Elementary geometry - Sholaster N.N. Signs of parallelism of two lines

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Signs of parallelism of two lines

Theorem 1. If at the intersection of two lines by a secant:

crossed 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 limit ourselves to proving case 1.

Let the intersecting lines a and b be crosswise and the angles AB be 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 external angle of triangle ABM. For definiteness, let ∠ 4 be the external angle of the triangle ABM, and ∠ 6 the internal one. From the theorem on the external angle of a triangle it follows that ∠ 4 is greater than ∠ 6, and this contradicts the condition, which means that lines a and 6 cannot intersect, so they are parallel.

Corollary 1. Two different 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 received its first name because at the beginning of the argument an assumption is made that is contrary (opposite) to what needs to be proven. It is called leading to absurdity due to the fact that, reasoning on the basis of the assumption made, we come to an absurd conclusion (to the absurd). Receiving such a conclusion forces us to reject the assumption made at the beginning and accept the one that needed to be proven.

Task 1. Construct a line passing through this point M and parallel to a given line a, not passing through the point M.

Solution. We draw a straight line p through the point M perpendicular to the straight line a (Fig. 3).

Then we draw a line b through point M perpendicular to the line p. Line b is parallel to line a according to the corollary of Theorem 1.

An important conclusion follows from the problem considered:
through a point not lying on a given line, it is always possible to draw a line parallel to the given one.

The main property of parallel lines is as follows.

Axiom of parallel lines. Through a given point that does not lie on a given line, there passes only one line parallel to the given one.

Let us consider some properties of parallel lines that follow from this axiom.

1) If a line intersects one of two parallel lines, then it also intersects the other (Fig. 4).

2) If two different lines are parallel to a third line, then they are parallel (Fig. 5).

The following theorem is also true.

Theorem 2. If two parallel lines are intersected by a transversal, then:

crosswise angles are equal;

corresponding angles are equal;

the sum of one-sided angles is 180°.

Corollary 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, that is, if a given theorem is true, then the inverse theorem may be false.

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 converse theorem would be: if two angles are equal, then they are vertical. And this, of course, is not true. Two equal angles don't 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 these angles.

Solution. Let Figure 6 meet the condition.

Each angle, depending on its size, has its own name:

Angle type	Size in degrees	Example
Spicy	Less than 90°
Straight	Equal to 90°. In a drawing, a right angle is usually denoted by a symbol drawn from one side of the angle to the other.
Blunt	More than 90° but less than 180°
Expanded	Equal to 180° Full angle equal to the sum two right angles, and a right angle is half a straight angle.
Convex	More than 180° but less than 360°
Full	Equal to 360°

The two angles are called adjacent, if they have one side in common, and the other two sides form a straight line:

Angles MOP And PON adjacent, since the beam OP - common side, and the other two sides - OM And ON make up a straight line.

The common side of adjacent angles is called oblique to straight, on which the other two sides lie, only in the case when adjacent angles are not equal to each other. If adjacent angles are equal, then their common side will be perpendicular.

The sum of adjacent angles is 180°.

The two angles are called vertical, if the sides of one angle complement the sides of the other angle to straight lines:

Angles 1 and 3, as well as angles 2 and 4, are vertical.

Vertical angles are equal.

Let us prove that the vertical angles are equal:

The sum of ∠1 and ∠2 is a straight angle. And the sum of ∠3 and ∠2 is a straight angle. So these two amounts are equal:

∠1 + ∠2 = ∠3 + ∠2.

In this equality, there is an identical term on the left and right - ∠2. Equality will not be violated if this term on the left and right is omitted. Then we get it.

Edited by Ivanitskaya V.P. - M.: State educational and pedagogical publishing house of the Ministry of Education of the RSFSR, 1959. - 272 p.
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If adjacent angles are equal, then each of them is called a right angle. Their common side is called perpendicular to the line formed by the other two sides. We can also say that the bisector of a reverse angle is perpendicular to the line formed by its sides.

Theorem. If the angles are equal, then the adjacent angles are equal.

Let (h, k) = ^. (I, m) and let ^ (h!, k) and ^ (/", t) be the corresponding adjacent angles (Fig. 20). Let, further, / be the movement at which ^ (h, k) is displayed in (I, tri). With this movement, the expanded ^ (h, K) will be mapped into the expanded (I, /"). It follows that ^(h", k) will be mapped into ^(V, m), i.e. ^(h!, k) = ^(V, m).

Theorem. There is a bisector of any angle and, moreover, a unique one.

Let ^(A, k) be different from the expanded one and let its interior region be convex. Let's lay out equal segments OA and OB on its sides from vertex O (Fig. 21, a) and connect points A and B. B isosceles triangle AOB A = ^B (§ 8). By connecting the middle C of the segment AB with the point O, we obtain triangles L OS and BOC that are equal in the first attribute. Therefore, AOC = BOC, and therefore the ray OS is a bisector (h, k).

If (h, k) is not convex (in the drawing its internal region is not shaded), then according to the previous

6}
t^

According to the theorem, its bisector is the ray m complementary to the ray /.

From the equality of triangles ACO and BCO it also follows that ^ ACO = BCO1 i.e. ray CO is the bisector of a reversed angle with sides CA and CB.

Let now be given an expanded ^(p,<7) (черт.21,6). Совершим движение, при котор ом р азвер нутый

ACB is displayed in

(p, q). The CO beam is mapped into the t beam. Since ^ (p, t) = ^lBCO , ^BCO= ^ACO and ^ACO= = (q, t), then (p, t) = = ^(q, t), i.e. t -bisector (p, q).

Let / be the bisector

(A, A), and Г is an arbitrary ray emerging from the vertex of the angle and lying in its interior region. If Γ lies in the inner region ^(A, /), then ^(A, /")<^ (А, /) и ^ (А, Г) >^ (A, /). Therefore, ^(A, G)<^ (А, /"). Отсюда следует, что угол имеет единственную биссектрису. Теорема доказана.

Corollary 1. There is one and only one perpendicular to a given line, emanating from a given point on it and lying in a given half-plane bounded by this line.

Corollary 2. Halves of equal angles are equal to each other.

Indeed, if ^(A, A) = ^(A", A"), then there is a movement / in which one of them is mapped into the other. According to the proven theorem, their bisectors / and Γ for a given motion should also be mapped into one another. Therefore ^(A, /) = ^(A", Г).

Since all straight angles are equal, a special case of Corollary 2 is the proposition: all right angles are equal to each other.

Straight lines a and A that form right angles when intersecting are called perpendicular (a ± b).

Reflection from a straight line. Let straight line a lie in plane a. The half-planes formed in this case will be denoted by X and p. (Figure 22). Let's take ray A on a straight line

emerging from point O. By the property of 6 motions (§ 7), there is a unique motion mapping the ray h into itself and the half-plane X into the half-plane jx. All points of this ray, according to the property of 5 movements, are mapped into themselves. All points of the ray k, complementary to the direct ray h, are also mapped onto themselves.

So, during the motion under consideration, all points of line a are mapped onto themselves. It is easy, further, to see that

Let us now take a point outside line a.

Theorem. Through any point not lying on a line there passes a single line perpendicular to the given line.

Proof. Let M be a point lying outside the straight line a (Fig. 23). Line a divides the plane defined by this line and

point M, into two half-planes: the half-plane X containing the point M, and the half-plane jx. When reflected from straight line a, point M is mapped to point M" of the half-plane jx. Since points M and M" lie in different half-planes,

ah, then straight MM" and Damn 23

intersect at some

point M0, which, when reflected, is mapped onto itself. It follows that the straight line MM" is mapped onto itself, and therefore the angles / and 2 formed by it with the straight line a (see Fig. 23) are mapped into one another.

The half-plane jx is mapped into the half-plane X.

The motion under consideration is called reflection from straight line a.

From the existence of the bisector of a reverse angle it follows that through any point lying on line a, it is always possible to draw a line b perpendicular to line a.

This means that these angles are equal, and since they are, in addition, adjacent, then MM" ± a. Now let another straight line be drawn through M, intersecting line a at some point Af0. It will be mapped into the line M"N0, a ^ MN0M0 will be mapped in M"N0M0. So, ^ 3 = ^i4. But by virtue of axiom 1 (§ 2), the points M1 N0 and M" do not lie on the same line, and therefore the sum of the angles 3 and 4, i.e. ^ MN0M", is not a reversed angle. It follows that angles 3 and 4 are different from a right angle and straight line MN0 will not be perpendicular to straight line a. Line MM" is, therefore, the only straight line perpendicular to a and passing through point M.