Research project remarkable triangle points. Research work “Remarkable points of the triangle

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 heights 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 has been 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 has been 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 has been 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.

Ministry of Education and Science of the Russian Federation Federal State Budgetary Educational Institution of 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, one should not limit oneself only to the remarkable points provided for in the school curriculum by the State educational standard, such as the center of the inscribed circle (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 to deeply penetrate into the nature of the triangle and comprehend its inexhaustibility, it is necessary to have ideas about as many remarkable points of the triangle as possible. 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.

The purpose of the course work is to study 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 has been 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 external 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 has been 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 be different 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 has been 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 angles, outside it - in obtuse angles or coincide with the vertex, in rectangular ones it coincides with the vertex at a 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 has been 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.

In this lesson we will look at four wonderful points of the triangle. Let us dwell on two of them in detail, recall the proofs of important theorems and solve the problem. Let us remember and characterize the remaining two.

Subject:Revision of the 8th grade geometry course

Lesson: Four Wonderful Points of a Triangle

A triangle is, first of all, three segments and three angles, therefore the properties of segments and angles are fundamental.

The segment AB is given. Any segment has a midpoint, and a perpendicular can be drawn through it - let’s denote it as p. Thus, p is the perpendicular bisector.

Theorem (main property of the perpendicular bisector)

Any point lying on the perpendicular bisector is equidistant from the ends of the segment.

Prove that

Proof:

Consider triangles and (see Fig. 1). They are rectangular and equal, because. have a common leg OM, and legs AO and OB are equal by condition, thus, we have two right triangles, equal in two legs. It follows that the hypotenuses of the triangles are also equal, that is, what was required to be proved.

Rice. 1

The converse theorem is true.

Theorem

Each point equidistant from the ends of a segment lies on the perpendicular bisector to this segment.

Given a segment AB, a perpendicular bisector to it p, a point M equidistant from the ends of the segment (see Fig. 2).

Prove that point M lies on the perpendicular bisector of the segment.

Rice. 2

Proof:

Consider a triangle. It is isosceles, as per the condition. Consider the median of a triangle: point O is the middle of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both an altitude and a bisector. It follows that . But line p is also perpendicular to AB. We know that at point O it is possible to draw a single perpendicular to the segment AB, which means that the lines OM and p coincide, it follows that the point M belongs to the straight line p, which is what we needed to prove.

If it is necessary to describe a circle around one segment, this can be done, and there are infinitely many such circles, but the center of each of them will lie on the perpendicular bisector to the segment.

They say that the perpendicular bisector is the locus of points equidistant from the ends of a segment.

A triangle consists of three segments. Let us draw bisectoral perpendiculars to two of them and obtain the point O of their intersection (see Fig. 3).

Point O belongs to the perpendicular bisector to side BC of the triangle, which means it is equidistant from its vertices B and C, let’s denote this distance as R: .

In addition, point O is located on the perpendicular bisector to segment AB, i.e. , at the same time, from here.

Thus, point O of the intersection of two midpoints

Rice. 3

perpendiculars of the triangle is equidistant from its vertices, which means it also lies on the third bisector perpendicular.

We have repeated the proof of an important theorem.

The three perpendicular bisectors of a triangle intersect at one point - the center of the circumcircle.

So, we looked at the first remarkable point of the triangle - the point of intersection of its bisectoral perpendiculars.

Let's move on to the property of an arbitrary angle (see Fig. 4).

The angle is given, its bisector is AL, point M lies on the bisector.

Rice. 4

If point M lies on the bisector of an angle, then it is equidistant from the sides of the angle, that is, the distances from point M to AC and to BC of the sides of the angle are equal.

Proof:

Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, and the angles are equal, since AL is the bisector of the angle. Thus, right triangles are equal in hypotenuse and acute angle, it follows that , which is what needed to be proved. Thus, a point on the bisector of an angle is equidistant from the sides of that angle.

The converse theorem is true.

Theorem

If a point is equidistant from the sides of an undeveloped angle, then it lies on its bisector (see Fig. 5).

An undeveloped angle is given, point M, such that the distance from it to the sides of the angle is the same.

Prove that point M lies on the bisector of the angle.

Rice. 5

Proof:

The distance from a point to a line is the length of the perpendicular. From point M we draw perpendiculars MK to side AB and MR to side AC.

Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, legs MK and MR are equal by condition. Thus, right triangles are equal in hypotenuse and leg. From the equality of triangles follows the equality of the corresponding elements; equal angles lie opposite equal sides, thus, Therefore, point M lies on the bisector of the given angle.

If you need to inscribe a circle in an angle, this can be done, and there are infinitely many such circles, but their centers lie on the bisector of a given angle.

They say that a bisector is the locus of points equidistant from the sides of an angle.

A triangle consists of three angles. Let's construct the bisectors of two of them and get the point O of their intersection (see Fig. 6).

Point O lies on the bisector of the angle, which means it is equidistant from its sides AB and BC, let’s denote the distance as r: . Also, point O lies on the bisector of the angle, which means it is equidistant from its sides AC and BC: , , from here.

It is easy to notice that the point of intersection of the bisectors is equidistant from the sides of the third angle, which means it lies on

Rice. 6

angle bisector. Thus, all three bisectors of the triangle intersect at one point.

So, we remembered the proof of another important theorem.

The bisectors of the angles of a triangle intersect at one point - the center of the inscribed circle.

So, we looked at the second remarkable point of the triangle - the point of intersection of the bisectors.

We examined the bisector of an angle and noted its important properties: the points of the bisector are equidistant from the sides of the angle, in addition, the tangent segments drawn to the circle from one point are equal.

Let us introduce some notation (see Fig. 7).

Let us denote equal tangent segments by x, y and z. The side BC lying opposite the vertex A is designated as a, similarly AC as b, AB as c.

Rice. 7

Problem 1: in a triangle, the semi-perimeter and length of side a are known. Find the length of the tangent drawn from the vertex A - AK, denoted by x.

Obviously, the triangle is not completely defined, and there are many such triangles, but it turns out that they have some elements in common.

For problems involving an inscribed circle, the following solution method can be proposed:

1. Draw bisectors and get the center of the inscribed circle.

2. From center O, draw perpendiculars to the sides and obtain points of tangency.

3. Mark equal tangents.

4. Write out the relationship between the sides of the triangle and the tangents.

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

Similarly, it is proved that

The theorem has been 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.

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 has been 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 has been proven.

Point of intersection of triangle altitudes

Theorem 6

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

Proof.

Figure 4. Triangle heights

Content

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

Chapter 1.

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

1.2. Medians of a triangle

1.4. Heights in a triangle

Conclusion

List of used literature

Booklet

Introduction

Geometry is a branch of mathematics that deals with various figures 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, a triangle is an atom of geometry.

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

Such points studied in a 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 bisector 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 about the triangle,its propertieswonderful points.

Target: exploration of the triangle to its remarkable points,studying themclassifications and properties.

Tasks:

1. Study the necessary literature

2. Study the classification of remarkable points of a triangle

3. Be able to construct remarkable triangle points.

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 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, Pappus, and 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 received 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 the “Euler straight line”.

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

IN A, B, C - vertices

AB, BC, SA - sides

A C

Each triangle has four points associated with it:

Intersection point of medians;

Intersection point of bisectors;

Point of intersection of heights.

The point of intersection of the perpendicular bisectors;

1.2. Medians of a triangle

Medina of a triangle - , connecting the vertex from the middle of the opposite side (Figure 1). The point where the median intersects the side of the triangle is called the base of the median.

Figure 1. Medians of a triangle

Let's construct the midpoints of the sides of the triangle and draw segments connecting each of the vertices with the midpoint of the opposite side. Such segments are called medians.

And again we observe that these segments intersect at one point. If we measure the lengths of the resulting median segments, we can check one more property: the intersection point of the medians divides all medians in a ratio of 2:1, counting from the vertices. And yet, the triangle, which rests on the tip of the needle at the point of intersection of the medians, is in equilibrium! A point with this property is called the center of gravity (barycenter). The center of equal mass 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 are divided by the intersection point in a ratio of 2:1, counting from the vertex.

1.3. Bisectors of a triangle

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

Figure 2. Triangle bisector

In an arbitrary triangle ABC we draw the bisectors of its angles. And again, with an 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 lowering the perpendiculars DA 1, DB 1 and DC1 to the sides of the triangle. All of them are equal to each other: DA1=DB1=DC1.

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

1.4. Heights in a triangle

Height of the triangle - , dropped from the 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 construct three altitudes in a triangle, then they will 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:

for an acute triangle - inside;

for a rectangular one - on the hypotenuse;

for an obtuse angle, it is on the outside.

Figure 4. Orthocenter of the triangle

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

1.5. Perpendicular bisectors 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's draw an arbitrary triangle ABC and draw perpendicular bisectors to its sides. If the construction is carried out accurately, then all 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 with a center at point O, passing through one of the vertices of the triangle, then it will also pass through its other two vertices.

A circle passing through all the vertices of a triangle is called circumscribed about it. 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. Study of the remarkable points of the triangle.

Study of height in triangles

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

The altitudes of an acute triangle are located strictly inside the triangle.

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

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

The altitude drawn to the hypotenuse lies inside the triangle.

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

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

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

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

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

The other two altitudes lie outside the triangle and are lowered to the continuation of the sides of the triangle.

AK is the height drawn to side BC.

BF - height drawn to the continuation of side AC.

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

The point of intersection of the altitudes 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 bisector of the angle of the triangle (ray) that is inside the triangle.

All three bisectors of a triangle intersect at one point.

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

Studying medians in a triangle

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

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

Study of perpendicular bisectors to a side of a triangle

Perpendicular bisector of a triangle is a perpendicular drawn to the middle of a side of a triangle.

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

The point of intersection of perpendicular bisectors in an acute triangle lies inside the triangle; in an obtuse angle - 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 assigned tasks were solved:

1). We studied the necessary literature;

2). We studied the classification of remarkable points of a triangle;

3). We learned how to build wonderful triangle points;

4). We summarized the studied material for the design of the booklet.

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

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

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

Bibliography

Textbook. L.S. Atanasyan “Geometry grades 7-9Mnemosyne, 2015.

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

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