How to draw 4 wonderful points of a triangle. Project "Remarkable Points of the Triangle"
Silchenkov Ilya
materials for the lesson, presentation with animation
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The midline of a triangle is a segment that connects the midpoints of two of its sides and is equal to half of this side. Also, according to the theorem middle line triangle is parallel to one of its sides and equal to half of that side.
If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
Remarkable triangle points
wonderful points triangle Intersection point of medians (centroid of triangle) ; The point of intersection of the bisectors, the center of the inscribed circle; The point of intersection of the perpendicular bisectors; Point of intersection of heights (orthocenter); Euler line and circle of nine points; Gergonne and Nagel points; Point Fermat-Torricelli;
Intersection point of medians
The median of a triangle is a line segment that connects the vertex of any angle of the triangle with the midpoint of the opposite side.
I. The medians of a triangle intersect at one point, which divides each median in a ratio of 2:1, counting from the top.
Proof:
A B C A 1 C 1 B 1 1 2 3 4 0 2. The segment A 1 B 1 is parallel to the side AB and 1/2 AB \u003d A 1 B 1 i.e. AB \u003d 2A1B1 (according to the triangle midline theorem), therefore 1 \u003d 4 and 3 \u003d 2 (because they internal cross-lying angles with parallel lines AB and A 1 B 1 and secant BB 1 for 1, 4 and AA 1 for 3, 2 3. Therefore, triangles AOB and A 1 OB 1 are similar in two angles, and, therefore, their sides are proportional , i.e. the ratios of the sides of AO and A 1 O, BO and B 1 O, AB and A 1 B 1 are equal. But AB = 2A 1 B 1, therefore AO \u003d 2A 1 O and BO \u003d 2B 1 O. Thus , the intersection point O of the medians BB 1 and AA 1 divides each of them in the ratio 2:1, counting from the top.
The center of mass is sometimes called the centroid. That is why they say that the point of intersection of the median is the centroid of the triangle. The center of mass of a homogeneous triangular plate is located at the same point. If a similar plate is placed on a pin so that the tip of the pin hits exactly the centroid of the triangle, then the plate will be in equilibrium. Also the point of intersection of the medians is the center of the incircle of its median triangle. An interesting property of the point of intersection of medians is connected with the physical concept of the center of mass. It turns out that if equal masses are placed at the vertices of a triangle, then their center will fall exactly at this point.
Intersection point of bisectors
Bisector of a triangle - a segment of the bisector of an angle connecting the vertex of one of the angles of the triangle with a point lying on the opposite side.
The bisectors of a triangle intersect at one point equidistant from its sides.
Proof:
C A B A 1 B 1 C 1 0 1. Denote by the letter O the intersection point of the bisectors AA 1 and BB 1 of the triangle ABC. 3. Let's use the fact that each point of the bisector of an unfolded angle is equidistant from its sides and vice versa: each point lying inside the angle and equidistant from the sides of the angle lies on its bisector. Then OK=OL and OK=OM. This means OM \u003d OL, i.e. point O is equidistant from the sides of the triangle ABC and, therefore, lies on the bisector CC1 of angle C. 4. Consequently, all three bisectors of the triangle ABC intersect at the point O. K L M The theorem is proved. 2. draw from this point the perpendiculars OK, OL and OM, respectively, to the straight lines AB, BC and CA.
Intersection point of perpendicular bisectors
The median perpendicular is a straight line passing through the midpoint of a given segment and perpendicular to it.
The perpendicular bisectors to the sides of the triangle intersect at one point equidistant from the vertices of the triangle.
Proof:
В С A m n 1. Denote by the letter O the point of intersection of the perpendicular bisectors m and n to the sides AB and BC of the triangle ABC. O 2. Using the theorem that each point of the perpendicular bisector to the segment is equidistant from the ends of this segment and vice versa: each point equidistant from the ends of the segment lies on the perpendicular bisector to it, we get that OB=OA and OB=OC. 3. 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 to this segment. 4. Therefore, all three perpendicular bisectors m, n and p to the sides of the triangle ABC intersect at the point O. The theorem is proved. R
Point of intersection of heights (or their extensions)
The height of a triangle is the perpendicular drawn from the vertex of any angle of the triangle to the line containing the opposite side.
The heights of a triangle or their extensions intersect at one point, which may lie in the triangle, or may be outside it.
Proof:
Let us prove that the lines AA 1 , BB 1 and CC 1 intersect at one point. B A C C2 C1 A1 A2 B 1 B 2 1. Draw a line through each vertex of triangle ABC parallel to the opposite side. We get a triangle A 2 B 2 C 2. 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 is perpendicular to A 2 B 2, AA 1 is perpendicular to B 2 C 2 and BB 1 is perpendicular to A 2 C 2 (from the corollary of the parallel lines and secant theorem). 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.
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 internal corners triangles 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 drawn Special attention, and since the 18th century they have been called "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 straight line, later called "Euler's line".
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.
IN 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 median segments, then we can check one more property: the median intersection point divides all medians in a 2:1 ratio, 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). 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 - height drawn from the top right angle And 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;
4). 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, 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 Russia http://cendomzn.ucoz.ru/index/0-15157
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.
Introduction
The objects of the world around us have certain properties, which are studied by various sciences.
Geometry is a branch of mathematics that considers various shapes and their properties, its roots go back to the distant past.
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. 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 "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 - "the geometry of a triangle" or "a new geometry of a triangle", 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 straight line, later called "Euler's line". In the twenties years XIX century, French mathematicians J. Poncelet, C. 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 established that the center of this circle lies on the Euler line.
“I think that we have never lived in such a geometrical period until now. Everything around is geometry. These words, spoken by the great French architect Le Corbusier at the beginning of the 20th century, very accurately characterize our time. The world in which we live is filled with the geometry of houses and streets, mountains and fields, the creations of nature and man.
We were interested in the so-called "wonderful points of the triangle."
After reading the literature on this topic, we fixed for ourselves the definitions and properties of the remarkable points of the triangle. But our work did not end there, and we wanted to explore these points ourselves.
That's why goal given work - the study of some wonderful points and lines of the triangle, the application of the knowledge gained to solving problems. In the process of achieving this goal, the following stages can be distinguished:
Selection and study educational material from various sources information, literature;
The study of the basic properties of the remarkable points and lines of the triangle;
Generalization of these properties and proof of necessary theorems;
Solving problems related to the remarkable points of the triangle.
ChapterI. Wonderful triangle dots and lines
1.1 Point of intersection of the midperpendiculars to the sides of a triangle
The perpendicular bisector is a straight line passing through the midpoint of a segment, perpendicular to it. We already know the theorem characterizing the property of the perpendicular bisector: each point of the perpendicular bisector to the segment is equidistant from its ends and vice versa, if the point is equidistant from the ends of the segment, then it lies on the perpendicular bisector.
The polygon is called inscribed into a circle if all its vertices belong to the circle. The circle is called circumscribed near the polygon.
A circle can be circumscribed around any triangle. Its center is the point of intersection of the medial perpendiculars to the sides of the triangle.
Let the point O be the point of intersection of the perpendicular bisectors to the sides of the triangle AB and BC.
Output: Thus, if the point O is the point of intersection of the midperpendiculars to the sides of the triangle, then OA = OS = OB, i.e. point O is equidistant from all vertices of the triangle ABC, which means that it is the center of the circumscribed circle.
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| acute-angled | obtuse | rectangular |
Consequences
sin γ \u003d c / 2R \u003d c / sin γ \u003d 2R.
It is proved similarly but/ sin α =2R, b/sin β =2R.
In this way:
This property is called the sine theorem.
In mathematics, it often happens that objects defined completely differently, turn out to match.
Example. Let A1, B1, C1 be the midpoints of the sides ∆ABS BC, AC, AB, respectively. Show that the circles circumscribed about the triangles AB1C1, A1B1C, A1BC1 intersect at one point. Moreover, this point is the center of the circumscribed about ∆ABS circle.
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| Consider the segment AO and construct a circle on this segment, as on a diameter. Points C1 and B1 fall on this circle, because are vertices of right angles based on AO. Points A, C1, B1 lie on a circle = this circle is circumscribed about ∆AB1C1. Similarly, we will draw a segment BO and construct a circle on this segment, as on a diameter. This will be a circle circumscribed about ∆BC1 A1. Let's draw a segment CO and build a circle on this segment, as on a diameter. This will be the circumscribed circle These three circles pass through the point O - the center of the circle circumscribed about ∆ABC. |
Generalization. If arbitrary points A 1 , B 1 , C 1 are taken on the sides ∆ABC AC, BC, AC, then the circles circumscribed about the triangles AB 1 C 1 , A 1 B 1 C, A 1 BC 1 intersect at one point.
1.2 Point of intersection of the bisectors of a triangle
The converse statement is also true: if a point is equidistant from the sides of an angle, then it lies on its bisector.
It is useful to mark the halves of one corner with the same letters:
OAF=OAD= α, OBD=OBE= β, OCE=OCF= γ.
Let the point O be the intersection point of the bisectors of angles A and B. By the property of a point lying on the bisector of angle A, OF=OD=r. By the property of a point lying on the bisector of angle B, OE=OD=r. Thus, OE=OD= OF=r= point O is equidistant from all sides of the triangle ABC, i.e. O is the center of the inscribed circle. (Point O is the only one).
Output: Thus, if the point O is the point of intersection of the bisectors of the angles of the triangle, then OE=OD= OF=r, i.e. point O is equidistant from all sides of the triangle ABC, which means that it is the center of the inscribed circle. The point O - the intersection of the bisectors of the angles of the triangle is a wonderful point of the triangle.
Consequences:
From the equality of triangles AOF and AOD (Figure 1) along the hypotenuse and sharp corner, follows that AF = AD . From the equality of triangles OBD and OBE it follows that BD = BE , It follows from the equality of triangles COE and COF that FROM F = CE . Thus, the segments of tangents drawn to the circle from one point are equal.
AF=AD= z, BD=BE= y, СF=CE= x
a=x+y (1), b= x+z (2), c= x+y (3).
+ (2) – (3), then we get: a+b-c=x+ y+ x+ z- z- y = a+b-c= 2x =
x=( b + c - a)/2
Similarly: (1) + (3) - (2), we get: y = (a + c -b)/2.
Similarly: (2) + (3) - (1), we get: z= (a +b - c)/2.
The angle bisector of a triangle divides the opposite side into segments proportional to the adjacent sides.
1.3 Point of intersection of the medians of a triangle (centroid)
Proof 1. Let A 1 , B 1 and C 1 be the midpoints of sides BC, CA and AB of triangle ABC, respectively (Fig. 4).
Let G be the intersection point of two medians AA 1 and BB 1 . Let us first prove that AG:GA 1 = BG:GB 1 = 2.
To do this, take the midpoints P and Q of segments AG and BG. According to the triangle midline theorem, the segments B 1 A 1 and PQ are equal to half of the side AB and are parallel to it. Therefore, the quadrilateral A 1 B 1 is a PQ-parallelogram. Then the intersection point G of its diagonals PA 1 and QB 1 bisects each of them. Therefore, the points P and G divide the median of AA 1 into three equal parts, and the points Q and G divide the median of BB 1 also into three equal parts. So, the point G of the intersection of the two medians of the triangle divides each of them in a ratio of 2:1, counting from the top.
The point of intersection of the medians of a triangle is called centroid or center of gravity triangle. This name is due to the fact that it is at this point that the center of gravity of a homogeneous triangular plate is located.
1.4 Point of intersection of the heights of the triangle (orthocenter)
1.5 Point Torricelli
The path given is triangle ABC. The Torricelli point of this triangle is such a point O, from which the sides of this triangle are visible at an angle of 120°, i.e. angles AOB, AOC and BOC are 120°.
Let us prove that if all the angles of the triangle are less than 120°, then the Torricelli point exists.
On the side AB of the triangle ABC, we construct an equilateral triangle ABC "(Fig. 6, a), and describe a circle around it. The segment AB subtends the arc of this circle with a value of 120 °. Therefore, the points of this arc, other than A and B, have the property that the segment AB is visible from them at an angle of 120 °. Similarly, on the AC side of the triangle ABC, we construct an equilateral triangle ACB "(Fig. 6, a), and describe a circle around it. Points of the corresponding arc, other than A and C, have the property that the segment AC is visible from them at an angle of 120°. In the case when the angles of the triangle are less than 120°, these arcs intersect at some interior point O. In this case, ∟AOB = 120°, ∟AOC = 120°. Therefore, ∟BOC = 120°. Therefore, the point O is the desired one.
In the case when one of the angles of the triangle, for example ABC, is equal to 120°, the point of intersection of the arcs of the circles will be point B (Fig. 6, b). In this case, the Torricelli point does not exist, since it is impossible to talk about the angles at which sides AB and BC are visible from this point.
In the case when one of the angles of the triangle, for example, ABC, is greater than 120° (Fig. 6, c), the corresponding arcs of the circles do not intersect, and the Torricelli point also does not exist.
Related to the Torricelli point is Fermat's problem (which we will consider in Chapter II) of finding the point from which the sum of the distances from which to three given points is the smallest.
1.6 Circle of nine points
Indeed, A 3 B 2 is the midline of triangle AHC and, consequently, A 3 B 2 || CC1. B 2 A 2 is the middle line of triangle ABC and, therefore, B 2 A 2 || AB. Since CC 1 ┴ AB, then A 3 B 2 A 2 = 90°. Similarly, A 3 C 2 A 2 = 90°. Therefore points A 2 , B 2 , C 2 , A 3 lie on the same circle with diameter A 2 A 3 . Since AA 1 ┴BC, the point A 1 also belongs to this circle. Thus, the points A 1 and A 3 lie on the circumcircle of the triangle A2B2C2. Similarly, it is shown that the points B 1 and B 3 , C 1 and C 3 lie on this circle. So all nine points lie on the same circle.
In this case, the center of the circle of nine points lies in the middle between the center of the intersection of the heights and the center of the circumscribed circle. Indeed, let in triangle ABC (Fig. 9), point O be the center of the circumscribed circle; G is the point of intersection of the medians. H point of intersection of heights. It is required to prove that the points O, G, H lie on the same straight line and the center of the circle of nine points N divides the segment OH in half.
Consider a homothety centered at G and with coefficient -0.5. Vertices A, B, C of triangle ABC will go to points A 2 , B 2 , C 2 respectively. The heights of the triangle ABC will go to the heights of the triangle A 2 B 2 C 2 and, consequently, the point H will go to the point O. Therefore, the points O, G, H will lie on one straight line.
Let us show that the midpoint N of the segment OH is the center of the circle of nine points. Indeed, C 1 C 2 is the nine-point chord of the circle. Therefore, the perpendicular bisector to this chord is the diameter and intersects OH at the midpoint of N. Similarly, the perpendicular bisector to the chord B 1 B 2 is the diameter and intersects OH at the same point N. Hence, N is the center of the circle of nine points. Q.E.D.
Indeed, let P be an arbitrary point lying on the circumcircle of triangle ABC; D, E, F are the bases of the perpendiculars dropped from the point P to the sides of the triangle (Fig. 10). Let us show that the points D, E, F lie on the same straight line.
Note that if AP passes through the center of the circle, then points D and E coincide with vertices B and C. Otherwise, one of the angles ABP or ACP is acute and the other is obtuse. It follows from this that points D and E will be located along different sides from the line BC and in order to prove that the points D, E and F lie on the same line, it suffices to check that ∟CEF =∟BED.
Let us describe a circle with diameter CP. Since ∟CFP = ∟CEP = 90°, the points E and F lie on this circle. Therefore, ∟CEF =∟CPF as inscribed angles based on one circular arc. Further, ∟CPF = 90°- ∟PCF = 90°- ∟DBP = ∟BPD. Let's describe a circle with diameter BP. Since ∟BEP = ∟BDP = 90°, the points F and D lie on this circle. Therefore, ∟BPD = ∟BED. Therefore, we finally obtain that ∟CEF =∟BED. So the points D, E, F lie on the same straight line.
ChapterIIProblem solving
Let's start with problems related to the location of bisectors, medians, and heights of a triangle. Their solution, on the one hand, allows you to recall the material covered earlier, and on the other hand, develops the necessary geometric representations, prepares more challenging tasks.
Task 1. At angles A and B of triangle ABC (∟A
Solution. Let CD be the height, CE the bisector, then
∟BCD = 90° - ∟B, ∟BCE = (180° - ∟A - ∟B):2.
Therefore, ∟DCE =.
Solution. Let O be the intersection point of the bisectors of triangle ABC (Fig. 1). Let's take advantage of the fact that a larger angle lies opposite the larger side of the triangle. If AB BC, then ∟A
Solution. Let O be the intersection point of the altitudes of triangle ABC (Fig. 2). If AC ∟B. A circle with diameter BC will pass through points F and G. Considering that the smaller of the two chords is the one on which the smaller inscribed angle rests, we get that CG
Proof. On sides AC and BC of triangle ABC, as on diameters, we construct circles. Points A 1 , B 1 , C 1 belong to these circles. Therefore, ∟B 1 C 1 C = ∟B 1 BC, as angles based on the same circular arc. ∟B 1 BC = ∟CAA 1 as angles with mutually perpendicular sides. ∟CAA 1 = ∟CC 1 A 1 as angles based on the same circular arc. Therefore, ∟B 1 C 1 C = ∟CC 1 A 1 , i.e. CC 1 is the bisector of angle B 1 C 1 A 1 . Similarly, it is shown that AA 1 and BB 1 are bisectors of angles B 1 A 1 C 1 and A 1 B 1 C 1 .
The considered triangle, whose vertices are the bases of the heights of a given acute-angled triangle, gives an answer to one of the classical extremal problems.
Solution. Let ABC be a given acute triangle. On its sides it is required to find such points A 1 , B 1 , C 1 for which the perimeter of the triangle A 1 B 1 C 1 would be the smallest (Fig. 4).
Let us first fix the point C 1 and look for the points A 1 and B 1 for which the perimeter of the triangle A 1 B 1 C 1 is the smallest (for the given position of the point C 1).
To do this, consider the points D and E symmetrical to the point C 1 with respect to the lines AC and BC. Then B 1 C 1 \u003d B 1 D, A 1 C 1 \u003d A 1 E and, therefore, the perimeter of the triangle A 1 B 1 C 1 will be equal to the length of the polyline DB 1 A 1 E. It is clear that the length of this polyline is the smallest if the points B 1 , A 1 lie on the line DE.
We will now change the position of the point C 1 and look for such a position at which the perimeter of the corresponding triangle A 1 B 1 C 1 is the smallest.
Since point D is symmetrical to C 1 with respect to AC, then CD = CC 1 and ACD=ACC 1 . Similarly, CE=CC 1 and BCE=BCC 1 . Therefore, triangle CDE is isosceles. Its side is equal to CC 1 . The base DE is equal to the perimeter P triangle A 1 B 1 C 1 . The angle DCE is equal to twice the angle ACB of the triangle ABC and, therefore, does not depend on the position of the point C 1 .
In an isosceles triangle with a given angle at the apex, the smaller the base, the smaller the side. Therefore, the smallest value of the perimeter P is achieved in the case of the smallest value of CC 1 . This value is taken if CC 1 is the height of triangle ABC. Thus, the required point C 1 on the side AB is the base of the height drawn from the top C.
Note that we could first fix not the point C 1 , but the point A 1 or the point B 1 and we would get that A 1 and B 1 are the bases of the corresponding altitudes of the triangle ABC.
From this it follows that the desired triangle, the smallest perimeter, inscribed in a given acute-angled triangle ABC is a triangle whose vertices are the bases of the altitudes of triangle ABC.
Solution. Let us prove that if the angles of the triangle are less than 120°, then the desired point in the Steiner problem is the Torricelli point.
Let's rotate the triangle ABC around the vertex C by an angle of 60°, fig. 7. Get the triangle A'B'C. Take an arbitrary point O in triangle ABC. When turning, it will go to some point O’. Triangle OO'C is equilateral since CO = CO' and ∟OCO' = 60°, hence OC = OO'. Therefore, the sum of the lengths of OA + OB + OC will be equal to the length of the polyline AO + OO’ + O’B’. It is clear that the length of this polyline takes on the smallest value if the points A, O, O', B' lie on the same straight line. If O is a Torricelli point, then it is. Indeed, ∟AOC = 120°, ∟COO" = 60°. Therefore, the points A, O, O' lie on the same straight line. Similarly, ∟CO'O = 60°, ∟CO"B" = 120°. Therefore, the points O, O', B' lie on the same line, which means that all the points A, O, O', B' lie on the same line.
Conclusion
The geometry of a triangle, along with other sections of elementary mathematics, makes it possible to feel the beauty of mathematics in general and can become for someone the beginning of the path to "big science".
Geometry - amazing science. Her history spans more than one millennium, but each meeting with her is able to endow and enrich (both the student and the teacher) with the exciting novelty of a small discovery, the amazing joy of creativity. Indeed, any problem of elementary geometry is, in essence, a theorem, and its solution is a modest (and sometimes huge) mathematical victory.
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 then become interesting and meaningful, only then can it become proper geometry, 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.
In this paper, the properties of bisectors, medians, perpendicular bisectors and altitudes of a triangle were considered, the number of remarkable points and lines of a triangle was expanded, theorems were formulated and proved. A number of problems on the application of these theorems have been solved.
The presented material can be used both in basic lessons and in optional classes, as well as in preparation for centralized testing and mathematics olympiads.
Bibliography
Berger M. Geometry in two volumes - M: Mir, 1984.
Kiselev A.P. Elementary geometry. – M.: Enlightenment, 1980.
Kokseter G.S., Greitzer S.L. New encounters with geometry. – M.: Nauka, 1978.
Latotin L.A., Chebotaravskiy B.D. Mathematics 9. - Minsk: Narodnaya Asveta, 2014.
Prasolov V.V. Problems in planimetry. - M.: Nauka, 1986. - Part 1.
Scanavi M. I. Mathematics. Problems with solutions. - Rostov-on-Don: Phoenix, 1998.
Sharygin I.F. Problems in geometry: Planimetry. – M.: Nauka, 1986.
