Segment midpoint of a segment construction of the midpoint of a segment. How to build the middle of a segment: school knowledge
Circle
A circle is a geometric figure consisting of all points on a plane that are at a given distance from a given point.
This point is called the center of the circle, and the given distance is called the radius of the circle.
The radius is a line segment that connects the center of the circle to any point on the circle. It follows from the definition that an infinite number of radii can be drawn and they all have the same length.
A segment that connects two points on a circle is called chord.
If the chord passes through the center of the circle, then it is called diameter circles.
Diameter is the longest chord.
An infinite number of diameters can also be drawn in a circle.
If you connect two points of a circle not by a segment, but by a curve passing along the circle itself, then the part of the circle between the two points is called arc .
If you mark two points on the circle, then you get two arcs. Therefore, for the name of the arc, three Latin letters are used, which can be both small and large.
In the picture above, we can name: arc \ (BDH \), arc \ (ACG \) and others.
The picture below shows: arc \ (AxB \) and arc \ (AyB \).
The part of the plane bounded by a circle is called a circle.
Building tasks
In tasks where it is necessary to perform constructions, are used compass and ruler.
It is very important to remember that in these tasks the ruler is not used as a measuring tool, but solely only to draw a straight line, ray or segment through two given points, that is, to draw a straight line. A compass is used to draw a circle or circular arc.
Consider the five main constructions in which we use the mentioned actions - building a straight line and a circle:
1. On a given ray from its beginning, lay a segment equal to the given one.
2. Construction of an angle equal to the given one.
3. Construction of the angle bisector.
4. Construction of perpendicular lines.
5. Construction of the midpoint of the segment.
1. On a given ray from its beginning, postpone a segment equal to the given.
Watch the video.
It is clear that in this way we got a segment equal to the given one. According to the definition of a circle, it consists of points located at a given distance (radius) from a certain point (center of the circle).
If the center is the starting point of the ray \ (C \), the radius is the given segment \ (AB \), then the intersection point of the circle and the ray \ (D \) is the desired endpoint of the segment \ (CD \), equal to the given segment \ (AB \).
2. Plotting an angle equal to a given one.
Watch the video.
Let us prove that the constructed angle \ (ECD \) is the desired angle equal to the given angle \ (AOB \).
If we have constructed a circle with center \ (C \) - the starting point of the ray and the same radius as the circle with center \ (O \), then \ (CD \) \ (= \) \ (OB \).
Conducted a beam \ (CE \). Obviously \ (OA \) \ (= \) \ (CE \).
This means that the triangles \ (AOB \) and \ (ECD \) are equal by the third sign of equality of triangles, they have equal angles, including the angle \ (ECD \) is equal to the angle \ (AOB \).
3. Plotting the bisector of an angle.
Watch the video.
To prove that \ (OC \) actually divides the angle \ (AOB \) in half, it suffices to consider the triangles \ (AOC \) and \ (BOC \).
Lesson number 2Topic : Draws the midpoint of the line. Drawing perpendicular lines
Goals:
educational: teach students to use a compass and a ruler to divide a segment in half; to form the skills and abilities of building perpendicular lines;
developing:
educational:
During the classes:
1. Updating the main theoretical concepts(5 minutes).
First, you can conduct a frontal survey on the following questions:
1. Give the definition of a circle. What are the center, radius, chord and diameter of a circle?
2. Which triangle is called isosceles? What are its sides called?
3. Which triangle is called equilateral?
4. What is called the midpoint of a segment?
Further suggestthe task: using a compass and a ruler, construct a bisector extending from the apex of an isosceles triangle. List its properties.
2. Learning new material ( practical work) (20 minutes)
Draw the midpoint of a line segment
When studying new material, table number 4 of Appendix 4 is used, according to which students make up a story about how to divide this segment in half. After that, the corresponding constructions are performed in the notebooks.
A task ... Build the middle of a given segment (explains the teacher with the help of students).
Solution ... Let AB be a given segment. Let's construct two circles with centers A and B of radius AB (Fig. 5).
Fig. 5.
They intersect at points P and Q. Draw a straight line PQ. The point O of the intersection of this line with the segment AB and the desired midpoint of the segment AB.
Indeed, triangles APQ and BPQ are equal on three sides, so 1 = 2.
Therefore, the segment RO is the bisector of the isosceles triangle ARV, and hence the median, i.e. point O - midpoint of segment AB.
Drawing perpendicular lines
It should be noted here that two cases are possible:
1. The point belongs to a straight line;
2. The point does not belong to a straight line.
After repetition, the teacher formulates the problem and explains the construction for the first case, while table number 3 of Appendix 4 can be used.
When considering the second case, students use Table 4 to build and prove independently.
A task ... Across this point Draw a straight line perpendicular to this straight line a (explains the teacher, after discussion with the students).
Solution ... Two cases are possible:
1) point O lies on the straight line a;
2) the point O does not lie on the straight line a.
Let's consider the first case (Fig. 6). From point O draw a circle with an arbitrary radius. It intersects the straight line a at two points: A and B. From points A and B we draw circles with a radius of AB. Let C be the point of their intersection. The desired line passes through points O and C.
Fig. 6.
The perpendicularity of the straight lines OS and AB follows from the equality of the angles at the vertex O of the triangles ACO and BCO.
These triangles are equal in the third criterion of equality of triangles.
Consider the construction and proof for the second case (Fig. 7).
Fig. 7.
From point O we draw a circle intersecting line a. Let A and B be the points of its intersection with the line a. Draw circles from points A and B with the same radius. Let O be the point of their intersection lying in a half-plane different from the one in which the point O lies. The sought line passes through the points O and O. Let us prove this. Let us denote by C the point of intersection of lines AB and OO. Triangles AOB and AOB are equal in the third attribute. Therefore, the OAC angle equal to the angle SLA. And then the triangles ОАС and ОАС are equal in the first sign. This means that their angles ACO and ACO are equal. And since they are adjacent, they are straight. Thus, OS is a perpendicular dropped from point O to line a.
3. Fastening (10 min)
A task. Build right triangle on his legs.
The student solves this problem at the blackboard, having previously analyzed it.
1. Analysis.
Fig. 8.
Let's make a drawing - a sketch (Fig. 8).
CA = b, CB = a, ACB =
2. Construction (fig. 9).
Fig. 9.
1. On the straight line, mark the point C and postpone the segment CB = a.
2. Construct a straight line passing through point C perpendicular to CB.
3. Set aside the segment CA = b
4. ABC - the desired one.
3. Proof.
In ABC, BC = a, CA = b, BDAC, therefore, the angle of BCA is 90є. So the triangle ABC is the desired one.
Also, for practicing skills and abilities, you can use tasks №154 (a, b) (see Appendix 1).
4. Summing up (3min)
1. During the lesson, we solved two building problems. Studied:
a) build the middle of the segment;
b) build perpendicular straight lines.
2. In the course of solving these problems:
a) remembered the signs of equality of triangles;
b) used the construction of circles, segments, rays.
5. Home (2min): # 153 (see Appendix 1).
Lesson number 3
Topic: Solving building problems
Goals:
educational: working out the skills and abilities of performing elementary constructions with the help of a compass and a ruler;
developing: development of spatial thinking, attention;
educational: education of diligence and accuracy.
During the classes:
1. Verification homework(10 min)
Check the completion of task # 153.
The check can be organized as follows: there are three students at the blackboard, they must build a straight line passing through point A perpendicular to the straight line a (Fig. 10).
Fig. 10.
The class at this time can complete the task: the triangle ABC is given. build the height AD. After completing the assignment, each step of the construction should be commented and justified.
Independent work is carried out in three ways and has a controlling character
1. Divide the segment into 4 equal parts.
2. Dan ABC. Construct the bisector VK.
3. The angle AOB is given. Construct the angle for which the OB ray is the bisector.
Geometry, 7-9, L.S. Atanasyan
Lesson topic: Constructing the midpoint of a line segment. Construction of perpendicular lines.
Goals: teach students to use a compass and a ruler to divide a segment in half; teach to build perpendicular straight lines.
Equipment: drawing tools; interactive whiteboard.
Educational task: teach to divide the segment in half; teach to build perpendicular straight lines.
I... Motivational and indicative part.
Organizing time : checking homework.
Knowledge update(test) (printouts of the test are given)
1) Write down the definition of the circle;
2) The diameter of the circle = this ...
a) a straight line passing through the center of the circle;
b) a chord passing through the center of the circle;
3) The center of the circle is ..
a) the middle of the circle;
b) the point where the leg of the compass is placed;
c) a point equidistant from all points of the circle;
4) What is the name of the segment connecting the center of the circle with any point on the circle?
a) circumference;
b) the radius of the circle;
c) half the diameter of the circle;
5) What triangle is called isosceles? (write definition)
6) What are the sides of an isosceles triangle called?
7) List the properties of an isosceles triangle?
8) Which triangle is called equilateral?
9) What is called the middle of a segment?
10) Using a compass and a ruler, draw an angle of 30 degrees.
Motivation: The art of building geometric shapes with the help of a compass and a ruler was in high degree developed in Ancient Greece... One of the most difficult construction tasks, which even then they knew how to perform, was the construction of a circle tangent to these three circles. This problem is called the Apollo problem - after the Greek geometer Apollonius of Perga (c. 200 BC)
However, ancient geometers did not manage to complete some constructions using only compasses and a ruler, and constructions made with other tools were not considered geometric. These problems include the so-called three famous classical problems of antiquity: squaring a circle, trisecting an angle, and doubling a cube.
These three problems have attracted the attention of eminent mathematicians for centuries, and only in the middle of the 19th century their undecidability was proved, i.e. the impossibility of these constructions only with the help of a compass and a ruler. These results were obtained by means not of geometry, but of algebra, which once again emphasized the unity of mathematics.
Today we will look at two new building problems.
So, let's write down the topic of the lesson: “ Creates the midpoint of a line segment. Construction of perpendicular lines ".(slide 1)
II... Content part.
One of the two tasks for building our today's lesson is the task of building the middle of a given segment. (slide 2)
Let's solve it:
Given: Construct: midpoint of segment AB.
Sharpening
1) let AB - this segment;
2) build two circles with centers A and B; They intersect at points P and Q.
3) draw a straight line PQ;
4) the point O of the intersection of this line with the segment AB is the desired midpoint of the segment AB.
Let us prove this: we connect the points A, B, P, Q by segments. 
(on three sides), therefore. Consequently, the segment RO is the bisector of the isosceles triangle ARV, and hence the median, that is, the point O is the midpoint of the segment AB. (slide 3)
So, we have solved the first problem.
Let's move on to problem number 2 of our topic.
A task: given a straight line and a point on it. Construct a straight line passing through a given point and perpendicular to a given straight line. (Slide 4)
Ano: Construct: a straight line passing through a given point and perpendicular to a given straight line.
Building
1) a straight line a is given and a given point M belongs to this straight line;
2) on the rays of the straight line a, outgoing from the point M, we postpone equal segments MA and MB;
3) construct two circles with centers A and B of radius AB. They intersect at two points: P and Q.
4) draw a straight line through the point M and one of these points, for example, the line MP.
Let us prove that the straight line МР а: since the median МР of the isosceles triangle PAB is also the height, then МР а. (slide 5)
So, we have solved two construction problems, let's fix it on the solution of the next problem.
Anchoring:(slide 6)
Task: Construct a right-angled triangle along its legs.
Ano: Construct: rectangular triangle.
Building
Teacher: Using the above solved building problems, where can we start?
Students: build a perpendicular to a line
Teacher: right, only here we will build a perpendicular to the beam
So let's write:
1) draw a ray O;
2) we build a perpendicular to the ray O
3) the point of intersection of the rays will be denoted by point A;
4) set aside from point A the leg equal to b, and the point of intersection of b and ray O will be point C.
5) put a leg equal to a upwards from point A, put point B.
6) connect points B and C, this is the hypotenuse;
7) the ABC triangle is the desired one.
III... Reflexively evaluative part.
Teacher: During the lesson, we solved two of the main building problems.
What have we learned?
Students: build the midpoint of a line segment, build perpendicular lines.
Teacher: in the course of solving these problems, what knowledge we learned earlier did we remember and use?
Pupils: We remembered the signs of equality of triangles; used the construction of circles, segments, rays.
Let's write down the assignment at home: No. 154 and paragraph 4 to repeat what we have gone through and studied again. Prepare for a little independent work. (Slide 7)
Building tasks
The main drawing tools with which geometric constructions are made are a ruler and a compass.
Using a compass, circles are drawn with a given center and a given radius. In particular, using a compass on the ray from its beginning, you can postpone a segment equal to the given one.
Problem 1 Using this figure, explain how to build a midpoint perpendicular to a given segment. AB .
Solution.
Describe circles with centers at points BUT and IN and a radius greater than half AB... We denote the points of their intersection lying on opposite sides of the straight line AB, across WITH 1 and C 2. Points WITH 1 and C 2 are equally distant from the ends of the segment AB... Therefore, they belong to the middle perpendicular to this segment. So straight C 1 WITH 2 will be the required perpendicular to the middle.
Task 2 Using this figure, explain how to build the middle of a given segment. AB .
We build the midpoint perpendicular to this segment and find its intersection point with this segment. It will be the desired middle.
Using this figure, explain how through this point O belonging to this line a, draw a straight line b perpendicular to the straight line a .
Centered at a point O draw a circle and denote A 1 , A 2 its points of intersection with a straight line a... Let's draw the middle perpendicular b to the segment A 1 A 2 . Straight b is the desired one.
Task 4. Using this figure, explain how from this point O not belonging to this line a , lower the perpendicular to this line.
Solution.
On a straight line a mark some point A... If the segment OA perpendicular a, then it is the desired one.
Otherwise, draw a circle centered at the point O and radius OA... She will cross the line a at the point A and some point B... As OA = OB then point O belongs to the middle perpendicular to the segment AB... The desired perpendicular will lie on the middle perpendicular to the segment AB... After that, you can use the construction of the median perpendicular.
Problem 5. Using this figure, explain how to build the bisector of a given angle.
Solution.
Describe a circle centered at the vertex O of a given corner, intersecting the sides of the corner in points BUT and IN... Then with the same solution of a compass with centers at points BUT and IN we describe two more circles. Their intersection point other than O, denote WITH. Let's pass the beam OS... Triangles SLA and OBC are equal in the third criterion of equality of triangles. Consequently, AOC = BOC, i.e. Ray OS is the required bisector.
Problem 6. Using this figure, explain how to build an angle equal to the given one, one of the sides of which coincides with the given ray.
Task 7.
Build a triangleABC on both sidesAB = c , AC = b and the corner between them.
Set aside segments on the sides of this corner AB = c and AC = b... Let's draw a segment BC ABC .
Task 8.
ABC on two given legs BC = a , AC = b .
C... Set aside segments on its sides BC = a and AC = b... Let's draw a segment AB... We get the required triangle ABC .
Problem 9.
Construct a right triangle ABC on the leg AC = b and hypotenuse AB = c .
Let's build a right angle with the vertex C... On one side of it, we postpone the segment AC = b... Centered at the point A c... We denote B its intersection point with the second side of the given corner. Let's draw a segment AB... We get the required triangle ABC... Note that a solution exists if c > b .
Problem 10.
Construct a right triangle ABC on hypotenuse AB = c and sharp corner A .
On one side of this corner, we postpone the segment AB = c From point B omit the perpendicular BC to the other side of the corner. We get the required triangle ABC .
Problem 11.
Build a triangle ABC on this side AB = c and two given angles A and B .
On a straight line, we postpone the segment AB = c... With the vertices at the ends of this segment to one side of the straight line, we postpone these angles A and B . We denote C their intersection point. The resulting triangle ABC will be the desired one. Note that a solution exists if the sides of the corners intersect.
Problem 12. Build a triangleABC on these three sidesAB = c , AC = b , AC = b .
On a straight line, we postpone the segment AB = c... Centered at a point A draw a circular arc of radius b... Centered at a point B draw a circular arc of radius a... We denote C their intersection point. Let's connect it with segments with points A and B... The resulting triangle will be the desired one. Note that a solution exists if a – b < c < a + b .
Putivskaya Julia
The task of constructing the midpoint of a segment defined by its ends using various tools
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THE PROBLEM OF CONSTRUCTING THE MIDDLE OF A INTERCEPT DEFINED BY ITS ENDS USING DIFFERENT TOOLS
Putivskaya Yulia Olegovna,
9th grade student of the Municipal Educational Institution "Zinaidinskaya
basic secondary school "
The first construction tasks arose in ancient times. They arose from the economic needs of man. Already the ancient architects and land surveyors had to solve the simplest construction problems associated with their profession. Solutions to the simplest geometric construction problems, which helped people in their economic life, were formulated in the form of "practical rules" based on clear considerations. It was these tasks that were the basis for the emergence of visual geometry, which found a fairly wide development among the ancient peoples of Egypt, Babylon, India, etc. Geometric constructions attracted the attention of ancient Greek mathematicians as early as the 6th-5th centuries. BC. The first Greek scientist to solve geometric construction problems was Thales of Miletus (624-547 BC). Almost all major Greek geometers were engaged in them: Pythagoras (6th century BC) and his students, Hippocrates (5th century BC), Euclid, Archimedes, Apollonius (3rd century BC), Pappus (III century AD) and many others.
Mathematicians from the school of Pythagoras have already managed to cope with such a relatively difficult task as building regular pentagon... In the V century. BC NS. the famous classical problems of squaring a circle, doubling a cube, and trisecting an angle arose (see Chapter VII). These problems, which, as it turned out later, were not solvable with the help of a compass and a ruler, for many centuries aroused the keen interest of various researchers. In the IV century. BC NS. Greek thinkers developed that general scheme solving a geometric construction problem (analysis - construction - proof - research), which we still use today.
The whole history of geometry and some other branches of mathematics is closely connected with the development of the theory of geometric constructions. The most important axioms of geometry, formulated by the founder of the scientific geometric system, Euclid, around 300. BC e., clearly show what role geometric constructions played in the formation of geometry. "From every point to every point, a straight line can be drawn", "A bounded straight line can be continued continuously", "A circle can be described from every center and every solution" - these postulates of Euclid clearly indicate the basic position of constructive methods in the geometry of the ancients.
Studying geometry in the 7th grade, I got acquainted with solving construction problems using a compass and a ruler. A question arose before me: "Is it possible to solve these simple tasks with the help of any other tools, and do they exist?" Turning this question to a mathematics teacher, I received the following book in return: Geometric constructions on a plane, B.I. Argunov and M.B. Balk - Uchpedgiz, 1955.
After reading many of its chapters, I learned that the most used geometric construction tools are: ruler (one-sided), compasses, two-sided ruler (with parallel edges), right angle and some others.
For constructive geometry it is necessary to have accurate and for mathematical purposes full description this or that instrument. This description is given in the form of axioms. These axioms in abstract mathematical form express the properties of real drawing tools that are used for geometric constructions.
I will formulate the corresponding axioms.
Ruler axiom.
The ruler allows you to perform the following geometric constructions:
a) build a segment connecting two constructed points;
b) build a straight line passing through two constructed points;
c) build a ray outgoing from the constructed point and passing through another constructed point.
Axiom of the compass.
The compass allows you to perform the following geometric constructions:
a) construct a circle if the center of the circle and the ends of a segment equal to the radius of the circle are constructed;
b) construct any of the two additional circular arcs if the center of the circle and the ends of the arc are constructed.
Axiom of a two-sided ruler.
The double-sided ruler allows you to:
a) perform any of the constructions listed in the ruler axiom;
b) in each of the half-planes defined by the constructed straight line, construct a straight line parallel to this straight line and passing from it at a distance h, where h - a segment fixed for a given ruler
(ruler width);
c) if two points A and B are constructed, then establish whether ABis greater than some fixed segment (the width of the ruler), and if AB> h, then construct two pairs of parallel lines passing through the points A, respectively and in and spaced apart from one another h.
Right angle axiom.
A right angle allows you to perform the following geometric constructions:
a) all constructions that can be performed with a one-sided ruler;
b) draw a straight line through a given point of the plane, perpendicular to some constructed straight line;
c) if the segment ABand some figure Ф, then establish whether figure Ф contains a point from which the segment is visible at a right angle, and if such a point exists, then build such a point.
The construction task consists in the fact that it is required to construct a certain figure in advance with the indicated tools, if some other figure is given and some ratios between the elements of the desired figure and the elements of this figure are indicated.
Each figure that satisfies the conditions of the problem is called a solution this task.
To find a solution to a construction problem means to reduce it to a finite number of basic constructions, that is, to indicate a finite sequence of basic constructions, after the execution of which the sought figure will already be considered constructed by virtue of the accepted axioms of constructive geometry. The list of admissible basic constructions, and, consequently, the course of solving the problem essentially depends on what tools are used for constructions.
As an example, I will consider the following problem:
Construct the midpoint of the segment defined by its ends A and B.
Let's find a solution to this problem using various tools.
1. Compass and ruler
(construction is studied in grade 7, p. 23 Examples of construction problems)
Let AB - this segment. Let's build two circles with centers A and B of radius AB ( ). They intersect at points P and Q. Draw the line PQ. Point O the intersection of this line with the segment AB is the desired midpoint of the segment AB.
Indeed, triangles APQ and BPQ equal on three sides, therefore 1 = 2 ().
Therefore, the segment RO is the bisector of an isosceles triangle ARV, and hence the median, i.e., the point O - the middle of the segment AB.
2. With a compass (picture below the description)
We build sequentially:
- circle with center B, radius VA;
- circle with center A and radius AB;
- common point C - the point of intersection of circles with center B with radius BA and with center A with radius AB;
- a circle with a center with a radius of CA;
- common point D - the point of intersection of circles with center B with radius BA and with center A with radius AC, different from point A;
- circle with center D with radius DB;
7) common point E - the point of intersection of circles with center B with radius BA and with center D with radius DB, other than point C;
Note that points A, B and Eare located on one straight line, with AE = 2AB. We build further:
8) a circle with a center E and a radius of EA;
9) a circle with center A with radius AB intersects a circle with center E with radius EA at points M and N
10) a circle with center M and radius MA;
11) a circle with center N and radius NA;
12) common point X - the point of intersection of circles with center M with radius MA and with center N with radius NA, different from A.
It is easy to see that point Xlocated on a straight line.
In addition, triangle AMX is similar to triangle AEM, since they are isosceles and have a common angle MAEat the base. Therefore, AX: AM = AM: AE or AX: AB = AB: 2AB, so
AX = AB and, therefore, the point X is the desired one.
3. Double-sided ruler (picture below description).
We build sequentially:
1) straight AB;
2) straight line a, parallel AB and passing at a distance h from her
(h is the width of the ruler);
3) straight line b parallel to a, spaced from it at a distance h and different from the straight line AB;
4) point C on line b;
5) direct AC and BC;
6) points D - the point of intersection of lines a and AC and E - the point of intersection of lines a and BC;
7) straight AE and BD;
8) point P - the point of intersection of straight lines AE and BD;
9) Direct CP;
10) point X - point of intersection of straight lines CP and AB.
Since DE - the middle line of the triangle ACB, then AE and BD is its median, and hence CP is the median, so that point X the desired one.
4. Right angle (picture below description)
1) We build a straight line AB;
2) we carry out straight AA "and BB ", perpendicular
straight AB;
3) choose on AA "arbitrary point C, different from A;
4) through point C we draw SS "AC.
Next, we build sequentially:
5) point D - intersection point of lines CC "and BB";
6) direct AD and BC;
7) point P - the point of intersection of lines AD and BC;
Point X is the desired one.
Thus, considered different ways solving the same building problem using different tools.
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