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 important role. In this chapter, you will learn about what the axioms of geometry are and what the axiom of parallel lines is, one of the most famous axioms of geometry.
In paragraph 1, we noted that two lines either have one common point, that is, intersect, or do not have one common point, i.e. they 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 paragraph 12, we established that such lines a and b do not intersect, i.e. they are parallel.
Rice. 98
Along with parallel lines, parallel segments are often considered. The two segments are called parallel, if they lie on parallel lines. In Figure 99, the segments AB and CD are parallel (AB || CD), but the segments MN and CD are not parallel. 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 similarly.
Rice. 99 Signs of parallelism of two lines
The straight line with is called secant in relation to straight lines a and b, if it intersects them at two points (Fig. 100). When lines a and b intersect with transversal c, eight angles are formed, which are indicated by numbers in Figure 100. Some pairs of these angles have special names:
crosswise angles: 3 and 5, 4 and 6;
one-sided angles: 4 and 5, 3 and 6;
corresponding angles
: 1 and 5, 4 and 8, 2 and 6, 3 and 7.
Rice. 100
Let's consider three signs of parallelism of two straight lines associated with these pairs of angles.
Theorem
Proof
Let the intersecting lines a and b crosswise the angles AB be equal: ∠1 = ∠2 (Fig. 101, a).
Let us prove that a || b. If angles 1 and 2 are right (Fig. 101, b), then lines a and b are perpendicular to line AB and, therefore, parallel.
Rice. 101
Let's consider the case when angles 1 and 2 are not right.
From the middle O of segment AB we draw a perpendicular OH to straight line a (Fig. 101, c). On straight line b from point B we will lay off the segment ВН 1, equal to the segment AH, as shown in Figure 101, c, and draw the segment OH 1. Triangles OHA and OH 1 B are equal on both sides and the angle between them (AO = VO, AN = BH 1, ∠1 = ∠2), therefore ∠3 = ∠4 and ∠5 = ∠6. From the equality ∠3 = ∠4 it follows that point H 1 lies on the continuation of the ray OH, i.e. points H, O and H 1 lie on the same straight line, and from the equality ∠5 = ∠6 it follows that 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 the corresponding angles be equal when lines a and b intersect with transversal c, for example ∠1 =∠2 (Fig. 102).
Rice. 102
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. The theorem has been proven.
Theorem
Proof
Let the intersection of straight lines a and b with transversal c sum the one-sided angles equal to 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, therefore lines a and b are parallel. The theorem has been proven.
Practical ways to construct parallel lines
The signs of parallel lines underlie the methods of constructing parallel lines using various tools used in practice. Consider, for example, the method of constructing parallel lines using a drawing square and a ruler. To construct a straight line passing through point M and parallel to a given line a, we apply a drawing square to straight line a, and a ruler to it as shown in Figure 103. Then, moving the square along the ruler, we will ensure that point M is on the side of the square , and draw straight line b. Straight lines a and b are parallel, since the corresponding angles, designated in Figure 103 by the letters α and β, are equal.
Rice. 103 Figure 104 shows a method for constructing parallel lines using a crossbar. This method is used in drawing practice.
Rice. 104 A similar method is used when performing carpentry work, where a block (two wooden planks fastened with a hinge, Fig. 105) is used to mark parallel lines.
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. Based on the data in Figure 107, prove that AB || D.E.
Rice. 107
188. Segments AB and CD intersect at their common midpoint. Prove that lines AC and BD are parallel.
189. Using the data in Figure 108, prove that BC || A.D.
Rice. 108
190. In Figure 109, AB = BC, AD = DE, ∠C = 70°, ∠EAC = 35°. Prove that DE || AC.
Rice. 109
191. Segment BK is the bisector of triangle ABC. A straight line is drawn through point K, intersecting side BC at 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 straight line AB.
193.V triangle ABC∠A = 40°, ∠B = 70°. A straight 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.
1. The first sign of parallelism.
If, when two straight lines intersect a third, the internal angles lying crosswise are equal, then these lines are parallel.
Let the lines AB and CD be intersected by the line EF and ∠1 = ∠2. Let's take point O - the middle of the segment KL of the secant EF (Fig.).
Let us lower the perpendicular OM from point O onto the line AB and continue it until it intersects 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 according to the theorem; ОK = ОL - by construction;
∠MOL = ∠NOK, like vertical angles. Thus, the side and two adjacent angles of one triangle are respectively equal to the side and two adjacent angles of another triangle; therefore, ΔMOL = ΔNOK, and hence ∠LMO = ∠KNO,
but ∠LMO is straight, which means ∠KNO is also straight. Thus, lines AB and CD are perpendicular to the same line MN, therefore, they are parallel, which was what needed to be proven.
Note. The intersection of straight lines MO and CD can be established by rotating the triangle MOL around point O by 180°.
2. The second sign of parallelism.
Let's see whether straight lines AB and CD are parallel if, when they intersect the third straight line EF, the corresponding angles are equal.
Let some corresponding angles be equal, for example ∠ 3 = ∠2 (Fig.);
∠3 = ∠1, as vertical angles; this means that ∠2 will be equal to ∠1. But angles 2 and 1 are intersecting interior angles, and we already know that if when two straight lines intersect the third, the intersecting interior angles are equal, then these lines are parallel. Therefore, AB || CD.
If, when two lines intersect a third, the corresponding angles are equal, then these two lines are parallel.
The construction of parallel lines using a ruler and a drawing triangle is based on this property. This is done as follows.
Let's attach the triangle to the ruler as shown in Fig. We will move the triangle so that one of its sides slides along the ruler, and we will draw several straight lines along some other side of the triangle. These lines will be parallel.
3. The third sign of parallelism.
Let us know that when two straight lines AB and CD intersect with a third straight line, the sum of any internal one-sided angles is equal to 2 d(or 180°). Will the straight lines AB and CD be parallel in this case (Fig.).
Let ∠1 and ∠2 be interior one-sided 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 lie crosswise. Therefore, AB || CD.
If, when two straight lines intersect a third, the sum of the internal one-sided angles is equal to 2 d (or 180°), then these two lines are parallel.
Signs of parallel lines:
1. If, when two lines intersect a third, the internal angles lying crosswise are equal, then these lines are parallel.2. If, when two lines intersect a third, the corresponding angles are equal, then these two lines are parallel.
3. If, when two lines intersect a third, the sum of the internal one-sided angles is 180°, then these two lines are parallel.
4. If two lines are parallel to a third line, then they are parallel to each other.
5. If two lines are perpendicular to a third line, then they are parallel to each other.
Euclid's Axiom of Parallelism
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 step (drawing 199).
We draw MN⊥AB and through 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 CD || AB.
2nd method (drawing 200).
We draw an MK intersecting AB at any angle α, and through the point M we draw a straight line EF, forming an angle EMK with the straight line MK, equal to angleα. Based on Theorem (), we conclude that EF || AB.
Having decided this task, we can consider it proven 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 are parallel to a given line and passing through this point, can exist?
The practice of construction allows us to assume that there is only one such straight line, since with a carefully executed drawing, straight lines drawn in different ways through the same point parallel to the same straight line merge.
In theory, the answer to the question posed is given by the so-called parallelism axiom of Euclid; it is formulated as follows:
Through a point taken outside a given line, only one line can be drawn parallel to this line.
In drawing 201, a straight line SC is drawn through point O, parallel to straight AB.
Any other line passing through point O will no longer be parallel to 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 straight line can be drawn parallel to this line, is called Euclid's axiom of parallelism.
More than two thousand years after Euclid, many mathematicians tried to prove this mathematical proposition, but their attempts were always unsuccessful. Only in 1826, the great Russian scientist, professor at Kazan University Nikolai Ivanovich Lobachevsky proved that using all the other axioms of Euclid, this mathematical proposition cannot be proven, that it should really be accepted as an axiom. N. I. Lobachevsky created new geometry, which, in contrast to Euclidean geometry, is called Lobachevsky geometry.
AB And WITHD crossed by the third straight 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 crosswise angles: 3 and 5, 4 and 6;
external crosswise angles: 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 according to what has been proven, ∠ 4 = ∠ 6.
Therefore, ∠ 2 =∠ 8.
3. Corresponding angles 2 and 6 are the same, since ∠ 2 = ∠ 4, and ∠ 4 = ∠ 6. Let’s 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. We 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 above proven justification we obtain converse theorems.
When, at the intersection of two lines with an arbitrary third line, we obtain that:
1. Internal crosswise angles are the same;
or 2. External crosswise angles are identical;
or 3. Corresponding angles are equal;
or 4. The sum of internal one-sided angles is 2d = 180 0;
or 5. The sum of external one-sided ones is 2d = 180 0 ,
then the first two lines are parallel.
CHAPTER III.
PARALLEL DIRECT
§ 35. SIGNS OF PARALLEL TWO LINES.
The theorem that two perpendiculars to one line are parallel (§ 33) gives a sign that two lines are parallel. You can withdraw more general signs parallelism of two lines.
1. The first sign of parallelism.
If, when two straight lines intersect a third, the internal angles lying crosswise are equal, then these lines are parallel.
Let straight lines AB and CD be intersected by straight line EF and / 1 = / 2. Take point O - the middle of the segment KL of the secant EF (Fig. 189).
Let us lower the perpendicular OM from point O onto the straight line AB and continue it until it intersects with the straight 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. In fact: /
1 = /
2 according to the conditions of the theorem; ОK = ОL - by construction;
/
MOL = /
NOK, like vertical angles. Thus, the side and two adjacent angles of one triangle are respectively equal to the side and two adjacent angles of another triangle; hence, /\
MOL = /\
NOK, and hence
/
LMO = /
KNO, but /
LMO is direct, which means /
KNO is also straight. Thus, the lines AB and CD are perpendicular to the same line MN, therefore, they are parallel (§ 33), which was what needed to be proven.
Note. The intersection of straight lines MO and CD can be established by rotating the triangle MOL around point O by 180°.
2. The second sign of parallelism.
Let's see whether straight lines AB and CD are parallel if, when they intersect the third straight line EF, the corresponding angles are equal.
Let some corresponding angles be equal, for example /
3 = /
2 (drawing 190);
/
3 = /
1, as the angles are vertical; Means, /
2 will be equal /
1. But angles 2 and 1 are intersecting interior angles, and we already know that if when two straight lines intersect the third, the intersecting interior angles are equal, then these lines are parallel. Therefore, AB || CD.
If, when two lines intersect a third, the corresponding angles are equal, then these two lines are parallel.
The construction of parallel lines using a ruler and a drawing triangle is based on this property. This is done as follows.
Let's 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 some other side of the triangle. These lines will be parallel.
3. The third sign of parallelism.
Let us know that when two straight lines AB and CD intersect with a third straight line, the sum of any internal one-sided angles is equal to 2 d(or 180°). Will the straight lines AB and CD be parallel in this case (Fig. 192).
Let /
1 and /
2 are interior one-sided angles and add up to 2 d.
But /
3 + /
2 = 2d as adjacent angles. Hence, /
1 + /
2 = /
3+ /
2.
From here / 1 = / 3, and these internal angles lie crosswise. Therefore, AB || CD.
If, when two straight lines intersect a third, the sum of the internal one-sided angles is equal to 2 d, then these two lines are parallel.
Exercise.
Prove that the lines are parallel:
a) if external crosswise angles are equal (Fig. 193);
b) if the sum of external one-sided angles equals 2 d(drawing 194).
