If 2 parallel lines intersect 3. Parallel lines, signs and conditions of parallel lines. crosswise angles are equal
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.
Definition:
Two direct na-zy-va-yut-sya pa-ral-lel-ny-mi, if they do not re-se-ka-yut-sya (Fig. 1). De-zna-cha-et-sya is like this:.
Through a point that does not lie on a given straight line, only one straight line passes, par-al-lel-naya given (Fig. 2) .
Consequences from the axiom
Consequence1:
If a straight line re-se-se-ka-et one of the para-l-lel-nyh straight lines, then it re-se-ka-et and the other.
Given:
.
Prove:.
Proof:
We will do-ka-zy-vat from against-against-no-go. Let's pretend that With do not re-se-ka-et straight b(Fig. 4).
Then: (according to the condition), (according to the pre-position). That is, through the point M pro-ho-dyat two straight lines ( a and c), pa-ral-lel-nye straight-my b. And this is pro-ti-vo-re-chit ak-sio-me. So, our assumption is wrong. Then straight c pe-re-se-even straight b.
Consequence 2:
If two straight lines are para-ral-lel-na of the third straight line, then they are par-ral-lel-na(Fig. 5) .
Given:.
Prove:.
Proof:
We will do-ka-zy-vat from against-against-no-go. Let's assume that we are direct a and b re-re-se-ka-yut-sya at some point M(Fig. 6).
In this way, in-lu-cha-em pro-ti-in-re-chie with ak-si-o-my: through the dot M pass two straight lines, one-but-time-men-but para-ral-lel-ny third straight-my.
Next-to-va-tel-but, our assumption is wrong. Then .
Theorems on the properties of parallel lines
Theo-re-ma 1:
If two straight lines are re-se-che-we se-ku-schey, then the cross-lying angles are equal(Fig. 7).
Given:.
Prove:.
Proof:
We will do-ka-zy-vat from against-against-no-go. Let's pretend that: .
Then from the beam MN you can from-lo-live a single corner ∠ PMN, someone will be equal to ∠ 2 ( Rice. 7). But then ∠ PMN and ∠ 2 - on the cross lying and equal. Then direct PM and b- pa-ral-lel-ny. Then through the point M pass two straight lines, para-ral-lel thirds. Namely:
Po-lu-cha-eat pro-ti-vo-re-chie with ak-si-o-my. So, our assumption is wrong. That is: .
Consequence:
If the line is per-pen-di-ku-lyar-on one of the parallel-lel-ny straight lines, then it is per-pen-di-ku-lyar-na and the second swarm.
Given:
Prove:
Proof:
1. With pe-re-se-ka-et a, which means, and pe-re-se-ka-et para-ral-lel-naya to her direct, that is b. Then With- se-ku-shaya from-no-she-niyu to a and b.
2. insofar as they are-la-yut-sya on the cross of le-zha-schi-mi. Then . That is .
Theo-re-ma 2:
If two parallel lines are re-se-che-we se-ku-schey, then the corresponding angles are equal.
Given:- se-ku-shaya.
Prove:(Fig. 9).
Proof:
If , then from the previous theo-re-we it follows that the cross-lying angles are equal. That is .
CHAPTER III.
PARALLEL LINES
§ 38. DEPENDENCE BETWEEN ANGLES,
FORMED BY TWO PARALLEL LINES AND A SECUTIVE.
We know that two lines are parallel if, at the intersection of their third line, the corresponding angles are equal, or internal or external cross-lying angles, or the sum of internal, or the sum of external one-sided angles is equal to 2 d. Let us prove that the converse theorems are also true, namely:
If two parallel lines are intersected by a third, then:
1) the corresponding angles are equal;
2) internal cross lying angles are equal;
3) external cross lying angles are equal;
4) the sum of internal one-sided angles is equal to 2
d
;
5) the sum of external one-sided angles is equal to 2
d
.
Let us prove, for example, that if two parallel lines are intersected by a third line, then the corresponding angles are equal.
Let the lines AB and CD be parallel, and MN be their secant (Fig. 202). Let us prove that the corresponding angles 1 and 2 are equal to each other.
Let's assume that / 1 and / 2 are not equal. Then, at point O, one can construct / IOC, corresponding and equal / 2 (dev. 203).
But if / IOC = / 2, then the line OK will be parallel to CD (§ 35).
We have obtained that two straight lines AB and OK are drawn through the point O and are parallel to the straight line CD. But this cannot be (§ 37).
We have arrived at a contradiction because we assumed that / 1 and / 2 are not equal. Therefore, our assumption is incorrect and / 1 must be equal / 2, i.e., the corresponding angles are equal.
Establish relationships between the remaining angles. Let the lines AB and CD be parallel, and MN be their secant (Fig. 204).
We have just proved that in this case the corresponding angles are equal. Let us suppose that any two of them have 119° each. Calculate the value of each of the remaining six angles. Based on the properties of adjacent and vertical angles, we get that four of the eight angles will have 119°, and the rest - 61°.
It turned out that both internal and external cross-lying angles are pairwise equal, and the sum of internal or external one-sided angles is equal to 180° (or 2 d).
The same will take place for any other value of equal corresponding angles.
Consequence 1. If each of the two lines AB and CD is parallel to the same third line MN, then the first two lines are parallel to each other (dev. 205).
Indeed, after drawing the secant EF (Fig. 206), we get:
a) /
1 = /
3, because AB || MN; b) /
2 = /
3, since CO || MN.
Means, / 1 = / 2, and these are the angles corresponding to the lines AB and CD and the secant EF, therefore, the lines AB and CD are parallel.
Consequence 2. If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other. (dev. 207).
Indeed, if EF _|_ AB, then / 1 = d; if AB || CD, then / 1 = / 2.
Consequently, / 2 = d i.e. EF _|_ CD .
Page 1 of 2
Question 1. Prove that two lines parallel to the third are parallel.
Answer. Theorem 4.1. Two lines parallel to a third are parallel.
Proof. Let lines a and b be parallel to line c. Assume that a and b are not parallel (Fig. 69). Then they do not intersect at some point C. Hence, two lines pass through the point C and are parallel to the line c. But this is impossible, since through a point that does not lie on a given line, at most one line parallel to the given line can be drawn. The theorem has been proven.
Question 2. Explain what angles are called internal one-sided. What angles are called internal cross lying?
Answer. Pairs of angles that are formed when lines AB and CD intersect AC have special names.
If the points B and D lie in the same half-plane relative to the straight line AC, then the angles BAC and DCA are called internal one-sided (Fig. 71, a).
If the points B and D lie in different half-planes relative to the line AC, then the angles BAC and DCA are called internal crosswise lying (Fig. 71, b).
Rice. 71
Question 3. Prove that if the internal cross-lying angles of one pair are equal, then the internal cross-lying angles of the other pair are also equal, and the sum of the internal one-sided angles of each pair is 180°.
Answer. The secant AC forms with lines AB and CD two pairs of internal one-sided and two pairs of internal cross-lying angles. The internal cross-lying corners of one pair, for example, angle 1 and angle 2, are adjacent to the internal cross-lying angles of another pair: angle 3 and angle 4 (Fig. 72).
Rice. 72
Therefore, if the internal cross-lying angles of one pair are equal, then the internal cross-lying angles of the other pair are also equal.
A pair of interior cross-lying corners, such as angle 1 and angle 2, and a pair of internal one-sided corners, such as angle 2 and angle 3, have one common angle, angle 2, and two other adjacent angles, angle 1 and angle 3.
Therefore, if the interior cross-lying angles are equal, then the sum of the interior angles is 180°. And vice versa: if the sum of interior cross-lying angles is equal to 180°, then the interior cross-lying angles are equal. Q.E.D.
Question 4. Prove the criterion for parallel lines.
Answer. Theorem 4.2 (test for parallel lines). If interior cross-lying angles are equal or the sum of interior one-sided angles is 180°, then the lines are parallel.
Proof. Let the lines a and b form equal internal crosswise lying angles with the secant AB (Fig. 73, a). Suppose the lines a and b are not parallel, which means they intersect at some point C (Fig. 73, b).
Rice. 73
The secant AB splits the plane into two half-planes. Point C lies in one of them. Let's construct a triangle BAC 1 , triangular ABC, with vertex C 1 in the other half-plane. By condition, the internal cross-lying angles for parallel a, b and secant AB are equal. Since the corresponding angles of triangles ABC and BAC 1 with vertices A and B are equal, they coincide with the internal cross-lying angles. Hence, line AC 1 coincides with line a, and line BC 1 coincides with line b. It turns out that two different lines a and b pass through the points C and C 1. And this is impossible. So lines a and b are parallel.
If lines a and b and secant AB have the sum of internal one-sided angles equal to 180°, then, as we know, the internal cross-lying angles are equal. Hence, by what was proved above, the lines a and b are parallel. The theorem has been proven.
Question 5. Explain what angles are called corresponding. Prove that if interior cross-lying angles are equal, then the corresponding angles are also equal, and vice versa.
Answer. If a pair of internal cross-lying angles has one angle replaced by a vertical one, then a pair of angles will be obtained, which are called the corresponding angles of the given lines with a secant. Which is what needed to be explained.
From the equality of internal cross-lying angles follows the equality of the corresponding angles, and vice versa. Let's say we have two parallel lines (because by condition the internal cross-lying angles are equal) and a secant, which form angles 1, 2, 3. Angles 1 and 2 are equal as internal cross-lying. And angles 2 and 3 are equal as vertical. We get: \(\angle\)1 = \(\angle\)2 and \(\angle\)2 = \(\angle\)3. By the property of transitivity of the equal sign, it follows that \(\angle\)1 = \(\angle\)3. The converse assertion is proved similarly.
This results in a sign of parallel lines at the corresponding angles. Namely, lines are parallel if the corresponding angles are equal. Q.E.D.
Question 6. Prove that through a point not lying on a given line, it is possible to draw a line parallel to it. How many lines parallel to a given line can be drawn through a point not on this line?
Answer. Problem (8). Given a line AB and a point C not lying on this line. Prove that through point C it is possible to draw a line parallel to line AB.
Solution. The straight line AC divides the plane into two half-planes (Fig. 75). Point B lies in one of them. From the half-line CA, let us plot the angle ACD equal to the angle CAB into the other half-plane. Then lines AB and CD will be parallel. Indeed, for these lines and the secant AC, the angles BAC and DCA are interior crosswise. And since they are equal, lines AB and CD are parallel. Q.E.D.
Comparing the statement of problem 8 and axiom IX (the main property of parallel lines), we come to an important conclusion: through a point that does not lie on a given line, one can draw a line parallel to it, and only one.
Question 7. Prove that if two lines intersect with a third line, then the interior cross-lying angles are equal, and the sum of interior one-sided angles is 180°.
Answer. Theorem 4.3(converse to Theorem 4.2). If two parallel lines intersect with a third line, then the interior cross-lying angles are equal, and the sum of the interior one-sided angles is 180°.
Proof. Let a and b be parallel lines and c be the line intersecting them at points A and B. Let us draw a line a 1 through point A so that the internal cross-lying angles formed by the secant c with lines a 1 and b are equal (Fig. 76).
By the criterion of parallelism of lines, lines a 1 and b are parallel. And since only one line passes through the point A, parallel to the line b, then the line a coincides with the line a 1 .
This means that the internal cross-lying angles formed by the secant with
parallel lines a and b are equal. The theorem has been proven.
Question 8. Prove that two lines perpendicular to a third are parallel. If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.
Answer. It follows from Theorem 4.2 that two lines perpendicular to a third are parallel.
Assume that any two lines are perpendicular to the third line. Hence, these lines intersect with the third line at an angle equal to 90°.
From the property of the angles formed at the intersection of parallel lines by a secant, it follows that if a line is perpendicular to one of the parallel lines, then it is also perpendicular to the other.
Question 9. Prove that the sum of the angles of a triangle is 180°.
Answer. Theorem 4.4. The sum of the angles of a triangle is 180°.
Proof. Let ABC be the given triangle. Draw a line through vertex B parallel to line AC. Mark a point D on it so that points A and D lie along different sides from straight line BC (Fig. 78).
Angles DBC and ACB are equal as internal crosswise, formed by the secant BC with parallel lines AC and BD. Therefore, the sum of the angles of the triangle at the vertices B and C is equal to the angle ABD.
And the sum of all three angles of a triangle is equal to the sum of angles ABD and BAC. Since these angles are internal one-sided for parallel AC and BD and secant AB, their sum is 180°. The theorem has been proven.
Question 10. Prove that any triangle has at least two acute angles.
Answer. Indeed, suppose that a triangle has only one acute angle or no acute angles at all. Then this triangle has two angles, each of which is at least 90°. The sum of these two angles is no less than 180°. But this is impossible, since the sum of all the angles of a triangle is 180°. Q.E.D.
Signs of parallelism of two lines
Theorem 1. If at the intersection of two lines of a secant:
diagonally lying angles are equal, or
corresponding angles are equal, or
the sum of one-sided angles is 180°, then
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 by a secant AB across the lying angles are equal. For example, ∠ 4 = ∠ 6. Let us prove that a || b.
Assume that lines a and b are not parallel. Then they intersect at some point M and, consequently, one of the angles 4 or 6 will be the external angle of the triangle ABM. Let, for definiteness, ∠ 4 be the outer corner of the triangle ABM, and ∠ 6 be the inner one. It follows from the theorem on the external angle of a triangle that ∠ 4 is greater than ∠ 6, and this contradicts the condition, which means that the lines a and 6 cannot intersect, therefore they are parallel.
Corollary 1. Two distinct lines in a plane perpendicular to the same line are parallel(Fig. 2).
Comment. The way we just proved case 1 of Theorem 1 is called the method of proof by 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 reduction to absurdity due to the fact that, arguing on the basis of the assumption made, we come to an absurd conclusion (absurdity). Receiving such a conclusion forces us to reject the assumption made at the beginning and accept the one that was required to be proved.
Task 1. Construct a line passing through a given point M and parallel to a given line a, not passing through the point M.
Solution. We draw a line p through the point M perpendicular to the line a (Fig. 3).
Then we draw a line b through the point M perpendicular to the line p. The line b is parallel to the line a according to the corollary of Theorem 1.
An important conclusion follows from the considered problem:
Through a point not on a given line, one can always draw a line parallel to the given line..
The main property of parallel lines is as follows.
Axiom of parallel lines. Through a given point not on a given line, there is only one line parallel to the given line.
Consider some properties of parallel lines that follow from this axiom.
1) If a line intersects one of the 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 crossed by a secant, then:
the lying angles are equal;
corresponding angles are equal;
the sum of one-sided angles is 180°.
Consequence 2. If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other.(see Fig.2).
Comment. Theorem 2 is called the inverse 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 an inverse, i.e. if this theorem is true, then converse theorem may be incorrect.
Let us explain this with 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 inverse theorem would be this: 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 internal one-sided angles is 30°. Find those angles.
Solution. Let figure 6 meet the condition.
