Corresponding angles are equal then the lines are parallel. Straight line

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 (CD) 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.

CHAPTER III.
PARALLEL LINES

§ 35. SIGNS OF PARALLELITY OF TWO DIRECT LINES.

The theorem that two perpendiculars to one line are parallel (§ 33) gives a sign that two lines are parallel. It is possible to derive more general signs of parallelism of two lines.

1. The first sign of parallelism.

If, at the intersection of two lines with a third, the interior angles lying across are equal, then these lines are parallel.

Let lines AB and CD intersect line EF and / 1 = / 2. Take the point O - the middle of the segment KL of the secant EF (Fig. 189).

Let us drop the perpendicular OM from the point O to the line AB and continue it until it intersects with the line CD, AB_|_MN. Let us prove that CD_|_MN.
To do this, consider two triangles: MOE and NOK. These triangles are equal to each other. Indeed: / 1 = / 2 by the condition of the theorem; OK = OL - by construction;
/ MOL = / NOK as vertical corners. Thus, the side and two angles adjacent to it of one triangle are respectively equal to the side and two angles adjacent to it of another triangle; Consequently, /\ MOL = /\ NOK, and hence
/ LMO = / kno but / LMO is direct, hence, and / KNO is also direct. Thus, the lines AB and CD are perpendicular to the same line MN, hence they are parallel (§ 33), which was to be proved.

Note. The intersection of the lines MO and CD can be established by rotating the triangle MOL around the point O by 180°.

2. The second sign of parallelism.

Let's see if the lines AB and CD are parallel if, at the intersection of their third line EF, the corresponding angles are equal.

Let some corresponding angles be equal, for example / 3 = / 2 (dev. 190);
/ 3 = / 1, as the corners are vertical; means, / 2 will be equal / 1. But angles 2 and 1 are internal crosswise angles, and we already know that if at the intersection of two straight lines by a third, the internal crosswise lying angles are equal, then these lines are parallel. Therefore, AB || CD.

If at the intersection of two lines of the third the corresponding angles are equal, then these two lines are parallel.

The construction of parallel lines with the help of a ruler and a drawing triangle is based on this property. This is done as follows.

Let us attach the triangle to the ruler as shown in drawing 191. We will move the triangle so that one of its sides slides along the ruler, and draw several straight lines along any other side of the triangle. These lines will be parallel.

3. The third sign of parallelism.

Let us know that at the intersection of two lines AB and CD by the third line, the sum of any internal one-sided angles is equal to 2 d(or 180°). Will the lines AB and CD be parallel in this case (Fig. 192).

Let / 1 and / 2 interior one-sided angles and add up to 2 d.
But / 3 + / 2 = 2d as adjacent angles. Consequently, / 1 + / 2 = / 3+ / 2.

From here / 1 = / 3, and these corners are internally lying crosswise. Therefore, AB || CD.

If at the intersection of two lines by a third, the sum of the interior one-sided angles is equal to 2 d, then the two lines are parallel.

An exercise.

Prove that the lines are parallel:
a) if the external cross-lying angles are equal (Fig. 193);
b) if the sum of external unilateral angles is 2 d(devil 194).

In this article, we will talk about parallel lines, give definitions, designate the signs and conditions of parallelism. For clarity of theoretical material, we will use illustrations and the solution of typical examples.

Definition 1

Parallel lines in the plane are two straight lines in the plane that do not have common points.

Definition 2

Parallel lines in 3D space- two straight lines in three-dimensional space that lie in the same plane and do not have common points.

It should be noted that in order to determine parallel lines in space, the clarification “lying in the same plane” is extremely important: two lines in three-dimensional space that do not have common points and do not lie in the same plane are not parallel, but intersecting.

To denote parallel lines, it is common to use the symbol ∥ . That is, if the given lines a and b are parallel, this condition should be briefly written as follows: a ‖ b . Verbally, the parallelism of lines is indicated as follows: lines a and b are parallel, or line a is parallel to line b, or line b is parallel to line a.

Let us formulate a statement that plays an important role in the topic under study.

Axiom

Through a point that does not belong to a given line, there is only one line parallel to the given line. This statement cannot be proved on the basis of the known axioms of planimetry.

In the case when it comes to space, the theorem is true:

Theorem 1

Through any point in space that does not belong to a given line, there will be only one line parallel to the given one.

This theorem is easy to prove on the basis of the above axiom (geometry program for grades 10-11).

The sign of parallelism is a sufficient condition under which parallel lines are guaranteed. In other words, the fulfillment of this condition is sufficient to confirm the fact of parallelism.

In particular, there are necessary and sufficient conditions for the parallelism of lines in the plane and in space. Let us explain: necessary means the condition, the fulfillment of which is necessary for parallel lines; if it is not satisfied, the lines are not parallel.

Summarizing, the necessary and sufficient condition for the parallelism of lines is such a condition, the observance of which is necessary and sufficient for the lines to be parallel to each other. On the one hand, this is a sign of parallelism, on the other hand, a property inherent in parallel lines.

Before giving a precise formulation of the necessary and sufficient conditions, we recall a few more additional concepts.

Definition 3

secant line is a line that intersects each of the two given non-coinciding lines.

Intersecting two straight lines, the secant forms eight non-expanded angles. To formulate the necessary and sufficient condition, we will use such types of angles as cross-lying, corresponding, and one-sided. Let's demonstrate them in the illustration:

Theorem 2

If two lines on a plane intersect a secant, then for the given lines to be parallel it is necessary and sufficient that the crosswise lying angles be equal, or the corresponding angles be equal, or the sum of one-sided angles be equal to 180 degrees.

Let us graphically illustrate the necessary and sufficient condition for parallel lines on the plane:

The proof of these conditions is present in the geometry program for grades 7-9.

In general, these conditions are also applicable for three-dimensional space, provided that the two lines and the secant belong to the same plane.

Let us point out a few more theorems that are often used in proving the fact that lines are parallel.

Theorem 3

In a plane, two lines parallel to a third are parallel to each other. This feature is proved on the basis of the axiom of parallelism mentioned above.

Theorem 4

In three-dimensional space, two lines parallel to a third are parallel to each other.

The proof of the attribute is studied in the 10th grade geometry program.

We give an illustration of these theorems:

Let us indicate one more pair of theorems that prove the parallelism of lines.

Theorem 5

In a plane, two lines perpendicular to a third are parallel to each other.

Let us formulate a similar one for a three-dimensional space.

Theorem 6

In three-dimensional space, two lines perpendicular to a third are parallel to each other.

Let's illustrate:

All the above theorems, signs and conditions make it possible to conveniently prove the parallelism of lines by the methods of geometry. That is, to prove the parallelism of lines, one can show that the corresponding angles are equal, or demonstrate the fact that two given lines are perpendicular to the third, and so on. But we note that it is often more convenient to use the coordinate method to prove the parallelism of lines in a plane or in three-dimensional space.

Parallelism of lines in a rectangular coordinate system

In a given rectangular coordinate system, a straight line is determined by the equation of a straight line on a plane of one of the possible types. Similarly, a straight line given in a rectangular coordinate system in three-dimensional space corresponds to some equations of a straight line in space.

Let us write the necessary and sufficient conditions for the parallelism of lines in a rectangular coordinate system, depending on the type of equation describing the given lines.

Let's start with the condition of parallel lines in the plane. It is based on the definitions of the direction vector of the line and the normal vector of the line in the plane.

Theorem 7

In order for two non-coincident lines to be parallel on a plane, it is necessary and sufficient that the direction vectors of the given lines be collinear, or the normal vectors of the given lines are collinear, or the direction vector of one line is perpendicular to the normal vector of the other line.

It becomes obvious that the condition of parallel lines on the plane is based on the condition of collinear vectors or the condition of perpendicularity of two vectors. That is, if a → = (a x , a y) and b → = (b x , b y) are the direction vectors of lines a and b ;

and n b → = (n b x , n b y) are normal vectors of lines a and b , then we write the above necessary and sufficient condition as follows: a → = t b → ⇔ a x = t b x a y = t b y or n a → = t n b → ⇔ n a x = t n b x n a y = t n b y or a → , n b → = 0 ⇔ a x n b x + a y n b y = 0 , where t is some real number. The coordinates of the directing or direct vectors are determined by the given equations of the lines. Let's consider the main examples.

1. The line a in a rectangular coordinate system is determined by the general equation of the line: A 1 x + B 1 y + C 1 = 0 ; line b - A 2 x + B 2 y + C 2 = 0 . Then the normal vectors of the given lines will have coordinates (A 1 , B 1) and (A 2 , B 2) respectively. We write the condition of parallelism as follows:

A 1 = t A 2 B 1 = t B 2

1. The straight line a is described by the equation of a straight line with a slope of the form y = k 1 x + b 1 . Straight line b - y \u003d k 2 x + b 2. Then the normal vectors of the given lines will have coordinates (k 1 , - 1) and (k 2 , - 1), respectively, and we write the parallelism condition as follows:

k 1 = t k 2 - 1 = t (- 1) ⇔ k 1 = t k 2 t = 1 ⇔ k 1 = k 2

Thus, if parallel lines on a plane in a rectangular coordinate system are given by equations with slope coefficients, then the slope coefficients of the given lines will be equal. And the converse statement is true: if non-coinciding lines on a plane in a rectangular coordinate system are determined by the equations of a line with the same slope coefficients, then these given lines are parallel.

1. The lines a and b in a rectangular coordinate system are given by the canonical equations of the line on the plane: x - x 1 a x = y - y 1 a y and x - x 2 b x = y - y 2 b y or the parametric equations of the line on the plane: x = x 1 + λ a x y = y 1 + λ a y and x = x 2 + λ b x y = y 2 + λ b y .

Then the direction vectors of the given lines will be: a x , a y and b x , b y respectively, and we write the parallelism condition as follows:

a x = t b x a y = t b y

Let's look at examples.

Example 1

Given two lines: 2 x - 3 y + 1 = 0 and x 1 2 + y 5 = 1 . You need to determine if they are parallel.

Solution

We write the equation of a straight line in segments in the form of a general equation:

x 1 2 + y 5 = 1 ⇔ 2 x + 1 5 y - 1 = 0

We see that n a → = (2 , - 3) is the normal vector of the line 2 x - 3 y + 1 = 0 , and n b → = 2 , 1 5 is the normal vector of the line x 1 2 + y 5 = 1 .

The resulting vectors are not collinear, because there is no such value of t for which the equality will be true:

2 = t 2 - 3 = t 1 5 ⇔ t = 1 - 3 = t 1 5 ⇔ t = 1 - 3 = 1 5

Thus, the necessary and sufficient condition of parallelism of lines on the plane is not satisfied, which means that the given lines are not parallel.

Answer: given lines are not parallel.

Example 2

Given lines y = 2 x + 1 and x 1 = y - 4 2 . Are they parallel?

Solution

Let's transform the canonical equation of the straight line x 1 \u003d y - 4 2 to the equation of a straight line with a slope:

x 1 = y - 4 2 ⇔ 1 (y - 4) = 2 x ⇔ y = 2 x + 4

We see that the equations of the lines y = 2 x + 1 and y = 2 x + 4 are not the same (if it were otherwise, the lines would be the same) and the slopes of the lines are equal, which means that the given lines are parallel.

Let's try to solve the problem differently. First, we check whether the given lines coincide. We use any point of the line y \u003d 2 x + 1, for example, (0, 1) , the coordinates of this point do not correspond to the equation of the line x 1 \u003d y - 4 2, which means that the lines do not coincide.

The next step is to determine the fulfillment of the parallelism condition for the given lines.

The normal vector of the line y = 2 x + 1 is the vector n a → = (2 , - 1) , and the direction vector of the second given line is b → = (1 , 2) . The scalar product of these vectors is zero:

n a → , b → = 2 1 + (- 1) 2 = 0

Thus, the vectors are perpendicular: this demonstrates to us the fulfillment of the necessary and sufficient condition for the original lines to be parallel. Those. given lines are parallel.

Answer: these lines are parallel.

To prove the parallelism of lines in a rectangular coordinate system of three-dimensional space, the following necessary and sufficient condition is used.

Theorem 8

For two non-coincident lines in three-dimensional space to be parallel, it is necessary and sufficient that the direction vectors of these lines be collinear.

Those. for given equations of lines in three-dimensional space, the answer to the question: are they parallel or not, is found by determining the coordinates of the direction vectors of the given lines, as well as checking the condition of their collinearity. In other words, if a → = (a x, a y, a z) and b → = (b x, b y, b z) are the direction vectors of the lines a and b, respectively, then in order for them to be parallel, the existence of such a real number t is necessary, so that equality holds:

a → = t b → ⇔ a x = t b x a y = t b y a z = t b z

Example 3

Given lines x 1 = y - 2 0 = z + 1 - 3 and x = 2 + 2 λ y = 1 z = - 3 - 6 λ . It is necessary to prove the parallelism of these lines.

Solution

The conditions of the problem are the canonical equations of one straight line in space and the parametric equations of another straight line in space. Direction vectors a → and b → given lines have coordinates: (1 , 0 , - 3) and (2 , 0 , - 6) .

1 = t 2 0 = t 0 - 3 = t - 6 ⇔ t = 1 2 , then a → = 1 2 b → .

Therefore, the necessary and sufficient condition for parallel lines in space is satisfied.

Answer: the parallelism of the given lines is proved.

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AB and FROMD crossed by the third line MN, then the angles formed in this case receive the following names in pairs:

corresponding angles: 1 and 5, 4 and 8, 2 and 6, 3 and 7;

internal cross-lying corners: 3 and 5, 4 and 6;

external cross-lying corners: 1 and 7, 2 and 8;

internal one-sided corners: 3 and 6, 4 and 5;

external one-sided corners: 1 and 8, 2 and 7.

So, ∠ 2 = ∠ 4 and ∠ 8 = ∠ 6, but by the proven ∠ 4 = ∠ 6.

Therefore, ∠ 2 = ∠ 8.

3. Respective angles 2 and 6 are the same, since ∠ 2 = ∠ 4, and ∠ 4 = ∠ 6. We also make sure that the other corresponding angles are equal.

4. Sum internal one-sided corners 3 and 6 will be 2d because the sum adjacent corners 3 and 4 is equal to 2d = 180 0 , and ∠ 4 can be replaced by the identical ∠ 6. Also make sure that sum of angles 4 and 5 is equal to 2d.

5. Sum external one-sided corners will be 2d because these angles are equal respectively internal one-sided corners like corners vertical.

From the justification proved above, we obtain inverse theorems.

When, at the intersection of two lines of an arbitrary third line, we obtain that:

1. Internal cross lying angles are the same;

or 2. External cross lying angles are the same;

or 3. The corresponding angles are the same;

or 4. The sum of internal one-sided angles is equal to 2d = 180 0 ;

or 5. The sum of the outer one-sided is 2d = 180 0 ,

then the first two lines are parallel.

1. The first sign of parallelism.

If, at the intersection of two lines with a third, the interior angles lying across are equal, then these lines are parallel.

Let lines AB and CD be intersected by line EF and ∠1 = ∠2. Let's take the point O - the middle of the segment KL of the secant EF (Fig.).

Let us drop the perpendicular OM from the point O to the line AB and continue it until it intersects with the line CD, AB ⊥ MN. Let us prove that CD ⊥ MN as well.

To do this, consider two triangles: MOE and NOK. These triangles are equal to each other. Indeed: ∠1 = ∠2 by the hypothesis of the theorem; OK = OL - by construction;

∠MOL = ∠NOK as vertical angles. Thus, the side and two angles adjacent to it of one triangle are respectively equal to the side and two angles adjacent to it of another triangle; therefore, ΔMOL = ΔNOK, and hence ∠LMO = ∠KNO,
but ∠LMO is direct, hence ∠KNO is also direct. Thus, the lines AB and CD are perpendicular to the same line MN, therefore, they are parallel, which was to be proved.

Note. The intersection of the lines MO and CD can be established by rotating the triangle MOL around the point O by 180°.

2. The second sign of parallelism.

Let's see if the lines AB and CD are parallel if, at the intersection of their third line EF, the corresponding angles are equal.

Let some corresponding angles be equal, for example ∠ 3 = ∠2 (Fig.);

∠3 = ∠1 as vertical angles; so ∠2 will be equal to ∠1. But angles 2 and 1 are internal crosswise angles, and we already know that if at the intersection of two lines by a third, the internal crosswise lying angles are equal, then these lines are parallel. Therefore, AB || CD.

If at the intersection of two lines of the third the corresponding angles are equal, then these two lines are parallel.

The construction of parallel lines with the help of a ruler and a drawing triangle is based on this property. This is done as follows.

Let us attach a triangle to the ruler as shown in Fig. We will move the triangle so that one side of it slides along the ruler, and draw several straight lines along any other side of the triangle. These lines will be parallel.

3. The third sign of parallelism.

Let us know that at the intersection of two lines AB and CD by the third line, the sum of any internal one-sided angles is equal to 2 d(or 180°). Will the lines AB and CD be parallel in this case (Fig.).

Let ∠1 and ∠2 be one-sided interior angles and add up to 2 d.

But ∠3 + ∠2 = 2 d as adjacent angles. Therefore, ∠1 + ∠2 = ∠3+ ∠2.

Hence ∠1 = ∠3, and these interior angles are crosswise. Therefore, AB || CD.

If at the intersection of two lines by a third, the sum of the interior one-sided angles is equal to 2 d (or 180°), then the two lines are parallel.

Signs of parallel lines:

1. If at the intersection of two straight lines by a third, the internal cross lying angles are equal, then these lines are parallel.

2. If at the intersection of two lines of the third, the corresponding angles are equal, then these two lines are parallel.

3. If at the intersection of two lines of the third, the sum of the internal one-sided angles is 180 °, then these two lines are parallel.

4. If two lines are parallel to the third line, then they are parallel to each other.

5. If two lines are perpendicular to the third line, then they are parallel to each other.

Euclid's axiom of parallelism

A task. Through a point M taken outside the line AB, draw a line parallel to the line AB.

Using the proven theorems on the signs of parallelism of lines, this problem can be solved in various ways,

Solution. 1st s o s o b (Fig. 199).

We draw MN⊥AB and through the point M we draw CD⊥MN;

we get CD⊥MN and AB⊥MN.

Based on the theorem ("If two lines are perpendicular to the same line, then they are parallel.") we conclude that СD || AB.

2nd s p o s o b (Fig. 200).

We draw a MK intersecting AB at any angle α, and through the point M we draw a straight line EF, forming an angle EMK with a straight line MK, equal to the angle α. Based on the theorem () we conclude that EF || AB.

Having solved this problem, we can consider it proved that through any point M, taken outside the line AB, it is possible to draw a line parallel to it. The question arises, how many lines parallel to a given line and passing through a given point can exist?

The practice of constructions allows us to assume that there is only one such line, since with a carefully executed drawing, lines drawn in various ways through the same point parallel to the same line merge.

In theory, the answer to this question is given by the so-called axiom of Euclid's parallelism; it is formulated like this:

Through a point taken outside a given line, only one line can be drawn parallel to this line.

In the drawing 201, a straight line SK is drawn through the point O, parallel to the straight line AB.

Any other line passing through the point O will no longer be parallel to the line AB, but will intersect it.

The axiom adopted by Euclid in his Elements, which states that on a plane through a point taken outside a given line, only one line can be drawn parallel to this line, is called Euclid's axiom of parallelism.

For more than two thousand years after Euclid, many mathematicians tried to prove this mathematical proposition, but their attempts were always unsuccessful. Only in 1826 did the great Russian scientist, professor of Kazan University Nikolai Ivanovich Lobachevsky prove that, using all other Euclid's axioms, this mathematical proposition cannot be proved, that it really should be taken as an axiom. N. I. Lobachevsky created a new geometry, which, in contrast to the geometry of Euclid, was called the geometry of Lobachevsky.