“Midline of the trapezoid” - The middle line of the trapezoid. A. MN is the midline of the trapezoid ABCD. In a triangle you can construct... middle lines. The middle line of the triangle has the property... MN = ? AB. Determination of the midline of the trapezoid. Theorem about the midline of a trapezoid. D. Continue the sentence: MN || AB.
“Equation of an ellipse” - Authors: Gololobova O. 9th grade Negrova O. 9th grade Dolgova K. 9th grade. Definition of an ellipse. How are the properties of the ellipse related to the properties of other “remarkable” curves? 2. We derived the canonical equation of the ellipse. Progress of the study. Results of the study: 4. Determine the main parameters of the ellipse: Purpose: Study the main parameters of the ellipse. 3. Constructed an ellipse.
“Thales’ Theorem” - It is believed that Thales was the first to study the movement of the Sun across the celestial sphere. Thales's theorem. A geometric theorem is named after Thales. Let us draw a line EF through point B2, parallel to line A1A3. Astronomy. Geometry. According to the property of a parallelogram, A1A2 = FB2, A2A3 = B2E. Milesian materialist. And since A1A2 = A2A3, then FB2 = B2E. Thales is widely known as a geometer.
“Problems about circle and circle” - 2. Answer: S=25? cm2; C=10? see Problem solving. 1. Circumference and area of a circle.
“Regular polygons geometry” - Around any regular polygon you can describe a circle, and only one. Let us derive a formula for calculating the angle аn of a regular n-gon. Take any three vertices of the polygon A1A2...An, for example A1, A2, A3. Let us now prove the uniqueness of such a circle. Center of a regular polygon. Theorem about the center of a regular polygon. The uniqueness of such a circle follows from the uniqueness of the circle circumscribed about the triangle.
“Motion geometry 9th grade” - Axial. Axial symmetry. Central and Axial symmetry. Theorem. Types of movements. Turn. Overlay. Any movement is an imposition. Axial symmetry Central symmetry Parallel translation Rotation. Parallel transfer. Movements. Central symmetry. Concept of movement. Geometry 9th grade. Central. When moving, a segment is mapped onto a segment.
Mapping the plane onto itself
Definition 1
Mapping the plane onto itself- this is a correspondence between each point of the plane and some point of the same plane, in which each point on the plane will be associated with some point.
Examples of mapping a plane onto itself can be axial symmetry (Fig. 1, a) and central symmetry (Fig. 1, b).
Figure 1. a) axial symmetry; b) central symmetry
Let us now introduce the definition of motion.
Definition 2
The movement of a plane is a mapping of the plane onto itself in which distances are preserved (Fig. 2).
Figure 2. Example of movement
Proof.
Let us be given a segment $MN$. Let, for a given motion of a plane, point $M$ be mapped to point $M_1$ of this plane, and point $N$ be mapped to point $N_1$ of this plane. Let's take an arbitrary point $P$ of the segment $MN$. Let it be mapped to point $\P_1$ of this plane (Fig. 3).
Figure 3. Mapping segment to segment while moving
Since the point $P$ belongs to the segment $MN$, then the equality
Since, by definition of motion, distances are conserved, then
Hence
This means that point $P_1$ lies on the segment $M_1N_1$. Due to the arbitrariness of the choice of point $P_1$, we obtain that the segment $MN$ during movement will be mapped onto the segment $M_1N_1$. The equality of these segments immediately follows from the definition of motion.
The theorem is proven.
Theorem 2
When moving, the triangle is mapped onto an equal triangle.
Proof.
Let us be given a triangle $ABC$. By Theorem 1, the segment $AB$ goes into the segment $A_1B_1$, the segment $AC$ goes into the segment $A_1C_1$, the segment $BC$ goes into the segment $B_1C_1$, and $(AB=A)_1B_1$, $(AC =A)_1C_1$, $(BC=B)_1C_1$. Consequently, according to the third criterion of equality of triangles, the triangle $ABC$ goes into the triangle $A_1B_1C_1$ equal to it.
The theorem is proven.
Similarly, it can be proven that ray is mapped to ray, angle is mapped to its equal angle.
To formulate the next theorem, we first introduce the following definition.
Definition 3
Overlay is called such a movement of the plane that has the following axioms:
Theorem 3
Any movement is an imposition.
Proof.
Consider the motion $g$ of triangle $ABC$. According to Theorem 2, when $g$ moves, the triangle $ABC$ transitions into the triangle $A_1B_1C_1$ equal to it. A-priory equal triangles we find that there is an overlay $f$ mapping points $A,B\ and\ C$ to points $A_1,B_1\ and\ C_1$, respectively. Let us prove that $g$ coincides with $f$.
Let's assume the opposite, that $g$ does not coincide with $f$. Then there is at least one point $M$, which, when $g$ moves, goes to the point $M_1$, and when $f$ is imposed, goes to the point $M_2$. Since the distances are preserved for $f$ and $g$, we have
That is, point $A_1$ is equidistant from points $M_1$ and $M_2$. Similarly, we find that points $B_1\ and\ C_1$ are equidistant from points $M_1$ and $M_2$. This means that points $A_1,B_1\ and\C_1$ lie on a line perpendicular to the segment $M_1M_2$ and passing through its center. This is not possible, since the points $A_1,B_1\ and\C_1$ do not lie on the same line. Therefore, the movement of $g$ coincides with the imposition of $f$.
The theorem is proven.
Example 1
Prove that when moving, an angle is mapped onto an angle equal to it.
Proof.
Let us be given an angle $AOB$. Let, for a given motion, points $A,\ O\ and\ B$ be mapped onto points $A_1,\ O_1\ and\ B_1$. By Theorem 2 we find that the triangle $AOB$ is mapped onto the triangle $A_1O_1B_1$, and these triangles are equal to each other. Therefore, $\angle AOB=\angle A_1O_1B_1$.
The word “movement” is familiar to you. But in geometry it has a special meaning. Which one you will learn about in this chapter. For now, let us note that with the help of movements it is possible to find beautiful solutions many geometric problems. You will find examples of such solutions in this chapter.
Let's imagine that each point of the plane is compared (put into correspondence) with some point of the same plane, and any point of the plane turns out to be associated with some point. Then they say that it is given mapping the plane onto itself.
In fact, we have already encountered mappings of a plane onto itself - let us remember axial symmetry (see paragraph 48). She gives us an example of such a mapping. In fact, let a be the axis of symmetry (Fig. 321). Let's take an arbitrary point M that does not lie on straight line a, and construct a point M 1 symmetrical to it relative to straight line a. To do this, you need to draw a perpendicular MR to straight line a and lay off on straight MR the segment RM 1, equal to the segment MR, as shown in Figure 321. Point M 1 will be the desired one. If the point M lies on the straight line a, then the point M 1 symmetrical to it coincides with the point M. We see that with the help of axial symmetry, each point M of the plane is associated with a point M of the same plane. In this case, any point M 1 turns out to be associated with some point M. This is clear from Figure 321.
Rice. 321
So, axial symmetry is a mapping of the plane onto itself.
Let us now consider the central symmetry of the plane (see paragraph 48). Let O be the center of symmetry. Each point M of the plane is associated with a point M 1, symmetrical to point M relative to point O (Fig. 322). Try to verify for yourself that the central symmetry of the plane is also a mapping of the plane onto itself.
Rice. 322
Axial symmetry has the following important property - is a mapping of the plane onto itself that preserves the distances between points.
Let's explain what this means. Let M and N be any points, and M 1 and N 1 be points symmetric to them relative to straight line a (Fig. 323). From points N and N 1 we draw perpendiculars NP and N 1 P 1 to the line MM 1. Right Triangles MNP and M 1 N 1 P 1 are equal on two legs: MP = M 1 P 1 and NP = N 1 P 1 (explain why these legs are equal). Therefore, the hypotenuses MN and M 1 N 1 are also equal.
Rice. 323
Hence, the distance between points M and N is equal to the distance between their symmetrical points M 1 and N 1. Consider other cases of the location of points M, N and M 1, N 1 yourself and make sure that in these cases MN = M 1 N 1 (Fig. 324). Thus, rotational symmetry is a mapping that preserves the distances between points. Any mapping that has this property is called motion (or translation).
Rice. 324
So, the movement of a plane is a mapping of the plane onto itself, preserving distances.
Why a mapping that preserves distances is called motion (or displacement) can be explained using the example of axial symmetry. It can be represented as a rotation of the plane in space by 180° around the a-axis. Figure 325 shows how this rotation occurs.
Rice. 325
Note that the central symmetry of the plane is also motion(using Figure 326, see this for yourself).
Rice. 326
Let us prove the following theorem:
Theorem
| When moving, the segment is mapped onto the segment. |
Proof
Let, for a given movement of the plane, the ends M and N of the segment MN be mapped to points M 1 and N 1 (Fig. 327). Let us prove that the entire segment MN is mapped onto the segment M 1 N 1 . Let P be an arbitrary point on the segment MN, P 1 be the point to which point P is mapped. Then MP + PN = MN. Since distances are conserved when moving, then
M 1 N 1 = MN, M 1 P 1 = MR and N 1 P 1 = NP. (1)
Rice. 327
From equalities (1) we obtain that M 1 P 1 + P 1 N 1 = M 1 N 1 , and, therefore, the point P 1 lies on the segment M 1 N 1 (if we assume that this is not the case, then the inequality M 1 P 1 +P 1 N 1 > M 1 N 1). So, the points of the segment MN are mapped to the points of the segment M 1 N 1 .
It is also necessary to prove that to each point P 1 of the segment M 1 N 1 some point P of the segment MN is mapped. Let's prove it. Let P 1 be an arbitrary point on the segment M 1 N 1, and point P, for a given movement, is mapped to point P 1. From relations (1) and the equality M 1 N 1 = M 1 P 1 + P 1 N 1 it follows that MR + PN = MN, and, therefore, point P lies on the segment MN. The theorem is proven.
Consequence
In fact, by virtue of the proven theorem, when moving, each side of the triangle is mapped onto an equal segment, therefore the triangle is mapped onto a triangle with, respectively equal sides, i.e. to an equal triangle.
Using the proven theorem, it is not difficult to verify that when moving, a straight line is mapped onto a straight line, a ray into a ray, and an angle into an angle equal to it.
Recall that in our geometry course, the equality of figures is determined using overlaps. We say that the figure Ф is equal to the figure Фп if the figure Ф can be combined by overlapping with the figure Ф 1. The concept of superposition in our course refers to the basic concepts of geometry, so the definition of superposition is not given. By superimposing the figure Φ onto the figure Φ 1, we mean a certain mapping of the figure Φ onto the figure Φ 1. Moreover, we believe that in this case not only the points of the figure Φ, but also any point on the plane are mapped to a certain point on the plane, i.e. overlay is the mapping of a plane onto itself.
However, we do not call every mapping of a plane onto itself an imposition. Impositions are those mappings of the plane onto itself that have properties expressed in axioms (see Appendix 1, axioms 7-13). These axioms allow us to prove all those properties of impositions that we imagine visually and which we use when proving theorems and solving problems. Let us prove, for example, that when superimposed, different points are mapped to different points.
In fact, let us assume that this is not the case, i.e., with some overlap, some two points A and B are mapped to the same point C. Then the figure Ф 1, consisting of points A and B, is equal to the figure Ф 2, consisting of one point C. It follows that Ф 2 = Ф 1 (axiom 12), i.e., with some overlap, the figure Ф 2 is mapped into the figure Ф 1. But this is impossible, since superimposition is a mapping, and with any mapping, point C is associated with only one point on the plane.
From the proven statement it follows that when superimposed, a segment is mapped onto an equal segment. Indeed, let, when superimposed, the ends A and B of the segment AB are mapped to points A 1 and B 1. Then the segment AB is mapped onto the segment A 1 B 1 (axiom 7), and, therefore, the segment AB is equal to the segment A 1 B 1. Because equal segments have equal lengths, then the superposition is a mapping of the plane onto itself, preserving distances, i.e. any overlap is a movement of the plane.
Let us prove that the converse is also true.
Theorem
Proof
Let's consider an arbitrary motion (denote it by the letter g) and prove that it is an imposition. Let's take some triangle ABC. When g moves, it is mapped onto an equal triangle A 1 B 1 C 1 . By the definition of congruent triangles, there is an overlap ƒ, in which points A, B and C are mapped to points A 1, B 1 and C 1, respectively.
Let us prove that the movement of g coincides with the imposition of ƒ. Let's assume that this is not the case. Then on the plane there is at least one such point M, which, when g moves, is mapped to point M„ and when ƒ is applied, to another point M2. Since distances are preserved when mapping ƒ u g, then AM = A 1 M 1, AM = A 1 M 2, therefore A 1 M 1 = A 1 M 2, i.e. point A 1 is equidistant from points M 1 and M 2 (Fig. 328). It is similarly proven that points B 1 and C 1 are equidistant from points M 1 and M 2. It follows that points A 1, B 1 and C 1 lie on the perpendicular bisector to the segment M 1 M 2. But this is impossible, since the vertices of triangle A 1 B 1 C 1 do not lie on the same straight line. Thus, the mappings ƒ u g coincide, i.e. the movement of g is an overlap. The theorem is proven.
Rice. 328
Consequence
1148. Prove that with axial symmetry of the plane:
a) a straight line parallel to the axis of symmetry is mapped onto a straight line parallel to the axis of symmetry;
b) a straight line perpendicular to the axis of symmetry is mapped onto itself.
1149. Prove that with central symmetry of the plane:
a) a straight line that does not pass through the center of symmetry is mapped onto a straight line parallel to it;
b) the line passing through the center of symmetry is mapped onto itself.
1150. Prove that when moving, an angle is mapped onto an angle equal to it.
Let, for a given movement, angle AOB be mapped onto angle A 1 O 1 B 1 , and points A, O, B are mapped to points A 1 , O 1 , B 1 , respectively. Since distances are maintained during movement, then OA = O 1 A 1, OB = O 1 B 1. If the angle AOB is not developed, then the triangles AOB and A 1 O 1 B 1 are equal on three sides, and, therefore, ∠AOB = ∠A 1 O 1 B 1. If angle AOB is reversed, then angle A 1 O 1 B 1 is reversed (prove this), so these angles are equal.
1151. Prove that when moving, parallel lines are mapped onto parallel lines.
1152. Prove that when moving: a) a parallelogram is mapped onto a parallelogram; b) trapezoid is mapped onto trapezoid; c) rhombus is mapped onto rhombus; d) a rectangle is mapped to a rectangle, and a square is mapped to a square.
1153. Prove that when moving, a circle is mapped onto a circle of the same radius.
1154. Prove that a plane mapping in which each point is mapped onto itself is an imposition.
1155. ABC and A 1 B 1 C 1 are arbitrary triangles. Prove that there is at most one motion in which points A, B and C are mapped to points A 1, B 1, C 1.
1156. In triangles ABC and A 1 B 1 C 1 AB = A 1 B 1, AC = A 1 C 1, BC = B 1 C 1. Prove that there is a movement in which points A, B and C are mapped to points A 1, B 1 and C 1, and only one.
According to the conditions of the problem, triangles ABC and A 1 B 1 C 1 are equal on three sides. Consequently, there is an overlap, that is, a movement in which points A, B and C are mapped to points A 1, B 1 and C 1, respectively. This movement is the only movement in which points A, B and C are mapped to points A 1, B 1 and C 1, respectively (problem 1155).
1157. Prove that two parallelograms are equal if the adjacent sides and the angle between them of one parallelogram are respectively equal to the adjacent sides and the angle between them of the other parallelogram.
1158. Given two straight lines a and b. Construct a line onto which line b is mapped with axial symmetry with the a axis.
1159. Given a line a and a quadrilateral ABCD. Construct a figure F onto which this quadrilateral is mapped with axial symmetry with the a axis. What does the F shape represent?
1160 Point O and line b are given. Construct a line onto which line b is mapped with central symmetry with center O.
1161 Point O and triangle ABC are given. Construct a figure F onto which triangle ABC is mapped with central symmetry to center O. What does figure F represent?
1151. Instruction. Prove by contradiction.
1154. Instruction. Use theorem 119.
1155. Instruction. The proof is carried out by contradiction (see the proof of the theorem, paragraph 119).
1157. Instruction. Use problems 1156 and 1051.
1158. Instruction. First, construct images of some two points of line b.
1159. F - quadrilateral.
1160. Instruction. The problem is solved similarly to problem 1158.
1161. F - triangle.
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