher additionally makes explanations.

"Theorists"

Task: define the basic concepts necessary when studying the topic “Four Remarkable Points of a Triangle” (height of a triangle, median of a triangle, bisector of a triangle, perpendicular bisector, inscribed circle, circumscribed circle), you can use the textbook; write the main concepts on a piece of paper.

"Historians"

bisectors center of the inscribed circle perpendiculars the center of the circumscribed circle. The Elements does not say that the three heights triangles intersect at one point called orthocenter medians center of gravity

In the 20s of the XIX century. French mathematicians J. Poncelet, Ch. Brianchon and others independently established the following theorem: the bases of the medians, the bases of the heights and the midpoints of the segments of the heights connecting the orthocenter with the vertices of the triangle lie on the same circle.

This circle is called the "circle of nine points", or the "circle of Feuerbach", or the "circle of Euler". K. Feuerbach found that the center of this circle lies on the "Euler line".

Task: analyze the article and fill in the table reflecting the studied material.

Point name

What intersects

"Creation"

Task: come up with a cinquain (s) on the topic “Four wonderful points of a triangle” (for example, a triangle, a point, a median, etc.)

Rule for writing cinquain:

In the first line, the topic is called by one word (usually a noun).

The second line is a description of the topic in a nutshell (2 adjectives).

The third line is a description of the action within the framework of this topic in three words (verbs, participles).

The fourth line is a 4-word phrase showing the attitude towards the topic.

The last line is a one-word synonym (metaphor) that repeats the essence of the topic.

"Logic constructors"

The median of a triangle is a segment that connects any vertex of the triangle with the midpoint of the opposite side. Any triangle has three medians.

A bisector is a segment of the bisector of any angle from the vertex to the intersection with the opposite side. Any triangle has three bisectors.

The altitude of a triangle is the perpendicular dropped from any vertex of the triangle to the opposite side or its extension. Any triangle has three altitudes.

The median perpendicular to a segment is a straight line passing through the midpoint of the given segment and perpendicular to it. Any triangle has three perpendicular bisectors.

Task: Using triangular sheets of paper, construct by bending the intersection points of medians, heights, bisectors, mid-perpendiculars. Explain this to the whole class.

"Practices"

In the fourth book of the "Beginnings", Euclid solves the problem "Inscribe a circle in a given triangle." It follows from the solution that three bisectors interior angles of a triangle intersect at one point center of the inscribed circle. From the solution of another Euclid problem, it follows that perpendiculars, restored to the sides of the triangle at their midpoints, also intersect at one point - the center of the circumscribed circle. The Elements does not say that the three heights of a triangle intersect at one point, called orthocenter(Greek word "orthos" means straight, right). This proposal was, however, known to Archimedes, Pappus, Proclus. The fourth singular point of the triangle is the point of intersection medians. Archimedes proved that she is center of gravity(barycenter) of the triangle. The above four points have received particular attention since the 18th century. They have been called "remarkable" or "singular points of the triangle".

The study of the properties of a triangle associated with these and other points served as the beginning for the creation of a new branch of elementary mathematics - "the geometry of a triangle", or "a new geometry of a triangle", one of the founders of which was Leonhard Euler.

Task: analyze the proposed material and come up with a diagram that reflects the semantic relationships between units, explain it, draw it on a piece of paper, and draw it on the board.

Remarkable points of the triangle

1.____________ 2.___________ 3.______________ 4.____________

Drawing 1 Drawing 2 Drawing 3 Drawing 4

____________ ___________ ______________ ____________

(explanation)

"Experts"

Task: make a table in which you evaluate the work of each group, select the parameters by which you will evaluate the work of the groups, determine the points.

The parameters can be as follows: the participation of each student in the work of his group, participation in the defense, an interesting presentation of the material, visualization is presented, etc.

In your presentation, you should note the positive and negative points in the activities of each group.

4. Performance of groups.(for 2-3 minutes)

The results of the work are posted on the board

5. Summing up the lesson.

Look at the goals you formulated at the beginning of the lesson. Did you manage to complete everything?

Do you agree with the epigraph that was chosen for today's lesson?

6. Homework.

1) Ensure that the triangle, which rests on the tip of the needle at a certain point, is in balance using the material of today's lesson.

2) Draw in different triangles all 4 wonderful points.

Let us first prove the angle bisector theorem.

Theorem

Proof

1) Take an arbitrary point M on the bisector of the angle BAC, draw perpendiculars MK and ML to the straight lines AB and AC and prove that MK = ML (Fig. 224). Consider right triangles AM K and AML. They are equal in hypotenuse and acute angle (AM - common hypotenuse, ∠1 = ∠2 by condition). Therefore, MK = ML.

2) Let the point M lie inside the angle BAC and be equidistant from its sides AB and AC. Let us prove that the ray AM is the bisector of the angle BAC (see Fig. 224). Draw perpendiculars MK and ML to straight lines AB and AC. Right-angled triangles AMK and AML are equal in hypotenuse and leg (AM - common hypotenuse, MK = ML by condition). Therefore, ∠1 = ∠2. But this means that the ray AM is the bisector of the angle BAC. The theorem has been proven.


Rice. 224

Corollary 1

Consequence 2

Indeed, let us denote by the letter O the point of intersection of the bisectors AA 1 and BB 1 of the triangle ABC and draw from this point the perpendiculars OK, OL and OM, respectively, to the lines AB, BC and CA (Fig. 225). By the proved theorem OK = OM and OK = OL. Therefore, OM \u003d OL, i.e., the point O is equidistant from the sides of the angle ACB and, therefore, lies on the bisector CC 1 of this angle. Consequently, all three bisectors of the triangle ABC intersect at the point O, which was to be proved.


Rice. 225

Properties of the perpendicular bisector to a line segment

The perpendicular bisector of a segment is a straight line passing through the midpoint of the given segment and perpendicular to it.


Rice. 226

Let us prove the theorem on the perpendicular bisector to a segment.

Theorem

Proof

Let the line m be the perpendicular bisector to the segment AB, the point O is the midpoint of this segment (Fig. 227, a).


Rice. 227

1) Consider an arbitrary point M of the line m and prove that AM = VM. If the point M coincides with the point O, then this equality is true, since O is the midpoint of the segment AB. Let M and O be distinct points. Right-angled triangles OAM and OBM are equal in two legs (OA = OB, OM - common leg), therefore AM = VM.

2) Consider an arbitrary point N, equidistant from the ends of the segment AB, and prove that the point N lies on the line m. If N is a point of the line AB, then it coincides with the midpoint O of the segment AB and therefore lies on the line m. If the point N does not lie on the line AB, then the triangle ANB is isosceles, since AN \u003d BN (Fig. 227, b). The segment NO is the median of this triangle, and hence the height. Thus, NO ⊥ AB; therefore, the lines ON and m coincide, i.e., N is a point of the line m. The theorem has been proven.

Corollary 1

Consequence 2

To prove this statement, consider the perpendicular bisectors m and n to the sides AB and BC of the triangle ABC (Fig. 228). These lines intersect at some point O. Indeed, if we assume the opposite, that is, that m || n, then the line BA, being perpendicular to the line m, would also be perpendicular to the line n parallel to it, and then two lines BA and BC, perpendicular to the line n, would pass through the point B, which is impossible.


Rice. 228

According to the proved theorem, OB = OA and OB = OS. Therefore, OA \u003d OC, that is, the point O is equidistant from the ends of the segment AC and, therefore, lies on the perpendicular bisector p to this segment. Therefore, all three perpendicular bisectors m, n and p to the sides of the triangle ABC intersect at point O.

Triangle intersection theorem

We have proved that the bisectors of a triangle intersect at one point, the perpendicular bisectors to the sides of the triangle intersect at one point. Earlier it was proved that the medians of a triangle intersect at one point (section 64). It turns out that the altitudes of a triangle have a similar property.

Theorem

Proof

Consider an arbitrary triangle ABC and prove that the lines AA 1 BB 1 and CC 1 containing its heights intersect at one point (Fig. 229).


Rice. 229

Draw a line through each vertex of triangle ABC parallel to the opposite side. We get a triangle A 2 B 2 C 2. Points A, B and C are the midpoints of the sides of this triangle. Indeed, AB \u003d A 2 C and AB \u003d CB 2 as opposite sides of the parallelograms ABA 2 C and ABCB 2, therefore A 2 C \u003d CB 2. Similarly, C 2 A \u003d AB 2 and C 2 B \u003d BA 2. In addition, as follows from the construction, CC 1 ⊥ A 2 B 2 , AA 1 ⊥ B 2 C 2 and BB 1 ⊥ A 2 C 2 . Thus, the lines AA 1, BB 1 and CC 1 are perpendicular bisectors to the sides of the triangle A 2 B 2 C 2. Therefore, they intersect at one point. The theorem has been proven.

So, four points are associated with each triangle: the point of intersection of the medians, the point of intersection of the bisectors, the point of intersection of the midperpendiculars to the sides, and the point of intersection of the heights (or their extensions). These four points are called wonderful points of the triangle.

Tasks

674. From the point M of the bisector of the non-expanded angle O, perpendiculars MA and MB are drawn to the sides of this angle. Prove that AB ⊥ OM.

675. The sides of angle O touch each of two circles having a common tangent at point A. Prove that the centers of these circles lie on the line O A.

676. The sides of angle A touch a circle with center O of radius r. Find: a) OA, if r = 5 cm, ∠A = 60°; b) d, if ОА = 14 dm, ∠A = 90°.

677. Bisectors of exterior angles at vertices B and C of triangle ABC intersect at point O. Prove that point O is the center of a circle tangent to lines AB, BC, AC.

678. Bisectors AA 1 and BB 1 of triangle ABC intersect at point M. Find the angles ACM and BCM if: a) ∠AMB = 136°; b) ∠AMB = 111°.

679. The perpendicular bisector to side BC of triangle ABC intersects side AC at point D. Find: a) AD and CD if BD = 5 cm, Ac = 8.5 cm; b) AC, if BD = 11.4 cm, AD = 3.2 cm.

680. The perpendicular bisectors to sides AB and AC of triangle ABC intersect at point D of side BC. Prove that: a) point D is the midpoint of side BC; b) ∠A - ∠B + ∠C.

681. The perpendicular bisector to side AB of an isosceles triangle ABC intersects side BC at point E. Find the base AC if the perimeter of triangle AEC is 27 cm and AB = 18 cm.

682. Isosceles triangles ABC and ABD have common ground AB. Prove that line CD passes through the midpoint of segment AB.

683. Prove that if the sides AB and AC in a triangle ABC are not equal, then the median AM of the triangle is not an altitude.

684. The angle bisectors at the base AB of an isosceles triangle ABC intersect at point M. Prove that line CM is perpendicular to line AB.

685. Altitudes AA 1 and BB 1 of an isosceles triangle ABC, drawn to the sides, intersect at point M. Prove that line MC is the perpendicular bisector to segment AB.

686. Construct the perpendicular bisector to the given segment.

Solution

Let AB be the given segment. Let's build two circles with centers at points A and B of radius AB (Fig. 230). These circles intersect at two points M 1 and M 2 . Segments AM 1 , AM 2 , VM 1 , VM 2 are equal to each other as the radii of these circles.


Rice. 230

Let's draw a straight line M 1 M 2. It is the required perpendicular bisector to segment AB. In fact, the points M 1 and M 2 are equidistant from the ends of the segment AB, so they lie on the perpendicular bisector to this segment. Hence, the line M 1 M 2 is the perpendicular bisector to the segment AB.

687. Given a line a and two points A and B lying on the same side of this line. On the line a, construct a point M, equidistant from points A to B.

688. An angle and a segment are given. Construct a point inside the given angle, equidistant from its sides and equidistant from the ends of the given segment.

Answers to tasks

    674. Instruction. First prove that triangle AOB is isosceles.

    676. a) 10 cm; b) 7√2 dm.

    678. a) 46° and 46°; b) 21° and 21°.

    679. a) AB = 3.5 cm, CD = 5 cm; b) AC = 14.6 cm.

    683. Instruction. Use the method of proof by contradiction.

    687. Instruction. Use the theorem of item 75.

    688. Instruction. Note that the desired point lies on the bisector of the given angle.

1 That is, it is equidistant from the lines containing the sides of the angle.

There are so-called four remarkable points in a triangle: the point of intersection of the medians. The point of intersection of the bisectors, the point of intersection of the heights and the point of intersection of the perpendicular bisectors. Let's consider each of them.

Point of intersection of the medians of a triangle

Theorem 1

On the intersection of the medians of a triangle: The medians of a triangle intersect at one point and divide the intersection point in a ratio of $2:1$ starting from the vertex.

Proof.

Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its median. Since the medians divide the sides in half. Consider the middle line $A_1B_1$ (Fig. 1).

Figure 1. Medians of a triangle

By Theorem 1, $AB||A_1B_1$ and $AB=2A_1B_1$, hence $\angle ABB_1=\angle BB_1A_1,\ \angle BAA_1=\angle AA_1B_1$. Hence the triangles $ABM$ and $A_1B_1M$ are similar according to the first triangle similarity criterion. Then

Similarly, it is proved that

The theorem has been proven.

Intersection point of the bisectors of a triangle

Theorem 2

On the intersection of the bisectors of a triangle: The bisectors of a triangle intersect at one point.

Proof.

Consider triangle $ABC$, where $AM,\ BP,\ CK$ are its bisectors. Let the point $O$ be the intersection point of the bisectors $AM\ and\ BP$. Draw from this point perpendicular to the sides of the triangle (Fig. 2).

Figure 2. Bisectors of a triangle

Theorem 3

Each point of the bisector of a non-expanded angle is equidistant from its sides.

By Theorem 3, we have: $OX=OZ,\ OX=OY$. Hence $OY=OZ$. Hence the point $O$ is equidistant from the sides of the angle $ACB$ and therefore lies on its bisector $CK$.

The theorem has been proven.

Intersection point of the perpendicular bisectors of a triangle

Theorem 4

The perpendicular bisectors of the sides of a triangle intersect at one point.

Proof.

Let a triangle $ABC$ be given, $n,\ m,\ p$ its perpendicular bisectors. Let the point $O$ be the intersection point of the perpendicular bisectors $n\ and\ m$ (Fig. 3).

Figure 3. Perpendicular bisectors of a triangle

For the proof we need the following theorem.

Theorem 5

Each point of the perpendicular bisector to a segment is equidistant from the ends of the given segment.

By Theorem 3, we have: $OB=OC,\ OB=OA$. Hence $OA=OC$. This means that the point $O$ is equidistant from the ends of the segment $AC$ and, therefore, lies on its perpendicular bisector $p$.

The theorem has been proven.

The point of intersection of the altitudes of the triangle

Theorem 6

The heights of a triangle or their extensions intersect at one point.

Proof.

Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its height. Draw a line through each vertex of the triangle parallel to the side opposite the vertex. We get a new triangle $A_2B_2C_2$ (Fig. 4).

Figure 4. Heights of a triangle

Since $AC_2BC$ and $B_2ABC$ are parallelograms with a common side, then $AC_2=AB_2$, that is, point $A$ is the midpoint of side $C_2B_2$. Similarly, we get that the point $B$ is the midpoint of the side $C_2A_2$, and the point $C$ is the midpoint of the side $A_2B_2$. From the construction we have that $(CC)_1\bot A_2B_2,\ (BB)_1\bot A_2C_2,\ (AA)_1\bot C_2B_2$. Hence $(AA)_1,\ (BB)_1,\ (CC)_1$ are the perpendicular bisectors of triangle $A_2B_2C_2$. Then, by Theorem 4, we have that the heights $(AA)_1,\ (BB)_1,\ (CC)_1$ intersect at one point.

© Kugusheva Natalya Lvovna, 2009 Geometry, Grade 8 TRIANGLES FOUR REMARKABLE POINTS

Intersection point of triangle medians Intersection point of triangle bisectors Intersection point of triangle heights Intersection point of perpendicular bisectors of a triangle

The median (BD) of a triangle is the line segment that connects the vertex of the triangle to the midpoint of the opposite side. A B C D Median

The medians of a triangle intersect at one point (the center of gravity of the triangle) and are divided by this point in a ratio of 2: 1, counting from the top. AM:MA 1 = VM:MV 1 = SM:MS 1 = 2:1. A A 1 B B 1 M C C 1

The bisector (A D) of a triangle is the segment of the bisector of the interior angle of the triangle.

Each point of the bisector of an unfolded angle is equidistant from its sides. Conversely, every point lying inside an angle and equidistant from the sides of the angle lies on its bisector. A M B C

All bisectors of a triangle intersect at one point - the center of the circle inscribed in the triangle. C B 1 M A B A 1 C 1 O The radius of the circle (OM) is a perpendicular dropped from the center (t.O) to the side of the triangle

HEIGHT The height (C D) of a triangle is the segment of the perpendicular dropped from the vertex of the triangle to the line containing the opposite side. A B C D

The heights of a triangle (or their extensions) intersect at one point. A A 1 B B 1 C C 1

MIDDLE PERPENDICULAR The perpendicular bisector (DF) is a line perpendicular to a side of a triangle and dividing it in half. A D F B C

A M B m O Each point of the perpendicular bisector (m) to a segment is equidistant from the ends of this segment. Conversely, each point equidistant from the ends of the segment lies on the perpendicular bisector to it.

All the perpendicular bisectors of the sides of a triangle intersect at one point - the center of the circle circumscribed about the triangle. A B C O The radius of the circumscribed circle is the distance from the center of the circle to any vertex of the triangle (OA). m n p

Student tasks Use a compass and straightedge to construct a circle inscribed in an obtuse triangle. To do this: Construct the bisectors of an obtuse triangle using a compass and straightedge. The intersection point of the bisectors is the center of the circle. Construct the radius of the circle: the perpendicular from the center of the circle to the side of the triangle. Construct a circle inscribed in a triangle.

2. Use a compass and straightedge to construct a circle circumscribed by obtuse triangle. To do this: Construct the perpendicular bisectors to the sides of an obtuse triangle. The point of intersection of these perpendiculars is the center of the circumscribed circle. The radius of a circle is the distance from the center to any vertex of the triangle. Construct a circle circumscribing a triangle.

Content

Introduction…………………………………………………………………………………………3

Chapter 1.

1.1 Triangle………………………………………………………………………………..4

1.2. Triangle medians

1.4. Heights in a triangle

Conclusion

List of used literature

Booklet

Introduction

Geometry is a branch of mathematics that deals with various shapes and their properties. Geometry starts with a triangle. For two and a half millennia, the triangle has been a symbol of geometry; but it is not only a symbol, the triangle is an atom of geometry.

In my work, I will consider the properties of the intersection points of the bisectors, medians and altitudes of a triangle, talk about their remarkable properties and the lines of the triangle.

These points studied in the school geometry course include:

a) the point of intersection of the bisectors (the center of the inscribed circle);

b) the point of intersection of the medial perpendiculars (the center of the circumscribed circle);

c) point of intersection of heights (orthocenter);

d) point of intersection of medians (centroid).

Relevance: expand your knowledge of the triangle,its propertieswonderful points.

Target: study of a triangle on its remarkable points,studying themclassifications and properties.

Tasks:

1. Study the necessary literature

2. Study the classification of the remarkable points of the triangle

3. Be able to build wonderful points of a triangle.

4. Summarize the studied material for the design of the booklet.

Project hypothesis:

the ability to find remarkable points in any triangle allows you to solve geometric construction problems.

Chapter 1. Historical information about the remarkable points of the triangle

In the fourth book of the "Beginnings" Euclid solves the problem: "Inscribe a circle in a given triangle." It follows from the solution that the three bisectors of the interior angles of a triangle intersect at one point - the center of the inscribed circle. From the solution of another problem of Euclid, it follows that the perpendiculars restored to the sides of the triangle at their midpoints also intersect at one point - the center of the circumscribed circle. The "Principles" does not say that the three heights of a triangle intersect at one point, called the orthocenter (the Greek word "orthos" means "straight", "correct"). This proposal was, however, known to Archimedes, Pappus, Proclus.

The fourth singular point of the triangle is the point of intersection of the medians. Archimedes proved that it is the center of gravity (barycenter) of the triangle. The above four points were given special attention, and since the 18th century they have been called the "remarkable" or "special" points of the triangle.

The study of the properties of a triangle associated with these and other points served as the beginning for the creation of a new branch of elementary mathematics - "triangle geometry" or "new triangle geometry", one of the founders of which was Leonhard Euler. In 1765, Euler proved that in any triangle, the orthocenter, barycenter, and center of the circumscribed circle lie on the same line, later called "Euler's line."

    1. Triangle

Triangle - geometric figure, consisting of three points that do not lie on the same straight line, and three segments connecting these points in pairs. Points -peaks triangles, line segmentssides triangle.

AT A, B, C - peaks

AB, BC, SA - sides

A C

Each triangle has four points associated with it:

    Intersection point of medians;

    Bisector intersection point;

    Height crossing point.

    The point of intersection of the perpendicular bisectors;

1.2. Triangle medians

Triangle medina - , connecting the top with the middle of the opposite side (Figure 1). The point of intersection of the median with the side of the triangle is called the base of the median.

Figure 1. Medians of a triangle

Let's build the midpoints of the sides of the triangle and draw a line segment connecting each of the vertices with the midpoint of the opposite side. Such segments are called the median.

And again we observe that these segments intersect at one point. If we measure the lengths of the resulting segments of the medians, then we can check one more property: the intersection point of the medians divides all medians in a ratio of 2: 1, counting from the vertices. And yet, the triangle, which rests on the tip of the needle at the point of intersection of the medians, is in equilibrium! A point with this property is called the center of gravity (barycenter). The center of equal masses is sometimes called the centroid. Therefore, the properties of the medians of a triangle can be formulated as follows: the medians of a triangle intersect at the center of gravity and the intersection point is divided in a ratio of 2:1, counting from the vertex.

1.3. Triangle bisectors

bisector called the bisector of an angle drawn from the vertex of the angle to its intersection with the opposite side. The triangle has three bisectors corresponding to its three vertices (Figure 2).

Figure 2. Bisector of a triangle

In an arbitrary triangle ABC, we draw the bisectors of its angles. And again, with exact construction, all three bisectors will intersect at one point D. Point D is also unusual: it is equidistant from all three sides of the triangle. This can be verified by dropping the perpendiculars DA 1, DB 1 and DC1 to the sides of the triangle. All of them are equal: DA1=DB1=DC1.

If you draw a circle centered at point D and radius DA 1, then it will touch all three sides of the triangle (that is, it will have only one common point with each of them). Such a circle is called inscribed in a triangle. So, the bisectors of the angles of a triangle intersect at the center of the inscribed circle.

1.4. Heights in a triangle

Triangle Height - , dropped from top to the opposite side or a straight line coinciding with the opposite side. Depending on the type of triangle, the height may be contained within the triangle (for triangle), coincide with its side (be triangle) or pass outside the triangle at an obtuse triangle (Figure 3).

Figure 3. Heights in triangles

    If you build three heights in a triangle, then they all intersect at one point H. This point is called the orthocenter. (Figure 4).

Using constructions, you can check that, depending on the type of triangle, the orthocenter is located differently:

    at an acute triangle - inside;

    in a rectangular one - on the hypotenuse;

    obtuse - outside.

Figure 4. Orthocenter of a triangle

Thus, we got acquainted with another remarkable point of the triangle and we can say that: the heights of the triangle intersect at the orthocenter.

1.5. Midperpendiculars to the sides of a triangle

The perpendicular bisector of a segment is a line perpendicular to the given segment and passing through its midpoint.

Let us draw an arbitrary triangle ABC and draw the perpendicular bisectors to its sides. If the construction is done exactly, then all the perpendiculars will intersect at one point - point O. This point is equidistant from all the vertices of the triangle. In other words, if you draw a circle centered at point O, passing through one of the vertices of the triangle, then it will pass through its other two vertices.

A circle passing through all the vertices of a triangle is called circumcircle. Therefore, the established property of a triangle can be formulated as follows: the perpendicular bisectors to the sides of the triangle intersect at the center of the circumscribed circle (Figure 5).

Figure 5. Triangle inscribed in a circle

Chapter 2

Exploring Height in Triangles

All three heights of a triangle intersect at one point. This point is called the orthocenter of the triangle.

The heights of an acute-angled triangle are located strictly inside the triangle.

Accordingly, the point of intersection of the heights is also inside the triangle.

In a right triangle, the two heights are the same as the sides. (These are the heights drawn from the vertices of acute angles to the legs).

The altitude drawn to the hypotenuse lies inside the triangle.

AC is the height drawn from vertex C to side AB.

AB is the height drawn from vertex B to side AC.

AK is the height drawn from the vertex of right angle A to the hypotenuse BC.

The heights of a right triangle intersect at the vertex of the right angle (A is the orthocenter).

In an obtuse triangle, there is only one height inside the triangle - the one drawn from the vertex of the obtuse angle.

The other two heights lie outside the triangle and are lowered to the extension of the sides of the triangle.

AK is the height drawn to side BC.

BF is the height drawn to the extension of side AC.

CD is the height drawn to the extension of side AB.

The intersection point of the heights of an obtuse triangle is also outside the triangle:

H is the orthocenter of triangle ABC.

Study of Bisectors in a Triangle

The bisector of a triangle is the part of the angle bisector of a triangle (a ray) that is inside the triangle.

All three bisectors of a triangle intersect at one point.


The intersection point of the bisectors in acute, obtuse and right triangles is the center of the circle inscribed in the triangle and is located inside.

Research medians in a triangle

Since a triangle has three vertices and three sides, there are also three segments connecting the vertex and the midpoint of the opposite side.


After examining these triangles, I realized that in any triangle the medians intersect at one point. This point is called center of gravity of the triangle.

Investigation of perpendicular bisectors to the side of a triangle

Midperpendicular A triangle is a perpendicular to the midpoint of a side of a triangle.

The three perpendicular bisectors of a triangle intersect at one point and are the center of the circumscribed circle.

The point of intersection of the perpendicular bisectors in an acute triangle lies inside the triangle; in obtuse - outside the triangle; in a rectangular one - in the middle of the hypotenuse.

Conclusion

In the course of the work done, we come to the following conclusions:

    Goal achieved:explored the triangle and found its remarkable points.

    The tasks set have been solved:

one). We studied the necessary literature;

2). Studied the classification of the remarkable points of the triangle;

3). Learned how to build wonderful points of a triangle;

four). Summarized the studied material for the design of the booklet.

The hypothesis that the ability to find the remarkable points of a triangle helps in solving construction problems has been confirmed.

The paper consistently outlines the techniques for constructing remarkable points of a triangle, provides historical information about geometric constructions.

Information from this work can be useful in geometry lessons in grade 7. The booklet can become a reference book on geometry on the topic presented.

Bibliography

    Textbook. L.S. Atanasyan "Geometry 7-9 gradesMnemosyne, 2015.

    Wikipediahttps://ru.wikipedia.org/wiki/Geometry#/media/File:Euclid%27s_postulates.png

    Portal Scarlet Sails

    Leading educational portal of Russia http://cendomzn.ucoz.ru/index/0-15157