If when crossing. N. Nikitin Geometry. criss-crossing angles are equal, or
Two corners are called vertical if the sides of one corner are a continuation of the sides of the other.
The angles in the picture 1 and 3 as well as the angles 2 and 4 - vertical. Injection 2 is contiguous as an angle 1 and with an angle 3. By property of adjacent corners 1 +2 = 180 0 and 3 +2 = 180 0. From here we get: 1=180 0 -2 , 3=180 0 -2. Thus, the degree measures of the angles 1 and 3 are equal. Hence 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 the two sides and the angle between them of another triangle, then such triangles are equal.
If a side and two adjacent angles of one triangle are respectively equal to the side and two adjacent angles of another triangle, then such triangles are equal.
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 = A 1 B 1, AC = 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 = (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 on the rays A 1 B 1 and A 1 C 1. Since AB = A 1 B 1, AC = A 1 C 1, then the AB side will be aligned with the A 1 B 1 side, and the AC side with the A 1 C 1 side; in particular, points B and B 1, C and C 1 will be combined. Consequently, the sides BC and B 1 C 1 will be combined. So, triangles ABC and A 1 B 1 C 1 will be completely combined, which means they are equal. CHTD
3. Theorem on the bisector of an isosceles triangle.
In an isosceles triangle, the bisector drawn to the base is the median and the height.
Let us refer to the figure in which ABC - isosceles triangle with the base ВС, АD is its bisector.
From the equality of triangles ABD and ACD (by 2 signs of equality of triangles: AD - common; angles 1 and 2 are equal since AD-bisector; AB = AC, since triangle is isosceles) it follows that BD = DC and 3 = 4. Equality ВD = DC means that point D is the middle of the side ВС and therefore АD is the median of triangle ABC. Since angles 3 and 4 are adjacent and equal to each other, they are straight. Consequently, the segment AO is also the height of the triangle ABC. CHTD.
4. If the lines are parallel -> angle…. (optional)
5. If the angle… ..-> straight lines are parallel (optional)
If at the intersection of two straight line secant the corresponding angles are equal, then the straight lines are parallel.
Let at the intersection of straight lines a and b secant c, the corresponding angles are equal, for example 1 = 2.
Since corners 2 and 3 are vertical, then 2 = 3. It follows from these two equalities that 1 = 3. But angles 1 and 3 are crosswise, therefore straight lines a and b are parallel. CHTD.
6. The theorem on the sum of the angles of a triangle.
The sum of the angles of a triangle is 180 0.
Consider arbitrary triangle ABC and prove that A + B + C = 180 0.
Let us draw a straight line a through vertex B, parallel to the AC side. Angles 1 and 4 are cross-lying angles at the intersection of parallel straight lines a and AC of the secant AB, and angles 3 and 5 are cross-lying angles at the intersection of the same parallel straight lines of the secant BC. Therefore (1) 4 = 1; 5 = 3.
Obviously, the sum of angles 4, 2 and 5 is equal to the extended angle with vertex B, i.e. 4 + 2 + 5 = 180 0. From here, taking into account equalities (1), we get: 1 + 2 + 3 = 180 0 or A + B + C = 180 0.
7. The sign of equality of right-angled triangles.
1. The first sign of parallelism.
If at the intersection of two straight lines the third internal cross lying angles are equal, then these straight lines are parallel.
Let lines AB and CD be intersected by line ЕF and ∠1 = ∠2. Take point O - the middle of the segment KL secant EF (Fig.).
Let us drop the perpendicular ОМ from the point O to the line AB and continue it to the intersection with the line CD, AB ⊥ МN. Let us prove that СD ⊥ МN.
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; ОK = ОL - by construction;
∠MOL = ∠NOK, as vertical angles. Thus, a side and two adjacent angles of one triangle are respectively equal to a side and two adjacent angles of another triangle; therefore, ΔMOL = ΔNOK, and hence ∠LMO = ∠KNO,
but ∠LMO is direct, hence ∠КNО is also direct. Thus, lines AB and CD are perpendicular to the same line МN, therefore, they are parallel, which was required to prove.
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 if the lines AB and CD will be parallel if the corresponding angles are equal at the intersection of their third line EF.
Let some corresponding angles be equal, for example ∠ 3 = ∠2 (fig.);
∠3 = ∠1, as vertical angles; hence ∠2 will be equal to ∠1. But angles 2 and 1 are cross-lying internal angles, and we already know that if at the intersection of two third straight lines, the cross-lying internal angles are equal, then these lines are parallel. Therefore, AB || CD.
If at the intersection of two straight lines the third corresponding angles are equal, then these two straight lines are parallel.
This property is based on the construction of parallel lines using a ruler and a drawing triangle. This is done as follows.
We attach the triangle to the ruler as shown in Fig. We will move the triangle so that one side of it slides along the ruler, and on some other side of the triangle we will draw several straight lines. These lines will be parallel.
3. The third sign of parallelism.
Suppose we know that at the intersection of two straight lines AB and CD of the third straight line, the sum of some internal one-sided angles is equal to 2 d(or 180 °). In this case, will the straight lines AB and CD be parallel (Fig.).
Let ∠1 and ∠2 be internal one-sided angles and add up to 2 d.
But ∠3 + ∠2 = 2 d as the corners are adjacent. Therefore, ∠1 + ∠2 = ∠3 + ∠2.
Hence, ∠1 = ∠3, and these internal angles lie crosswise. Therefore, AB || CD.
If at the intersection of two straight lines the third, the sum of the inner one-sided angles is 2 d (or 180 °), then these two lines are parallel.
Signs of parallel lines:
1. If at the intersection of two straight lines the third internal cross lying angles are equal, then these straight lines are parallel.2. If at the intersection of two straight lines the third corresponding angles are equal, then these two straight lines are parallel.
3. If at the intersection of two straight lines the third sum of the inner one-sided angles is 180 °, then these two straight 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.
Euclidean parallelism axiom
Task. Through point M, taken outside the straight line AB, draw a straight line parallel to the straight line AB.
Using the proven theorems on criteria for parallelism of lines, this problem can be solved in various ways,
Solution. 1st with p about with about b (Fig. 199).
Draw МN⊥АВ and draw СD⊥МN through point М;
we get СD⊥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 with about with about b (Fig. 200).
We draw the MK that intersects AB at any angle α, and through the point M we draw a straight line EF, which forms an angle EMC with the straight MK equal to the angle α. Based on theorem (), we conclude that ЕF || AB.
Having solved this problem, we can consider it proved that through any point M, taken outside the straight line AB, it is possible to draw a straight line parallel to it. The question arises, how many straight 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 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 this question is given by the so-called Euclidean parallelism axiom; it is formulated as follows:
Through a point taken outside this straight line, you can draw only one straight line parallel to this straight line.
In drawing 201, a straight line CK is drawn through the point O, parallel to the straight line AB.
Any other straight line passing through point O will no longer be parallel to line AB, but will intersect it.
The axiom adopted by Euclid in his "Principles", which states that on a plane through a point taken outside a given straight line, only one straight line parallel to this straight line can be drawn, is called Euclid's axiom of parallelism.
For more than two millennia after Euclid, many mathematicians tried to prove this mathematical proposition, but their attempts were always unsuccessful. Only in 1826 the great Russian scientist, professor of Kazan University Nikolai Ivanovich Lobachevsky proved that using all other Euclidean axioms, this mathematical proposition cannot be proved, that it really should be taken as an axiom. NI Lobachevsky created a new geometry, which, in contrast to the geometry of Euclid, is called Lobachevsky's geometry.
Signs of parallelism of two straight lines
Theorem 1. If at the intersection of two secant lines:
criss-crossing angles are equal, or
the corresponding angles are equal, or
the sum of the one-sided angles is 180 °, then
straight 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 secant AB, the intersecting angles are equal. For example, ∠ 4 = ∠ 6. Let us prove that a || b.
Suppose that lines a and b are not parallel. Then they intersect at some point M and, therefore, one of the angles 4 or 6 will be the outer corner of the triangle ABM. Let, for definiteness, ∠ 4 be the outer corner of the triangle ABM, and ∠ 6 - the inner one. From the theorem about outer corner the triangle implies that ∠ 4 is greater than ∠ 6, and this contradicts the condition, which means that the lines a and 6 cannot intersect, so they are parallel.
Corollary 1. Two different straight lines in a plane perpendicular to the same straight line are parallel(fig. 2).
Comment. The way in which we have just proved case 1 of Theorem 1 is called 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 a reduction to absurdity due to the fact that, arguing on the basis of the assumption made, we come to an absurd conclusion (to an absurdity). The receipt of such a conclusion forces us to reject the assumption made at the beginning and accept the one that was required to be proved.
Objective 1. Construct a straight line passing through a given point M and parallel to a given straight line a, not passing through M.
Solution. Draw a straight line p through the point M perpendicular to the straight line a (Fig. 3).
Then we draw a straight line b through point M perpendicular to a straight line p. Line b is parallel to line a according to the corollary to Theorem 1.
An important conclusion follows from the considered problem:
through a point not lying on a given straight line, you can always draw a straight line parallel to a given.
The main property of parallel lines is as follows.
Axiom of parallel lines. Through a given point, which does not lie on a given straight line, only one straight line, parallel to the given one, passes.
Consider some properties of parallel lines that follow from this axiom.
1) If a line intersects one of 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 intersected by a secant, then:
criss-cross corners are equal;
the corresponding angles are equal;
the sum of the one-sided angles is 180 °.
Corollary 2. If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other(see fig. 2).
Comment. Theorem 2 is called the converse 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 the converse, that is, if this theorem is true, then converse theorem may be wrong.
Let us explain this using 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 theorem converse to it would be as follows: 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 inner one-sided angles is 30 °. Find these corners.
Solution. Let Figure 6 meet the condition.
This chapter is devoted to the study of parallel lines. This is the name of two straight lines on a plane that do not intersect. Sections of parallel lines we see in the environment - these are two edges of a rectangular table, two edges of a book cover, two trolley 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, that is, they intersect, or do not have a single point in common, that is, they do not intersect.
Definition
The parallelism of straight lines a and b is denoted as follows: a || b.
Figure 98 shows straight 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, parallel lines are often considered. The two segments are called parallel if they lie on parallel lines. In Figure 99, and segments AB and CD are parallel (AB || CD), and 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 straight lines
Straight with is called secant with respect to straight lines a and b, if it intersects them at two points (Fig. 100). At the intersection of straight lines a and b secant c, eight corners are formed, 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 straight lines associated with these pairs of angles.
Theorem
Proof
Suppose that at the intersection of lines a and b secant AB, the intersecting angles are equal: ∠1 = ∠2 (Fig. 101, a).
Let us prove that a || b. If angles 1 and 2 are straight (Fig. 101, b), then straight lines a and b are perpendicular to line AB and, therefore, are parallel.
Rice. 101
Consider the case where corners 1 and 2 are not straight.
From the middle O of the segment AB we draw the perpendicular OH to the straight line a (Fig. 101, c). On the straight line b from point B, we postpone the segment BH 1, equal to the segment AH, as shown in Figure 101, c, and draw the segment OH 1. The triangles ОНА and ОН 1 В are equal on two sides and the angle between them (AO = BO, AH = BH 1, ∠1 = ∠2), therefore ∠3 = ∠4 and ∠5 = ∠6. From the equality ∠3 = ∠4 it follows that the point H 1 lies on the extension of the ray OH, that is, the points H, O and H 1 lie on one straight line, and from the equality ∠5 = ∠6 it follows that the angle 6 is a straight line (since angle 5 is straight). So, straight lines a and b are perpendicular to the straight line HH 1, so they are parallel. The theorem is proved.
Theorem
Proof
Let at the intersection of the straight lines a and b secant with the corresponding angles are equal, for example, ∠1 = ∠2 (Fig. 102).
Rice. 102
Since corners 2 and 3 are vertical, then ∠2 = ∠3. It follows from these two equalities that ∠1 = ∠3. But angles 1 and 3 are crosswise, so lines a and b are parallel. The theorem is proved.
Theorem
Proof
Let at the intersection of straight lines a and b secant with the sum of one-sided angles equal to 180 °, for example ∠1 + ∠4 = 180 ° (see Fig. 102).
Since angles 3 and 4 are adjacent, ∠3 + ∠4 = 180 °. From these two equalities it follows that the cross-lying angles 1 and 3 are equal, therefore the straight lines a and b are parallel. The theorem is proved.
Practical ways to build parallel lines
The signs of parallelism of straight lines underlie the methods of constructing parallel straight lines using 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 point M and parallel to a given straight line a, we apply a drawing square to a straight line a, and a ruler to it as shown in Figure 103. Then, moving the square along the ruler, we will achieve that point M is on the side of the square , and draw a line b. 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 way to construct parallel straight lines using a flight bus. This method is used in drawing practice.
Rice. 104 A similar method is used when performing carpentry work, where a malka is used to mark parallel straight lines (two wooden planks, fastened with a hinge, Fig. 105).
Rice. 105
Tasks
186. In Figure 106, straight lines a and b are intersected by straight 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 || DE.
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 ВС || AD.
Rice. 108
190. In Figure 109 AB = BC, AD = DE, ∠C = 70 °, ∠EAC = 35 °. Prove that DE || AC.
Rice. 109
191. Segment BK - bisector of triangle ABC. A straight line is drawn through point K, intersecting the side BC at point M so that BM = MK. Prove that lines KM and AB are parallel.
192. In triangle ABC, angle A is equal to 40 °, and angle BCE adjacent to angle ACB is equal to 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 the AC side. Through point D, using a drawing square and a ruler, draw straight lines parallel to the other two sides of the triangle.
CHAPTER III.
PARALLEL LINE
§ 35. SIGNS OF PARALLELITY OF TWO LINE.
The theorem that two perpendiculars to one straight line are parallel (§ 33) gives a criterion for the parallelism of two straight lines. You can deduce more general signs of parallelism of two straight lines.
1. The first sign of parallelism.
If at the intersection of two straight lines the third internal cross lying angles are equal, then these straight lines are parallel.
Let lines AB and CD be intersected by line EF and / 1 = / 2. Take point O - the middle of the segment KL secant EF (Fig. 189).
Let us drop the perpendicular ОМ from point O to the line AB and continue it to the intersection with the line CD, AB_ | _МN. Let us prove that СD_ | _МN.
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; ОK = ОL - by construction;
/
MOL = /
NOK like vertical corners. Thus, a side and two adjacent angles of one triangle are respectively equal to a side and two adjacent angles of another triangle; hence, /\
MOL = /\
NOK, and hence
/
LMO = /
KNO, but /
LMO is straight, which means that /
KNO is also straightforward. Thus, lines AB and CD are perpendicular to the same line MN, therefore, they are parallel (§ 33), as required.
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 if the lines AB and CD will be parallel if the corresponding angles are equal at the intersection of their third line EF.
Let some corresponding angles be equal, for example /
3 = /
2 (Fig. 190);
/
3 = /
1, as the corners are vertical; means, /
2 will be equal /
1. But angles 2 and 1 are cross-lying internal angles, and we already know that if at the intersection of two third straight lines the internal cross-lying angles are equal, then these lines are parallel. Therefore, AB || CD.
If at the intersection of two straight lines the third corresponding angles are equal, then these two straight lines are parallel.
This property is based on the construction of parallel lines using a ruler and a drawing triangle. This is done as follows.
Let us apply the triangle to the ruler as shown in drawing 191. We will move the triangle so that one side of it slides along the ruler, and on some other side of the triangle we draw several straight lines. These lines will be parallel.
3. The third sign of parallelism.
Suppose we know that at the intersection of two straight lines AB and CD of the third straight line, the sum of some internal one-sided angles is equal to 2 d(or 180 °). Will in this case straight lines AB and CD be parallel (Fig. 192).
Let be /
1 and /
2 are inner one-sided corners and add up to 2 d.
But /
3 + /
2 = 2d as the corners are adjacent. Hence, /
1 + /
2 = /
3+ /
2.
From here / 1 = / 3, and these internal angles lie crosswise. Therefore, AB || CD.
If at the intersection of two straight lines the third, the sum of the inner one-sided angles is 2 d, then these two lines are parallel.
The exercise.
Prove that the lines are parallel:
a) if the outer angles lying crosswise are equal (Fig. 193);
b) if the sum of the outer one-sided corners is 2 d(Fig. 194).
