Signs of parallelism of two straight lines. Properties of parallel straight lines. Straight line. Parallel straight. Basic concepts

Chapter 1 Care with the power line (attack line) This principle is expressed by the folk wisdom: "Do not climb on the dog". Rogshon is a number for which a fool goes directly, that is, in the forehead. In general, in the life of the frontal attack, in the literal and figurative sense, the case is ungrateful and the domestic traumatic. For

Determining parallel lines. Parallel are called two straight lines lying in the same plane and not intersecting throughout their entire.

Direct AB and CD (Damn 57) will be parallel. The fact that they are parallel, sometimes express writing: AB || CD.

Theorem 34.. Two straight, perpendicular to the same third, parallel.

Dana straight CD and EF perpendicular to AB (damn 58)

CD ⊥ AB and EF ⊥ AB.

It is required to prove that CD || EF.

Evidence. If direct CD and EF were parallel to, they would cross in some kind of point M. In this case, two perpendicular would be omitted from point M, which is impossible (theorem 11), hence the direct CD || EF (CHDD).

Theorem 35.. Two straight, of which one is perpendicular, and the other inclined to the third, always intersect.

Two straight lines EF and CG are given, of which EF ⊥ AB, and Cg is tilted to AB (damn 59).

It is required to prove that CG will meet with the EF line or that CG is not parallel to EF.

Evidence. From the point C, restore to the AB line perpendicular CD, then at point C is formed the DCG angle, which will repeat as many times so that the CK line falls below the AB line. We put that we for this corner DCG will repeat n times like that

Similarly, we will postpone directly direct CE on direct AB, too N time so CN \u003d NCE.

From the points C, E, L, M, N, restore perpendicular LL ", MM", NN. "The space contained between two parallel segments of CD, NN" and CN segment will be in N times more space concluded between the two perpendicular CDs, EF and CE segment, so DCNN "\u003d NDCEF.

The space consistent with DCK contains the DCNN space in itself, "therefore

DCK\u003e CDNN "or
ndcg\u003e ndcef, from where
DCG\u003e DCEF.

The last inequality can only take place when direct CG will be released with its continuation from the limits of the DCEF space, i.e. when direct CG will meet with a straight EF, therefore direct CG is not parallel to CF (CF).

Theorem 36.. Direct, perpendicular to one of the parallel, perpendicular to the other.

Two parallel straight lines AB and CD and direct EF perpendicular to CD (damn 60).

AB || CD, EF ⊥ CD

It is required to prove that EF ⊥ AB.

Evidence. If the direct AB was inclined to EF, then two direct CD and AB would cross it, for CD ⊥ EF and AB inclined to EF (Theorem 35), and direct AB and CD would not be parallel, which would contradict this conditionTherefore, straight EF perpendicular to CD (CHDD).

The angles formed by the intersection of two direct Third Direct. When crossing two direct AB and CDs, the third straight EF (damn 61) is formed eight angles α, β, γ, δ, λ, μ, ν, ρ. These angles receive special names.

Four angle α, β, ν and ρ are called external.

Four angle γ, δ, λ, μ are called internal.

Four angle β, γ, μ, ν and four angles α, δ, λ, ρ are called one-sidedbecause they lie on one side of the straight EF.

In addition, the angles, being taken in pairs, receive the following names:

The angles β and μ are called respective . In addition to this pair, there will be a pair of corners with the same corresponding angles:γ and ν, α and λ, δ and ρ.

N's angles δ and μ, as well as γ and λ are called internal cross-lying .

Pairs of angles β and ρ, as well as α and ν are called external pass-lying .

Pairs of angles γ and μ, as well as δ and λ are called internal one-sided .

Couples of the angles β and ν, as well as α and ρ are called external one-sided .

Conditions of parallelism of two straight lines

Theorem 37.. Two straight parallels, if they are in crossing their third, they are equal: 1) corresponding angles, 2) internal closures-lying, 3) external cross-lying, and, finally, if 4) the sum of the internal one-sided is equal to two straight, 5) the amount of external One-way is equal to two straight.

Let us prove each of these parts theorem separately.

1st case. Corresponding angles are equal (Damn 62).

Dano. The angles β and μ are equal.

Evidence. If the AB and CD lines were intersected at the point q, the GQH triangle would have turned out, in which the β β would be equal to inner corner μ, which would contradict theorem 22, therefore, direct AB and CDs do not intersect or AB || CD (CHDD).

2nd case. Internal cross-lying angles are equal, that is, Δ \u003d μ.

Evidence. Δ \u003d β as vertical, δ \u003d μ under the condition, therefore, β \u003d μ. That is, corresponding angles are equal, and in this case the lines are parallel (1st case).

3rd case. External cross-lying angles are equal, that is, β \u003d ρ.

Evidence. β \u003d ρ under the condition, μ \u003d ρ as vertical, therefore, β \u003d μ, because the corresponding angles are equal. Hence it follows that AB || CD (1st case).

4th case. The sum of the internal one-sided is equal to two direct or γ + μ \u003d 2D.

Evidence. β + γ \u003d 2D as the sum of adjacent, γ + μ \u003d 2D by condition. Consequently, β + γ \u003d γ + μ, from where β \u003d μ. Corresponding angles are equal, therefore, AB || CD.

5th case. The sum of external one-sided is equal to two direct, that is, β + ν \u003d 2D.

Evidence. μ + ν \u003d 2D as the sum of adjacent, β + ν \u003d 2d by condition. Consequently, μ + ν \u003d β + ν, from where μ \u003d β. Corresponding angles are equal, therefore, AB || CD.

Thus, in all cases AB || CD (CHDD).

Theorem 38. (inverse 37). If two straight parallels, then when crossing their third straight will be equal to: 1) internal passage-lying angles, 2) outer cross-lying, 3) corresponding angles and are equal to two direct 4) the sum of the internal one-sided and 5) the amount of external one-sided corners.

There are two parallel straight AB and CDs, that is, AB || CD (Damn 63).

It is required to prove that all the above conditions are performed.

1st case. We cross two parallel direct AB and CD third inclined direct EF. Denote by g and n point of intersection of direct AB and CD direct EF. From the point o of the middle of the direct GH, we lower the perpendicular to the direct CD and continue it to intersee with a straight AB at P. P. Direct OQ perpendicular to CD perpendicular to AB (Theorem 36). Rectangular triangle OPG and OHQ are equal, for OG \u003d OH on construction, ∠ HOQ \u003d. ∠ POG as vertical angles, therefore, Op \u003d OQ.

It follows that δ \u003d μ, i.e. internal cross-lying angles are equal.

2nd case. If AB || CD, then δ \u003d μ, and since δ \u003d β, and μ \u003d ρ, then β \u003d ρ, that is, external cross-lying angles are equal.

3rd case. If AB || CD, then δ \u003d μ, and since δ \u003d β, then β \u003d μ, therefore, corresponding angles are equal.

4th case. If AB || CD, then δ \u003d μ, and since Δ + γ \u003d 2d, then μ + γ \u003d 2d, that is, the sum of the internal one-sided is equal to two direct.

5th case. If AB || CD, then Δ \u003d μ.

Since μ + ν \u003d 2d, μ \u003d Δ \u003d β, therefore, ν + β \u003d 2d, that is, the sum of external one-sided is equal to two direct.

From these theorems follows corollary. After a point you can spend only one straight, parallel to another straight line.

Theorem 39.. Two straight, parallel to the third, parallel between themselves.

Three straight lines (damn 64) AB, CD and EF, of which AB || EF, CD || EF.

It is required to prove that AB || CD.

Evidence. We cross these direct fourth direct GH.

If AB || EF, T. ∠ α = ∠ γ how appropriate. If CD || EF, T. ∠ β = ∠ γ also as appropriate. Hence, ∠ α = ∠ β .

If the corresponding angles are equal, then straight parallel, therefore, AB || CD (CHDD).

Theorem 40.. The same angles with parallel sides are equal.

The same names are given (both sharp or both stupid) ABC and DEF angles, their parties are parallel, i.e. AB || DE, BC || EF (Damn 65).

It is required to prove that ∠ B \u003d. ∠ E.

Evidence. Continue the side of DE to crossing it with a straight BC at the point G, then

∠ E \u003d. ∠ G As corresponding to the intersection of the sides of parallel BC and EF Third Direct DG.

∠ B \u003d ∠ G as appropriate from the intersection of parallel sides of AB and DG direct BC, therefore,

∠ E \u003d. ∠ B (CHDD).

Theorem 41.. Multimame corners with parallel sides complement each other to two straight lines.

There are two different ABC and DEF angle (damn 66) with parallel sides, therefore AB || DE and BC || EF.

It is required to prove that ABC + DEF \u003d 2D.

Evidence. We continue directly to intersections with a direct BC at the point G.

∠ B +. ∠ DGB \u003d 2D as the sum of the internal one-sided angles formed by the intersection of parallel AB and DG of the Third Direct BC.

∠ DGB \u003d. ∠ DEF as appropriate, therefore,

∠ B +. ∠ Def \u003d 2D (CTD).

Theorem 42.. The same angles with perpendicular sides are equal and the variemen complement each other to two straight lines.

Consider two cases: when a) the angles of the same name and when b) they are different.

1st case. The sides of the two ADF and ABC angles of the same name (damn 67) are perpendicular, i.e. DE ⊥ AB, EF ⊥ Bc.

It is required to prove that ∠ DEF \u003d ∠ ABC.

Evidence. Spend from point b straight BM and BN parallel direct DE and EF so that

Bm || DE, BN || EF.

These are also perpendicular to the sides of this angle of ABC, i.e.

BM ⊥ AB and BN ⊥ BC.

As ∠ nbc \u003d d, ∠ mba \u003d d, then

∠ NBC \u003d. ∠ MBA (A)

Lying from both parts of equality (a) on the corner of NBA, we find

∠ MBN \u003d ∠ ABC

Since MBN and DEF angles of the same name and parallel sides are equal (Theorem 40).

∠ MBN \u003d ∠ DEF (B)

Equality (a) and (b) implies equality

∠ ABC \u003d ∠ DEF (CTD).

2nd case. GED and ABC corners with perpendicular sides of the variemen.

It is required to prove that ∠ GED + ∠ ABC \u003d 2D (damn 67).

Evidence. The sum of the angles of GED and DEF is equal to two straight.

GED + DEF \u003d 2D
DEF \u003d ABC, therefore,
GED + ABC \u003d 2D (CHDD).

Theorem 43.. Parts of parallel direct between other parallel are equal.

Four direct AB, BD, CD, AC (damn 68) are given, of which AB || CD and BD || AC.

It is required to prove that AB \u003d CD and BD \u003d AC.

Evidence. Connecting a point C with a point B with a bc segment, we get two equal triangles ABC and BCD, for

BC - Total side,

∠ α \u003d ∠ β (as internal cross-lying on the intersection of parallel direct AB and CDs of the Third Direct BC),

∠ γ \u003d ∠ Δ (as internal cross-lying on the intersection of parallel direct BD and AC direct BC).

Thus, triangles are on the equal side and on two equal corners lying on it.

Against equal angles α and β, there are equal part of AC and BD, and against equal angles γ and δ - equal parties AB and CD, therefore,

AC \u003d BD, AB \u003d CD (CHDD).

Theorem 44.. Parallel straight lines in the whole distance are at an equal distance from each other.

The distance of the point from the straight line is determined by the perpendicular length, lowered from the point to the straight line. To determine the distance of any two points A and B parallel to AB from CD, from the points A and B, to lower the perpendicular of AC and BD.

Dana direct AB parallel CD, AC and BD segments perpendicular to a straight line CD, i.e. AB || CD, AC ⊥ DC, BD ⊥ CD (Damn 69).

It is required to prove that AC \u003d BD.

Evidence. Direct AC and BD, being both perpendicular to CD, are parallel, and therefore AC and BD as part of parallel between parallel are equal, that is, AC \u003d BD (CTD).

Theorem 45. (inverse 43). If the opposite parts of the four intersecting straight lines are equal, these parts are parallel.

Four intersecting straight lines, opposite parts of which are equal to: AB \u003d CD and BD \u003d AC (damn 68).

It is required to prove that AB || CD and BD || AC.

Evidence. Connect points b and c direct BC. ABC and BDC triangles are equal, for

BC - general side,
AB \u003d CD and BD \u003d AC by condition.

From here

∠ α = ∠ β , ∠ γ = ∠ δ

Hence,

AC || BD, AB || CD (CHDD).

Theorem 46.. The sum of the corners of the triangle is equal to two direct.

Dan is a triangle ABC (Damn 70).

It is required to prove that A + B + C \u003d 2D.

Evidence. We carry out from the point C direct CF parallel side AB. At point C, three angles of BCA, α and β are formed. The amount of them is equal to two direct:

BCA +. α + β \u003d 2D

α \u003d B (as internal cross-lying angles when crossing parallel direct AB and CF direct BC);

β = A (as corresponding angles with the intersection of direct AB and CF direct AD).

Replacing the angles α and β their values, we get:

BCA + A + B \u003d 2D (CTD).

The following consequences flow from this theorem:

Corollary 1.. The outer angle of the triangle is equal to the sum of the internal non-adjacent with it.

Evidence. Indeed, from the drawing 70,

∠ BCD \u003d. ∠ α + ∠ β

Since ∠ α \u003d ∠ b, ∠ β \u003d ∠ A, then

∠ BCD \u003d. ∠ A + ∠ B.

Corollary 2.. IN rectangular triangle The sum of sharp corners is equal to direct.

Indeed, in a rectangular triangle (damn 40)

A + b + c \u003d 2d, a \u003d d, therefore,
B + C \u003d D.

Corollary 3.. The triangle cannot be more than one direct or one stupid angle.

Corollary 4.. In the equilateral triangle, each angle is 2/3 D .

Indeed, in the equilateral triangle

A + B + C \u003d 2D.

As a \u003d b \u003d c, then

3A \u003d 2D, a \u003d 2/3 d.

This article is about parallel direct and on direct parallelism. At first, the definition of parallel direct on the plane and in space, the notation was introduced, examples and graphic illustrations of parallel straight lines are given. Next disassemble the signs and conditions of the parallelism of direct. The conclusion shows the solutions of characteristic tasks on the proof of the parallelism of direct, which are set by some equations direct in the rectangular coordinate system on the plane and in three-dimensional space.

Navigating page.

Parallel direct - basic information.

Definition.

Two straight planes are called parallelIf they do not have common points.

Definition.

Two straight in three-dimensional space are called parallelIf they lie in the same plane and do not have common points.

Please note that the reservation "If they lie in the same plane" in the definition of parallel direct in space is very important. Let us explain this moment: two straight in three-dimensional space that do not have common points and do not lie in the same plane are not parallel, but are crossing.

Here are some examples of parallel straight lines. The opposite edges of the notebook leaf lie on parallel straight lines. Direct, by which the plane of the wall of the house crosses the plane of the ceiling and floor, are parallel. Railway rails on a flat terrain can also be viewed as parallel straight.

To indicate parallel direct use the symbol "". That is, if direct a and b are parallel, then you can briefly record a b.

Please note: if direct a and b are parallel, then we can say that direct A is parallel to the straight line B, and also that straight b is parallel to direct a.

Let's voice the statement that plays an important role in the study of parallel straight lines on the plane: through a point that does not lie on this direct, the only straight line, parallel to this. This statement is adopted as a fact (it cannot be proven on the basis of the planimetry by the axis), and it is called axiom of parallel straight lines.

For the case in space, the theorem is valid: through any point of space that does not lie on a given straight line, the only straight line, parallel to this. This theorem is easily proved with the help of the above axiom parallel direct (its proof you can find in the geometry textbook 10-11, which is specified at the end of the article on the literature).

For the case in space, the theorem is valid: through any point of space that does not lie on a given straight line, the only straight line, parallel to this. This theorem is easily proved using the above axiom parallel straight lines.

Parallelism of direct - features and conditions of parallelism.

A sign of direct parallelism It is a sufficient condition for parallelism of direct, that is, such a condition, the execution of which guarantees the parallelity of direct. In other words, the implementation of this condition is enough to state the fact of parallelism of direct.

There are also necessary and sufficient conditions for parallelism of direct on the plane and in three-dimensional space.

Let us explain the meaning of the phrase "necessary and sufficient condition of parallelism of direct".

With a sufficient condition of parallelism, we have already figured out. And what is " prerequisite parallelism direct "? By the name "necessary" it is clear that the execution of this condition is necessary for the parallelism of direct. In other words, if the required condition of direct parallelism is not fulfilled, then the straight lines are not parallel. In this way, required and sufficient condition parallelism - This condition, the execution of which is both necessary and enough for the parallelism of the straight lines. That is, on the one hand, this is a sign of direct parallelism, and on the other hand, this is a property that has parallel straight.

Before formulating the necessary and sufficient condition for parallelism of direct, it is advisable to remind several auxiliary definitions.

Singing straight - This is a straight line that crosses each of the two defined uncompanying straight lines.

When crossing two direct secant, eight non-verminated. In the wording of the necessary and sufficient condition of the parallelism of direct participate so-called led, respectively and one-sided corners. Show them in the drawing.

Theorem.

If two direct on the plane are crossed by the unit, then for their parallel, it is necessary and enough so that the underlying angles are equal to, or the corresponding angles were equal, or the sum of one-sided corners was 180 degrees.

We show the graphic illustration of this necessary and sufficient condition of parallelism directly on the plane.

The evidence of these conditions of parallelism direct can be found in geometry textbooks for 7 -9 classes.

Note that these conditions can also be used in three-dimensional space - the main thing is that two straight and secant lay in the same plane.

We give a few theorems that are often used in the proof of parallelism of direct.

Theorem.

If two straight line on the plane are parallel to the third straight, then they are parallel. The proof of this feature follows from the axiom of parallel direct.

There is a similar condition for parallelism of direct in three-dimensional space.

Theorem.

If the two straight in the space are parallel to the third straight, then they are parallel. The proof of this feature is considered in the lessons of geometry in the 10th grade.

We illustrate the voiced theorems.

We give another theorem that allows you to prove the parallelism of direct on the plane.

Theorem.

If two straight ones are perpendicular to the third straight, then they are parallel.

There is a similar theorem for direct in space.

Theorem.

If two direct in three-dimensional space are perpendicular to one plane, then they are parallel.

I will depict the drawings corresponding to these theorems.

All the theorems formulated above, signs and necessary and sufficient conditions are perfectly suitable for evidence of parallelism of direct geometry methods. That is, to prove the parallelism of the two specified directs to show that they are parallel to the third straight, or to show the equality of the passage of lying angles, etc. Many such tasks are solved in the lessons of geometry in high school. However, it should be noted that in many cases it is convenient to use the coordinate method to proof the parallelism of direct on the plane or in three-dimensional space. We formulate the necessary and sufficient conditions for the parallelism of direct, which are specified in the rectangular coordinate system.

Parallelism of direct in the rectangular coordinate system.

In this paragraph of the article we will formulate required and sufficient conditions of direct parallelism In the rectangular coordinate system, depending on the type of equations that define these direct, and also give detailed solutions Characteristic tasks.

Let's start with the condition of the parallelism of two direct on the plane in the rectangular Oxy coordinate system. The basis of its proof is the definition of the guide vector direct and the definition of the normal vectors on the plane.

Theorem.

For the parallelity of the two inconsistent straight lines on the plane, it is necessary and enough that the guide vectors of these lines were collinear, or the normal vectors of these straight lines were collinear, or the director of one straight was perpendicular to the normal vector of the second direct.

Obviously, the condition of parallelism of two straight lines on the plane is reduced to (guide vectors of direct or normal vectors of straight lines) or K (guide vector of one straight and normal vector second line). Thus, if both - direct vectors of direct a and b, and and - Normal vectors of straight lines a and b, respectively, then the necessary and sufficient condition of parallelism of direct a and b will be recorded as , or , or, where T is some valid number. In turn, the coordinates of the guide and (or) normal vectors of straight lines A and B are located according to the well-known equations of direct.

In particular, if direct A in the rectangular system of Oxy coordinates on the plane sets the general equation of direct type , and straight b - , The normal vectors of these directs have coordinates and, accordingly,, and the condition of parallelism of direct a and b will be recorded as.

If the direct A corresponds to the equation of a straight line with an angular coefficient of the species, and direct b -, then the normal vectors of these directs have coordinates and, and the condition of the parallelism of these direct will take the form . Therefore, if direct on the plane in the rectangular coordinate system is parallel and can be set by equations of direct with angular coefficients, the corner coefficients will be equal. And back: if the coordinate-straight lines on the plane in the rectangular coordinate system can be given by the equations of direct with equal angular coefficients, then such direct are parallel.

If direct a and straight b in a rectangular coordinate system define canonical equations direct on the plane of the species and , or parametric equations direct on the plane of the species and Accordingly, the guide vectors of these directs have coordinates and, and the condition of parallelism of direct a and b is recorded as.

We will analyze the solutions of several examples.

Example.

Whether straight lines are parallel and?

Decision.

I rewrite the equation is straight in segments in the form of a general direct equation: . Now it can be seen that - normal vector straight , and - normal vector straight. These vectors are not collinear, since there is no such valid number T for which the equality is true ( ). Therefore, the necessary and sufficient condition of parallelism of direct on the plane is not performed, therefore, the specified straight lines are not parallel.

Answer:

No, straight is not parallel.

Example.

Are the straight and parallel?

Decision.

We give the canonical equation direct to the equation direct with the angular coefficient :. It is obvious that the equations of direct and not the same (in this case, the specified straight lines would be coinciding) and the angular coefficients of the direct are equal, therefore, the initial straight parallels.

Which lie in the same plane or coincide, or do not intersect. In some school definitions, the coinciding straight lines are not considered parallel, this definition is not considered.

Properties

1. Parallelism is a binary equivalence ratio, therefore, it breaks all the many direct lines in parallel between themselves.
2. Through any point you can spend exactly one straight, parallel to this. This is a distinctive feature of the Euclidean geometry, in other geometries, the number 1 is replaced by others (in the geometry of Lobachevsky such direct minimum two)
3. 2 parallel straight lines in space lie in the same plane.
4. When crossing 2 parallel straight third, called sale:
   1. The sequential necessarily crosses both straight.
   2. With the intersection, 8 angles are formed, some of whose characteristic pairs have special names and properties:
      1. Low Lying Corners are equal.
      2. Respective Corners are equal.
      3. Unilateral Corners in sum are 180 °.

In the geometry of Lobachevsky

In the geometry of Lobachevsky in the plane through the point It is impossible to disassemble the expression (lexical error): C out of this direct $AB$

There is an infinite set of direct, non-crossing A.B. . Of these, parallel to A.B. only two are called.

Straight C.E. called an equiliburnal (parallel) direct A.B. in the direction Ot A. to B. , if a:

1. points B. and E. lie on one side from the straight A.C. ;
2. straight C.E. Does not cross straight A.B. but every ray passing inside the angle A.C.E. , crosses the ray A.B. .

Similarly, a straight line, equating A.B. in the direction Ot B. to A. .

All other straight, non-crossing this, are called ultraparallel or discussion.

see also

Wikimedia Foundation. 2010.

• Straight crossing
• Nesterichin, Yuri Efremovich

Watch what is "parallel straight lines" in other dictionaries:

Parallel straight - Parallel straight, non-vigorous straight, lying in the same plane ... Modern encyclopedia

Parallel straight Big Encyclopedic Dictionary

Parallel straight - Parallel straight, non-destroyed straight, lying in the same plane. ... Illustrated Encyclopedic Dictionary

Parallel straight - In the Euclidean geometry, direct, which lie in the same plane and do not intersect. In absolute geometry (see absolute geometry) through a point that does not lie on this direct, at least one direct, which does not cross this one. IN… … Great Soviet Encyclopedia

parallel straight - Insecumbered straight, lying in the same plane. * * * Parallel straight parallel straight, non-destroyed straight, lying in the same plane ... encyclopedic Dictionary

Parallel straight - In the Euclidean geometry, the straight lines are lying in the same plane and do not intersect. In absolute geometry through a point that does not lie on this line, at least one direct, which does not cross this is passed. In the Euclidean geometry there is only one ... ... Mathematical encyclopedia

Parallel straight - Non-destroyed straight, lying in the same plane ... Natural science. encyclopedic Dictionary

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