Two angles are called vertical if the sides of one angle are an extension of the sides of the other.
The figure shows the corners 1 and 3 , as well as angles 2 and 4 - vertical. Injection 2 is adjacent to both angle 1 , and with the angle 3. According to the property of adjacent angles 1 +2 =180 0 and 3 +2 =1800. From here we get: 1=180 0 -2 , 3=180 0 -2. Thus, the degree measures of the angles 1 and 3 are equal. 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 two sides and the angle between them of another triangle, then such triangles are congruent.
If a side and two adjacent angles of one triangle are respectively equal to a side and two adjacent angles of another triangle, then such triangles are congruent.
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 \u003d A 1 B 1, AC \u003d 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 \u003d (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, respectively, on the rays A 1 B 1 and A 1 C 1 . Since AB \u003d A 1 B 1, AC \u003d A 1 C 1, then side AB will be combined with side A 1 B 1, and side AC - with side A 1 C 1; in particular, points B and B 1 , C and C 1 will coincide. Therefore, the sides BC and B 1 C 1 will be aligned. So, the triangles ABC and A 1 B 1 C 1 are completely compatible, which means they are equal. CTD
3. The theorem on the bisector of an isosceles triangle.
In an isosceles triangle, the bisector drawn to the base is the median and height.
Let us turn to the figure, in which ABC is an isosceles triangle with base BC, AD is its bisector.
From the equality of triangles ABD and ACD (according to the 2nd criterion for the equality of triangles: AD is common; angles 1 and 2 are equal because the AD-bisector; AB=AC, because the triangle is isosceles) it follows that BD = DC and 3 = 4. The equality BD = DC means that the point D is the midpoint of the side BC and therefore AD is the median of the triangle ABC. Since angles 3 and 4 are adjacent and equal to each other, they are right angles. Therefore, segment AO is also the height of triangle ABC. CHTD.
4. If the lines are parallel -> angle…. (optional)
5. If the angle ... ..-> lines are parallel (optional)
If at the intersection of two lines of a secant the corresponding angles are equal, then the lines are parallel.
Let at the intersection of lines a and b of the secant with the corresponding angles be equal, for example 1=2.
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. CHTD.
6. Theorem on the sum of the angles of a triangle.
The sum of the angles of a triangle is 180 0.
Consider an arbitrary triangle ABC and prove that A+B+C=180 0 .
Let us draw a straight line a through the vertex B, parallel to the side AC. Angles 1 and 4 are crosswise lying angles at the intersection of parallel lines a and AC by the secant AB, and angles 3 and 5 are crosswise lying angles at the intersection of the same parallel lines by the secant BC. Therefore (1)4=1; 5=3.
Obviously, the sum of angles 4, 2 and 5 is equal to the straight angle with vertex B, i.e. 4+2+5=1800 . Hence, taking into account equalities (1), we obtain: 1+2+3=180 0 or A+B+C=180 0 .
7. Sign of equality of right triangles.
1. The first sign of parallelism.
If, at the intersection of two lines with a third, the interior angles lying across are equal, then these lines are parallel.
Let lines AB and CD be intersected by line EF and ∠1 = ∠2. Let's take the point O - the middle of the segment KL of the secant EF (Fig.).
Let us drop the perpendicular OM from the point O to the line AB and continue it until it intersects with the line CD, AB ⊥ MN. Let us prove that CD ⊥ MN as well.
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; OK = OL - by construction;
∠MOL = ∠NOK as vertical angles. Thus, the side and two angles adjacent to it of one triangle are respectively equal to the side and two angles adjacent to it of another triangle; therefore, ΔMOL = ΔNOK, and hence ∠LMO = ∠KNO,
but ∠LMO is direct, hence ∠KNO is also direct. Thus, the lines AB and CD are perpendicular to the same line MN, therefore, they are parallel, which was to be proved.
Note. The intersection of the lines MO and CD can be established by rotating the triangle MOL around the point O by 180°.
2. The second sign of parallelism.
Let's see if the lines AB and CD are parallel if, at the intersection of their third line EF, the corresponding angles are equal.
Let some corresponding angles be equal, for example ∠ 3 = ∠2 (Fig.);
∠3 = ∠1 as vertical angles; so ∠2 will be equal to ∠1. But angles 2 and 1 are internal crosswise angles, and we already know that if at the intersection of two lines by a third, the internal crosswise lying angles are equal, then these lines are parallel. Therefore, AB || CD.
If at the intersection of two lines of the third the corresponding angles are equal, then these two lines are parallel.
The construction of parallel lines with the help of a ruler and a drawing triangle is based on this property. This is done as follows.
Let us attach a triangle to the ruler as shown in Fig. We will move the triangle so that one side of it slides along the ruler, and draw several straight lines along any other side of the triangle. These lines will be parallel.
3. The third sign of parallelism.
Let us know that at the intersection of two lines AB and CD by the third line, the sum of any internal one-sided angles is equal to 2 d(or 180°). Will the lines AB and CD be parallel in this case (Fig.).
Let ∠1 and ∠2 be one-sided interior 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 are crosswise. Therefore, AB || CD.
If at the intersection of two lines by a third, the sum of the interior one-sided angles is equal to 2 d (or 180°), then the two lines are parallel.
Signs of parallel lines:
1. If at the intersection of two straight lines by a third, the internal cross lying angles are equal, then these lines are parallel.2. If at the intersection of two lines of the third, the corresponding angles are equal, then these two lines are parallel.
3. If at the intersection of two lines of the third, the sum of the internal one-sided angles is 180 °, then these two 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.
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,
Decision. 1st s o s o b (Fig. 199).
We draw MN⊥AB and through the 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 СD || AB.
2nd s p o s o b (Fig. 200).
We draw a MK intersecting AB at any angle α, and through the point M we draw a straight line EF, forming an angle EMK with a straight line MK, equal to the angle α. Based on the theorem () we conclude that EF || AB.
Having solved this problem, we can consider it proved 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 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 line, since with a carefully executed drawing, lines drawn in various ways through the same point parallel to the same line merge.
In theory, the answer to this question is given by the so-called axiom of Euclid's parallelism; it is formulated like this:
Through a point taken outside a given line, only one line can be drawn parallel to this line.
In the drawing 201, a straight line SK is drawn through the point O, parallel to the straight line AB.
Any other line passing through the point O will no longer be parallel to the 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 line can be drawn parallel to this line, is called Euclid's axiom of parallelism.
For more than two thousand years after Euclid, many mathematicians tried to prove this mathematical proposition, but their attempts were always unsuccessful. Only in 1826 did the great Russian scientist, professor of Kazan University Nikolai Ivanovich Lobachevsky prove that, using all other Euclid's axioms, this mathematical proposition cannot be proved, that it really should be taken as an axiom. N. I. Lobachevsky created a new geometry, which, in contrast to the geometry of Euclid, was called the geometry of Lobachevsky.
AB and WithD crossed by the third 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 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.
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.
Decision. 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 a given theorem is true, then the inverse theorem may be false.
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.
Decision. Let figure 6 meet the condition.
