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Question 1. Prove that two lines parallel to the 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. Assume that a and b are not parallel (Fig. 69). Then they do not intersect at some point C. Hence, two lines pass through the point C and are parallel to the line c. But this is impossible, since through a point that does not lie on a given line, at most one line parallel to the given line can be drawn. The theorem has been proven.

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

Rice. 71

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

Rice. 72

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

Question 4. Prove the criterion for parallel lines.
Answer. Theorem 4.2 (test for parallel lines). If interior cross-lying angles are equal or the sum of interior one-sided angles is 180°, then the lines are parallel.
Proof. Let the lines a and b form equal internal crosswise lying angles with the secant AB (Fig. 73, a). Suppose the lines a and b are not parallel, which means they intersect at some point C (Fig. 73, b).

Rice. 73

The secant AB splits the plane into two half-planes. Point C lies in one of them. Let's construct triangle BAC 1 , equal to triangle ABC, with vertex C 1 in the other half-plane. By condition, the internal cross-lying angles for parallel a, b and 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 cross-lying angles. Hence, line AC 1 coincides with line a, and line BC 1 coincides with line b. It turns out that two different lines a and b pass through the points C and C 1. And this is impossible. So lines a and b are parallel.
If lines a and b and secant AB have the sum of internal one-sided angles equal to 180°, then, as we know, the internal cross-lying angles are equal. Hence, by what was proved above, the lines a and b are parallel. The theorem has been proven.

Question 5. Explain what angles are called corresponding. Prove that if interior cross-lying angles are equal, then the corresponding angles are also equal, and vice versa.

Answer. If a pair of internal cross-lying angles has one angle replaced by a vertical one, then a pair of angles will be obtained, which are called the corresponding angles of the given lines with a secant. Which is what needed to be explained.
From the equality of internal cross-lying angles follows the equality of the corresponding angles, and vice versa. Let's say we have two parallel lines (because by condition the internal cross-lying angles are equal) and a secant, which form angles 1, 2, 3. Angles 1 and 2 are equal as internal cross-lying. And angles 2 and 3 are equal as vertical. We get: \(\angle\)1 = \(\angle\)2 and \(\angle\)2 = \(\angle\)3. By the property of transitivity of the equal sign, it follows that \(\angle\)1 = \(\angle\)3. The converse assertion is proved similarly.
This results in a sign of parallel lines at the corresponding angles. Namely, 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, it is possible to draw a line parallel to it. How many lines parallel to a given line can be drawn through a point not on this line?

Answer. Problem (8). Given a line AB and a point C not lying on this line. Prove that through point C it is possible to draw a line parallel to line AB.
Solution. The straight line AC divides the plane into two half-planes (Fig. 75). Point B lies in one of them. From the half-line CA, let us plot the angle ACD equal to the angle CAB into the other half-plane. Then lines AB and CD will be parallel. Indeed, for these lines and the secant AC, the angles BAC and DCA are interior crosswise. And since they are equal, 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 that does not lie on a given line, one can draw a line parallel to it, and only one.

Question 7. Prove that if two lines intersect with a third line, then the interior cross-lying angles are equal, and the sum of interior one-sided angles is 180°.

Answer. Theorem 4.3(converse to Theorem 4.2). If two parallel lines intersect with a third line, then the interior cross-lying angles are equal, and the sum of interior one-sided angles is 180°.
Proof. Let a and b be parallel lines and c be the line that intersects them at points A and B. Draw a line a 1 through point A so that the internal cross-lying angles formed by the secant c with lines a 1 and b are equal (Fig. 76).
By the criterion of parallelism of lines, lines a 1 and b are parallel. And since only one line passes through the point A, parallel to the line b, then the line a coincides with the line a 1 .
This means that the internal cross-lying angles formed by the secant 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. It follows from Theorem 4.2 that two lines perpendicular to a third are parallel.
Assume that any two lines are perpendicular to the third line. Hence, these lines intersect with the third line at an angle equal to 90°.
From the property of the angles formed at the intersection of parallel lines by a secant, 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. Draw a line through vertex B parallel to line AC. Mark point D on it so that points A and D lie on opposite sides of the line BC (Fig. 78).
Angles DBC and ACB are equal as internal crosswise, formed by the secant BC with parallel lines AC and BD. Therefore, the sum of the angles of the triangle at the vertices B and C is equal to the angle ABD.
And the sum of all three angles of a triangle is equal to the sum of angles ABD and BAC. Since these angles are internal one-sided 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, suppose that a 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.

Signs of parallelism of two lines

Theorem 1. If at the intersection of two lines of a secant:

diagonally lying angles are equal, or

corresponding angles are equal, or

the sum of one-sided angles is 180°, then

lines are parallel(Fig. 1).

Proof. We restrict ourselves to the proof of case 1.

Suppose that at the intersection of lines a and b by a secant AB across the lying angles are equal. For example, ∠ 4 = ∠ 6. Let us prove that a || b.

Assume that lines a and b are not parallel. Then they intersect at some point M and, consequently, one of the angles 4 or 6 will be the external angle of the triangle ABM. Let, for definiteness, ∠ 4 be the outer corner of the triangle ABM, and ∠ 6 be the inner one. It follows from the theorem on the external angle of a triangle that ∠ 4 is greater than ∠ 6, and this contradicts the condition, which means that the lines a and 6 cannot intersect, therefore they are parallel.

Corollary 1. Two distinct lines in a plane perpendicular to the same line are parallel(Fig. 2).

Comment. The way we just proved case 1 of Theorem 1 is called the method of proof by contradiction or reduction to absurdity. This method got its first name because at the beginning of the reasoning, an assumption is made that is opposite (opposite) to what is required to be proved. It is called reduction to absurdity due to the fact that, arguing on the basis of the assumption made, we come to an absurd conclusion (absurdity). Receiving such a conclusion forces us to reject the assumption made at the beginning and accept the one that was required to be proved.

Task 1. Construct a line passing through a given point M and parallel to a given line a, not passing through the point M.

Solution. We draw a line p through the point M perpendicular to the line a (Fig. 3).

Then we draw a line b through the point M perpendicular to the line p. The line b is parallel to the line a according to the corollary of Theorem 1.

An important conclusion follows from the considered problem:
Through a point not on a given line, one can always draw a line parallel to the given line..

The main property of parallel lines is as follows.

Axiom of parallel lines. Through a given point not on a given line, there is only one line parallel to the given line.

Consider some properties of parallel lines that follow from this axiom.

1) If a line intersects one of the two parallel lines, then it intersects the other (Fig. 4).

2) If two different lines are parallel to the third line, then they are parallel (Fig. 5).

The following theorem is also true.

Theorem 2. If two parallel lines are crossed by a secant, then:

the lying angles are equal;

corresponding angles are equal;

the sum of one-sided angles is 180°.

Consequence 2. If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.(see Fig.2).

Comment. Theorem 2 is called the inverse of Theorem 1. The conclusion of Theorem 1 is the condition of Theorem 2. And the condition of Theorem 1 is the conclusion of Theorem 2. Not every theorem has an inverse, i.e. if a given theorem is true, then the inverse theorem may be false.

Let us explain this with 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 inverse theorem would be this: 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 at all.

Example 1 Two parallel lines are crossed by a third. It is known that the difference between two internal one-sided angles is 30°. Find those angles.

Solution. Let figure 6 meet the condition.

They do not intersect, no matter how long they continue. The parallelism of lines in writing is indicated as follows: AB|| FROME

The possibility of the existence of such lines is proved by a theorem.

Theorem.

Through any point taken outside a given line, one can draw a parallel to this line..

Let AB this line and FROM some point taken outside of it. It is required to prove that FROM you can draw a straight line parallelAB. Let's drop on AB from a point FROM perpendicularFROMD and then we will FROME^ FROMD, what is possible. Straight CE parallel AB.

For the proof, we assume the opposite, i.e., that CE intersects AB at some point M. Then from the point M to a straight line FROMD we would have two different perpendiculars MD and MS, which is impossible. Means, CE cannot intersect with AB, i.e. FROME parallel AB.

Consequence.

Two perpendiculars (CEandD.B.) to one straight line (СD) are parallel.

Axiom of parallel lines.

Through the same point it is impossible to draw two different lines parallel to the same line.

So if a straight line FROMD, drawn through the point FROM parallel to a straight line AB, then any other line FROME through the same point FROM, cannot be parallel AB, i.e. she continues intersect With AB.

The proof of this not quite obvious truth turns out to be impossible. It is accepted without proof as a necessary assumption (postulatum).

Consequences.

1. If straight(FROME) intersects with one of parallel(SW), then it intersects with the other ( AB), because otherwise through the same point FROM two different straight lines, parallel AB, which is impossible.

2. If each of the two direct (AandB) are parallel to the same third line ( FROM) , then they are parallel between themselves.

Indeed, if we assume that A and B intersect at some point M, then two different straight lines, parallel to each other, would pass through this point. FROM, which is impossible.

Theorem.

If a straight line is perpendicular to one of the parallel lines, then it is perpendicular to the other parallel.

Let AB || FROMD and EF ^ AB.It is required to prove that EF ^ FROMD.

PerpendicularEF, intersecting with AB, will certainly intersect and FROMD. Let the point of intersection be H.

Suppose now that FROMD not perpendicular to EH. Then some other line, for example HK, will be perpendicular to EH and hence through the same point H two straight parallel AB: one FROMD, by condition, and the other HK as proven before. Since this is impossible, it cannot be assumed that SW was not perpendicular to EH.

Class: 2

The purpose of the lesson:

• form the concept of parallelism of 2 lines, consider the first sign of parallel lines;
• develop the ability to apply the sign in solving problems.

Tasks:

1. Educational: repetition and consolidation of the studied material, the formation of the concept of parallelism of 2 lines, proof of the 1st sign of parallelism of 2 lines.
2. Educational: to cultivate the ability to accurately keep notes in a notebook and follow the rules for constructing drawings.
3. Developmental tasks: development of logical thinking, memory, attention.

Lesson equipment:

• multimedia projector;
• screen, presentations;
• drawing tools.

During the classes

I. Organizational moment.

Greetings, checking readiness for the lesson.

II. Preparation for active UPD.

Stage 1.

In the first lesson of geometry, we considered the relative position of 2 lines on the plane.

Question. How many common points can two lines have?
Answer. Two lines can either have one common point, or not have more than one common point.

Question. How will the 2 lines be located relative to each other if they have one common point?
Answer. If lines have one common point, then they intersect

Question. How are the 2 lines located relative to each other if they do not have common points?
Answer. In this case, the lines do not intersect.

Stage 2.

In the last lesson, you were given the task of making a presentation where we meet with non-intersecting lines in our life and in nature. Now we will look at these presentations and choose the best of them. (The jury included students who, due to low intelligence, find it difficult to create their own presentations.)

Viewing presentations made by students: "Parallelism of lines in nature and life", and choosing the best of them.

III. Active UPD (explanation of new material).

Stage 1.

Picture 1

Definition. Two lines in a plane that do not intersect are called parallel.

This table shows various cases of arranging 2 parallel lines on a plane.

Consider which segments will be parallel.

Figure 2

1) If line a is parallel to b, then segments AB and CD are also parallel.

2) A line segment can be parallel to a straight line. So the segment MN is parallel to the line a.

Figure 3

3) Segment AB is parallel to ray h. Ray h is parallel to beam k.

4) If line a is perpendicular to line c, and line b is perpendicular to line c, then lines a and b are parallel.

Stage 2.

Angles formed by two parallel lines and a transversal.

Figure 4

Two parallel lines intersect a third line at two points. In this case, eight corners are formed, indicated in the figure by numbers.

Some pairs of these angles have special names (see Figure 4).

Exists three signs, parallelism of two lines associated with these angles. In this lesson, we'll look at first sign.

Stage 3.

Let us repeat the material needed to prove this feature.

Figure 5

Question. What are the names of the corners shown in Figure 5?
Answer. Angles AOC and COB are called adjacent.

Question. What angles are called adjacent? Give a definition.
Answer. Two angles are called adjacent if they have one side in common and the other two are extensions of each other.

Question. What are the properties of adjacent angles?
Answer. Adjacent angles add up to 180 degrees.
AOC + COB = 180°

Question. What are angles 1 and 2 called?
Answer. Angles 1 and 2 are called vertical.

Question. What are the properties of vertical angles?
Answer. The vertical angles are equal to each other.

Stage 4.

Proof of the first sign of parallelism.

Theorem. If at the intersection of two lines by a transversal, the lying angles are equal, then the lines are parallel.

Figure 6

Given: a and b are straight
AB - secant
1 = 2
Prove: a//b.

1st case.

Figure 7

If 1 and 2 are straight lines, then a is perpendicular to AB, and b is perpendicular to AB, then a//b.

2nd case.

Figure 8

Consider the case when 1 and 2 are not straight lines. We divide the segment AB in half by the point O.

Question. What will be the length of the segments AO and OB?
Answer. Segments AO and OB are equal in length.

1) From the point O we draw a perpendicular to the line a, OH is perpendicular to a.

Question. What will angle 3 be?
Answer. Corner 3 will be right.

2) From point A on the straight line b, we set aside the segment AH 1 = BH with a compass.

3) Let's draw a segment OH 1.

Question. What triangles were formed as a result of the proof?
Answer. Triangle ONV and triangle OH 1 A.

Let's prove that they are equal.

Question. What angles are equal according to the hypothesis of the theorem?
Answer. Angle 1 is equal to angle 2.

Question. Which sides are equal in construction.
Answer. AO = OB and AN 1 = VN

Question. On what basis are triangles congruent?
Answer. Triangles are equal in two sides and the angle between them (the first sign of equality of triangles).

Question. What property do congruent triangles have?
Answer. Equal triangles have equal angles opposite equal sides.

Question. What angles will be equal?
Answer. 5 = 6, 3 = 4.

Question. What are 5 and 6 called?
Answer. These angles are called vertical.

From this it follows that the points: H 1 , O, H lie on one straight line.
Because 3 is straight, and 3 = 4, then 4 is straight.

Question. How are lines a and b located with respect to line HH 1 if angles 3 and 4 are right?
Answer. Lines a and b are perpendicular to HH 1 .

Question. What can we say about two perpendiculars to one straight line?
Answer. Two perpendiculars of one line are parallel.

So a//b. The theorem has been proven.

Now I will repeat all the proof from the beginning, and you will listen to me carefully and try to understand everything to remember.

IV. Consolidation of new material.

Work in groups with different levels of intelligence, followed by a check on the screen and on the board. 3 students work at the blackboard (one from each group).

№1 (for students with a reduced level of intellectual development).

Given: a and b are straight
c - secant
1 = 37°
7 = 143°
Prove: a//b.

Solution.

7 = 6 (vertical) 6 = 143°
1 + 4 = 180° (adjacent) 4 =180° – 37° = 143°
4 \u003d 6 \u003d 143 °, and they lie crosswise a//b 5 \u003d 48 °, 3 and 5 are cross-lying angles, they are equal to a//b.

Figure 11

V. Summary of the lesson.

The result of the lesson is carried out using figures 1-8.

The activity of students in the lesson is assessed (each student receives an appropriate emoticon).

Homework: teach - pp. 52-53; solve No. 186 (b, c).

Parallelism is a very useful property in geometry. In real life, parallel sides allow you to create beautiful, symmetrical things that are pleasing to any eye, so geometry has always needed ways to check this parallelism. We will talk about the signs of parallel lines in this article.

Definition for parallelism

Let us single out the definitions that you need to know to prove the signs of parallelism of two lines.

Lines are called parallel if they have no points of intersection. In addition, in solutions, parallel lines usually go in conjunction with a secant line.

A secant line is a line that intersects both parallel lines. In this case, lying, corresponding and one-sided angles are formed crosswise. The pairs of angles 1 and 4 will be lying across; 2 and 3; 8 and 6; 7 and 5. Corresponding will be 7 and 2; 1 and 6; 8 and 4; 3 and 5.

Unilateral 1 and 2; 7 and 6; 8 and 5; 3 and 4.

When properly formatted, it is written: “Cross-lying angles with two parallel lines a and b and a secant c”, because for two parallel lines there can be an infinite number of secants, so you need to specify which secant you mean.

Also, for the proof, we need the theorem on the external angle of a triangle, which states that the external angle of a triangle is equal to the sum of two angles of a triangle that are not adjacent to it.

signs

All signs of parallel lines are tied to the knowledge of the properties of angles and the theorem on the external angle of a triangle.

Feature 1

Two lines are parallel if the intersecting angles are equal.

Consider two lines a and b with a secant c. Crosswise lying angles 1 and 4 are equal. Assume that the lines are not parallel. This means that the lines intersect and there should be an intersection point M. Then a triangle AVM is formed with an external angle of 1. The external angle must be equal to the sum of angles 4 and AVM as non-adjacent to it according to the theorem on the external angle in a triangle. But then it turns out that angle 1 is greater than angle 4, and this contradicts the condition of the problem, which means that the point M does not exist, the lines do not intersect, that is, they are parallel.

Rice. 1. Drawing for proof.

Feature 2

Two lines are parallel if the corresponding secant angles are equal.

Consider two lines a and b with a secant c. The corresponding angles 7 and 2 are equal. Let's pay attention to angle 3. It is vertical for angle 7. Therefore, angles 7 and 3 are equal. So angles 3 and 2 are also equal, since<7=<2 и <7=<3. А угол 3 и угол 2 являются накрест лежащими. Следовательно, прямые параллельны, что и требовалось доказать.

Rice. 2. Drawing for proof.

Feature 3

Two lines are parallel if the sum of one-sided angles is 180 degrees.

Rice. 3. Drawing for proof.

Consider two lines a and b with a secant c. The sum of one-sided angles 1 and 2 is 180 degrees. Let's pay attention to angles 1 and 7. They are adjacent. That is:

$$<1+<7=180$$

$$<1+<2=180$$

Subtract the second from the first expression:

$$(<1+<7)-(<1+<2)=180-180$$

$$(<1+<7)-(<1+<2)=0$$

$$<1+<7-<1-<2=0$$

$$<7-<2=0$$

$<7=<2$ - а они являются соответственными. Значит, прямые параллельны.

What have we learned?

We analyzed in detail what angles are obtained when cutting parallel lines with a third line, identified and described in detail the proof of three signs of parallelism of lines.

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