By studying the properties of geometric figures, we proved a number of theorems. In doing so, we relied, as a rule, on previously proven theorems. What are the proofs of the very first theorems of geometry based on? The answer to this question is this: certain statements about the properties of geometric figures are accepted as starting points, on the basis of which further theorems are proved and, in general, all geometry is constructed. Such initial positions are called axioms.
Some axioms were formulated back in the first chapter (although they were not called axioms there). For example, it is an axiom that
Many other axioms, although not particularly emphasized, were actually used in our reasoning. Thus, we compared two segments by superimposing one segment on another. The possibility of such an overlap follows from the following axiom:
Comparison of two angles is based on a similar axiom:
All these axioms are clearly obvious and beyond doubt. The word “axiom” itself comes from the Greek “axios”, which means “valuable, worthy.” We provide a complete list of planimetry axioms adopted in our geometry course at the end of the textbook.
This approach to the construction of geometry, when the initial positions - axioms - are first formulated, and then other statements are proven on their basis through logical reasoning, originated in ancient times and was outlined in the famous work “Principles” by the ancient Greek scientist Euclid. Some of Euclid's axioms (some of them he called postulates) and are now used in geometry courses, and the geometry itself, presented in the “Elements”, is called Euclidean geometry. In the next paragraph we will get acquainted with one of the most famous axioms of geometry.
Axiom of parallel lines
Let us consider an arbitrary straight line a and a point M that does not lie on it (Fig. 110, a). Let us prove that through the point M it is possible to draw a line parallel to the line a. To do this, draw two straight lines through point M: first straight line c perpendicular to straight line a, and then straight line b perpendicular to straight line c (Fig. 110, (b). Since straight lines a and b are perpendicular to straight line c, they are parallel.
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So, through point M there passes a line b parallel to line a. The following question arises: is it possible to draw another line through point M, parallel to straight line a?
It seems to us that if straight line b is “turned” even by a very small angle around point M, then it will intersect straight line a (line b" in Figure 110.6). In other words, it seems to us that it is impossible to draw another straight line through point M ( different from b), parallel to line a. Is it possible to prove this statement?
This question has a long history. Euclid’s “Elements” contains a postulate (Euclid’s fifth postulate), from which it follows that through a point not lying on a given line, only one straight line can be drawn parallel to the given one. Many mathematicians, starting from ancient times, have attempted to prove Euclid's fifth postulate, that is, to derive it from other axioms. However, these attempts were unsuccessful every time. And only in the last century it was finally clarified that the statement about the uniqueness of a line passing through a given point parallel to a given line cannot be proven on the basis of the remaining axioms of Euclid, but is itself an axiom.
The great Russian mathematician Nikolai Ivanovich Lobachevsky (1792-1856) played a huge role in solving this difficult issue.
So, as another starting point we accept axiom of parallel lines.
Statements that are derived directly from axioms or theorems are called consequences. For example, statements 1 and 2 (see p. 35) are consequences of the theorem on the bisector of an isosceles triangle.
Let's consider some corollaries from the axiom of parallel lines.
Indeed, let straight lines a and b be parallel and straight line c intersect straight line a at point M (Fig. 111, a). Let us prove that line c also intersects line b. If line c did not intersect line b, then two lines (lines a and c) parallel to line b would pass through point M (Fig. 111, b). But this contradicts the axiom of parallel lines, and, therefore, line c intersects line b.
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Indeed, let straight lines a and b be parallel to straight line c (Fig. 112, a). Let us prove that a || b. Let us assume that lines a and b are not parallel, that is, they intersect at some point M (Fig. 112.6). Then two lines pass through point M (lines a and b), parallel to line c.
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But this contradicts the axiom of parallel lines. Therefore, our assumption is incorrect, which means that lines a and b are parallel.
Theorems on angles formed by two parallel lines and a transversal
Every theorem has two parts: condition And conclusion. The condition of the theorem is what is given, and the conclusion is what needs to be proven.
Let us consider, for example, a theorem expressing the criterion for the parallelism of two straight lines: if, when two straight lines intersect with a transversal, the lying angles are equal, then the straight lines are parallel.
In this theorem, the condition is the first part of the statement: “when two lines intersect crosswise, the lying angles are equal” (this is given), and the conclusion is the second part: “the lines are parallel” (this needs to be proven).
The converse of this theorem, is a theorem in which the condition is the conclusion of the theorem, and the conclusion is the condition of the theorem. Let us prove the theorems converse to the three theorems in paragraph 25.
Theorem
Proof
Let parallel lines a and b be intersected by the secant MN. Let us prove that the angles lying crosswise, for example 1 and 2, are equal (Fig. 113).
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Let's assume that angles 1 and 2 are not equal. Let us subtract from the ray MN an angle PMN equal to angle 2, so that ∠PMN and ∠2 are crosswise angles at the intersection of lines MR and b by the secant MN. By construction, these crossed angles are equal, so MR || b. We found that two lines pass through point M (lines a and MP), parallel to line b. But this contradicts the axiom of parallel lines. This means that our assumption is incorrect and ∠1 = ∠2. The theorem is proven.
Comment
In proving this theorem, we used a method of reasoning called by proof by contradiction.
We assumed that when parallel lines a and b intersect the secant MN crosswise, the lying angles 1 and 2 are not equal, i.e. we assumed the opposite of what needs to be proven. Based on this assumption, through reasoning we came to a contradiction with the axiom of parallel lines. This means that our assumption is incorrect and therefore ∠1 = ∠2.
This way of reasoning is often used in mathematics. We used it earlier, for example, in paragraph 12 when proving that two lines perpendicular to a third do not intersect. We used the same method in paragraph 28 to prove corollaries 1 0 and 2 0 from the axiom of parallel lines.
Consequence
Indeed, let a || b, c ⊥ a, i.e. ∠1 = 90° (Fig. 114). Line c intersects line a, so it also intersects line b. When parallel lines a and b intersect with a transversal c, equal crosswise angles are formed: ∠1=∠2. Since ∠1 = 90°, then ∠2 = 90°, i.e., c ⊥ b, which is what needed to be proved.
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Theorem
Proof
Let parallel lines a and b be intersected by a secant c. Let us prove that the corresponding angles, for example 1 and 2, are equal (see Fig. 102). Since a || b, then the crosswise angles 1 and 3 are equal.
Angles 2 and 3 are equal as vertical. From the equalities ∠1 = ∠3 and ∠2 = ∠3 it follows that ∠1 = ∠2. The theorem is proven.
Theorem
Proof
Let parallel lines a and b be intersected by a secant c (see Fig. 102). Let us prove, for example, that ∠1 + ∠4 = 180°. Since a || b, then the corresponding angles 1 and 2 are equal. Angles 2 and 4 are adjacent, so ∠2 + ∠4 = 180°. From the equalities ∠1 = ∠2 and ∠2 + ∠4 = 180° it follows that ∠1 + ∠4 = 180°. The theorem is proven.
Comment
If a certain theorem is proven, then the converse statement does not follow. Moreover, the converse is not always true. Let's give a simple example. We know that if the angles are vertical, then they are equal. The converse statement: “if the angles are equal, then they are vertical” is, of course, false.
Angles with respectively parallel or perpendicular sides
Let us prove the theorem about angles with correspondingly parallel sides.
Theorem
Proof
Let ∠AOB and ∠A 1 O 1 B 1 be the given angles and OA || O 1 A 1 , OB || About 1 In 1. If angle AOB is developed, then angle A 1 O 1 B 1 is also developed (explain why), so these angles are equal. Let ∠AOB be an undeveloped angle. Possible cases of the location of the angles AOB and A 1 O 1 B 1 are shown in Figure 115, a and b. The straight line O 1 B 1 intersects the line O 1 A 1 and, therefore, intersects the line OA parallel to it at some point M. The parallel lines OB and O 1 B 1 are intersected by the secant OM, therefore one of the angles formed at the intersection of the straight lines O 1 B 1 and OA (angle 1 in Figure 115), is equal to angle AOB (like crosswise angles). Parallel lines OA and O 1 A 1 are intersected by the secant O 1 M, therefore either ∠1 = ∠A 1 O 1 B 1 (Fig. 115, a), or ∠1 + ∠A 1 O 1 B 1 = 180° (Fig. . 115, b). From the equality ∠1 = ∠AOB and the last two equalities it follows that either ∠AOB = ∠A 1 O 1 B 1 (see Fig. 115, a), or ∠AOB + ∠A 1 O 1 B 1 = 180° ( see Fig. 115, b). The theorem is proven.
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Let us now prove the theorem about angles with correspondingly perpendicular sides.
Theorem
Proof
Let ∠AOB and ∠A 1 O 1 B 1 be given angles, OA ⊥ O 1 A 1 , OB ⊥ O 1 B 1 . If angle AOB is reversed or straight, then angle A 1 O 1 B 1 is reversed or straight (explain why), so these angles are equal. Let ∠AOB< 180°, О ∉ О 1 А 1 , О ∉ О 1 В 1 (случаи О ∈ O 1 А 1 , О ∈ О 1 В 1 рассмотрите самостоятельно).
Two cases are possible (Fig. 116).
1 0 . ∠AOB< 90° (см. рис. 116, а). Проведём луч ОС так, чтобы прямые ОА и ОС были взаимно перпендикулярными, а точки В и С лежали по разные стороны от прямой О А. Далее, проведём луч OD так, чтобы прямые ОВ и OD были взаимно перпендикулярными, а точки С и D лежали по одну сторону от прямой О А. Поскольку ∠AOB = 90° - ∠AOD и ∠COD = 90° - ∠AOD, то ∠AOB = ∠COD. Стороны угла COD соответственно параллельны сторонам угла А 1 О 1 В 1 (объясните почему), поэтому либо ∠COD = ∠A 1 O 1 B 1 , либо ∠COD + ∠A 1 O 1 B 1 = 180°. Следовательно, либо ∠AOB = ∠A 1 O 1 B 1 , либо ∠AOB + ∠A 1 O 1 B 1 = 180°.
2 0 . ∠AOB > 90° (see Fig. 116, b). Let us draw the ray OS so that the angle AOS is adjacent to the angle AOB. Angle AOC is acute, and its sides are correspondingly perpendicular to the sides of angle A 1 O 1 B 1 . Therefore, either ∠AOC + ∠A 1 O 1 B 1 = 180°, or ∠AOC = ∠A 1 O 1 B 1 . In the first case, ∠AOB = ∠A 1 O 1 B 1, in the second case, ∠AOB + ∠A 1 O 1 B 1 = 180°. The theorem is proven.
Tasks
196. Given a triangle ABC. How many lines parallel to side AB can be drawn through vertex C?
197. Four straight lines are drawn through a point not lying on the line p. How many of these lines intersect line p? Consider all possible cases.
198. Lines a and b are perpendicular to line p, line c intersects line a. Does line c intersect line b?
199. Line p is parallel to side AB of triangle ABC. Prove that lines BC and AC intersect line r.
200. In Figure 117 AD || p and PQ || Sun. Prove that line p intersects lines AB, AE, AC, BC and PQ.
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201. The sum of crosswise angles when two parallel lines intersect with a transversal is equal to 210°. Find these angles.
202. In Figure 118, lines a, b and c are intersected by line d, ∠1 = 42°, ∠2 = 140°, ∠3 = 138°. Which of the lines a, b and c are parallel?
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203. Find all the angles formed when two parallel lines a and b intersect with a transversal c, if:
a) one of the angles is 150°;
b) one of the angles is 70° greater than the other.
204. The ends of the segment AB lie on parallel lines a and b. The straight line passing through the middle O of this segment intersects lines a and b at points C and D. Prove that CO = OD.
205. Using the data in Figure 119, find ∠1.
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206. ∠ABC = 70°, and ABCD = 110°. Can direct AB and CD be:
a) parallel;
b) intersecting?
207. Answer the questions in Problem 206 if ∠ABC = 65° and ∠BCD = 105°.
208. The difference between two one-sided angles when two parallel lines intersect with a transversal is 50°. Find these angles.
209. In figure 120 a || b, c || d, ∠4 = 45°. Find angles 1, 2 and 3.
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210. Two bodies P 1 and P 2 are suspended at the ends of a thread thrown over blocks A and B (Fig. 121). The third body P 3 is suspended from the same thread at point C and balances the bodies P 1 and P 2. (In this case, AP 1 || BP 2 || CP 3 .) Prove that ∠ACB = ∠CAP 1 + ∠CBP 2 .
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211. Two parallel lines are intersected by a transversal. Prove that: a) bisectors of opposite angles are parallel; b) bisectors of one-sided angles are perpendicular.
212. Straight lines containing altitudes AA 1 and BB 1 of triangle ABC intersect at point H, angle B is obtuse, ∠C = 20°. Find the angle ABB.
Answers to problems
196. One straight line.
197. Three or four.
201. 105°, 105°.
203. b) Four angles are 55°, four other angles are 125°.
206. a) Yes; b) yes.
207. a) No; b) yes.
208. 115° and 65°.
209. ∠1 = 135°, ∠2 = 45°, ∠3=135°.
210. Instruction. Consider the continuation of the beam CP 3.
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Fig.1-2
For example, the task is given to draw two parallel lines, and so that through a given point M at least one of the straight lines passed. Thus, through a given point M draw mutually perpendicular lines MN And CD . And through the point N let's draw a second straight line AB , it must be perpendicular to the line MN .
Let's conclude: straight AB perpendicular to the line MN and straight CD is also perpendicular to the line MN , and since these lines are parallel to one line, then, as a consequence, the line CD parallel AB . So, through the point M there is a straight line CD , which is parallel to the line AB . Let's find out: is it possible to draw another straight line through the point? M so that it is parallel to the line AB ?
This statement is the answer to our question: through a point on the plane that does not lie on a given line, you can draw only one straight line, which will be parallel to the given line. Such a rejection in a different formulation without evidence was accepted in ancient times by the scientist Euclid. It is known that such statements, accepted without proof, are called axioms.
The above statement is called the parallel lines axiom. This axiom of Euclid is of great importance for the proof of many theorems.
Let's consider the converse theorem. If a straight line intersects parallel lines, then the angles lying crosswise at parallel lines are correspondingly equal.
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3
Proof: suppose that AC And ВD are parallel lines, then the line AB is their secant line. We need to prove that РСАВ =Р АВD .
We need to draw a straight line like this AC1 , to РС1АВ=РАВD . In accordance with the axiom of parallel lines AC1||ВD , in the condition we have AC||ВD . And this means that through this point A two lines pass through, and they are parallel to the line ВD . This results in a contradiction to the axiom of parallel lines, which means that the straight line AC1 carried out incorrectly.
It will be correct if РСАВ=РАВD . Let us conclude: in the case when a given straight line is perpendicular to one of the parallel lines, then it will be perpendicular to the second line.
It turns out if (MN)^(CD) And (CD)||(AB) , That р1=р2=90о . And this means: (MN)^(AB) (Fig. 1) .
Let's prove the theorem: if two lines are parallel to the third, then they will be parallel to the second.
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Let it be straight a parallel to the line With and straight b also parallel to the line With (Fig. 4 a) . We need to prove that a||b .
Let's assume that straight lines a And b are not parallel, but they intersect at a point M (Fig. 4 b) . And this means that two straight lines a And b , which are parallel to the line with pass through one point, and this is a complete contradiction to the axiom of parallel lines. So ours are direct a And b parallel.
