Location of three points on a straight line. The relative position of a line and a point

The article talks about the concept of a straight line on a plane. Let's look at the basic terms and their designations. Let's work with the relative position of a line and a point and two lines on a plane. Let's talk about axioms. Finally, we will discuss methods and methods for defining a straight line on a plane.

Straight line on a plane - concept

First you need to have a clear understanding of what a plane is. Any surface of something can be classified as a plane, only it differs from objects in its boundlessness. If we imagine that the plane is a table, then in our case it will not have boundaries, but will be infinitely huge.

If you touch the table with a pencil, a mark will remain, which can be called a “dot”. Thus, we get an idea of ​​a point on the plane.

Let's consider the concept of a straight line on a plane. If you draw a straight line on a sheet, it will appear on it with a limited length. We did not get the entire straight line, but only part of it, since in fact it does not have an end, just like a plane. Therefore, the depiction of lines and planes in the notebook is formal.

We have an axiom:

Definition 1

Points can be marked on each straight line and in each plane.

Points are designated in both large and small Latin letters. For example, A and D or a and d.

For a point and a line, only two possible locations are known: a point on a line, in other words, that the line passes through it, or a point not on a line, that is, the line does not pass through it.

To indicate whether a point belongs to a plane or a point to a line, use the sign “∈”. If the condition is given that the point A lies on the line a, then it has the following form of writing A ∈ a. In the case when point A does not belong, then another entry A ∉ a.

Fair judgment:

Definition 2

Through any two points located in any plane, there is a single straight line that passes through them.

This statement is considered an akisoma, and therefore does not require proof. If you consider this yourself, you can see that with two existing points there is only one option for connecting them. If we have two given points A and B, then the line passing through them can be called by these letters, for example, line A B. Consider the figure below.

A straight line located on a plane has a large number of points. This is where the axiom comes from:

Definition 3

If two points of a line lie in a plane, then all other points of this line belong to the plane.

The set of points located between two given points is called a straight segment. It has a beginning and an end. A two-letter designation has been introduced.

If it is given that points A and P are the ends of a segment, then its designation will take the form P A or A P. Since the designations of a segment and a line coincide, it is recommended to add or finish the words “segment”, “straight line”.

A shorthand notation for membership involves the use of the signs ∈ and ∉. In order to fix the location of a segment relative to a given line, use ⊂. If the condition states that the segment A P belongs to the line b, then the entry will look like this: A P ⊂ b.

The case where three points simultaneously belong to one line occurs. This is true when one point lies between two others. This statement is considered to be an axiom. If points A, B, C are given, which belong to the same line, and point B lies between A and C, it follows that all given points lie on the same line, since they lie on both sides of point B.

A point divides a line into two parts, called rays. We have an axiom:

Definition 4

Any point O located on a straight line divides it into two rays, with any two points of one ray lying on one side of the ray relative to point O, and others on the other side of the ray.

The arrangement of straight lines on a plane can take the form of two states.

Definition 5

coincide.

This opportunity arises when straight lines have common points. Based on the axiom written above, we have that a straight line passes through two points and only one. This means that when 2 straight lines pass through given 2 points, they coincide.

Definition 6

Two straight lines on a plane can cross.

This case shows that there is one common point, which is called the intersection of lines. The intersection is designated by the sign ∩. If there is a notation form a ∩ b = M, then it follows that the given lines a and b intersect at the point M.

When straight lines intersect, we deal with the resulting angle. The section where straight lines intersect on a plane to form an angle of 90 degrees, that is, a right angle, is subject to separate consideration. Then the lines are called perpendicular. The form of writing two perpendicular lines is as follows: a ⊥ b, which means that line a is perpendicular to line b.

Definition 7

Two straight lines on a plane can be parallel.

Only if two given lines have no common intersections, and therefore no points, are they parallel. A notation is used that can be written for a given parallelism of lines a and b: a ∥ b.

A straight line on a plane is considered together with vectors. Particular importance is attached to zero vectors that lie on a given line or on any of the parallel lines; they are called direction vectors of a line. Consider the figure below.

Non-zero vectors located on lines perpendicular to a given one are otherwise called normal line vectors. There is a detailed description in the article of the normal vector of a line on a plane. Consider the picture below.

If there are 3 lines on a plane, their location can be very different. There are several options for their location: the intersection of all, parallelism, or the presence of different intersection points. The figure shows the perpendicular intersection of two lines relative to one.

To do this, we present the necessary factors that prove their relative position:

• if two lines are parallel to a third, then they are all parallel;
• if two lines are perpendicular to a third, then these two lines are parallel;
• If on a plane a straight line intersects one parallel line, then it will also intersect another.

Let's look at this in the pictures.

A straight line on a plane can be specified in several ways. It all depends on the conditions of the problem and on what its solution will be based. This knowledge can help for the practical arrangement of straight lines.

Definition 8

The straight line is defined using the specified two points located in the plane.

From the considered axiom it follows that through two points it is possible to draw a straight line and, moreover, only one single one. When a rectangular coordinate system specifies the coordinates of two divergent points, then it is possible to fix the equation of a straight line passing through the two given points. Consider a drawing where we have a line passing through two points.

Definition 9

A straight line can be defined through a point and a line to which it is parallel.

This method exists because through a point it is possible to draw a straight line parallel to a given one, and only one. The proof is already known from a school course in geometry.

If a line is given relative to a Cartesian coordinate system, then it is possible to construct an equation for a line passing through a given point parallel to a given line. Let's consider the principle of defining a straight line on a plane.

Definition 10

The straight line is specified through the specified point and the direction vector.

When a straight line is specified in a rectangular coordinate system, it is possible to compose canonical and parametric equations on the plane. Let us consider in the figure the location of the straight line in the presence of a direction vector.

The fourth point in specifying a straight line makes sense when the point through which it should be drawn and the straight line perpendicular to it are indicated. From the axiom we have:

Definition 11

Through a given point located on a plane, only one straight line will pass, perpendicular to the given one.

And the last point related to specifying a line on a plane is given the specified point through which the line passes, and in the presence of a normal vector of the line. Given the known coordinates of a point located on a given line and the coordinates of the normal vector, it is possible to write down the general equation of the line.

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Straight line on a plane - necessary information.

In this article we will dwell in detail on one of the primary concepts of geometry - the concept of a straight line on a plane. First, let's define the basic terms and designations. Next, we will discuss the relative position of a line and a point, as well as two lines on a plane, and present the necessary axioms. In conclusion, we will consider ways to define a straight line on a plane and provide graphic illustrations.

Page navigation.

• A straight line on a plane is a concept.
• The relative position of a straight line and a point.
• The relative position of lines on a plane.
• Methods for defining a straight line on a plane.

A straight line on a plane is a concept.

Before giving the concept of a straight line on a plane, you should clearly understand what a plane is. Concept of a plane allows you to get, for example, a flat surface on a table or a wall at home. It should, however, be borne in mind that the dimensions of the table are limited, and the plane extends beyond these boundaries to infinity (as if we had an arbitrarily large table).

If we take a well-sharpened pencil and touch its tip to the surface of the “table”, we will get an image of a point. This is how we get representation of a point on a plane.

Now you can move on to the concept of a straight line on a plane.

Place a sheet of clean paper on the table surface (on a plane). In order to draw a straight line, we need to take a ruler and draw a line with a pencil as far as the size of the ruler and sheet of paper we are using allows us to do. It should be noted that in this way we will only get part of the line. We can only imagine an entire straight line extending into infinity.

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The relative position of a straight line and a point.

We should start with the axiom: on every straight line and in every plane there are points.

Points are usually denoted in capital Latin letters, for example, points A And F. In turn, straight lines are denoted in small Latin letters, for example, straight lines a And d.

Possible two options for the relative position of a line and a point on a plane: either the point lies on the line (in this case it is also said that the line passes through the point), or the point does not lie on the line (it is also said that the point does not belong to the line or the line does not pass through the point).

To indicate that a point belongs to a certain line, the symbol “ ” is used. For example, if the point A lies on a straight line A, then we can write . If the point A does not belong to the line A, then write it down.

The following statement is true: there is only one straight line passing through any two points.

This statement is an axiom and should be accepted as a fact. In addition, this is quite obvious: we mark two points on paper, apply a ruler to them and draw a straight line. A straight line passing through two given points (for example, through points A And IN), can be denoted by these two letters (in our case, the straight line AB or VA).

It should be understood that on a straight line defined on a plane there are infinitely many different points, and all these points lie in the same plane. This statement is established by the axiom: if two points of a line lie in a certain plane, then all points of this line lie in this plane.

The set of all points located between two points given on a line, together with these points, is called straight line segment or simply segment. The points limiting the segment are called the ends of the segment. A segment is denoted by two letters corresponding to the endpoints of the segment. For example, let the points A And IN are the ends of a segment, then this segment can be denoted AB or VA. Please note that this designation for a segment coincides with the designation for a straight line. To avoid confusion, we recommend adding the word “segment” or “straight” to the designation.

To briefly record whether a certain point belongs or does not belong to a certain segment, the same symbols and are used. To show that a certain segment lies or does not lie on a line, use the symbols and, respectively. For example, if the segment AB belongs to the line A, can be briefly written down.

We should also dwell on the case when three different points belong to the same line. In this case, one, and only one point, lies between the other two. This statement is another axiom. Let the points A, IN And WITH lie on the same straight line, and the point IN lies between the points A And WITH. Then we can say that the points A And WITH are on opposite sides of the point IN. It can also be said that the points IN And WITH the points lie on one side A, and the points A And IN lie on one side of the point WITH.

To complete the picture, we note that any point on a line divides this line into two parts - two beam. For this case the axiom is given: arbitrary point ABOUT, belonging to a line, divides this line into two rays, and any two points of one ray lie on the same side of the point ABOUT, and any two points of different rays are on opposite sides of the point ABOUT.

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Test questions for §1.

Basic properties of the simplest geometric figures.

Question 1. Give examples of geometric shapes.
Answer: Examples of geometric shapes: triangle, square, circle.

Question 2. Name the basic geometric shapes on a plane.
Answer: The main geometric figures on a plane are a point and a straight line.

Question 3. How are points and lines designated?
Answer: Points are designated in capital Latin letters: A, B, C, D, …. Direct lines are designated by lowercase Latin letters: a, b, c, d, ….
A straight line can be denoted by two points lying on it. For example, line a in Figure 4 can be labeled AC, and line b can be labeled BC. Fig.4

Question 4. Formulate the basic properties of membership of points and lines.
Answer: Whatever the line, there are points that belong to this line and points that do not belong to it.
Through any two points you can draw a straight line, and only one.
Question 5. Explain what a line segment with ends at these points is.
Answer: A segment is a part of a line that consists of all points of this line lying between two given points. These points are called the ends of the segment. A segment is indicated by indicating its ends. When they say or write: “segment AB,” they mean a segment with ends at points A and B.

Question 6. State the basic property of the location of points on a straight line.
Answer: Of the three points on a line, one and only one lies between the other two.

Question 7. Formulate the basic properties of measuring segments.
Answer: Each segment has a certain length greater than zero. The length of a segment is equal to the sum of the lengths of the parts into which it is divided by any of its points.
Question 8. What is the distance between two given points?
Answer: The length of segment AB is called the distance between points A and B.
Question 9. What properties does the division of a plane into two half-planes have?
Answer: Partitioning a plane into two half-planes has the following property. If the ends of a segment belong to the same half-plane, then the segment does not intersect the line. If the ends of a segment belong to different half-planes, then the segment intersects a line.

Question 10. Formulate the basic property of the location of points relative to a straight line on a plane.
Answer: A straight line divides a plane into two half-planes.

Question 11. What is a half-line or ray? Which half-lines are called complementary?
Answer: A half-line or ray is a part of a line that consists of all points of this line lying on one side of a given point. This point is called the starting point of the half-line. Different half-lines of the same line that have a common starting point are called complementary.

Question 12. How are half-lines designated?
Answer: Semi-straight lines, like straight lines, are designated in lowercase Latin letters.
Question 13. What figure is called an angle?
Answer: An angle is a figure that consists of a point - the vertex of the angle - and two different half-lines emanating from this point - the sides of the angle.

Question 14. How is the angle indicated?
Answer: An angle is designated either by indicating its vertex, or by indicating its sides, or by indicating three points: the vertex and two points on the sides of the angle. The word "angle" is sometimes replaced by a sign.
Question 15. Which angle is called a straight angle?
Answer: If the sides of an angle are additional half-lines of one straight line, then the angle is called developed.

Question 16. Explain what the expression means: “A half-line passes between the sides of an angle.”
Answer: We will say that a ray passes between the sides of a given angle if it comes from its vertex and intersects some segment with ends on the sides of the angle.
Question 17. In what units are angles measured and with what tool? Explain how the measurement is carried out.
Answer: Angles are measured in degrees using a protractor.

Question 18. Formulate the basic properties of measuring angles.
Answer: Each angle has a certain degree measure greater than zero. The rotated angle is 180°. The degree measure of an angle is equal to the sum of the degree measures of the angles into which it is divided by any ray passing between its sides.
Question 19. Formulate the basic properties of laying out segments and angles.
Answer: On any half-line from its starting point, you can plot a segment of a given length, and only one. From any half-line, into a given half-plane, you can put an angle with a given degree measure less than 180°, and only one.
Question 20. What is a triangle?
Answer: A triangle is a figure that consists of three points that do not lie on the same line, and three segments connecting these points in pairs. The points are called the vertices of the triangle, and the segments are called the sides.

Question 21. What is the angle of a triangle at a given vertex?
Answer: The angle of a triangle ABC at vertex A is the angle formed by half lines AB and AC. The angles of the triangle at vertices B and C are also determined.

Question 22. Which segments are called equal?
Answer: Segments are called equal if their lengths are equal.
Question 23. What angles are called equal?
Answer: Angles are called equal if their degree measures are equal.
Question 24. Which triangles are called equal?
Answer: Triangles are called congruent if their corresponding sides are equal and their corresponding angles are equal. In this case, the corresponding angles must lie opposite the corresponding sides.
Question 25. How are the corresponding sides and angles marked in the figure for equal triangles?
Answer: In the drawing, equal segments are usually marked with one, two or three lines, and equal angles with one, two or three arcs.

Question 26. Using Figure 23, explain the existence of a triangle equal to this one.
Answer: Let us have a triangle ABC and a ray a (Fig. 23, a). Let's move triangle ABC so that its vertex A is aligned with the beginning of ray a, vertex B is on ray a, and vertex C is in a given half-plane relative to ray a and its extension. We will denote the vertices of our triangle in this new position as A 1, B 1, C 1 (Fig. 23, b).
Triangle A 1 B 1 C 1 is equal to triangle ABC.
Question 27. Which lines are called parallel? What sign is used to indicate parallel lines?
Answer: Two lines are called parallel if they do not intersect. To indicate parallelism of lines, the sign || is used. a||b.

Question 28. State the main property of parallel lines.
Answer: Through a point not lying on a given line, it is possible to draw on the plane at most one straight line parallel to the given one.
Question 29. Give an example of the theorem.
Answer: If a line that does not pass through any of the vertices of a triangle intersects one of its sides, then it intersects only one of the other two sides.

Test questions for §2. Adjacent and vertical angles.

Question 1. What angles are called adjacent?
Answer: Two angles are called adjacent if they have one side in common, and the other sides of these angles are complementary half-lines.
In Figure 31, the angles (a 1 b) and (a 2 b) are adjacent. They have side b in common, and sides a 1 and a 2 are additional half-lines.

Question 2. Prove that the sum of adjacent angles is 180°.
Answer: Theorem 2.1. The sum of adjacent angles is 180°.
Proof. Let angle (a 1 b) and angle (a 2 b) be given adjacent angles (see Fig. 31). Ray b passes between sides a 1 and a 2 of a straight angle. Therefore, the sum of the angles (a 1 b) and (a 2 b) is equal to the unfolded angle, i.e. 180°. Q.E.D.

Question 3. Prove that if two angles are equal, then their adjacent angles are also equal.
Answer:

From the theorem 2.1 It follows that if two angles are equal, then their adjacent angles are equal.
Let's say the angles (a 1 b) and (c 1 d) are equal. We need to prove that the angles (a 2 b) and (c 2 d) are also equal. The sum of adjacent angles is 180°. It follows from this that a 1 b + a 2 b = 180° and c 1 d + c 2 d = 180°. Hence, a 2 b = 180° - a 1 b and c 2 d = 180° - c 1 d. Since the angles (a 1 b) and (c 1 d) are equal, we get that a 2 b = 180° - a 1 b = c 2 d. By the property of transitivity of the equal sign it follows that a 2 b = c 2 d. Q.E.D.

Question 4. What angle is called right (acute, obtuse)?
Answer: An angle equal to 90° is called a right angle. An angle less than 90° is called an acute angle. An angle greater than 90° and less than 180° is called obtuse.

Question 5. Prove that an angle adjacent to a right angle is a right angle.
Answer: From the theorem on the sum of adjacent angles it follows that an angle adjacent to a right angle is a right angle: x + 90° = 180°, x = 180° - 90°, x = 90°.

Question 6. What angles are called vertical?
Answer: Two angles are called vertical if the sides of one angle are complementary half-lines of the sides of the other.

Question 7. Prove that the vertical angles are equal.
Answer: Theorem 2.2. Vertical angles are equal.
Proof. Let (a 1 b 1) and (a 2 b 2) be the given vertical angles (Fig. 34). Angle (a 1 b 2) is adjacent to angle (a 1 b 1) and to angle (a 2 b 2). From here, using the theorem on the sum of adjacent angles, we conclude that each of the angles (a 1 b 1) and (a 2 b 2) complements the angle (a 1 b 2) to 180°, i.e. angles (a 1 b 1) and (a 2 b 2) are equal. Q.E.D.

Question 8. Prove that if, when two lines intersect, one of the angles is right, then the other three angles are also right.
Answer: Suppose lines AB and CD intersect each other at point O. Suppose angle AOD is 90°. Since the sum of adjacent angles is 180°, we get that AOC = 180° - AOD = 180° - 90° = 90°. Angle COB is vertical to angle AOD, so they are equal. That is, angle COB = 90°. Angle COA is vertical to angle BOD, so they are equal. That is, angle BOD = 90°. Thus, all angles are equal to 90°, that is, they are all right angles. Q.E.D.

Question 9. Which lines are called perpendicular? What sign is used to indicate perpendicularity of lines?
Answer: Two lines are called perpendicular if they intersect at right angles. The perpendicularity of lines is indicated by the sign ⊥.. Recording a⊥b reads: “Line a is perpendicular to line b.”

Question 10. Prove that through any point on a line you can draw a line perpendicular to it, and only one.
Answer: Theorem 2.3. Through each line you can draw a line perpendicular to it, and only one.
Proof. Let a be a given line and A a given point on it. Let us denote by a 1 one of the half-lines of the straight line a with the starting point A (Fig. 38). Let us subtract an angle (a 1 b 1) equal to 90° from the half-line a 1. Then the straight line containing the ray b 1 will be perpendicular to the straight line a.

Let us assume that there is another line, also passing through point A and perpendicular to line a. Let us denote by c 1 the half-line of this line lying in the same half-plane with the ray b 1 . Angles (a 1 b 1) and (a 1 c 1), each equal to 90°, are laid out in one half-plane from the half-line a 1. But from the half-line a 1 only one angle equal to 90° can be put into a given half-plane. Therefore, there cannot be another line passing through point A and perpendicular to line a. The theorem has been proven.

Question 11. What is perpendicular to a line?
Answer: A perpendicular to a given line is a segment of a line perpendicular to a given line, which has one of its ends at their intersection point. This end of the segment is called basis perpendicular.

Question 12. Explain what proof by contradiction consists of.
Answer: The proof method we used in Theorem 2.3 is called proof by contradiction. This method of proof consists of first making an assumption opposite to what the theorem states. Then, by reasoning, relying on axioms and proven theorems, we come to a conclusion that contradicts either the conditions of the theorem, or one of the axioms, or a previously proven theorem. On this basis, we conclude that our assumption was incorrect, and therefore the statement of the theorem is true.

Question 13. What is the bisector of an angle?
Answer: The bisector of an angle is a ray that emanates from the vertex of the angle, passes between its sides and divides the angle in half.

Test questions for § 3.Signs of equality of triangles.

Question 1. Prove the first sign that triangles are equal. What axioms are used in the proof of Theorem 3.1?
Answer: The first sign of equality of triangles is Theorem 3.1. (a sign of equality of triangles on two sides and the angle between them). If two sides and the angle between them of one triangle are equal, respectively, to two sides and the angle between them of another triangle, then such triangles are congruent.
Proof. Let triangles ABC and A 1 B 1 C 1 have angle A= angle A 1, AB=A 1 B 1, AC=A 1 C 1 (Fig. 44). Rice. 44.
Let us prove that the triangles are congruent. Let A 1 B 2 C 2 be a triangle equal to triangle ABC, with vertex B 2 on ray A 1 B 1 and vertex C 2 in the same half-plane relative to straight line A 1 B 1, where vertex C 1 lies (Fig. 45, a ). Since A 1 B 1 = A 1 B 2, the vertex B 2 coincides with the vertex B 1 (Fig. 45, b). Since angle B 1 A 1 C 1 = angle B 2 A 1 C 2, then ray A 1 C 2 coincides with ray A 1 C 1 (Fig. 45, c). Since A 1 C 1 = A 1 C 2, the vertex C 2 coincides with the vertex C 1 (Fig. 45, d).
So, triangle A 1 B 1 C 1 coincides with triangle A 1 B 2 C 2, which means it is equal to triangle ABC. The theorem has been proven.
At the beginning of the proof, draw a triangle A 1 B 2 C 2 equal to triangle ABC with vertex B 2 on ray A 1 B 1 and vertex C 2 in the same half-plane relative to straight line A 1 B 1, where vertex C 1 lies (Fig. 45, a ). Such a triangle exists according to the axiom about the existence of a triangle equal to the given one (whatever the triangle is, there is a triangle equal to it in a given location relative to a given half-line).
Then the coincidence of vertices B 1 and B 2 is asserted on the basis that A 1 B 1 = A 1 B 2. The axiom of delaying segments is used here (on any half-line from its starting point, you can set aside a segment of a given length, and only one).
Next, the coincidence of rays A 1 C 2 and A 1 C 1 is asserted on the basis that ∠B 2 A 1 C 1 = ∠B 2 A 1 C 2 . Here we use the axiom of laying off angles (from any half-line into a given half-plane you can put off an angle with a given degree measure less than 180°, and only one). Finally, the coincidence of vertices C 1 and C 2 is confirmed, since A 1 C 1 = A 2 C 2. Here again the axiom of delaying segments is used (on any half-line from its starting point you can set aside a segment of a given length, and only one).
So, in the proof of Theorem 3.1, the axioms for setting aside segments and angles and the axiom about the existence of a triangle equal to the given one are used.

Question 2. Formulate and prove the second criterion for the equality of triangles.
Answer: The second criterion for the equality of triangles is Theorem 3.2 (a criterion for the equality of triangles along a side and adjacent angles). If a side and its adjacent angles of one triangle are equal, respectively, to a side and its adjacent angles of another triangle, then such triangles are congruent.
Proof. Let ABC and A 1 B 1 C 1 be two triangles in which AB = A 1 B 1 , angle A = angle A 1 and angle B = angle B 1 (Fig. 47). Let us prove that the triangles are congruent.
Let A 1 B 2 C 2 be a triangle equal to triangle ABC, with vertex B 2 on ray A 1 B 1 and vertex C 2 in the same half-plane relative to the line A 1 B 1 where vertex C 1 lies.
Since A 1 B 2 =A 1 B 1, then the vertex B 2 coincides with the vertex B 1. Since angle B 1 A 1 C 2 = angle B 1 A 1 C 1 and angle A 1 B 1 C 2 = angle A 1 B 1 C 1, then ray A 1 C 2 coincides with ray A 1 C 1, and ray B 1 C 2 coincides with the ray B 1 C 1. It follows that vertex C 2 coincides with vertex C 1 .
So, triangle A 1 B 1 C 1 coincides with triangle A 1 B 2 C 2, and therefore is equal to triangle ABC. The theorem has been proven.

Question 3. Which triangle is called isosceles? Which sides of an isosceles triangle are called lateral sides? Which side is called the base?
Answer: A triangle is called isosceles if its two sides are equal. These equal sides are called the sides, and the third side is called the base of the triangle.

Question 4. Prove that in an isosceles triangle the base angles are equal.
Answer: Theorem 3.3 (property of the angles of an isosceles triangle). In an isosceles triangle, the base angles are equal.
Proof. Let ABC be an isosceles triangle with base AB (Fig. 48). Let's prove that his angle A = angle B.

Triangle CAB is equal to triangle CBA according to the first criterion of equality of triangles. Indeed, CA= CB, CB= CA, angle C= angle C. From the equality of triangles it follows that angle A= angle B. The theorem is proven.

Question 5. Which triangle is called equilateral?
Answer: A triangle in which all sides are equal is called equilateral.

Question 6. Prove that if two angles in a triangle are equal, then it is isosceles.
Answer: Theorem 3.4 (test of an isosceles triangle). If two angles in a triangle are equal, then it is isosceles.
Proof. Let ABC be a triangle in which angle A = angle B (Fig. 50). Let us prove that it is isosceles with base AB.

Triangle ABC is equal to triangle BAC according to the second criterion for the equality of triangles. Indeed, AB=BA, angle B= angle A, angle A= angle B. From the equality of triangles it follows that AC= BC. So, by definition, triangle ABC is isosceles. The theorem has been proven.

Question 7. Explain what the converse theorem is. Give an example. Is the converse true for every theorem?
Answer: Theorem 3.4 is called the converse of Theorem 3.3. The conclusion of Theorem 3.3 is a condition of Theorem 3.4. And the condition of Theorem 3.3 is the conclusion of Theorem 3.4. Not every theorem has a converse, that is, if a given theorem is true, then the converse theorem may be false. Let us explain this using the example of the theorem on vertical angles. This theorem can be formulated as follows: if two angles are vertical, then they are equal. The converse theorem would be: if two angles are equal, then they are vertical. And this, of course, is not true. Two equal angles do not have to be vertical.

Question 8. What is the height of a triangle?
Answer:Height of a triangle dropped from a given vertex is called a perpendicular drawn from this vertex to a straight line that contains the opposite side of the triangle (Fig. 51, a-b).

Question 9. What is the bisector of a triangle?
Answer:Bisector of a triangle drawn from a given vertex is called the segment of the bisector of the angle of the triangle, connecting this vertex with a point on the opposite side (Fig. 52, a).

Question 10. What is the median of a triangle?
Answer:Median of a triangle drawn from a given vertex is called a segment connecting this vertex with the middle of the opposite side of the triangle (Fig. 52, b).

Question 11. Prove that in an isosceles triangle the median drawn to the base is the bisector and the altitude.
Answer: Theorem 3.5 (property of the median of an isosceles triangle). In an isosceles triangle, the median drawn to the base is the bisector and the altitude.
Proof. Let ABC be a given isosceles triangle with base AB and CD the median drawn to the base (Fig. 53). Triangles CAD and CBD are equal according to the first sign of equality of triangles. (Their sides AC and BC are equal because triangle ABC is isosceles. Angles CAD and CBD are equal as the base angles of an isosceles triangle. Sides AD and BD are equal because D is the midpoint of segment AB.)
From the equality of triangles follows the equality of angles: angle ACD = angle BCD, angle ADC = angle BDC. Since angles ACD and BCD are equal, CD is a bisector. Since angles ADC and BDC are adjacent and equal, they are right angles, so CD is the altitude of the triangle.

Question 12. Prove the third criterion for the equality of triangles.
Answer: The third criterion for the equality of triangles is Theorem 3.6 (test for the equality of triangles on three sides). If three sides of one triangle are equal to three sides of another triangle, then such triangles are congruent.

Proof. Let ABC and A 1 B 1 C 1 be two triangles in which AB = A 1 B 1, AC = A 1 C 1, BC = B 1 C 1 (Fig. 55). You need to prove that the triangles are congruent.
Let's say the triangles are not equal. Then their angle A does not = angle A 1, angle B does not = angle B 1, angle C does not = angle C 1. Otherwise they would be equal on the first basis.
Let A 1 B 1 C 2 be a triangle equal to triangle ABC, whose vertex C 2 lies in the same half-plane with vertex C 1 relative to the straight line A 1 B 1 (see Fig. 55).
Let D be the midpoint of the segment C 1 C 2 . Triangles A 1 C 1 C 2 and B 1 C 1 C 2 are isosceles with a common base C 1 C 2. Therefore their medians A 1 D and B 1 D are heights. This means that lines A 1 D and B 1 D are perpendicular to line C 1 C 2. Lines A 1 D and B 1 D do not coincide, since points A 1, B 1, D do not lie on the same line. But through point D of straight line C 1 C 2 only one straight line perpendicular to it can be drawn. We have arrived at a contradiction. The theorem has been proven.

Test questions for §4.Sum of triangle angles.

Question 1. Prove that two lines parallel to a third are parallel.
Answer: Theorem 4.1. Two lines parallel to a third are parallel.
Proof. Let lines a and b be parallel to line c. Let us assume that a and b are not parallel (Fig. 69). Then they do not intersect at some point C. This means that two lines pass through point C parallel to line c. But this is impossible, since through a point not lying on a given line, you can draw at most one straight line parallel to the given one. The theorem has been proven.

Question 2. Explain which angles are called one-sided interior angles. What angles are called internal cross-lying ones?
Answer: Pairs of angles that are formed when lines AB and CD intersect with secant AC have special names.
If points B and D lie in the same half-plane relative to straight line AC, then angles BAC and DCA are called one-sided internal angles (Fig. 71, a).
If points B and D lie in different half-planes relative to straight line AC, then angles BAC and DCA are called internal cross-lying angles (Fig. 71, b).

Question 3. Prove that if the interior angles of one pair are equal, then the interior angles of the other pair are also equal, and the sum of the interior angles of each pair is 180°.
Answer: The secant AC forms with the straight lines AB and CD two pairs of internal one-sided angles and two pairs of internal cross-lying angles. The internal cross-lying angles of one pair, for example, angle 1 and corner 2, are adjacent to the internal cross-lying angles of another pair: angle 3 and angle 4 (Fig. 72). Rice. 72

Therefore, if the interior angles of one pair are congruent, then the interior angles of the other pair are also equal.
A pair of internal cross-lying angles, for example angle 1 and angle 2, and a pair of internal one-sided angles, for example angle 2 and angle 3, have one angle in common - angle 2, and two other angles are adjacent: angle 1 and angle 3.
Therefore, if internal crosswise angles are equal, then the sum of the internal angles is 180°. And vice versa: if the sum of internal intersecting angles is equal to 180°, then intersecting internal angles are equal. Q.E.D.

Question 4. Prove a test for parallel lines.
Answer: Theorem 4.2 (test for parallel lines). If internal crosswise angles are equal or the sum of internal one-sided angles is equal to 180°, then the lines are parallel.
Proof. Let the straight lines a and b form equal internal crosswise angles with the secant AB (Fig. 73, a). Let's say that lines a and b are not parallel, which means they intersect at some point C (Fig. 73, b). Rice. 73

The secant AB divides the plane into two half-planes. Point C lies in one of them. Let's construct a triangle BAC 1, equal to triangle ABC, with vertex C 1 in another half-plane. By condition, internal crosswise angles for parallel a, b and the secant AB are equal. Since the corresponding angles of triangles ABC and BAC 1 with vertices A and B are equal, they coincide with the internal angles lying crosswise. This means that line AC 1 coincides with line a, and line BC 1 coincides with line b. It turns out that two different straight lines a and b pass through points C and C 1. And this is impossible. This means that lines a and b are parallel.
If the lines a and b and the transversal AB have the sum of the internal one-sided angles equal to 180°, then, as we know, the internal angles lying crosswise are equal. This means, according to what was proven above, lines a and b are parallel. The theorem has been proven.

Question 5. Explain which angles are called corresponding angles. Prove that if internal crosswise angles are equal, then the corresponding angles are also equal, and vice versa.

Answer: If for a pair of internal crosswise angles one angle is replaced by a vertical one, then we get a pair of angles that are called the corresponding angles of these lines with a transversal. Which is what needed to be explained.
From the equality of interior angles lying crosswise follows the equality of the corresponding angles, and vice versa. Let's say we have two parallel lines (since by condition, internal angles lying across each other are equal) and a transversal, which form angles 1, 2, 3. Angles 1 and 2 are equal as internal angles lying across each other. And angles 2 and 3 are equal as vertical. We get: ∠1 = ∠2 and ∠2 = ∠3. By the property of transitivity of the equal sign it follows that ∠1 = ∠3.

3. The converse statement is proved in a similar way.
From this we get the sign that straight lines are parallel at the corresponding angles. Namely: straight lines are parallel if the corresponding angles are equal. Q.E.D.

Question 6. Prove that through a point not lying on a given line you can draw a line parallel to it. How many lines parallel to a given line can be drawn through a point not lying on this line?

Answer: Problem (8). Given a line AB and a point C that does not lie on this line. Prove that through point C you can draw a line parallel to line AB.
Solution. Line AC divides the plane into two half-planes (Fig. 75). Point B lies in one of them. Let us add angle ACD from the half-line CA to another half-plane, equal to angle CAB. Then the lines AB and CD will be parallel. In fact, for these lines and the secant AC, the interior angles BAC and DCA lie crosswise. And since they are equal, the lines AB and CD are parallel. Q.E.D. Comparing the statement of problem 8 and axiom IX (the main property of parallel lines), we come to an important conclusion: through a point not lying on a given line, it is possible to draw a line parallel to it, and only one.

Question 7. Prove that if two lines are intersected by a third line, then the intersecting interior angles are equal, and the sum of the interior one-sided angles is 180°.

Answer: Theorem 4.3(the converse of Theorem 4.2). If two parallel lines intersect with a third line, then the intersecting internal angles are equal, and the sum of the internal one-sided angles is 180°.
Proof. Let a and b be parallel lines and c be a line intersecting them at points A and B. Let us draw a line a 1 through point A so that the internal crosswise angles formed by the transversal c with the lines a 1 and b are equal (Fig. 76). According to the principle of parallelism of lines, lines a 1 and b are parallel. And since only one line passes through point A, parallel to line b, then line a coincides with line a 1. This means that the internal crosswise angles formed by a transversal with parallel lines a and b are equal. The theorem has been proven.

Question 8. Prove that two lines perpendicular to a third are parallel. If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
Answer: From Theorem 4.2 it follows that two lines perpendicular to a third are parallel.
Suppose that any two lines are perpendicular to a third line. This means that these lines intersect with the third line at an angle equal to 90°. From the property of angles formed when parallel lines intersect with a transversal, it follows that if a line is perpendicular to one of the parallel lines, then it is also perpendicular to the other.

Question 9. Prove that the sum of the angles of a triangle is 180°.

Answer: Theorem 4.4. The sum of the angles of a triangle is 180°.
Proof. Let ABC be the given triangle. Let us draw a line through vertex B parallel to line AC. Let's mark point D on it so that points A and D lie on opposite sides of straight line BC (Fig. 78). Angles DBC and ACB are congruent as internal cross-lying ones formed by the transversal BC with parallel lines AC and BD. Therefore, the sum of the angles of a triangle at vertices B and C is equal to angle ABD.
And the sum of all three angles of a triangle is equal to the sum of angles ABD and BAC. Since these are one-sided interior angles for parallel AC and BD and secant AB, their sum is 180°. The theorem has been proven.

Question 10. Prove that any triangle has at least two acute angles.
Answer: Indeed, let us assume that the triangle has only one acute angle or no acute angles at all. Then this triangle has two angles, each of which is at least 90°. The sum of these two angles is no less than 180°. But this is impossible, since the sum of all the angles of a triangle is 180°. Q.E.D.

Question 11. What is an exterior angle of a triangle?
Answer: The external angle of a triangle at a given vertex is the angle adjacent to the angle of the triangle at this vertex (Fig. 79).

Question 12. Prove that an exterior angle of a triangle is equal to the sum of two interior angles that are not adjacent to it.

Answer: Theorem 4.5. An exterior angle of a triangle is equal to the sum of two interior angles that are not adjacent to it.
Proof. Let ABC be the given triangle (Fig. 80). By the theorem on the sum of angles of a triangle, ∠A + ∠B + ∠C = 180°. It follows that ∠A + ∠B = 180° - ∠C. On the right side of this equality is the degree measure of the external angle of the triangle at vertex C. The theorem is proven.

Question 13. Prove that an exterior angle of a triangle is greater than any interior angle not adjacent to it.
Answer: From Theorem 4.5 it follows that the exterior angle of a triangle is greater than any interior angle not adjacent to it.

Question 14. Which triangle is called a right triangle?
Answer: A triangle is called right-angled if it has a right angle.

Question 15. What is the sum of the acute angles of a right triangle?
Answer: Since the sum of the angles of a triangle is 180°, a right triangle has only one right angle. The other two angles of a right triangle are acute. The sum of the acute angles of a right triangle is 180° - 90° = 90°.

Question 16. Which side of a right triangle is called the hypotenuse? Which sides are called legs?

Answer: The side of a right triangle opposite the right angle is called the hypotenuse, the other two sides are called legs (Fig. 82).

Question 17. Formulate a test for the equality of right triangles along the hypotenuse and leg.

Answer: If the hypotenuse and leg of one right triangle are respectively equal to the hypotenuse and leg of another triangle, then such triangles are congruent.

Question 18. Prove that from any point not lying on a given line, you can drop a perpendicular to this line, and only one.

Answer: Theorem 4.6. From any point not lying on a given line, you can drop a perpendicular to this line, and only one.
Proof. Let a be a given line and A a point not lying on it (Fig. 85). Let us draw a perpendicular line through some point of line a. Now let’s draw a line b parallel to it through point A. It will be perpendicular to line a, since line a, being perpendicular to one of the parallel lines, is also perpendicular to the other. The segment AB of line b is the perpendicular drawn from point A to line a.
Let us prove the uniqueness of the perpendicular AB. Let's say there is another perpendicular AC. Then triangle ABC will have two right angles. And this, as we know, is impossible. The theorem has been proven.

Question 19. What is the distance from a point to a line called?
Answer: The length of a perpendicular drawn from a given point to a straight line is called the distance from the point to the straight line.

Question 20. Explain what the distance between parallel lines is.
Answer: The distance between parallel lines is the distance from any point on one line to another line.

Test questions for §5. Geometric constructions.

Question 1. What is a circle, center of a circle, radius?
Answer: A circle is a figure that consists of all points of the plane that are equidistant from a given point. This point is called the center of the circle. The distance from the points of a circle to its center is called the radius. The radius is also called any segment connecting a point on a circle with its center.

Question 2. What is a chord of a circle? Which chord is called diameter?
Answer: A segment connecting two points on a circle is called a chord. The chord passing through the center is called the diameter.

Question 3. What circle is called the circumscribed circle of a triangle?
Answer: A circle is called circumscribed about a triangle if it passes through all its vertices.

Question 4. Prove that the center of a circle circumscribed about a triangle lies at the intersection of the perpendicular bisectors to the sides of the triangle.
Answer: Theorem 5.1. The center of a circle circumscribed about a triangle is the point of intersection of perpendiculars to the sides of the triangle drawn through the midpoints of these sides.
Proof. Let ABC be the given triangle and O be the center of the circle circumscribed around it (Fig. 93). Triangle AOC is isosceles: its sides OA and OC are equal as radii. The median OD of this triangle is also its height. Therefore, the center of the circle lies on a line perpendicular to side AC and passing through its midpoint. In the same way, it is proved that the center of the circle lies at perpendiculars to the other two sides of the triangle. The theorem has been proven.

Question 5. Which line is called tangent to a circle?
Answer: A straight line passing through a point on a circle perpendicular to the radius drawn to this point is called a tangent. In this case, this point on the circle is called the point of tangency.

Question 6. What does it mean: the circles touch at a given point?
Answer: Two circles having a common point are said to touch at this point if they have a common tangent at this point (Fig. 97).

Question 7. Which contact of the circles is called external and which is called internal?
Answer: The tangency of the circles is called internal if the centers of the circles lie on one side of their common tangent (Fig. 97, a). The tangent of the circles is called external if the centers of the circles lie on opposite sides of their common tangent (Fig. 97, b).

Question 8. What circle is called inscribed in a triangle?
Answer: A circle is said to be inscribed in a triangle if it touches all its sides.

Question 9. Prove that the center of a circle inscribed in a triangle lies at the intersection of its bisectors.
Answer: Theorem 5.2. The center of a circle inscribed in a triangle is the intersection point of its bisectors.
Proof. Let ABC be the given triangle, O the center of the circle inscribed in it, D, E and F the points of contact of the circle with the sides (Fig. 98). Right triangles AOD and AOE are equal in hypotenuse and leg. They have a common hypotenuse AO, and the legs OD and OE are equal as radii. From the equality of triangles it follows that the angles OAD and OAE are equal. This means that point O lies on the bisector of the triangle drawn from vertex A. It is proved in the same way that point O lies on the other two bisectors of the triangle. The theorem has been proven.

Question 10. Explain how to construct a triangle using three sides.
Answer: Problem 5.1. Construct a triangle with given sides a, b, c (Fig. 99, a).
Solution. Using a ruler, draw an arbitrary straight line and mark an arbitrary point B on it (Fig. 99, b). Using a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. Now, with a compass opening equal to c, we describe a circle from center B, and with a compass opening equal to b, we describe a circle from center C. Let A be the intersection point of these circles. Let's draw segments AB and AC. Triangle ABC has sides equal to a, b, c. Which is what needed to be explained.

Question 11. Explain how to plot an angle equal to a given angle from a given half-line into a given half-plane.
Answer: Problem 5.2. Subtract from a given half-line into a given half-plane an angle equal to a given angle.
Solution. Let's draw an arbitrary circle with its center at vertex A of a given angle (Fig. 100, a). Let B and C be the points of intersection of the circle with the sides of the angle. With radius AB we draw a circle with the center at point O - the starting point of this half-line (Fig. 100, b). Let us denote the point of intersection of this circle with this half-line as B 1 . Let us describe a circle with center B 1 and radius BC. Point C 1 of intersection of the constructed circles in the indicated half-plane lies on the side of the desired angle. To prove it, it is enough to note that triangles ABC and OB 1 C 1 are congruent as triangles with respectively equal sides. Angles A and O are the corresponding angles of these triangles. Which is what needed to be explained.

Question 12. Explain how to divide this angle in half.
Answer: Problem 5.3. Construct the bisector of the given angle.
Solution. From vertex A of a given angle, as from the center, we describe a circle of arbitrary radius (Fig. 101). Let B and C be the points of its intersection with the sides of the angle. From points B and C we describe circles with the same radius. Let D be their intersection point, different from A. Draw a half-line AD. Ray AD is a bisector because it bisects angle BAC. This follows from the equality of triangles ABD and ACD, whose angles DAB and DAC are corresponding. Which is what needed to be explained.

Question 13. Explain how to divide a line segment in half.
Answer: Problem 5.4. Divide the segment in half.
Solution. Let AB be the given segment (Fig. 102). From points A and B with radius AB we describe circles. Let C and C 1 be the intersection points of these circles. They lie in different half-planes relative to straight line AB. Segment CC 1 intersects line AB at some point O. This point is the midpoint of segment AB. Indeed, triangles CAC 1 and CBC 1 are equal according to the third criterion for the equality of triangles. This implies that the angles ACO and BCO are equal. Triangles ACO and BCO are equal according to the first criterion of equality of triangles. The sides AO and BO of these triangles are corresponding and therefore they are equal. Thus, O is the midpoint of segment AB. Which is what needed to be explained.

Question 14. Explain how to draw a line perpendicular to a given line through a given point.
Answer: Problem 5.5. Through a given point O draw a line perpendicular to a given line a.
Solution. There are two possible cases:
1) point O lies on line a;
2) point O does not lie on line a.
Let's consider the first case (Fig. 103).
From point O we draw a circle of arbitrary radius. It intersects line a at two points: A and B. From points A and B we draw circles of radius AB. Let C be their intersection point. The required line passes through points O and C.
The perpendicularity of the lines OC and AB follows from the equality of the angles at the vertex O of the triangles ACO and BCO. These triangles are wounds according to the third sign of equality of triangles.
Let's consider the second case (Fig. 104).
From point O we draw a circle intersecting line a. From points A and B we draw circles with the same radius. Let O1 be the point of their intersection, lying in a half-plane different from the one in which point O lies. The desired straight line passes through points O and O1. Let's prove it.
Let us denote by C the point of intersection of lines AB and OO1. Triangles AOB and AO1B are equal according to the third criterion. Therefore, angle OAC is equal to angle O1AC. And then the triangles OAC and O1AC are equal according to the first criterion. This means that their angles ACO and ACO1 are equal. And since they are adjacent, they are straight. Thus, OC is a perpendicular dropped from point O to straight line a. Which is what needed to be explained.

Question 15. What is the locus of points equidistant from two given points?
Answer: Theorem 5.3. The locus of points equidistant from two given points is a straight line perpendicular to the segment connecting these points and passing through its midpoint.
Proof. Let A and B be given points, a be a straight line passing through the midpoint O of segment AB perpendicular to it (Fig. 105). We must prove that:
1) Each point of line a is equidistant from points A and B;
2) Each point D of the plane, equidistant from points A and B, lies on line a.
The fact that each point C of line a is at the same distance from points A and B follows from the equality of triangles AOC and BOC. These triangles have right angles at vertex O, side OC is common, and AO = OB, since O is the midpoint of segment AB.
Let us now show that every point D of the plane, equidistant from points A and B, lies on line a. Consider triangle ADB. It is isosceles, since AD=BD. In it, DO is the median. According to the property of an isosceles triangle, the median drawn to the base is the height. This means that point D lies on line a. The theorem has been proven.