Wonderful points of the circle. The Wonderful Points of the Triangle Project
FOUR WONDERFUL POINTS
TRIANGLE
Geometry
8th grade
Sakharova Natalia Ivanovna
MBOU SOSH №28 of Simferopol
- Intersection point of triangle medians
- Intersection point of triangle bisectors
- Intersection point of triangle heights
- The point of intersection of the median perpendiculars of the triangle
Median
Median (BD) A triangle is a segment that connects the apex of the triangle to the middle of the opposite side.
Medians triangles intersect at one point (center of gravity triangle) and are divided by this point in a ratio of 2: 1, counting from the top.
BISECTOR
Bisector (AD) of a triangle is the segment of the bisector of the inner corner of the triangle. ∟ BAD = ∟ CAD.
Every point bisectors an undeveloped corner is equidistant from its sides.
Back: each point lying inside the corner and equidistant from the sides of the corner lies on its the bisector.
All bisectors triangles intersect at one point - center inscribed into a triangle circles.
Circle radius (OM) - perpendicular dropped from the center (point O) to the side of the triangle
HEIGHT
Height (CD) a triangle is a segment of a perpendicular dropped from the apex of a triangle to a straight line containing the opposite side.
Heights triangles (or their extensions) intersect one point.
MIDDLE PERPENDICULAR
Center perpendicular (DF) is called a straight line perpendicular to the side of the triangle and dividing it in half.
Every point middle perpendicular(m) to the segment equidistant from the ends of this segment.
Back: each point equidistant from the ends of the segment lies on the middle perpendicular to him.
All the median perpendiculars of the sides of the triangle intersect at one point - center described near triangle circles .
The radius of the circumscribed circle is the distance from the center of the circle to any vertex of the triangle (OA).
P. 177 No. 675 (picture)
Homework
Page 173 § 3 definitions and theorems page 177 No. 675 (finish)
Ministry of Education and Science Russian Federation Federal State Budgetary educational institution higher vocational education
Magnitogorsk State University»
Faculty of Physics and Mathematics
Department of Algebra and Geometry
Course work
Wonderful points of the triangle
Completed: student of 41 groups
Vakhrameeva A.M
supervisor
A.S. Velikikh
Magnitogorsk 2014
Introduction
Historically, geometry began with a triangle, therefore, 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 exactly a triangle can be considered an atom of geometry? Because the preceding concepts - point, line and angle - are obscure 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 a deep and comprehensive study of the triangle appears in it.
Surprisingly, the triangle, despite its apparent simplicity, is an inexhaustible object of study - no one, even in our time, dares to say that he has studied and knows all the properties of 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, a source different techniques its study - it is necessary to select and develop material for the study of the geometry of the wonderful points of the triangle. Moreover, when choosing this material, one should not be limited only to the remarkable points provided in school curriculum State educational standards, such as the center of the inscribed circle (the point of intersection of the bisectors), the center of the inscribed circle (the point of intersection of the median perpendiculars), the point of intersection of the medians, the point of intersection of heights. But for a deep penetration into the nature of the triangle and comprehending its inexhaustibility, it is necessary to have ideas about how more wonderful 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 method of studying the triangle can be constructed so that all the other properties of the triangle can be strung on this basis. In other words, no matter where you start the study of the triangle, you can always go to any depths of this amazing figure. But then - as an option - you can start studying the triangle by examining its wonderful points.
Target term paper is the study of the wonderful points of the triangle. To achieve this goal, it is necessary to solve the following tasks:
· To study the concepts of bisector, median, height, bisector and their properties.
· Consider Gergonne's point, Euler's circle and Euler's line, not studied in school.
CHAPTER 1. Bisector of a triangle, the center of the incircle of a triangle. Properties of the bisector of a triangle. Gergonne point
1 Inscribed circle center of a triangle
Wonderful 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 segment of the bisector of the angle of the triangle that connects the vertex to a point on the opposite side.
Theorem. Each point of the bisector of an undeveloped angle is equidistant (i.e. equidistant from the straight lines containing the sides of the triangle) from its sides. Conversely: each point that lies inside the corner and equidistant from the sides of the corner lies on its bisector.
Proof. 1) Take an arbitrary point M on the bisector of the angle BAC, draw perpendiculars MK and ML to lines AB and AC and prove that MK = ML. Consider right triangles ?AMK and ?AML. They are equal in hypotenuse and sharp corner(AM - general hypotenuse, 1 = 2 by condition). Therefore, MK = ML.
) Let point M lie inside BAC and equidistant from its sides AB and AC. Let us prove that the ray AM is the bisector of BAC. Let's draw perpendiculars MK and ML to straight lines AB and AC. Rectangular triangles AKM and ALM are equal in hypotenuse and leg (AM is the general hypotenuse, MK = ML by condition). Therefore, 1 = 2. But this also means that the beam AM is the bisector of BAC. The theorem is proved.
Consequence. The bisectors of a triangle intersect at one point, (the center of the inscribed circle 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: each point lying inside the corner 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 of ACB and, therefore, lies on the bisector CC1 of this angle. Therefore, all three bisectors ?ABCs intersect at point O, as required.
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.
It is required to prove that if ABD = DBC, then AD: DC = AB: BC.
We will carry out the CE || BD before crossing at point E with the continuation of side AB. Then, according to the theorem on the proportionality of segments formed on straight lines intersected by several parallel straight lines, we will have the proportion: AD: DC = AB: BE. To go from this proportion to the one that is required to prove, it is enough to find that BE = BC, i.e., that ?EVERYTHING is isosceles. In this triangle, E = ABD (as angles corresponding to parallel lines) and BCE = DBC (as angles lying crosswise at the same parallel lines).
But ABD = DBC by condition; hence, E = ALL, and therefore the sides BE and BC are equal, lying opposite to equal angles.
Now, replacing BE in the above proportion with BC, we get the proportion that you want to prove.
20 The bisectors of the inner and adjacent corner of the triangle are perpendicular.
Proof. Let BD be the bisector of ABC (Figure 1.2), and 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 the outer corner of a triangle divides the opposite side outwardly 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 corner of a triangle divides the opposite side into segments proportional to the adjacent sides of the triangle.
Proof. Consider ?ABC. For definiteness, let the bisector CAB intersect the side BC at point D (Figure 1.4). Let us show that BD: DC = AB: AC. To do this, draw a line through point C parallel to line AB, and denote by E the point of intersection of this line AD. Then DAB = DEC, ABD = ECD and therefore ?DAB ~ ?DEC for the first sign of similarity of triangles. Further, since ray AD is the bisector of CAD, then CAE = EAB = AEC and, therefore, ?ECA is isosceles. Hence AC = CE. But in this case, from the similarity ?DAB and ?DEC implies that BD: DC = AB: CE = AB: AC, and this is what was required to prove.
If the bisector of the outer corner of a triangle intersects the continuation 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. Consider ?ABC. Let F be a point on the extension of the side CA, D be the intersection point of the bisector of the outer BAF triangle with the extension of the side CB (Fig. 1.5). Let us show that DC: DB = AC: AB. Indeed, let us draw a line through point C parallel to line AB, and denote by E the point of intersection of this line with line DA. Then the triangle ADB ~ ?EDC and therefore DC: DB = EC: AB. And since ?EAC = ?BAD = ?CEA, the isosceles ?CEA side AC = EC and thus DC: DB = AC: AB, as required.
3 Solving problems on the application of the properties of the bisector
Problem 1. Let O be the center of the circle inscribed in ?ABC, CAB = ?... Prove COB = 900 +? /2.
Solution. Since O is the center inscribed in ?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, as required.
Problem 2. Let O be the center of the described ?ABC of the circle, H - base of the height drawn to side BC. Prove that the bisector CAB is also a bisector? OAH.
Let AD be the bisector of CAB, AE the diameter of the circumscribed ?ABC circle (fig. 1.7,1.8). If ?ABC - acute-angled (fig. 1.7) and, therefore, ABC<900, то так как ABC = AEC= ½ arc AC, and ?BHA and ?ECA rectangular (BHA = ECA = 900), then ?BHA ~ ?ECA and therefore CAO = CAE = HAB. Further, BAD and CAD are conditionally equal, therefore HAD = BAD - BAH = CAD - CAE = EAD = OAD. Now let ABC = 900. In this case, the height AH coincides with the side AB, then the point O will belong to the hypotenuse AC and therefore the validity of the statement of the problem is obvious.
Consider the case when ABC> 900 (Figure 1.8). Here the quadrilateral ABCE is inscribed in a circle and, therefore, AEC = 1800 - ABC. On the other hand, ABH = 1800 - ABC, i.e. AEC = ABH. And since ?BHA and ?ECA - rectangular and, therefore, HAB = 900 - ABH = 900 - AEC = EAC, then HAD = HAB + BAD = EAC + CAD = EAD = OAD. Cases where BAC and ACB are dumb are treated similarly. ?
4 Point Gergonne
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 to these vertices and the circle inscribed in the triangle.
Let point O be the incircle center of triangle ABC. Let the incircle touch the sides of the triangle BC, AC and AB at points D, E and F, respectively. Gergonne's point is the intersection of line 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. Gergonne's point is the intersection of line segments AD, BE, and CF.
Let us prove that these three segments do indeed intersect at one point. Note that the center of the inscribed circle is the intersection point of the bisectors of the angles ?ABC, and the radii of the inscribed circle 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 Cheva'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 meet at one point. Then
(go around the triangle clockwise)), the segments intersect at one point.
Inscribed circle properties:
A circle is called inscribed in a triangle if it touches all its sides.
You can inscribe a circle in any triangle.
Given: ABC is a given triangle, O is the point of intersection of the bisectors, M, L and K are 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 from point O the perpendiculars OK, OL and ОМ, respectively, to the sides AB, BC and CA (Figure 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. Hence, a circle with center O of radius OK is inscribed in triangle ABC. The theorem is proved.
The center of a circle inscribed in a triangle is the point of intersection of its bisectors.
Let ABC be given, O - the center of the inscribed circle, D, E and F - points of tangency of the circle with the sides (Figure 1.12). ? AEO =? AOD by hypotenuse and leg (EO = OD - as radius, AO - total). From the equality of the triangles it follows that? OAD =? OAE. Hence AO is the bisector of the angle EAD. In the same way, it is proved that the 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 circle (O; R) be the given circle (Fig. 1.13), the line a touches it at the point P. Let the radius OP be not perpendicular to a. Draw from point O the perpendicular OD to the tangent line. By the definition of a tangent, all its points other than the point P, and, in particular, the 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 wonderful points triangle, Euler's circle, Euler's line.
1 Center of the circumscribed circle of a triangle
The midpoint perpendicular to a segment is a straight line passing through the midpoint of the segment and perpendicular to it.
Theorem. Each point of the midpoint perpendicular to the segment is equidistant from the ends of this segment. Conversely: each point equidistant from the ends of a line segment lies on the perpendicular to it.
Proof. Let the line m be the midpoint perpendicular to the segment AB, point O - the midpoint of the segment.
Consider an arbitrary point M of the line m and prove that AM = BM. If the point M coincides with the point O, then this equality is true, since O is the midpoint of the segment AB. Let M and O be different points. Rectangular ?OAM and ?OBM are equal in two legs (OA = OB, OM - common leg), therefore AM = VM.
) Consider an arbitrary point N equidistant from the ends of the segment AB and prove that the point N lies on the line m. If N is a point of the straight line AB, then it coincides with the midpoint O of the segment AB and therefore lies on the straight line m. If point N does not lie on line AB, then consider ?АNB, which is isosceles, since АN = BN. The segment NO is the median of this triangle, and hence 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 proved.
Consequence. The mid-perpendiculars to the sides of the triangle intersect at one point, (the center of the circumscribed circle).
We denote by O, the point of intersection of the mid-perpendiculars m and n to the sides AB and BC ?ABC. According to the theorem (each point of the median perpendicular to the segment is equidistant from the ends of this segment. Conversely: each point equidistant from the ends of the segment lies on the median perpendicular 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 middle perpendicular p to this segment. Therefore, all three perpendiculars m, n, and p to the sides ?ABC intersect at point O.
In an acute-angled triangle, this point lies inside, in an obtuse-angled triangle, outside the triangle, in a rectangular triangle, in the middle of the hypotenuse.
Mid perpendicular property of a triangle:
The straight lines on which the bisectors of the inner and outer corners triangles, emerging from one vertex, intersect with the middle to the opposite side of the perpendicular from the diametrically opposite points of the circle circumscribed about the triangle.
Proof. For example, let the bisector ABC intersect the described one about ?ABC circle at point D (fig. 2.1). Then since the inscribed ABD and DBC are equal, then AD = arc DC. But the middle perpendicular to the AC side also divides the arc AC in half, so point D will also belong to this middle perpendicular. Further, since, according to property 30 from item 1.3, the bisector of BD is ABC adjacent to ABC, the latter will intersect the circle at a point diametrically opposite to point D, since the inscribed right angle always rests on the diameter.
2 Orthocenter of a circle of a triangle
Height is a perpendicular drawn from the apex of a triangle to a straight line containing the opposite side.
The heights of the triangle (or their extensions) intersect at one point, (orthocenter).
Proof. Consider an arbitrary ?ABC and prove that the straight lines AA1, BB1, CC1, containing its heights, intersect at one point. Let's draw 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 the sides of this triangle. Indeed, AB = A2C and AB = CB2 as opposite sides of the 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, straight lines AA1, BB1 and CC1 are perpendiculars to the sides ?A2B2C2. Therefore, they intersect at one point.
Depending on the type of triangle, the orthocenter can be inside a triangle in acute-angled, outside it - in obtuse-angled or coincide with the vertex, in rectangular ones - coincide with the vertex at a right angle.
Triangle Height Properties:
A segment connecting the bases of two heights of an acute-angled triangle cuts off a triangle similar to this one from it, with a similarity coefficient equal to cosine general angle.
Proof. Let AA1, BB1, CC1 be the heights of an acute-angled triangle ABC, and ABC = ?(fig. 2.2). Right-angled 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, and the similarity coefficient is cos ?... It is proved in a similar way that ?A1CB1 ~ ?ABC with the similarity coefficient cos BCA, and ?B1AC1 ~ ?ABC with the coefficient of similarity cos CAB.
The height, lowered by the hypotenuse of a right-angled triangle, divides it into two similar to each other and similar to the original triangle, triangles.
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 the similarity of right-angled triangles in the proportionality of two legs, since AD / CD = CD / DB. Each of the right-angled triangles ADC and BDC is similar to the original right-angled triangle, if only on the basis of the similarity feature in two angles.
Solving problems on the application of height properties
Problem 1. Prove that a triangle, one of the vertices of which is the vertex of a given obtuse triangle, and the other two vertices are the bases of the heights of an obtuse triangle, omitted from its other two vertices, is similar to this triangle with a similarity coefficient equal to the modulus of the cosine of the angle at the first vertex ...
Solution. Consider obtuse ?ABC with blunt 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 (Figure 2.4) with a similarity coefficient k = cos ?, completely repeats the reasoning carried out in the proof of property 1, section 2.2.
Let us prove that ?A1CB ~ ?ABC (Fig. 2.5) with the similarity coefficient k1 = cos ?, a ?B1AC1 ~ ?ABC (Fig. 2.6) with the similarity coefficient k2 = | cos? |.
Indeed, right-angled triangles CA1A and CB1B have a common angle ?and therefore are similar. It follows that B1C / BC = A1C / AC = cos ?and therefore B1C / A1C = BC / AC = cos ?, i.e. sides in triangles A1CB1 and ABC forming a common ??are proportional. And then on the second basis of the similarity of triangles ?A1CB ~ ?ABC, and the similarity coefficient k1 = cos ?... As for the last case (Figure 2.6), then from the consideration of right-angled triangles ?BB1A and ?CC1A with equal vertical corners BAB1 and C1AC it follows that they are similar and, therefore, B1A / BA = C1A / CA = cos (1800 - ?) = | cos ?|, since ??- stupid. Hence B1A / C1A = BA / CA = | cos ?| and thus in triangles ?B1AC1 and ?ABC sides forming equal angles are proportional. This means that ?B1AC1 ~ ?ABC with the similarity coefficient k2 = | cos? |.
Problem 2. Prove that if point O is the intersection point of the heights of an acute-angled 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 other two formulas are proved in a similar way. So, let ABC = ?, AOC = ?... A1, B1 and C1 are the bases of the heights of the triangle drawn from the vertices A, B and C, respectively (Figure 2.7). Then from the right-angled triangle BC1C it follows that BCC1 = 900 - ?and thus, in the right-angled triangle OA1C, the angle COA1 is ?... But the sum of the angles AOC + COA1 = ? + ?gives an unfolded angle, and therefore AOC + COA1 = AOC + ABC = 1800, which is what was required to prove.
Problem 3. Prove that the heights of an acute-angled triangle are the bisectors of the angles of the triangle, the vertices of which are the bases of the heights of this triangle.
Figure 2.8
Solution. Let AA1, BB1, CC1 be the heights of an acute-angled triangle ABC and let CAB = ?(Figure 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 also means that AA1 is the bisector of the angle C1A1B1. Similarly, it is proved that the other two heights of the triangle ABC are the bisectors of the other two corresponding angles of the triangle A1B1C1.
3 Center of gravity of the circle of the triangle
The median of a triangle is the segment that connects any vertex of the triangle to the middle of the opposite side.
Theorem. The median of the triangle intersect at one point, (center of gravity).
Proof. Consider an arbitrary one? ABC.
Let us denote by the letter O the point of intersection of the medians AA1 and BB1 and draw the middle line A1B1 of this triangle. The segment A1B1 is parallel to the side AB, therefore 1 = 2 and 3 = 4. Therefore, ?AOB and ?A1OB1 are similar in two angles, and, therefore, their sides are proportional: AO: A1O = BO: B1O = AB: A1B1. But AB = 2A1B1, therefore AO = 2A1O and BO = 2B1O. Thus, the point O of the intersection of the medians AA1 and BB1 divides each of them in a ratio of 2: 1, counting from the top.
Similarly, it is proved that the point of intersection of the medians BB1 and CC1 divides each of them in a ratio of 2: 1, counting from the top, and, therefore, coincides with the point O and divides it in a ratio of 2: 1, counting from the top.
Median triangle properties:
10 The medians of the triangle intersect at one point and are divided by the point of intersection in a ratio of 2: 1, counting from the vertex.
Given: ?ABC, AA1, BB1 - medians.
Prove: AO: OA1 = BO: OV1 = 2: 1
Proof. Let's draw the middle line A1B1 (Figure 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 straight lines A1B1 and AB and secant BB1.
Hence, ?AOB ~ ?A1OB1 by the equality of two angles, which means that the sides are proportional: AO / A1O = OB / OB1 = AB / A1B = 2/1, AO / A1O = 2/1; OB / OB1 = 2/1.
The median splits a triangle into two triangles of equal area.
Proof. BD - median ?ABC (Figure 2.11), BE is its height. Then ?ABD and ?DBCs are equal in size, since they have equal bases AD and DC, respectively, and a total height BE.
The entire triangle is divided by its medians into six equal triangles.
If, on the continuation of the median of the triangle, we postpone a segment equal in length to the median 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 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 halved, it follows from Property 13.4 that the quadrilateral ABEC is a parallelogram.
Solving problems on the use of properties of medians:
Problem 1. Prove that if O is the point of intersection of the medians ?ABC then ?AOB, ?BOC and ? AOC are equal.
Solution. Let AA1 and BB1 be the medians ?ABC (fig.2.13). Consider ?AOB and ?BOC. Obviously, 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, whence it follows that S ?AOB = S ?BOC. The equality S ? AOB = S ?AOC.
Problem 2. Prove that if point O lies inside ?ABC and ?AOB, ?BOC and ?AOC are equal, then O is the point of intersection of the medians? ABC.
Solution. Consider ?ABC (2.14) and suppose 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 the point O belongs to two other medians ?ABC. From this it follows that the point O is really the point of intersection of the 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 circle and extend its bisector BD to the intersection with the circle at point K. Through point K there will be a perpendicular midpoint to the segment AC (property 1, from item 2.1), which has common point M. But since the segments BH and MK are parallel, and the points B and K lie along different sides from the line AC, then the intersection point of the segments BK and AC belongs to the segment HM, and this proves the required result.
Problem 4. In ?ABC The median BM is half the size of side AB and makes an angle of 400 with it. Find ABC.
Solution. Let us extend the median BM beyond the point M by its length and get the point D (Fig. 2.15). Since AB = 2BM, then AB = BD, that is, triangle ABD is isosceles. Therefore, BAD = BDA = (180o - 40o): 2 = 70o. Quadrilateral ABCD is a parallelogram since its diagonals are halved by the intersection point. So 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. (Figure 2.16).
Solution. Let us double the median by completing? ABC to the parallelogram ACVP, and apply Theorem 8 to this parallelogram. We obtain: CP2 + AB2 = 2AC2 + 2BC2, i.e. (2mc) 2 + c2 = 2b2 + 2a2, whence we find:
2.4 Euler's circle. Euler's line
Theorem. The bases of the medians, heights of an arbitrary triangle, as well as the midpoints of the segments connecting the vertices of the triangle with its orthocenter, lie on one 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. Take the median? MNL (Fig. 2.17) and describe a circle around it W. The segment LQ is the median in the rectangular? AQB, therefore LQ = 1 / 2AB. Segment MN = 1 / 2AB, because MN- middle line? ABC. It follows that the trapezoid QLMN is isosceles. Since the circle W passes through 3 vertices of the isosceles trapezoid L, M, N, it also passes 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 middle line? AHB. BH is perpendicular to AC and since AC is parallel to LM, BH is perpendicular to LM. Therefore, XLM = P / 2. Similarly, XNM = P / 2.
In the quadrilateral LXNM, two opposite corners are straight lines, 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 coefficient of similarity is 2. Therefore, the radius of the circle of nine points is equal to R / 2.
Euler circle properties:
The radius of a circle of nine points is equal to half the radius of a circle circumscribed about? ABC.
The circle of nine points is homothetic to the circle circumscribed about? ABC, with the coefficient ½ and the center of homothety at point H.
Theorem. The orthocenter, centroid, the center of the circumcircle and the center of the circle of nine points lie on the same straight line. Euler's line.
Proof. Let H be the orthocenter? ABC (Fig.2.18) and O - the center of the circumscribed circle. By construction, the median perpendiculars? ABC contain the heights of the median? MNL, i.e. O is simultaneously the orthocenter? LMN. ? LMN ~? ABC, their similarity coefficient is 2, therefore 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 fulfilled.
Let further TF be the median perpendicular? MNL and F - the point of intersection of this perpendicular with the line HO. Consider similar? TGF and? NGO. Point G is the centroid? MNL; therefore, the similarity coefficient? TGF and? NGO is 2. Hence OG = 2GF and since HG = 2GO, then HF = FO and F is the middle of the HO segment.
If we carry out the same reasoning about the midpoint perpendicular to the other side? MNL, then it must also go through the middle of the segment HO. But this means that point F is the point of the mid-perpendiculars? MNL. Such a point is the center of Euler's circle. The theorem is proved.
CONCLUSION
In this paper, we examined 4 wonderful points of the triangle studied in school and their properties, on the basis of which we can solve many problems. Gergonne's point, Euler's circle and Euler's line were also considered.
LIST OF USED SOURCES
1.Geometry 7-9. Textbook for secondary schools // Atanasyan L.S., Butuzov V.F. and others - M .: Education, 1994.
2.V.V. Amelkin Geometry on a plane: Theory, tasks, solutions: Textbook. A manual on mathematics // V.V. Amelkin, V.L. Rabtsevich, V.L. Timokhovich - Minsk: "Asar", 2003.
.V.S. Bolodurin, O.A. Vakhmyanina, T.S. Izmailova // Manual on elementary geometry. Orenburg, OGPI, 1991.
.V.G. Prasolov Planimetry tasks. - 4th ed., Supplemented - M .: Publishing house of the Moscow Center for Continuous Mathematical Education, 2001.
Content
Introduction …………………………………………………………………………………… 3
Chapter 1.
1.1 Triangle …………………………………………………………………………… ..4
1.2. Triangle medians
1.4. Heights in a triangle
Conclusion
List of used literature
Booklet
Introduction
Geometry is a branch of mathematics that deals with various shapes and their properties. Geometry starts with a triangle. For two and a half millennia, the triangle has been a symbol of geometry; but it is not only a symbol, the triangle is an atom of geometry.
In my work, I will consider the properties of the intersection points of the bisectors, medians and heights of the triangle, I will talk about their wonderful properties and the lines of the triangle.
These points studied in the school geometry course include:
a) the point of intersection of the bisectors (the center of the inscribed circle);
b) the point of intersection of the median perpendiculars (the center of the circumscribed circle);
c) the point of intersection of heights (orthocenter);
d) the point of intersection of the medians (centroid).
Relevance: expand your knowledge of the triangle,properties of itswonderful points.
Target: exploration of the triangle for its remarkable points,studying themclassifications and properties.
Tasks:
1. Study the necessary literature
2. Learn the classification of the wonderful points of the triangle
3. Be able to build wonderful triangle points.
4. Summarize the material studied for the design of the booklet.
Project hypothesis:
the ability to find wonderful 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 "Principles" Euclid solves the problem: "Inscribe a circle in a given triangle." It follows from the solution that the three bisectors inner corners triangles intersect at one point - the center of the inscribed circle. It follows from the solution of another Euclidean problem that the perpendiculars restored to the sides of the triangle in their midpoints also intersect at one point - the center of the circumscribed circle. In the "Beginnings" it is not said that the three heights of the triangle intersect at one point, called the orthocenter (the Greek word "orthos" means "straight", "correct"). This proposal was, however, known to Archimedes, Pappus, Proclus.
The fourth singular point of the triangle is the intersection of the medians. Archimedes proved that it is the center of gravity (barycenter) of the triangle. The above four points were drawn Special attention, and since the 18th century they have been called "remarkable" or "special" points of the triangle.
The study of the properties of the 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 Leonard Euler. In 1765, Euler proved that in any triangle the orthocenter, barycenter and the center of the circumscribed circle lie on one straight line, later called "Euler's line".
Triangle
Triangle - geometric figure, consisting of three points that do not lie on one straight line, and three segments connecting these points in pairs. Points -tops triangle, segments -parties triangle.
V A, B, C - peaks
AB, BC, CA - sides
A C
There are four points associated with each triangle:
Intersection point of medians;
Intersection point of bisectors;
Intersection point of heights.
The point of intersection of the midpoint perpendiculars;
1.2. Triangle medians
Medina of the triangle - connecting the vertex with the middle of the opposite side (Figure 1). The point of intersection of the median with the side of the triangle is called the base of the median.
Figure 1. Medians of a triangle
Construct the midpoints of the sides of the triangle and draw segments connecting each of the vertices with the midpoint of the opposite side. These segments are called the median.
And again we observe that these segments also intersect at one point. If we measure the lengths of the resulting segments of the medians, then we can check one more property: the point of intersection of the medians divides all the medians in a ratio of 2: 1, counting from the vertices. And also, 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 the intersection point is divided in a ratio of 2: 1, counting from the vertex.
1.3. Bisectors of a triangle
Bisector called the bisector of an angle, drawn from the apex of the angle to its intersection with the opposite side. A triangle has three bisectors corresponding to its three vertices (Figure 2).
Figure 2. Bisector of a triangle
In an arbitrary triangle ABC, draw the bisectors of its angles. And again, with an accurate 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 seen by dropping the perpendiculars DA 1, DB 1 and DC1 to the sides of the triangle. They are all equal to each other: DA1 = DB1 = DC1.
If you draw a circle centered at point D and radius DA 1, then it will touch all three sides of the triangle (that is, it will have only one common point with each of them). Such a circle is called inscribed in a triangle. So, the bisectors of the angles of the triangle intersect at the center of the inscribed circle.
1.4. Heights in a triangle
The height of the triangle - dropped from the top to the opposite side or a straight line that coincides 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 three heights are constructed in a triangle, then they all intersect at one point H. This point is called the orthocenter. (Figure 4).
With the help of constructions, you can check that, depending on the type of triangle, the orthocenter is located in different ways:
in an acute-angled triangle - inside;
for a rectangular one - on the hypotenuse;
the obtuse - outside.
Figure 4. Orthocenter of a triangle
Thus, we got acquainted with one more remarkable point of the triangle and we can say that: the heights of the triangle intersect at the orthocenter.
1.5. Mid perpendiculars to the sides of the triangle
The midpoint perpendicular to a line segment is a line perpendicular to that line segment and passing through its midpoint.
Let's draw an arbitrary triangle ABC and draw perpendiculars to its sides. If the construction is performed exactly, then all perpendiculars will intersect at one point - point O. This point is equidistant from all vertices of the triangle. In other words, if you draw a circle centered at point O, passing through one of the vertices of the triangle, then it will pass through its other two vertices.
A circle passing through all the vertices of a triangle is called circumscribed about it. Therefore, the established property of a triangle can be formulated as follows: the median perpendiculars to the sides of the triangle intersect at the center of the circumscribed circle (Figure 5).
Figure 5. Triangle inscribed in a circle
Chapter 2. Exploring the remarkable points of the triangle.
Examining Height in Triangles
All three heights of the triangle intersect at one point. This point is called the orthocenter of the triangle.
The heights of an acute-angled triangle are located strictly inside the triangle.
Accordingly, the intersection point of the heights is also inside the triangle.
In a right-angled triangle, two heights coincide with the sides. (These are heights drawn from the tops of sharp corners to the legs).
The height drawn to the hypotenuse lies within the triangle.
AC - the height drawn from the top of the C to the side AB.
AB is the height drawn from the top of B to the side of AC.
AK - the height drawn from the top right angle And to the hypotenuse BC.
The heights of a right-angled triangle intersect at the apex of a right angle (A - orthocenter).
In an obtuse triangle, there is only one height inside the triangle - the one that is drawn from the apex of the obtuse angle.
The other two heights lie outside the triangle and are lowered to the extension of the sides of the triangle.
AK is the height drawn to the BC side.
BF is the height drawn to the extension of the AC side.
CD is the height drawn to the extension of the AB side.
The intersection of the heights of an obtuse triangle is also outside the triangle:
H is the orthocenter of triangle ABC.
Examining bisectors in a triangle
The bisector of a triangle is the part of the bisector of the angle of a triangle (ray) that is inside the triangle.
All three bisectors of the triangle intersect at one point.
Intersection point of bisectors in acute-angled, obtuse-angled and right-angled triangles, is the center of the inscribed circle and is inside.
Examining 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.
After examining these triangles, I realized that in any triangle, the medians intersect at one point. This point is called the center of gravity of the triangle.
Examining the perpendiculars to the side of the triangle
Mid perpendicular of a triangle is a perpendicular drawn to the middle of a side of a triangle.
The three perpendiculars of the triangle intersect at one point, they are the center of the circumscribed circle.
The point of intersection of the mid-perpendiculars in an acute-angled triangle lies inside the triangle; in obtuse - outside the triangle; in a rectangular one - in the middle of the hypotenuse.
Conclusion
In the course of the work done, we come to the following conclusions:
The goal is achieved:explored the triangle and found its remarkable points.
The tasks have been solved:
1). We studied the necessary literature;
2). Learned the classification of the remarkable points of the triangle;
3). Learned how to build wonderful points of the triangle;
4). We summarized the material studied for the design of the booklet.
The hypothesis that the ability to find wonderful points of a triangle helps in solving construction problems was confirmed.
In the work, the methods of constructing remarkable points of the triangle are consistently presented, historical information about geometric constructions.
Information from this work can be useful in geometry lessons in grade 7. The booklet can serve as a guide to geometry on a given topic.
Bibliography
Textbook... L.S. Atanasyan "Geometry grades 7-9Mnemosina, 2015.
Wikipediahttps: //ru.wikipedia.org/wiki/Geometry#/media/File: Euclid% 27s_postulates.png
Scarlet Sails Portal
Leading educational portal Russia http://cendomzn.ucoz.ru/index/0-15157
Introduction
The objects of the world around us have certain properties, which are studied by various sciences.
Geometry is a branch of mathematics that considers various shapes and their properties, with its roots going back to the distant past.
In the fourth book of the "Elements" Euclid solves the problem: "Inscribe a circle in a given triangle." From the solution it follows that the three bisectors of the inner angles of the triangle intersect at one point - the center of the inscribed circle. It follows from the solution of another Euclidean problem that the perpendiculars restored to the sides of the triangle in their midpoints also intersect at one point - the center of the circumscribed circle. In the "Beginnings" it is not said that the three heights of the triangle intersect at one point, called the orthocenter (the Greek word "orthos" means "straight", "correct"). This proposal was, however, known to Archimedes. The fourth singular point of the triangle is the intersection of the medians. Archimedes proved that it is the center of gravity (barycenter) of the triangle.
Special attention was paid to the above four points, and since the 18th century they have been called "remarkable" or "special" points of the triangle. The study of the properties of the 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 Leonard Euler.
In 1765, Euler proved that in any triangle the orthocenter, barycenter and the center of the circumscribed circle lie on the same straight line, later called "Euler's line". In the twenties years XIX century, the French mathematicians J. Poncelet, C. Brianchon and others independently established the following theorem: the bases of the medians, the bases of the heights and the midpoints of the segments of heights connecting the orthocenter with the vertices of the triangle lie on the same circle. This circle is called the "circle of nine points" or "Feuerbach's circle" or "Euler's circle". K. Feuerbach established that the center of this circle lies on the Euler line.
“I think that never until now 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, the creations of nature and man.
We were interested in the so-called "wonderful points of the triangle".
After reading the literature on this topic, we fixed for ourselves the definitions and properties of the wonderful points of the triangle. But this was not the end of our work, and we wanted to explore these points ourselves.
That's why goal given work - the study of some wonderful points and lines of the triangle, the application of the knowledge gained to solving problems. In the process of achieving this goal, the following stages can be distinguished:
Selection and study teaching material from various sources information, literature;
Study of the basic properties of the wonderful points and lines of the triangle;
Generalization of these properties and proof of necessary theorems;
Solving problems related to wonderful points of a triangle.
ChapterI... Wonderful points and lines of the triangle
1.1 The point of intersection of the perpendiculars to the sides of the triangle
The midpoint is a straight line passing through the midpoint of a line segment, perpendicular to it. We already know the theorem that characterizes the property of the midpoint perpendicular: each point of the mid-perpendicular to the segment is equidistant from its ends and back, if the point is equidistant from the ends of the segment, then it lies on the mid-perpendicular.
The polygon is called inscribed into a circle if all its vertices belong to the circle. In this case, 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 perpendiculars to the sides of the triangle.
Let point O be the point of intersection of the perpendiculars to the sides of the triangle AB and BC.
Output: thus, if point O is the point of intersection of the middle perpendiculars to the sides of the triangle, then OA = OC = OB, i.e. point O is equidistant from all vertices of the triangle ABC, which means that it is the center of the circumscribed circle.
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| acute-angled | obtuse | rectangular |
Consequences
sin γ = c / 2R = c / sin γ = 2R.
It is similarly proved a/ sin α = 2R, b / sin β = 2R.
Thus:
This property is called the sine theorem.
In mathematics, it often happens that objects that are completely defined differently, turn out to be the same.
Example. Let A1, B1, C1 be the midpoints of the sides ∆ABS BC, AC, AB, respectively. Show that the circles circumscribed about the triangles AB1C1, A1B1C, A1BC1 intersect at one point. Moreover, this point is the center of a circle described about ∆ABS.
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| Consider a segment AO and construct a circle on this segment, as on a diameter. Points C1 and B1 fall on this circle, since are the vertices of right angles based on AO. Points A, C1, B1 lie on a circle = this circle is circumscribed about ∆АВ1С1. Similarly, we draw a segment VO and construct a circle on this segment, as on a diameter. This will be a circle around ∆BC1 A1. Let's draw a segment of CO and construct a circle on this segment, as on a diameter. This will be the circle around These three circles pass through the point O - the center of the circle circumscribed about ∆ABS. |
Generalization. If on the sides of ∆АВС АС, ВС, АС we take arbitrary points А 1, В 1, С 1, then the circles described around triangles AB 1 С 1, А 1 В 1 С, А 1 ВС 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 point of intersection of the bisectors of angles A and B. By the property of a point lying on the bisector of angle A, OF = OD = r. By the property of a point lying on the bisector of angle B, OE = OD = r. Thus, OЕ = OD = OF = r = point O is equidistant from all sides of the triangle ABC, i.e. O is the center of the inscribed circle. (Point O is the only one).
Output: thus, if point O is the intersection point of the bisectors of the angles of the triangle, then OЕ = OD = OF = r, i.e. point O is equidistant from all sides of the triangle ABC, which means that it is the center of the inscribed circle. The point O-intersection of the bisectors of the angles of the triangle is a wonderful point of the triangle.
Consequences:
From the equality of triangles AOF and AOD (Figure 1) for the hypotenuse and acute angle, it follows that AF = AD ... It follows from the equality of the OBD and OBE triangles that BD = BE , It follows from the equality of the triangles COE and COF that WITH F = CE ... Thus, the segments of the tangents drawn to the circle from one point are equal.
AF = AD = z, BD = BE = y, СF = CE = x
a = x + y (1), b= x +z (2), c = x + y (3).
+ (2) - (3), then we get: a +b-c =x+ y+ x+ z- z- y = a +b-c = 2x =
x = ( b + c - a) / 2
Similarly: (1) + (3) - (2), 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 splits the opposite side into segments proportional to the adjacent sides.
1.3 Intersection point 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. First, let us 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 median line of a triangle, the segments B 1 A 1 and PQ are equal to half of the side AB and are parallel to it. Therefore, the quadrilateral A 1 B 1 is a PQ-parallelogram. Then the intersection point G of its diagonals PA 1 and QB 1 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 divide the median BB 1 also into three equal parts. So, the intersection point G of the two medians of the triangle divides each of them in a ratio of 2: 1, counting from the vertex.
The point of intersection of the medians of the 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 Intersection point of triangle heights (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 of this triangle are visible at an angle of 120 °, i.e. angles AOB, AOC and BOC are equal to 120 °.
Let us prove that if all the angles of the triangle are less than 120 °, then the Torricelli point exists.
On the side AB of triangle ABC, construct an equilateral triangle ABC "(Fig. 6, a), and describe a circle around it. The segment AB contracts an arc of this circle of 120 °. Consequently, the points of this arc, other than A and B, have the property that the segment AB is visible from them at an angle of 120 °. Similarly, on the side AC of triangle ABC, construct an equilateral triangle ACB "(Fig. 6, a), and describe a circle around it. Points of the corresponding arc, other than A and C, have the property that the segment AC is seen 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 the triangle, for example ABC, is equal to 120 °, the point of intersection of the arcs of the circles will be point B (Fig. 6, b). In this case, the Torricelli point does not exist, since it is impossible to talk about the angles at which the sides AB and BC are visible from this point.
In the case when one of the angles of the triangle, for example ABC, is greater than 120 ° (Fig. 6, c), the corresponding arcs of the circles do not intersect, and the Torricelli point also does not exist.
Associated with the Torricelli point is Fermat's problem (which we will consider in Chapter II) of finding a point whose sum of distances to three given points is the smallest.
1.6 Circle of nine points
Indeed, A 3 B 2 is the midline of triangle AHC and, 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 °. Similarly, A 3 C 2 A 2 = 90 °. Therefore, points A 2, B 2, C 2, A 3 lie on the same circle with diameter A 2 A 3. Since AA 1 ┴BC, the point A 1 also belongs to this circle. Thus, points A 1 and A 3 lie on a circle circumscribed about 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 heights and the center of the circumscribed circle. Indeed, let in triangle ABC (Fig. 9), point O - the center of the circumscribed circle; G is the point of intersection of the medians. H is the point of intersection of heights. It is required to prove that the points O, G, H lie on one straight line and the center of the circle of nine points N divides the segment OH in half.
Consider a homothety centered at G and with a coefficient of -0.5. The vertices A, B, C of the triangle ABC will move, respectively, to points A 2, B 2, C 2. The heights of the triangle ABC will go to the heights of the triangle A 2 B 2 C 2 and, therefore, the point H will go to the point O. Therefore, the points O, G, H will lie on one straight line.
Let us show that the midpoint N of the segment OH is the center of the circle of nine points. Indeed, C 1 C 2 is a chord of a circle of nine points. Therefore, the mid-perpendicular to this chord is the diameter and intersects OH in the middle of N. Similarly, the mid-perpendicular to the chord B 1 B 2 is the 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 a circle circumscribed about a triangle ABC; D, E, F - the bases of the perpendiculars dropped from the point P to the sides of the triangle (Fig. 10). Let us show that points D, E, F are collinear.
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 sharp and the other is obtuse. This implies that points D and E will be located on opposite sides of line BC, and in order to prove that points D, E and F lie on the same line, it suffices to verify that ∟CEF = ∟BED.
Let us describe a circle with a diameter CP. Since ∟CFP = ∟CEP = 90 °, points E and F lie on this circle. Therefore, ∟CEF = ∟CPF as inscribed angles supported by one circular arc. Further, ∟CPF = 90 ° - ∟PCF = 90 ° - ∟DBP = ∟BPD. Let's describe a circle with a diameter BP. Since ∟BEP = ∟BDP = 90 °, points F and D lie on this circle. Therefore, ∟BPD = ∟BED. Therefore, we finally obtain that ∟CEF = ∟BED. Hence the points D, E, F lie on one straight line.
ChapterIISolving problems
Let's start with the problems related to the location of the bisectors, medians and heights of the triangle. Their solution, on the one hand, allows you to recall the previously covered material, and on the other hand, develops the necessary geometric representations, prepares more complex tasks.
Objective 1. At angles A and B of triangle ABC (∟A
Solution. Let CD be the height, CE be the bisector, then
∟BCD = 90 ° - ∟B, ∟BCE = (180 ° - ∟A - ∟B): 2.
Therefore, ∟DCE =.
Solution. Let O be the point of intersection of the bisectors of triangle ABC (Fig. 1). Let's take advantage of the fact that there is a larger angle opposite the larger side of the triangle. If AB BC, then ∟A
Solution. Let O be the point of intersection of the heights of the triangle ABC (Fig. 2). If AC ∟B. The circle with the diameter BC will pass through the points F and G. Taking into account that of the two chords the smaller is the one on which the smaller inscribed angle rests, we obtain that CG
Proof. On the sides AC and BC of the triangle ABC, as on the diameters, draw 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 corners based on the same circular arc. Therefore, ∟B 1 C 1 C = ∟CC 1 A 1, that is, 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 heights of this acute-angled triangle, gives an answer to one of the classical extremal problems.
Solution. Let ABC be a given acute-angled triangle. On its sides it is required to find such points A 1, B 1, C 1, for which the perimeter of the triangle A 1 B 1 C 1 would be the smallest (Fig. 4).
First we fix the point C 1 and look for the points A 1 and B 1 for which the perimeter of the triangle A 1 B 1 C 1 is the smallest (for a given position of the point C 1).
To do this, consider points D and E symmetric to point C 1 relative to 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 the 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 with respect to AC, then CD = CC 1 and ACD = ACC 1. Likewise, CE = CC 1 and BCE = BCC 1. Therefore, triangle CDE is isosceles. Its side is equal to CC 1. The base of DE is equal to the perimeter P triangle A 1 B 1 C 1. The angle DCE is equal to twice the angle ACB of triangle ABC and, therefore, does not depend on the position of point C 1.
In an isosceles triangle with a given apex angle, the smaller the side, the smaller the base. Therefore, the smallest perimeter value P is reached in the case of the smallest CC 1. This value is assumed if CC 1 is the height of triangle ABC. Thus, the desired point C 1 on the side AB is the base of the height drawn from the vertex C.
Note that we could fix first not point C 1, but point A 1 or point B 1 and would get that A 1 and B 1 are the bases of the corresponding heights of triangle ABC.
It follows from this that the required triangle, the smallest perimeter, inscribed in a given acute-angled triangle ABC is a triangle whose vertices are the bases of the heights 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 the triangle ABC around the vertex C at an angle of 60 °, Fig. 7. We get a triangle A'B'C. Take an arbitrary point O in triangle ABC. When turning, it will move to some point O '. The OO'C triangle 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 polyline takes the smallest value if the points A, O, O ', B' lie on one straight line. If O is the Torricelli point, then it is. Indeed, ∟AOC = 120 °, ∟COO "= 60 °. Consequently, the points A, O, O 'lie on one straight line. Similarly, ∟CO'O = 60 °, ∟CO" B "= 120 °. Therefore, points O, O ', B' lie on one straight line, so all points A, O, O ', B' lie on one straight line.
Conclusion
The geometry of a triangle, along with other sections of elementary mathematics, makes it possible to feel the beauty of mathematics in general and can become for someone the beginning of the path to "big science".
Geometry - amazing science... Its history goes back more than one millennium, but each meeting with it is able to bestow and enrich (both the student and the teacher) with an exciting novelty of a small discovery, an amazing joy of creativity. Indeed, any problem in elementary geometry is, in essence, a theorem, and its solution is a modest (and sometimes huge) mathematical victory.
Historically, geometry began with a triangle, therefore, for two and a half millennia, the triangle has been a symbol of geometry. School geometry can only become interesting and meaningful, only then can it become geometry proper when a deep and comprehensive study of the triangle appears in it. Surprisingly, the triangle, despite its apparent simplicity, is an inexhaustible object of study - no one, even in our time, dares to say that he has studied and knows all the properties of the triangle.
In this paper, the properties of bisectors, medians, perpendiculars and heights of a triangle were considered, the number of remarkable points and lines of a triangle was expanded, and theorems were formulated and proved. A number of problems on the application of these theorems have been solved.
The presented material can be used both in basic lessons and in optional classes, as well as in preparation for centralized testing and olympiads in mathematics.
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