Ministry of General and vocational education Sverdlovsk region.
Municipal Educational Institution of Yekaterinburg.
Educational institution – MOUSOSH No. 212 “Ekaterinburg Cultural Lyceum”
Educational field – mathematics.
Subject - geometry.
Remarkable points of the triangle
Referent: 8th grade student
Selitsky Dmitry Konstantinovich.
Scientific adviser:
Rabkanov Sergey Petrovich.
Ekaterinburg, 2001
Introduction 3
Descriptive part:
Orthocenter 4
Icenter 5
Center of gravity 7
Circumcenter 8
Euler line 9
Practical part:
Orthocentric triangle 10
Conclusion 11
References 11
Introduction.
Geometry starts with a triangle. For two and a half millennia, the triangle has been a symbol of geometry. Its new properties are constantly being discovered. To talk about all the known properties of a triangle will take a lot of time. I was interested in the so-called " Wonderful points triangle." An example of such points is the intersection point of bisectors. The remarkable thing is that if you take three arbitrary points in space, construct a triangle from them and draw bisectors, then they (the bisectors) will intersect at one point! It would seem that this is not possible, because we took arbitrary points, but this rule always applies. Other “remarkable points” have similar properties.
After reading the literature on this topic, I fixed for myself the definitions and properties of five wonderful points and a triangle. But my work didn’t end there; I wanted to explore these points myself.
That's why target This work is a study of some remarkable properties of a triangle, and a study of an orthocentric triangle. In the process of achieving this goal, the following stages can be distinguished:
Selection of literature, with the help of a teacher
Studying the basic properties of the remarkable points and lines of a triangle
Generalization of these properties
Drawing up and solving a problem involving an orthocentric triangle
I presented the results obtained in this research work. I made all the drawings using computer graphics (vector graphics editor CorelDRAW).
Orthocenter. (Point of intersection of heights)
Let us prove that the heights intersect at one point. Let's take you through the peaks A, IN And WITH triangle ABC straight lines parallel to opposite sides. These lines form a triangle A 1 IN 1 WITH 1 . height of the triangle ABC are the perpendicular bisectors to the sides of the triangle A 1 IN 1 WITH 1 . therefore, they intersect at one point - the center of the circumcircle of the triangle A 1 IN 1 WITH 1 . The point of intersection of the altitudes of a triangle is called the orthocenter ( H).
Icentre is the center of the inscribed circle.
(Point of intersection of bisectors)
Let us prove that the bisectors of the angles of a triangle ABC intersect at one point. Consider the point ABOUT angle bisector intersections A And IN. any points of the bisector of angle A are equidistant from the lines AB And AC, and any point of the angle bisector IN equidistant from straight lines AB And Sun, so point ABOUT equidistant from straight lines AC And Sun, i.e. it lies on the bisector of the angle WITH. dot ABOUT equidistant from straight lines AB, Sun And SA, which means there is a circle with center ABOUT, tangent to these lines, and the points of tangency lie on the sides themselves, and not on their extensions. In fact, the angles at the vertices A And IN triangle AOB sharp therefore projection point ABOUT directly AB lies inside the segment AB.
For parties Sun And SA the proof is similar.
The icenter has three properties:
If the continuation of the angle bisector WITH intersects the circumcircle of a triangle ABC at the point M, That MA=MV=MO.
If AB- base of an isosceles triangle ABC, then the circle tangent to the sides of the angle DIA at points A And IN, passes through the point ABOUT.
If a line passing through a point ABOUT parallel to the side AB, crosses the sides Sun And SA at points A 1 And IN 1 , That A 1 IN 1 =A 1 IN+AB 1 .
Center of gravity. (Point of intersection of medians)
Let us prove that the medians of a triangle intersect at one point. For this, consider the point M, at which the medians intersect AA 1 And BB 1 . let's draw in a triangle BB 1 WITH midline A 1 A 2 , parallel BB 1 . Then A 1 M:AM=IN 1 A 2 :AB 1 =IN 1 A 2 :IN 1 WITH=VA 1 :SUN=1:2, i.e. median intersection point BB 1 And AA 1 divides the median AA 1 in a ratio of 1:2. Similarly, the intersection point of the medians SS 1 And AA 1 divides the median AA 1 in a ratio of 1:2. Therefore, the intersection point of the medians AA 1 And BB 1 coincides with the intersection point of the medians AA 1 And SS 1 .
If the intersection point of the medians of a triangle is connected to the vertices, then the triangles will be divided into three triangles of equal area. Indeed, it is enough to prove that if R– any point of the median AA 1 in a triangle ABC, then the areas of the triangles AVR And ASR are equal. After all, medians AA 1 And RA 1 in triangles ABC And RVS cut them into triangles of equal area.
The converse statement is also true: if for some point R, lying inside the triangle ABC, area of triangles AVR, ON WEDNESDAY And SAR are equal, then R– point of intersection of medians.
The intersection point has one more property: if you cut a triangle out of any material, draw medians on it, attach a rod at the intersection point of the medians and secure the suspension on a tripod, then the model (triangle) will be in a state of equilibrium, therefore, the intersection point is nothing more than the center of gravity of the triangle.
Center of the circumscribed circle.
Let us prove that there is a point equidistant from the vertices of the triangle, or, in other words, that there is a circle passing through the three vertices of the triangle. The locus of points equidistant from points A And IN, is perpendicular to the segment AB, passing through its middle (the perpendicular bisector to the segment AB). Consider the point ABOUT, in which the bisectors of the perpendiculars to the segments intersect AB And Sun. Dot ABOUT equidistant from points A And IN, as well as from points IN And WITH. therefore it is equidistant from the points A And WITH, i.e. it also lies on the perpendicular bisector to the segment AC.
Center ABOUT the circumcircle lies inside a triangle only if the triangle is acute. If the triangle is right-angled, then the point ABOUT coincides with the middle of the hypotenuse, and if the angle at the vertex WITH blunt then straight AB separates the points ABOUT And WITH.
In mathematics, it often happens that objects defined in completely different ways turn out to be the same. Let's show this with an example.
Let A 1 , IN 1 ,WITH 1 – midpoints of the sides Sun,SA and AB. It can be proven that circumscribed circles of triangles AB 1 WITH, A 1 Sun 1 And A 1 IN 1 WITH 1 intersect at one point, and this point is the circumcenter of the triangle ABC. So, we have two seemingly completely different points: the point of intersection of the perpendicular bisectors to the sides of the triangle ABC and the intersection point of the circumcircles of the triangles AB 1 WITH 1 , A 1 Sun And A 1 IN 1 WITH 1 . but it turns out that these two points coincide.
Euler's straight line.
The most amazing property The remarkable points of the triangle is that some of them are connected to each other by certain relationships. For example, center of gravity M, orthocenter N and the center of the circumcircle ABOUT lie on the same straight line, and the point M divides the segment OH so that the relation is valid OM:MN=1:2. This theorem was proven in 1765 by the Swiss scientist Leonardo Euler.
Orthocentric triangle.
Orthocentric triangle(orthotriangle) is a triangle ( MNTO), the vertices of which are the bases of the altitudes of this triangle ( ABC). This triangle has many interesting properties. Let's give one of them.
Property.
Prove:
Triangles AKM, CMN And BKN similar to a triangle ABC;
Angles of an orthotriangle MNK are: L KNM = π - 2 L A,LKMN = π – 2 L B, L MNK = π - - 2 L C.
Proof:
We have AB cos A, A.K. cos A. Hence, A.M./AB = A.K./A.C..
Because at triangles ABC And AKM corner A– common, then they are similar, from which we conclude that the angle L AKM = L C. That's why L BKM = L C. Next we have L MKC= π/2 – L C, L NKC= π/2 – - - L C, i.e. SK– angle bisector MNK. So, L MNK= π – 2 L C. The remaining equalities are proved similarly.
Conclusion.
At the end of this research work, the following conclusions can be drawn:
The notable points and lines of the triangle are:
orthocenter of a triangle is the point of intersection of its heights;
andcentre triangle is the point of intersection of the bisectors;
center of gravity of a triangle is the point of intersection of its medians;
circumcenter– is the point of intersection of the bisector perpendiculars;
Euler's straight line- this is the straight line on which the center of gravity, the orthocenter and the center of the circumscribed circle lie.
An orthocentric triangle divides a given triangle into three similar ones.
After doing this work, I learned a lot about the properties of a triangle. This work was relevant for me from the point of view of developing my knowledge in the field of mathematics. In the future, I intend to develop this interesting topic.
Bibliography.
Kiselyov A.P. Elementary geometry. – M.: Education, 1980.
Coxeter G.S., Greitzer S.L. New encounters with geometry. – M.: Nauka, 1978.
Prasolov V.V. Problems in planimetry. – M.: Nauka, 1986. – Part 1.
Sharygin I.F. Geometry problems: Planimetry. – M.: Nauka, 1986.
Scanavi M.I. Mathematics. Problems with solutions. – Rostov-on-Don: Phoenix, 1998.
Berger M. Geometry in two volumes - M: Mir, 1984.
Introduction
Objects of the world around us have certain properties, the study of which is carried out by various sciences.
Geometry is a branch of mathematics that examines various figures and their properties; its roots go back to the distant past.
In the fourth book of Elements, Euclid solves the problem: “To inscribe a circle in a given triangle.” It follows from the solution that the three bisectors of the interior angles of the triangle intersect at one point - the center of the inscribed circle. From the solution of another Euclidean problem 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 Elements does not say that the three altitudes of the triangle intersect at one point, called the orthocenter (the Greek word “orthos” means “straight”, “correct”). This proposal, however, was 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 addressed 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 circumcenter lie on the same straight line, later called “Euler’s straight line.” In the twenties of the 19th century, French mathematicians J. Poncelet, C. Brianchon and others independently established the following theorem: the bases of medians, the bases of altitudes and the midpoints of segments of altitudes connecting the orthocenter with the vertices of a triangle lie on the same circle. This circle is called the “circle of nine points”, or “Feuerbach circle”, or “Euler circle”. K. Feuerbach established that the center of this circle lies on the Euler straight line.
“I think that never before have we lived in such a geometric period. 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, creations of nature and man.
We were interested in the so-called “remarkable points of the triangle”.
After reading the literature on this topic, we fixed for ourselves the definitions and properties of the remarkable points of a triangle. But our work did not end there, and we wanted to explore these points ourselves.
That's why target given work – studying some remarkable points and lines of a triangle, applying the acquired knowledge to solving problems. In the process of achieving this goal, the following stages can be distinguished:
Selection and study educational material from various sources of information, literature;
Studying the basic properties of remarkable points and lines of a triangle;
Generalization of these properties and proof of the necessary theorems;
Solving problems involving remarkable points of a triangle.
ChapterI. Remarkable triangle points and lines
1.1 The point of intersection of the perpendicular bisectors to the sides of the triangle
A perpendicular bisector is a line passing through the middle 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 a segment is equidistant from its ends and vice versa; if a 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 about the polygon.
A circle can be described around any triangle. Its center is the point of intersection of the perpendicular bisectors to the sides of the triangle.
Let point O be the intersection point of the perpendicular bisectors to the sides of the triangle AB and BC.
Conclusion: thus, if point O is the point of intersection of the perpendicular bisectors to the sides of the triangle, then OA = OC = OB, i.e. point O is equidistant from all vertices of triangle ABC, which means it is the center of the circumscribed circle.
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Consequences
sin γ = c/2R = c/sin γ =2R.
It is proved in a similar way A/ sin α =2R, b/ sin β =2R.
Thus:
This property is called the theorem of sines.
In mathematics it often happens that objects that are completely defined differently, turn out to be identical.
Example. Let A1, B1, C1 be the midpoints of the sides ∆ABC BC, AC, AB, respectively. Show that the circles described around the triangles AB1C1, A1B1C, A1BC1 intersect at one point. Moreover, this point is the center of a circle circumscribed about ∆ABC.
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| Let's 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 the vertices of right angles based on AO. Points A, C1, B1 lie on a circle = this circle is circumscribed about ∆AB1C1. Let us similarly draw the segment BO and construct a circle on this segment, as on a diameter. This will be a circle circumscribed about ∆ВС1 А1. Let's draw a segment CO and construct a circle on this segment, as on a diameter. This will be a circle circumscribed by about These three circles pass through point O - the center of the circle circumscribed about ∆ABC. |
Generalization. If on the sides ∆ABC AC, BC, AC we take arbitrary points A 1, B 1, C 1, 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 Intersection point of triangle bisectors
The converse 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 point O be the intersection point of the bisectors of angles A and B. By the property of the point lying on the bisector of angle A, OF=OD=r. According to the property of the point lying on the bisector of angle B, OE=OD=r. Thus, OE=OD= OF=r= point O is equidistant from all sides of triangle ABC, i.e. O is the center of the inscribed circle. (Point O is the only one).
Conclusion: thus, if point O is the point of intersection of the bisectors of the angles of a triangle, then OE=OD= OF=r, i.e. point O is equidistant from all sides of triangle ABC, which means it is the center of the inscribed circle. The O-point of intersection of the bisectors of the angles of a triangle is a remarkable point of the triangle.
Consequences:
From the equality of triangles AOF and AOD (Figure 1) along the hypotenuse and acute angle, it follows that A.F. = AD . From the equality of triangles OBD and OBE it follows that BD = BE , From the equality of triangles COE and COF it follows that WITH F = C.E. . Thus, the tangent segments drawn to the circle from one point are equal.
AF=AD= z, BD=BE= y, CF=CE= x
a=x+y (1), b= x+z (2), c= x+y (3).
+ (2) – (3), then we get: a+b-с=x+ y+ x+ z- z- y = a+b-с= 2x =
x=( b + c - a)/2
Similarly: (1) + (3) – (2), then we get: y = (a + c –b)/2.
Similarly: (2) + (3) – (1), then 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 triangle medians (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 the segments AG and BG. By the theorem on the midline of a triangle, the segments B 1 A 1 and PQ are equal to half of side AB and parallel to it. Therefore, the quadrilateral A 1 B 1 is a PQ parallelogram. Then the point G of the intersection of its diagonals PA 1 and QB 1 divides each of them in half. Therefore, points P and G divide the median AA 1 into three equal parts, and points Q and G also divide the median BB 1 into three equal parts. So, the point G of the intersection of two medians of a triangle divides each of them in the ratio 2:1, counting from the vertex.
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 triangle altitudes (orthocenter)
1.5 Torricelli point
The path is given by triangle ABC. The Torricelli point of this triangle is a point O from which the sides given triangle visible at an angle of 120°, i.e. angles AOB, AOC and BOC are equal to 120°.
Let us prove that if all angles of a triangle are less than 120°, then the Torricelli point exists.
On side AB of triangle ABC we construct an equilateral triangle ABC" (Fig. 6, a), and describe a circle around it. The segment AB subtends an arc of this circle measuring 120°. Consequently, 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 side AC of triangle ABC we will construct an equilateral triangle ACB" (Fig. 6, a), and describe a circle around it. Points of the corresponding arc, different from 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 internal point O. In this case, ∟AOB = 120°, ∟AOC = 120°. Therefore, ∟BOC = 120°. Therefore, point O is the desired one.
In the case when one of the angles of a triangle, for example ABC, is equal to 120°, the point of intersection of the circular arcs will be point B (Fig. 6, b). In this case, Torricelli's 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 a triangle, for example ABC, is greater than 120° (Fig. 6, c), the corresponding arcs of circles do not intersect, and Torricelli’s point also does not exist.
The Torricelli point is associated with Fermat’s problem (which we will consider in Chapter II) of finding the point whose sum of distances to three given points is the smallest.
1.6 Nine-point circle
Indeed, A 3 B 2 is the midline of triangle AHC and, therefore, A 3 B 2 || CC 1. 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°. Likewise, 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, then point A 1 also belongs to this circle. Thus, points A 1 and A 3 lie on the circumcircle of triangle A2B2C2. Similarly, it is shown that points B 1 and B 3, C 1 and C 3 lie on this circle. This means that 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 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 – point of intersection of medians. H is the point where the heights intersect. You need to prove that points O, G, H lie on the same line and the center of the circle of nine points N divides the segment OH in half.
Consider a homothety with center at point G and coefficient -0.5. Vertices A, B, C of triangle ABC will go, respectively, to points A 2, B 2, C 2. The altitudes of triangle ABC will go into the altitudes of triangle A 2 B 2 C 2 and, therefore, point H will go to point O. Therefore, points O, G, H will lie on the same 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 a chord of the circle of nine points. Therefore, the perpendicular bisector of this chord is a diameter and intersects OH at the middle of N. Similarly, the perpendicular bisector of the chord B 1 B 2 is a diameter and intersects OH at the same point N. So 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 – the bases of the perpendiculars dropped from point P to the sides of the triangle (Fig. 10). Let us show that points D, E, F lie on the same 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 line BC and in order to prove that points D, E and F lie on the same line, it is enough to check that ∟CEF =∟BED.
Let us describe a circle with diameter CP. Since ∟CFP = ∟CEP = 90°, then points E and F lie on this circle. Therefore, ∟CEF =∟CPF as inscribed angles subtended by one arc of a circle. Next, ∟CPF = 90°- ∟PCF = 90°- ∟DBP = ∟BPD. Let us describe a circle with diameter BP. Since ∟BEP = ∟BDP = 90°, then points F and D lie on this circle. Therefore ∟BPD =∟BED. Therefore, we finally get that ∟CEF =∟BED. This means points D, E, F lie on the same line.
ChapterIIProblem solving
Let's start with problems related to the location of bisectors, medians and altitudes of a triangle. Solving them, on the one hand, allows you to remember previously covered material, and on the other hand, develops the necessary geometric concepts, prepares you for solving more complex tasks.
Task 1. At angles A and B of triangle ABC (∟A
Solution. Let CD be the height and CE be 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 the larger angle lies opposite the larger side of the triangle. If AB BC, then ∟A
Solution. Let O be the point of intersection 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 obtain that CG
Proof. On the sides AC and BC of triangle ABC, as on the 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 arc of a circle. ∟B 1 BC = ∟CAA 1 as angles with mutually perpendicular sides. ∟CAA 1 = ∟CC 1 A 1 as angles subtended by the same arc of a circle. 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 the bisectors of the angles B 1 A 1 C 1 and A 1 B 1 C 1 .
The considered triangle, the vertices of which are the bases of the altitudes of a given acute triangle, provides an answer to one of the classical extremal problems.
Solution. Let ABC be the given acute triangle. On its sides, you need to find 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's first fix point C 1 and look for points A 1 and B 1 for which the perimeter of triangle A 1 B 1 C 1 is the smallest (for a given position of point C 1).
To do this, consider points D and E symmetrical to point C 1 relative to straight lines AC and BC. Then B 1 C 1 = B 1 D, A 1 C 1 = A 1 E and, therefore, the perimeter of the triangle A 1 B 1 C 1 will be equal to the length of the broken line DB 1 A 1 E. It is clear that the length of this broken line is the smallest if the points B 1, A 1 lie on line DE.
We will now change the position of point C 1 and look for 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 relative to AC, then CD = CC 1 and ACD = ACC 1. Likewise, CE=CC 1 and BCE=BCC 1. Therefore, triangle CDE is isosceles. Its lateral side is equal to CC 1. The base DE is equal to the perimeter P triangle A 1 B 1 C 1. Angle DCE is equal to double angle ACB of triangle ABC and, therefore, does not depend on the position of point C 1.
In an isosceles triangle with a given angle at the vertex, the smaller the side, the smaller the base. That's why smallest value perimeter P is achieved in the case of the lowest CC 1 value. This value is taken if CC 1 is the height of triangle ABC. Thus, the required point C 1 on side AB is the base of the altitude drawn from the vertex C.
Note that we could first fix not point C 1, but point A 1 or point B 1 and would obtain that A 1 and B 1 are the bases of the corresponding altitudes of triangle ABC.
It follows from this that the required triangle of the smallest perimeter inscribed in a given acute 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 required point in the Steiner problem is the Torricelli point.
Let's rotate triangle ABC around vertex C by an angle of 60°, Fig. 7. We get triangle A’B’C. Let's 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°, therefore OC = OO'. Therefore, the sum of the lengths OA + OB + OC will be equal to the length of the broken line AO + OO’ + O’B’. It is clear that the length of this broken line takes the smallest value if points A, O, O’, B’ lie on the same straight line. If O is a Torricelli point, then this is so. Indeed, ∟AOC = 120°, ∟COO" = 60°. Therefore, points A, O, O' lie on the same straight line. Similarly, ∟CO'O = 60°, ∟CO"B" = 120°. Therefore, points O, O', B' lie on the same line, which means that all 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 is an amazing science. Its history goes back more than one thousand years, but every meeting with it can gift and enrich (both student and teacher) with the exciting novelty of a small discovery, the amazing joy of creativity. Indeed, any problem in elementary geometry is essentially 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 become interesting and meaningful, only then can it become geometry proper when it includes a deep and comprehensive study of the triangle. Surprisingly, the triangle, despite its apparent simplicity, is an inexhaustible object of study - no one, even in our time, dares to say that they have studied and know all the properties of the triangle.
In this work, the properties of bisectors, medians, perpendicular bisectors and altitudes of a triangle were considered, the number of remarkable points and lines of the triangle was expanded, and theorems were formulated and proven. 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 elective classes, as well as in preparation for centralized testing and mathematics Olympiads.
Bibliography
Berger M. Geometry in two volumes - M: Mir, 1984.
Kiselyov A.P. Elementary geometry. – M.: Education, 1980.
Coxeter G.S., Greitzer S.L. New encounters with geometry. – M.: Nauka, 1978.
Latotin L.A., Chebotaravsky 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. Geometry problems: Planimetry. – M.: Nauka, 1986.
Ministry of Education and Science Russian Federation Federal state budget educational institution higher professional education
"Magnitogorsk State University»
Faculty of Physics and Mathematics
Department of Algebra and Geometry
Course work
Remarkable points of the triangle
Completed by: student of group 41
Vakhrameeva A.M.
Scientific director
Velikikh A.S.
Magnitogorsk 2014
Introduction
Historically, geometry began with a triangle, so for two and a half millennia the triangle has been, as it were, a symbol of geometry; but he is not only a symbol, he is an atom of geometry.
Why can a triangle be considered an atom of geometry? Because the previous concepts - point, straight line and angle - are vague and intangible abstractions along with an associated set of theorems and problems. Therefore, today school geometry can only become interesting and meaningful, only then can it become geometry proper when it includes a deep and comprehensive study of the triangle.
Surprisingly, the triangle, despite its apparent simplicity, is an inexhaustible object of study - no one, even in our time, dares to say that they have studied and know all the properties of the triangle.
This means that the study of school geometry cannot be carried out without a deep study of the geometry of the triangle; in view of the diversity of the triangle as an object of study - and, therefore, the source of various methods for studying it - it is necessary to select and develop material for studying the geometry of the remarkable points of the triangle. Moreover, when selecting this material, you should not limit yourself only to the remarkable points provided for in school curriculum State educational standard, such as the center of the incircle (the point of intersection of bisectors), the center of the circumcircle (the point of intersection of bisectors), the point of intersection of medians, the point of intersection of heights. But for deep penetration into the nature of the triangle and comprehension of its inexhaustibility, it is necessary to have as many ideas as possible about more remarkable points of the triangle. In addition to the inexhaustibility of the triangle as a geometric object, it is necessary to note the most amazing property of the triangle as an object of study: the study of the geometry of a triangle can begin with the study of any of its properties, taking it as a basis; then the methodology for studying the triangle can be constructed in such a way that all other properties of the triangle can be strung on this basis. In other words, no matter where you start studying the triangle, you can always reach any depth of this amazing figure. But then - as an option - you can start studying the triangle by studying its remarkable points.
Target course work consists of studying the remarkable points of a triangle. To achieve this goal, it is necessary to solve the following tasks:
· Study the concepts of bisector, median, height, perpendicular bisector and their properties.
· Consider the Gergonne point, Euler circle and Euler line, which are not studied at school.
CHAPTER 1. Bisector of a triangle, center of the inscribed circle of a triangle. Properties of the bisector of a triangle. Gergonna point
1 Center of the inscribed circle of a triangle
Remarkable points of a triangle are points whose location is uniquely determined by the triangle and does not depend on the order in which the sides and vertices of the triangle are taken.
The bisector of a triangle is the bisector segment of an angle of a triangle connecting a vertex to a point on the opposite side.
Theorem. Each point of the bisector of an undeveloped angle is equidistant (that is, equidistant from the lines containing the sides of the triangle) from its sides. Conversely: every point lying inside an angle and equidistant from the sides of the angle lies on its bisector.
Proof. 1) Take an arbitrary point M on the bisector of the angle BAC, draw perpendiculars MK and ML to straight lines AB and AC and prove that MK = ML. Consider right triangles ?AMK and ?AML. They are equal in hypotenuse and acute angle (AM - common hypotenuse, 1 = 2 by convention). Therefore, MK=ML.
) Let point M lie inside YOU and be equidistant from its sides AB and AC. Let us prove that the ray AM is the bisector BAC. Let us draw perpendiculars MK and ML to straight lines AB and AC. Right triangles AKM and ALM are equal in hypotenuse and leg (AM is the common hypotenuse, MK = ML by convention). Therefore, 1 = 2. But this means that the ray AM is the bisector of BAC. The theorem is proven.
Consequence. The bisectors of a triangle intersect at one point (the center of the incircle and the center).
Let us denote by the letter O the point of intersection of the bisectors AA1 and BB1 of the triangle ABC and draw from this point the perpendiculars OK, OL and OM, respectively, to the straight lines AB, BC and CA. According to the theorem (Each point of the bisector of an undeveloped angle is equidistant from its sides. Conversely: every point lying inside the angle and equidistant from the sides of the angle lies on its bisector) we say that OK = OM and OK = OL. Therefore, OM = OL, that is, point O is equidistant from the sides ACB and, therefore, lies on the bisector CC1 of this angle. Therefore, all three bisectors ?ABC intersect at point O, which is what needed to be proven.
circle bisector triangle line
1.2 Properties of the bisector of a triangle
Bisector BD (Fig. 1.1) of any angle ?ABC divides the opposite side into parts AD and CD proportional to the adjacent sides of the triangle.
We need to prove that if ABD = DBC, then AD: DC = AB: BC.
Let's carry out CE || BD to the intersection at point E with the continuation of side AB. Then, according to the theorem on the proportionality of segments formed on lines intersected by several parallel lines, we will have the proportion: AD: DC = AB: BE. In order to move from this proportion to the one that needs to be proved, it is enough to discover that BE = BC, i.e. that ?ALL isosceles. In this triangle E = ABD (as corresponding angles with parallel lines) and ALL = DBC (as crosswise angles with the same parallel lines).
But ABD = DBC by condition; this means E = ALL, and therefore the sides BE and BC lying opposite equal angles are equal.
Now, replacing BE in the proportion written above with BC, we obtain the proportion that needs to be proved.
20 The bisectors of the internal and adjacent angles of a triangle are perpendicular.
Proof. Let BD be the bisector of ABC (Fig. 1.2), and BE be the bisector of the external CBF adjacent to the specified internal angle, ?ABC. Then if we denote ABD = DBC = ?, CBE = EBF = ?, then 2 ? + 2?= 1800 and thus ?+ ?= 900. And this means that BD? BE.
30 The bisector of an exterior angle of a triangle divides the opposite side externally into parts proportional to the adjacent sides.
(Fig.1.3) AB: BC = AD: DC, ?AED~ ?CBD, AE/BC = AD/DC = AE/BC.
40 The bisector of any angle of a triangle divides the opposite side into segments proportional to the adjacent sides of the triangle.
Proof. Let's consider ?ABC. For definiteness, let the bisector CAB intersect side BC at point D (Fig. 1.4). Let us show that BD: DC = AB: AC. To do this, draw a line parallel to line AB through point C, and denote by E the intersection point of this line AD. Then DAB=DEC, ABD=ECD and therefore ?DAB ~ ?DEC based on the first criterion of similarity of triangles. Further, since the ray AD is a bisector CAD, then CAE = EAB = AEC and, therefore, ?ECA isosceles. Hence AC=CE. But in this case, from the similarity ?DAB and ?DEC follows that BD: DC=AB: CE =AB: AC, and this was what needed to be proven.
If the bisector of an external angle of a triangle intersects the extension of the side opposite the vertex of this angle, then the segments from the resulting intersection point to the ends of the opposite side are proportional to the adjacent sides of the triangle.
Proof. Let's consider ?ABC. Let F be a point on the extension of side CA, D be the point of intersection of the bisector of the outer triangle BAF with the extension of side CB (Fig. 1.5). Let us show that DC:DB=AC:AB. Indeed, let us draw a line parallel to line AB through point C, and denote by E the point of intersection of this line with line DA. Then triangle ADB ~ ?EDC and therefore DC:DB=EC:AB. And since ?EAC= ?BAD= ?CEA, then in isosceles ?CEA side AC=EC and, thus, DC:DB=AC:AB, which is what needed to be proven.
3 Solving problems using the properties of the bisector
Problem 1. Let O be the center of a circle inscribed in ?ABC, CAB = ?. Prove that COB = 900 + ? /2.
Solution. Since O is the center of the inscribed ?ABC of a circle (Figure 1.6), then rays BO and CO are bisectors ABC and BCA, respectively. And then COB = 1800 - (OBC + BCO) = 1800 - (ABC + BCA)/2 = 1800 -(1800 - ?)/2 = 900 + ?/2, which is what needed to be proven.
Problem 2. Let O be the center of the described about ?ABC of a circle, H is the base of the altitude drawn to side BC. Prove that the bisector CAB is also the bisector ? OAH.
Let AD be the bisector of CAB, AE be the diameter of the circumscribed ?ABC of a circle (Fig. 1.7, 1.8). If ?ABC is acute (Fig. 1.7) and, therefore, ABC<900, то так как ABC = AEC= ½ AC arcs, and ?BHA and ?ECA rectangular (BHA =ECA = 900), then ?BHA~ ?ECA and therefore CAO = CAE =HAB. Further, BAD and CAD are equal by condition, so HAD = BAD - BAH =CAD - CAE = EAD = OAD. Let now ABC = 900. In this case, the height AH coincides with the side AB, then point O will belong to the hypotenuse AC and therefore the validity of the problem statement is obvious.
Let's consider the case when ABC > 900 (Fig. 1.8). Here the quadrilateral ABCE is inscribed in a circle and hence AEC = 1800 - ABC. On the other hand, ABH = 1800 - ABC, i.e. AEC = ABH. And since ?BHA and ?ECA are rectangular and, therefore, HAB = 900 - ABH = 900 - AEC = EAC, then HAD = HAB +BAD = EAC + CAD = EAD = OAD. Cases where BAC and ACB are obtuse are treated similarly. ?
4 Point Gergonna
The Gergonne point is the point of intersection of the segments that connect the vertices of the triangle with the points of tangency of the sides opposite these vertices and the inscribed circle of the triangle.
Let point O be the center of the incircle of triangle ABC. Let the incircle touch the sides of the triangle BC, AC and AB at points D,E and F respectively. The Gergonne point is the intersection point of segments AD, BE and CF. Let point O be the center of the inscribed circle ?ABC. Let the incircle touch the sides of the triangle BC, AC and AB at points D, E and F respectively. The Gergonne point is the intersection point of segments AD, BE and CF.
Let us prove that these three segments actually intersect at one point. Note that the center of the incircle is the point of intersection of the angle bisectors ?ABC, and the radii of the incircle are OD, OE and OF ?sides of the triangle. Thus, we have three pairs of equal triangles (AFO and AEO, BFO and BDO, CDO and CEO).
Works AF?BD ? CE and AE ? BE? CF are equal, since BF = BD, CD = CE, AE = AF, therefore, the ratio of these products is equal, and by Ceva’s theorem (Let the points A1, B1, C1 lie on the sides BC, AC and AB ?ABC, respectively. Let the segments AA1 , BB1 and CC1 intersect at one point. Then
(we go around the triangle clockwise)), the segments intersect at one point.
Properties of the inscribed circle:
A circle is said to be inscribed in a triangle if it touches all its sides.
A circle can be inscribed in any triangle.
Given: ABC - this triangle, O - the point of intersection of the bisectors, M, L and K - the points of tangency of the circle with the sides of the triangle (Fig. 1.11).
Prove: O is the center of a circle inscribed in ABC.
Proof. Let us draw perpendiculars OK, OL and OM from point O to the sides AB, BC and CA, respectively (Fig. 1.11). Since point O is equidistant from the sides of triangle ABC, then OK = OL = OM. Therefore, a circle with center O of radius OK passes through points K, L, M. The sides of triangle ABC touch this circle at points K, L, M, since they are perpendicular to the radii OK, OL and OM. This means that a circle with center O of radius OK is inscribed in triangle ABC. The theorem is proven.
The center of a circle inscribed in a triangle is the intersection point of its bisectors.
Let ABC be given, O be the center of the circle inscribed in it, D, E and F be the points of contact of the circle with the sides (Fig. 1.12). ? AEO = ? AOD on the hypotenuse and leg (EO = OD - as radius, AO - total). From the equality of triangles, what follows? OAD = ? O.A.E. So AO is the bisector of angle EAD. It is proved in the same way that point O lies on the other two bisectors of the triangle.
The radius drawn to the tangent point is perpendicular to the tangent.
Proof. Let the surrounding (O; R) be a given circle (Fig. 1.13), straight line a touches it at point P. Let the radius OP not be perpendicular to a. Let us draw a perpendicular OD from point O to the tangent. By the definition of a tangent, all its points other than point P, and in particular point D, lie outside the circle. Therefore, the length of the perpendicular OD is greater than the length R of the oblique OP. This contradicts the oblique property, and the resulting contradiction proves the statement.
CHAPTER 2. 3 remarkable points of the triangle, Euler's circle, Euler's straight line.
1 Center of the circumcircle of a triangle
A perpendicular bisector to a segment is a line passing through the middle of the segment and perpendicular to it.
Theorem. Each point of the perpendicular bisector of a segment is equidistant from the ends of that segment. Conversely: every point equidistant from the ends of a segment lies on the perpendicular bisector to it.
Proof. Let straight line m be the perpendicular bisector to segment AB, and point O be the midpoint of the segment.
Let's consider an arbitrary point M on a straight line m and prove that AM=BM. If point M coincides with point O, then this equality is true, since O is the midpoint of segment AB. Let M and O - various points. Rectangular ?OAM and ?OBM are equal on two legs (OA = OB, OM is the common leg), therefore AM = VM.
) Consider an arbitrary point N, equidistant from the ends of the segment AB, and prove that point N lies on line m. If N is a point on line AB, then it coincides with the midpoint O of segment AB and therefore lies on line m. If point N does not lie on line AB, then consider ?ANB, which is isosceles, since AN=BN. The segment NO is the median of this triangle, and therefore the height. Thus, NO is perpendicular to AB, therefore lines ON and m coincide, and, therefore, N is a point of line m. The theorem is proven.
Consequence. The perpendicular bisectors to the sides of the triangle intersect at one point (the center of the circumcircle).
Let us denote O, the point of intersection of the bisectoral perpendiculars m and n to the sides AB and BC ?ABC. According to the theorem (each point of the perpendicular bisector to a segment is equidistant from the ends of this segment. Conversely: every point equidistant from the ends of the segment lies on the perpendicular bisector to it.) we conclude that OB = OA and OB = OC therefore: OA = OC, That is, 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 bisectors m, n and p to the sides ?ABC intersect at point O.
For an acute triangle this point lies inside, for an obtuse triangle it lies outside the triangle, for a right triangle it lies in the middle of the hypotenuse.
Property of the perpendicular bisector of a triangle:
The lines on which the bisectors of the internal and external angles of the triangle lie, coming out from one vertex, intersect with the perpendicular midway to the opposite side from diametrically opposite points of the circle circumscribed about the triangle.
Proof. Let, for example, the bisector ABC intersect the one described about ?ABC circle at point D (Fig. 2.1). Then since the inscribed ABD and DBC are equal, then AD = arc DC. But the perpendicular bisector to side AC also bisects arc AC, so point D will also belong to this perpendicular bisector. Further, since by property 30 from paragraph 1.3 the bisector BD ABC adjacent to ABC, the latter will intersect the circle at a point diametrically opposite to point D, since an inscribed right angle always rests on the diameter.
2 Orthocenter of the circle of a triangle
Height is a perpendicular drawn from the vertex of a triangle to a straight line containing the opposite side.
The altitudes of a triangle (or their extensions) intersect at one point (orthocenter).
Proof. Consider an arbitrary ?ABC and prove that the lines AA1, BB1, CC1 containing its heights intersect at one point. Let's go through each vertex ?ABC is a straight line parallel to the opposite side. We get ?A2B2C2. Points A, B and C are the midpoints of this triangle. Indeed, AB=A2C and AB=CB2 are like opposite sides of parallelograms ABA2C and ABCB2, therefore A2C=CB2. Similarly C2A=AB2 and C2B=BA2. In addition, as follows from the construction, CC1 is perpendicular to A2B2, AA1 is perpendicular to B2C2 and BB1 is perpendicular to A2C2. Thus, lines AA1, BB1 and CC1 are perpendicular bisectors to the sides ?A2B2C2. Therefore, they intersect at one point.
Depending on the type of triangle, the orthocenter can be inside the triangle in acute-angled ones, outside it - in obtuse-angled ones or coincide with the vertex, in rectangular ones - coincides with the vertex at right angle.
Properties of the altitude of a triangle:
A segment connecting the bases of two altitudes of an acute triangle cuts off from it a triangle similar to the given one, with a similarity coefficient equal to the cosine of the common angle.
Proof. Let AA1, BB1, CC1 be the heights of acute triangle ABC, and ABC = ?(Fig. 2.2). Right triangles BA1A and CC1B have a common ?, so they are similar, which means BA1/BA = BC1/BC = cos ?. It follows that BA1/BC1=BA/BC = cos ?, i.e. V ?C1BA1 and ?ABC sides adjacent to the common ??C1BA1~ ?ABC, with the similarity coefficient equal to cos ?. In a similar way it is proved that ?A1CB1~ ?ABC with similarity coefficient cos BCA, and ?B1AC1~ ?ABC with similarity coefficient cos CAB.
The altitude dropped to the hypotenuse of a right triangle divides it into two triangles similar to each other and similar to the original triangle.
Proof. Consider a rectangular ?ABC, which has ?BCA = 900, and CD is its height (Fig. 2.3).
Then the similarity ?ADC and ?BDC follows, for example, from the sign of similarity of right triangles by the proportionality of two legs, since AD/CD = CD/DB. Each of the right triangles ADC and BDC is similar to the original right triangle, at least based on the similarity at two angles.
Solving problems involving the use of elevation properties
Problem 1. Prove that a triangle, one of the vertices of which is the vertex of the given obtuse triangle, and the other two vertices are the bases of the altitudes of the obtuse triangle, omitted from its two other vertices, is similar to the given triangle with a similarity coefficient equal to the modulus of the cosine of the angle at the first vertex .
Solution. Consider an obtuse ?ABC with dumb CAB. Let AA1, BB1, CC1 be its heights (Fig. 2.4, 2.5, 2.6) and let CAB = ?, ABC = ? , BCA = ?.
Proof of the fact that ?C1BA1~ ?ABC (Fig. 2.4) with similarity coefficient k = cos ?, completely repeats the reasoning carried out in the proof of property 1, paragraph 2.2.
Let's prove that ?A1CB~ ?ABC (Fig. 2.5) with similarity coefficient k1= cos ?, A ?B1AC1~ ?ABC (Fig. 2.6) with similarity coefficient k2 = |cos? |.
Indeed, right triangles CA1A and CB1B have a common angle ?and therefore similar. It follows that B1C/ BC = A1C / AC= cos ?and, therefore, B1C/ A1C = BC / AC = cos ?, i.e. in triangles A1CB1 and ABC the sides forming a common ??, are proportional. And then, according to the second criterion of similarity of triangles ?A1CB~ ?ABC, with similarity coefficient k1= cos ?. As for the last case (Fig. 2.6), then from the consideration of right triangles ?BB1A and ?CC1A with equal vertical angles BAB1 and C1AC it follows that they are similar and therefore B1A / BA = C1A / CA = cos (1800 - ?) = |cos ?|, since ??- blunt. Hence B1A / C1A = BA /CA = |cos ?| and thus in triangles ?B1AC1 and ?ABC sides forming equal angles, are proportional. And this means that ?B1AC1~ ?ABC with similarity coefficient k2 = |cos? |.
Problem 2. Prove that if point O is the point of intersection of the altitudes of an acute triangle ABC, then ABC + AOC = 1800, BCA + BOA = 1800, CAB + COB = 1800.
Solution. Let us prove the validity of the first of the formulas given in the problem statement. The validity of the remaining two formulas is proven similarly. So let ABC = ?, AOC = ?. A1, B1 and C1 are the bases of the altitudes of the triangle drawn from vertices A, B and C, respectively (Fig. 2.7). Then from the right triangle BC1C it follows that BCC1 = 900 - ?and thus in right triangle OA1C angle COA1 is equal to ?. But the sum of angles AOC + COA1 = ? + ?gives a straight angle and therefore AOC + COA1 = AOC + ABC = 1800, which is what needed to be proved.
Problem 3. Prove that the altitudes of an acute triangle are the bisectors of the angles of a triangle whose vertices are the bases of the altitudes of this triangle.
is.2.8
Solution. Let AA1, BB1, CC1 be the altitudes of acute triangle ABC and let CAB = ?(Fig. 2.8). Let us prove, for example, that the height AA1 is the bisector of the angle C1A1B1. Indeed, since triangles C1BA1 and ABC are similar (property 1), then BA1C1 = ?and, therefore, C1A1A = 900 - ?. From the similarity of triangles A1CB1 and ABC it follows that AA1B1 = 900 - ?and therefore C1A1A = AA1B1= 900 - ?. But this means that AA1 is the bisector of angle C1A1B1. Similarly, it is proved that the other two altitudes of triangle ABC are the bisectors of the other two corresponding angles of triangle A1B1C1.
3 Center of gravity of the circle of a triangle
The median of a triangle is a segment connecting any vertex of the triangle to the midpoint of the opposite side.
Theorem. The median of the triangle intersect at one point (the center of gravity).
Proof. Let's consider arbitrary? ABC.
Let us denote the intersection point of the medians AA1 and BB1 with the letter O and draw the middle line A1B1 of this triangle. The segment A1B1 is parallel to side AB, therefore 1 = 2 and 3 = 4. Therefore, ?AOB and ?A1OB1 are similar at two angles, and, therefore, their sides are proportional: AO:A1O=BO:B1O=AB:A1B1. But AB=2A1B1, so AO=2A1O and BO=2B1O. Thus, point O of the intersection of medians AA1 and BB1 divides each of them in a ratio of 2:1, counting from the vertex.
It is similarly proven that the intersection point of the medians BB1 and CC1 divides each of them in the ratio 2:1, counting from the vertex, and, therefore, coincides with point O and is divided by it in the ratio 2:1, counting from the vertex.
Properties of the median of a triangle:
10 The medians of a triangle intersect at one point and are divided by the intersection point in a ratio of 2:1, counting from the vertex.
Given: ?ABC, AA1, BB1 - medians.
Prove: AO:OA1=VO:OB1=2:1
Proof. Let's draw the middle line A1B1 (Fig. 2.10), according to the property of the middle line A1B1||AB, A1B1=1/2 AB. Since A1B1 || AB, then 1 = 2 lying crosswise with parallel lines AB and A1B1 and secant AA1. 3 = 4 lying crosswise with parallel lines A1B1 and AB and secant BB1.
Hence, ?AOB ~ ?A1OB1 by the equality of two angles, which means the sides are proportional: AO/A1O = OB/OB1 = AB/A1B = 2/1, AO/A1O = 2/1; OB/OB1 = 2/1.
The median divides a triangle into two triangles of equal area.
Proof. BD - median ?ABC (Fig. 2.11), BE - its height. Then ?ABD and ?DBC are equal in size because they have equal bases AD and DC respectively and a common height BE.
The entire triangle is divided by its medians into six equal triangles.
If, on the continuation of the median of the triangle, a segment equal in length to the median is laid off from the middle of the side of the triangle, then the end point of this segment and the vertices of the triangle are the vertices of the parallelogram.
Proof. Let D be the midpoint of side BC ?ABC (Fig. 2.12), E is a point on the line AD such that DE=AD. Then, since the diagonals AE and BC of the quadrilateral ABEC at the point D of their intersection are bisected, it follows from property 13.4 that the quadrilateral ABEC is a parallelogram.
Solving problems using the properties of medians:
Problem 1. Prove that if O is the intersection point of the medians ?ABC then ?A.O.B. ?BOC and ?AOC are equal in size.
Solution. Let AA1 and BB1 be medians ?ABC(Fig. 2.13). Let's consider ?AOB and ?BOC. It is obvious that S ?AOB = S ?AB1B-S ?AB1O,S ?BOC=S ?BB1C-S ?OB1C. But by property 2 we have S ?AB1B=S ?BB1C, S ?AOB = S ?OB1C, which means that S ?AOB = S ?BOC. The equality S ?AOB = S ?AOC.
Problem 2. Prove that if point O lies inside ?ABC and ?A.O.B. ?BOC and ?AOC are equal in area, then O is the point of intersection of the medians? ABC.
Solution. Let's consider ?ABC (2.14) and assume that point O does not lie on the median BB1. Then since OB1 is the median ?AOC then S ?AOB1 = S ?B1OC , and since by condition S ?AOB = S ?BOC, then S ?AB1OB = S ?BOB1C. But this cannot be, since S ?ABB1 = S ?B1BC. The resulting contradiction means that point O lies on the median BB1. Similarly, it is proved that point O belongs to two other medians ?ABC. It follows that point O is really the point of intersection of three medians? ABC.
Problem 3. Prove that if in ?ABC sides AB and BC are not equal, then its bisector BD lies between the median BM and the height BH.
Proof. Let's describe about ?ABC is a circle and extend its bisector BD until it intersects the circle at point K. The perpendicular midpoint to segment AC will pass through point K (property 1, from paragraph 2.1), which has a common point M with the median. But since segments BH and MK are parallel, and points B and K lie on opposite sides of the line AC, then the intersection point of the segments BK and AC belong to the segment HM, and this proves the required.
Problem 4. B ?ABC median BM is half the size of side AB and forms an angle of 400 with it. Find ABC.
Solution. Let's extend the median BM beyond point M by its length and get point D (Fig. 2.15). Since AB = 2BM, then AB = BD, that is, the triangle ABD is isosceles. Therefore, BAD = BDA = (180o - 40o) : 2 = 70o. Quadrilateral ABCD is a parallelogram because its diagonals are bisected by their intersection point. This means CBD = ADB = 700. Then ABC = ABD + CBD =1100. The answer is 1100.
Problem 5. The sides?ABC are equal to a, b, c. Calculate the median mc drawn to side c (Fig. 2.16).
Solution. Let's double the median by building ?ABC to the parallelogram ACBP, and apply Theorem 8 to this parallelogram. We get: CP2+AB2 = 2AC2+2BC2, i.e. (2mc)2+c2= 2b2+2a2, from where we find:
2.4 Euler circle. Euler's line
Theorem. The bases of the medians, altitudes of an arbitrary triangle, as well as the midpoints of the segments connecting the vertices of the triangle with its orthocenter lie on the same circle, the radius of which is equal to half the radius of the circle circumscribed about the triangle. This circle is called the nine-point circle or Euler's circle.
Proof. Let's take the middle?MNL (Fig. 2.17) and describe a circle W around it. The segment LQ is the median in the rectangular?AQB, so LQ=1/2AB. The segment MN=1/2AB, because MN - middle line?ABC. It follows that the trapezoid QLMN is isosceles. Since circle W passes through 3 vertices of an isosceles trapezoid L, M, N, it will also pass through the fourth vertex Q. Similarly, it is proved that P belongs to W, R belongs to W.
Let's move on to points X, Y, Z. The segment XL is perpendicular to BH as the midline?AHB. The segment BH is perpendicular to AC and since AC is parallel to LM, then BH is perpendicular to LM. Therefore, XLM=P/2. Likewise, XNM= P/2.
In the quadrilateral LXNM, two opposite angles are right angles, so a circle can be described around it. This will be the circle W. So X belongs to W, similarly Y belongs to W, Z belongs to W.
The middle?LMN is similar to?ABC. The similarity coefficient is 2. Therefore, the radius of the circle of nine points is R/2.
Properties of the Euler circle:
The radius of the circle of nine points is equal to half the radius of the circle circumscribed about?ABC.
The circle of nine points is homothetic to the circle circumscribed about? ABC, with the coefficient ½ and the homothety center at point H.
Theorem. The orthocenter, centroid, circumcenter, and nine-point circle center lie on the same straight line. Euler's straight line.
Proof. Let H be the orthocenter? ABC (Fig. 2.18) and O be the center of the circumscribed circle. By construction, the perpendicular bisectors?ABC contain the heights of the median?MNL, i.e. O is simultaneously the orthocenter?LMN. ?LMN ~ ?ABC, their similarity coefficient is 2, so BH=2ON.
Let's draw a straight line through points H and O. We get two similar triangles?NOG and?BHG. Since BH=2ON, then BG=2GN. The latter means that point G is the centroid?ABC. For point G the ratio HG:GO=2:1 is satisfied.
Let further TF be the perpendicular bisector? MNL and F be the point of intersection of this perpendicular with the line HO. Let's consider the similar ?TGF and ?NGO. Point G is the centroid of?MNL, so the similarity coefficient of?TGF and?NGO is equal to 2. Hence OG=2GF and since HG=2GO, then HF=FO and F is the middle of the segment HO.
If we carry out the same reasoning regarding the perpendicular bisector to the other side?MNL, then it must also pass through the middle of the segment HO. But this means that point F is the point of the perpendicular bisectors?MNL. This point is the center of the Euler circle. The theorem is proven.
CONCLUSION
In this work, we looked at 4 wonderful points of a triangle, studied at school, and their properties, on the basis of which we can solve many problems. The Gergonne point, Euler circle and Euler straight line were also considered.
LIST OF SOURCES USED
1.Geometry 7-9. Textbook for secondary schools // Atanasyan L.S., Butuzov V.F. and others - M.: Education, 1994.
2.Amelkin V.V. Geometry on the plane: Theory, problems, solutions: Proc. A manual on mathematics // V.V. Amelkin, V.L. Rabtsevich, V.L. Timokhovich - Mn.: “Asar”, 2003.
.V.S. Bolodurin, O.A. Vakhmyanina, T.S. Izmailova // Manual on elementary geometry. Orenburg, OGPI, 1991.
.Prasolov V.G. Problems in planimetry. - 4th ed., supplemented - M.: Publishing house of the Moscow Center for Continuing Mathematical Education, 2001.
Baranova Elena
This work examines the remarkable points of the triangle, their properties and patterns, such as the nine-point circle and the Euler straight line. Given historical reference discovery of Euler's straight line and the nine-point circle. The practical direction of application of my project is proposed.
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"WONDERFUL POINTS OF A TRIANGLE." (Applied and fundamental questions of mathematics) Elena Baranova 8th grade, MKOU “Secondary School No. 20” Pos. Novoizobilny, Tatyana Vasilievna Dukhanina, mathematics teacher of MKOU "Secondary School No. 20" Novoizobilny village 2013. Municipal government educational institution"Secondary school No. 20"
Goal: study the triangle for its remarkable points, study their classifications and properties. Tasks: 1.Study necessary literature 2. Study the classification of remarkable points of a triangle 3.. Get acquainted with the properties of remarkable points of a triangle 4. Be able to construct remarkable points of a triangle. 5. Explore the scope of remarkable points. Object of study - section of mathematics - geometry Subject of study - triangle Relevance: expand your knowledge about the triangle, the properties of its remarkable points. Hypothesis: connection between triangle and nature
The point of intersection of the perpendicular bisectors. It is equidistant from the vertices of the triangle and is the center of the circumscribed circle. Circles circumscribed about triangles, the vertices of which are the midpoints of the sides of the triangle and the vertices of the triangle intersect at one point, which coincides with the point of intersection of the perpendicular bisectors.
Intersection point of bisectors The intersection point of bisectors of a triangle is equidistant from the sides of the triangle. OM=OA=OB
Point of intersection of altitudes The point of intersection of the bisectors of a triangle, the vertices of which are the bases of the heights, coincides with the point of intersection of the altitudes of the triangle.
Intersection point of medians The medians of a triangle intersect at one point, which divides each median in a ratio of 2:1, counting from the vertex. If the intersection point of the medians is connected to the vertices, then the triangle will be divided into three triangles of equal area. An important property of the intersection point of the medians is the fact that the sum of vectors, the beginning of which is the intersection point of the medians, and the ends are the vertices of the triangles, is equal to zero M1 N C B A m2 m3 M1 N C B A m2 m3 M1 N C B A m2 m3 M1 N C B A m2 m3
Torricelli point Note: A Torricelli point exists if all angles of the triangle are less than 120.
Circle of nine points B1, A1, C1 – bases of heights; A2, B2, C2 – the midpoints of the corresponding sides; A3, B3, C3, are the midpoints of segments AN, VN and CH.
Euler's straight line The point of intersection of medians, the point of intersection of heights, the center of a circle of nine points lie on one straight line, which is called Euler's straight line in honor of the mathematician who determined this pattern.
A little from the history of the discovery of remarkable points In 1765, Euler discovered that the midpoints of the sides of a triangle and the bases of its altitudes lie on the same circle. The most amazing property of the remarkable points of a triangle is that some of them are connected to each other by a certain ratio. The point of intersection of the medians M, the point of intersection of the heights H, and the center of the circumscribed circle O lie on the same straight line, and the point M divides the segment OH so that the relation OM: OH = 1: 2 is valid. This theorem was proven by Leonhard Euler in 1765.
The connection between geometry and nature. In this position, the potential energy has the smallest value and the sum of the segments MA+MB+MC will be the smallest, and the sum of the vectors lying on these segments with the beginning at the Torricelli point will be equal to zero.
Conclusions I learned that in addition to the wonderful points of intersection of heights, medians, bisectors and perpendicular bisectors that I know, there are also wonderful points and lines of a triangle. I can use the knowledge gained on this topic in my educational activities, independently apply theorems to certain problems, apply the learned theorems in a real situation. I believe that using the wonderful points and lines of a triangle in learning mathematics is effective. Knowing them significantly speeds up the solution of many tasks. The proposed material can be used both in mathematics lessons and in extracurricular activities for students in grades 5-9.
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There are so-called four remarkable points in a triangle: the point of intersection of the medians. The point of intersection of bisectors, the point of intersection of altitudes and the point of intersection of perpendicular bisectors. Let's look at each of them.
Intersection point of triangle medians
Theorem 1
On the intersection of medians of a triangle: The medians of a triangle intersect at one point and are divided by the intersection point in the ratio $2:1$ starting from the vertex.
Proof.
Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ are its medians. Since medians divide the sides in half. Let's consider the middle line $A_1B_1$ (Fig. 1).
Figure 1. Medians of a triangle
By Theorem 1, $AB||A_1B_1$ and $AB=2A_1B_1$, therefore, $\angle ABB_1=\angle BB_1A_1,\ \angle BAA_1=\angle AA_1B_1$. This means that triangles $ABM$ and $A_1B_1M$ are similar according to the first criterion of similarity of triangles. Then
Similarly, it is proved that
The theorem is proven.
Intersection point of triangle bisectors
Theorem 2
On the intersection of 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 $AM\ and\BP$. Let us draw perpendiculars from this point to the sides of the triangle (Fig. 2).
Figure 2. Triangle bisectors
Theorem 3
Each point of the bisector of an undeveloped angle is equidistant from its sides.
By Theorem 3, we have: $OX=OZ,\ OX=OY$. Therefore, $OY=OZ$. This means that the point $O$ is equidistant from the sides of the angle $ACB$ and, therefore, lies on its bisector $CK$.
The theorem is proven.
The point of intersection of the perpendicular bisectors of a triangle
Theorem 4
The perpendicular bisectors to 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 bisectoral perpendiculars $n\ and\ m$ (Fig. 3).
Figure 3. Perpendicular bisectors of a triangle
To prove it, we need the following theorem.
Theorem 5
Each point of the perpendicular bisector to a segment is equidistant from the ends of the segment.
By Theorem 3, we have: $OB=OC,\ OB=OA$. Therefore, $OA=OC$. 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 is proven.
Point of intersection of triangle altitudes
Theorem 6
The altitudes of a triangle or their extensions intersect at one point.
Proof.
Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its altitude. Let us draw a straight 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. Triangle heights
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 find that point $B$ is the midpoint of side $C_2A_2$, and point $C$ is the midpoint of side $A_2B_2$. From the construction we have that $(CC)_1\bot A_2B_2,\ (BB)_1\bot A_2C_2,\ (AA)_1\bot C_2B_2$. Therefore, $(AA)_1,\ (BB)_1,\ (CC)_1$ 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.
