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. You can withdraw more common signs 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 lying 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 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).

Two angles are called vertical if the sides of one angle are an extension of the sides of the other.

The figure shows the corners 1 and 3 , as well as angles 2 and 4 - vertical. Corner 2 is adjacent to both angle 1 , and with the angle 3. By property adjacent corners 1 +2 =180 0 and 3 +2 =1800. From here we get: 1=180 0 -2 , 3=180 0 -2. Thus, the degree measures of the angles 1 and 3 are equal. It follows that the angles themselves are equal. So the vertical angles are equal.

2. Signs of equality of triangles.

If two sides and the angle between them of one triangle are respectively equal to two sides and the angle between them of another triangle, then such triangles are congruent.

If a side and two adjacent angles of one triangle are respectively equal to a side and two adjacent angles of another triangle, then such triangles are congruent.

3. If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are equal.

1 sign of equality of triangles:

Consider triangles ABC and A 1 B 1 C 1, in which AB \u003d A 1 B 1, AC \u003d A 1 C 1, angles A and A 1 are equal. Let us prove that ABC=A 1 B 1 C 1 .
Since (y) A \u003d (y) A 1, then the triangle ABC can be superimposed on the triangle A 1 B 1 C 1 so that the vertex A is aligned with the vertex A1, and the sides AB and AC are superimposed, respectively, on the rays A 1 B 1 and A 1 C 1 . Since AB \u003d A 1 B 1, AC \u003d A 1 C 1, then side AB will be combined with side A 1 B 1, and side AC - with side A 1 C 1; in particular, points B and B 1 , C and C 1 will coincide. Therefore, the sides BC and B 1 C 1 will be aligned. So, the triangles ABC and A 1 B 1 C 1 are completely compatible, which means they are equal. CTD

3. The theorem on the bisector of an isosceles triangle.

In an isosceles triangle, the bisector drawn to the base is the median and height.

Let us turn to the figure, in which ABC is an isosceles triangle with base BC, AD is its bisector.

From the equality of triangles ABD and ACD (according to the 2nd criterion for the equality of triangles: AD is common; angles 1 and 2 are equal because the AD-bisector; AB=AC, because the triangle is isosceles) it follows that BD = DC and 3 = 4. The equality BD = DC means that the point D is the midpoint of the side BC and therefore AD is the median of the triangle ABC. Since angles 3 and 4 are adjacent and equal to each other, they are right angles. Therefore, segment AO is also the height of triangle ABC. CHTD.

4. If the lines are parallel -> angle…. (optional)

5. If the angle ... ..-> lines are parallel (optional)

If at the intersection of two lines of a secant the corresponding angles are equal, then the lines are parallel.

Let at the intersection of lines a and b of the secant with the corresponding angles be equal, for example 1=2.

Since angles 2 and 3 are vertical, then 2=3. From these two equalities it follows that 1=3. But angles 1 and 3 are crosswise, so lines a and b are parallel. CHTD.

6. Theorem on the sum of the angles of a triangle.

The sum of the angles of a triangle is 180 0.

Consider an arbitrary triangle ABC and prove that A+B+C=180 0 .

Let us draw a straight line a through the vertex B, parallel to the side AC. Angles 1 and 4 are crosswise lying angles at the intersection of parallel lines a and AC by the secant AB, and angles 3 and 5 are crosswise lying angles at the intersection of the same parallel lines by the secant BC. Therefore (1)4=1; 5=3.

Obviously, the sum of angles 4, 2 and 5 is equal to the straight angle with vertex B, i.e. 4+2+5=1800 . Hence, taking into account equalities (1), we obtain: 1+2+3=180 0 or A+B+C=180 0 .

7. Sign of equality of right triangles.

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 a. 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 new geometry, which, in contrast to Euclid's geometry, is called Lobachevsky's geometry.

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.

This chapter is devoted to the study of parallel lines. This is the name given to two straight lines in a plane that do not intersect. We see segments of parallel lines in the environment - these are two edges of a rectangular table, two edges of a book cover, two trolleybus bars, etc. Parallel lines play a very important role in geometry. In this chapter, you will learn about what the axioms of geometry are and what the axiom of parallel lines consists of - one of the most famous axioms of geometry.

In Section 1, we noted that two lines either have one common point, i.e., intersect, or have neither common point, i.e., do not intersect.

Definition

The parallelism of lines a and b is denoted as follows: a || b.

Figure 98 shows lines a and b perpendicular to line c. In Section 12 we established that such lines a and b do not intersect, that is, they are parallel.

Rice. 98

Along with parallel lines, one often considers parallel segments. The two segments are called parallel if they lie on parallel lines. In figure 99, and the segments AB and CD are parallel (AB || CD), and the segments MN and CD are not parallel. Similarly, the parallelism of a segment and a straight line (Fig. 99, b), a ray and a straight line, a segment and a ray, two rays (Fig. 99, c) is determined.

Rice. 99 Signs of parallelism of two lines

Direct with is called secant with respect to lines a and b, if it intersects them at two points (Fig. 100). At the intersection of lines a and b, the secant c forms eight angles, which are indicated by numbers in Figure 100. Some pairs of these angles have special names:

criss-cross corners: 3 and 5, 4 and 6;
one-sided corners: 4 and 5, 3 and 6;
corresponding angles: 1 and 5, 4 and 8, 2 and 6, 3 and 7.

Rice. 100

Consider three signs of parallelism of two lines associated with these pairs of angles.

Theorem

Proof

Suppose that at the intersection of lines a and b by a secant AB, the lying angles are equal: ∠1 = ∠2 (Fig. 101, a).

Let us prove that a || b. If angles 1 and 2 are right (Fig. 101, b), then the lines a and b are perpendicular to the line AB and, therefore, parallel.

Rice. 101

Consider the case when angles 1 and 2 are not right.

From the middle O of the segment AB, draw a perpendicular OH to the straight line a (Fig. 101, c). On the line b from point B, we set aside the segment VH 1, equal to the segment AH, as shown in Figure 101, c, and draw the segment OH 1. Triangles ONA and OH 1 V are equal in two sides and the angle between them (AO = BO, AN = VN 1, ∠1 = ∠2), therefore ∠3 = ∠4 and ∠5 = ∠6. From the equality ∠3 = ∠4 it follows that the point H 1 lies on the continuation of the ray OH, i.e. the points H, O and H 1 lie on the same straight line, and from the equality ∠5 = ∠6 it follows that the angle 6 is a straight line (since angle 5 is a right angle). So lines a and b are perpendicular to line HH 1, so they are parallel. The theorem has been proven.

Theorem

Proof

Let at the intersection of lines a and b the secant with the corresponding angles be equal, for example ∠1 = ∠2 (Fig. 102).

Rice. 102

Since angles 2 and 3 are vertical, then ∠2 = ∠3. These two equalities imply that ∠1 = ∠3. But angles 1 and 3 are crosswise, so lines a and b are parallel. The theorem has been proven.

Theorem

Proof

Let, at the intersection of lines a and b, the secant with the sum of one-sided angles be 180°, for example ∠1 + ∠4 = 180° (see Fig. 102).

Since angles 3 and 4 are adjacent, then ∠3 + ∠4 = 180°. From these two equalities it follows that the crosswise angles 1 and 3 are equal, so the lines a and b are parallel. The theorem has been proven.

Practical Ways to Draw Parallel Lines

The signs of parallel lines underlie the ways of constructing parallel lines with the help of various tools used in practice. Consider, for example, a method for constructing parallel lines using a drawing square and a ruler. To build a straight line passing through the point M and parallel to the given line a, we apply a drawing square to the straight line a, and a ruler to it as shown in Figure 103. Then, moving the square along the ruler, we will ensure that the point M is on the side of the square , and draw a line b. The lines a and b are parallel, since the corresponding angles, denoted in Figure 103 by the letters α and β, are equal.

Rice. 103 Figure 104 shows a method for constructing parallel lines using a T-square. This method is used in drawing practice.

Rice. 104 A similar method is used when performing carpentry work, where a bevel is used to mark parallel lines (two wooden planks fastened with a hinge, Fig. 105).

Rice. 105

Tasks

186. In Figure 106, lines a and b are intersected by line c. Prove that a || b if:

a) ∠1 = 37°, ∠7 = 143°;
b) ∠1 = ∠6;
c) ∠l = 45°, and angle 7 is three times larger than angle 3.

Rice. 106

187. According to figure 107 prove that AB || D.E.

Rice. 107

188. Segments AB and CD intersect in their common middle. Prove that lines AC and BD are parallel.

189. Using the data in Figure 108, prove that BC || AD.

Rice. 108

190. In Figure 109 AB = BC, AD = DE, ∠C = 70°, ∠EAC = 35°. Prove that DE || AS.

Rice. 109

191. The segment VK is the bisector of the triangle ABC. A straight line is drawn through the point K, intersecting the side BC at the point M so that BM = MK. Prove that lines KM and AB are parallel.

192. In triangle ABC, angle A is 40°, and angle ALL adjacent to angle ACB is 80°. Prove that the bisector of angle ALL is parallel to line AB.

193. In triangle ABC ∠A = 40°, ∠B = 70°. Line BD is drawn through vertex B so that ray BC is the bisector of angle ABD. Prove that lines AC and BD are parallel.

194. Draw a triangle. Through each vertex of this triangle, using a drawing square and a ruler, draw a straight line parallel to the opposite side.

195. Draw triangle ABC and mark point D on side AC. Through point D, using a drawing square and a ruler, draw straight lines parallel to the other two sides of the triangle.