1 sign of equality of triangles is called. The third sign of equality of triangles

Among huge amount polygons, which are essentially a closed non-intersecting polyline, a triangle is the figure with the fewest angles. In other words, this is the simplest polygon. But, despite all its simplicity, this figure is fraught with many mysteries and interesting discoveries that are illuminated special section mathematics - geometry. This discipline in schools begins to be taught from the seventh grade, and the topic "Triangle" is given here Special attention. Children not only learn the rules about the figure itself, but also compare them, studying 1, 2 and 3 sign of equality of triangles.

First meeting

One of the first rules that students learn is something like this: the sum of the values of all the angles of a triangle is 180 degrees. To confirm this, it is enough to measure each of the vertices with the help of a protractor and add up all the resulting values. Based on this, with two known values, it is easy to determine the third. for example: In a triangle, one of the angles is 70° and the other is 85°, what is the value of the third angle?

180 - 85 - 70 = 25.

Answer: 25°.

Tasks can be even more complex if only one value of the angle is indicated, and the second value is said only by how much or how many times it is larger or smaller.

In a triangle, to determine one or another of its features, special lines can be drawn, each of which has its own name:

height - a perpendicular line drawn from the top to the opposite side;
all three heights drawn simultaneously intersect in the center of the figure, forming the orthocenter, which, depending on the type of triangle, can be both inside and outside;
median - a line connecting the top with the middle of the opposite side;
the intersection of the medians is the point of its gravity, located inside the figure;
bisector - a line passing from the vertex to the point of intersection with the opposite side, the point of intersection of three bisectors is the center of the inscribed circle.

Simple Truths About Triangles

Triangles, like, in fact, all shapes, have their own characteristics and properties. As already mentioned, this figure is the simplest polygon, but with its own characteristic features:

opposite the longest side there is always an angle with a larger value, and vice versa;
equal angles lie against equal sides, an example of this is an isosceles triangle;
the sum of the interior angles is always 180°, which has already been demonstrated with an example;
when one side of a triangle is extended beyond its limits, an external angle is formed, which will always be is equal to the sum corners not adjacent to it;
either side is always less than the sum of the other two sides, but greater than their difference.

Types of triangles

The next stage of acquaintance is to determine the group to which the presented triangle belongs. Belonging to a particular species depends on the magnitude of the angles of the triangle.

Isosceles - with two equal sides, which are called lateral, the third in this case acts as the base of the figure. The angles at the base of such a triangle are the same, and the median drawn from the vertex is the bisector and height.
correct, or equilateral triangle, is one in which all its sides are equal.
Rectangular: one of its angles is 90°. In this case, the side opposite this angle is called the hypotenuse, and the other two are the legs.
Acute triangle - all angles are less than 90°.
Obtuse - one of the angles is greater than 90°.

Equality and similarity of triangles

In the learning process, they not only consider a single figure, but also compare two triangles. And this, it would seem, simple theme has a lot of rules and theorems by which you can prove that the figures under consideration are equal triangles. Triangles are equal if their corresponding sides and angles are the same. With this equality, if you put these two figures on top of each other, all their lines will converge. Also, the figures can be similar, in particular, this applies in practice identical figures, differing only in magnitude. In order to make such a conclusion about the triangles presented, one of the following conditions must be met:

two angles of one figure are equal to two angles of another;
two sides of one are proportional to two sides of the second triangle, and the angles formed by the sides are equal;
the three sides of the second figure are the same as those of the first.

Of course, for indisputable equality, which will not cause the slightest doubt, it is necessary to have the same values \u200b\u200bof all the elements of both figures, however, using theorems, the task is greatly simplified, and only a few conditions are allowed to prove the equality of triangles.

The first sign of equality of triangles

Problems on this topic are solved on the basis of the proof of the theorem, which sounds like this: "If two sides of a triangle and the angle that they form are equal to two sides and an angle of another triangle, then the figures are also equal to each other."

How does the proof of the theorem about the first criterion for the equality of triangles sound? Everyone knows that two segments are equal if they have the same length, or circles are equal if they have the same radius. And in the case of triangles, there are several signs, having which, we can assume that the figures are identical, which is very convenient to use when solving various geometric problems.

How the theorem “The first sign of equality of triangles” sounds is described above, but here is its proof:

Suppose triangles ABC and A 1 B 1 C 1 have the same sides AB and A 1 B 1 and, accordingly, BC and B 1 C 1, and the angles that are formed by these sides have the same value, that is, they are equal. Then, by superimposing △ ABC on △ A 1 B 1 C 1, we get the coincidence of all lines and vertices. It follows from this that these triangles are absolutely identical, which means they are equal to each other.

The theorem "The first criterion for the equality of triangles" is also called "By two sides and an angle." Actually, this is its essence.

Second feature theorem

The second sign of equality is proved similarly, the proof is based on the fact that when the figures are superimposed on each other, they completely coincide in all vertices and sides. And the theorem sounds like this: "If one side and two angles in the formation of which it participates correspond to the side and two angles of the second triangle, then these figures are identical, that is, equal."

Third Sign and Proof

If both 2 and 1 sign of equality of triangles concerned both the sides and corners of the figure, then the 3rd applies only to the sides. So, the theorem has the following formulation: "If all sides of one triangle are equal to three sides of the second triangle, then the figures are identical."

To prove this theorem, we need to delve into the very definition of equality in more detail. Essentially, what does the expression "triangles are equal" mean? Identity says that if you superimpose one figure on another, all their elements will coincide, this can only be the case when their sides and angles are equal. At the same time, the angle opposite one of the sides, which is the same as that of the other triangle, will be equal to the corresponding vertex of the second figure. It should be noted that at this point the proof can be easily translated to 1 criterion for the equality of triangles. In the event that such a sequence is not observed, the equality of triangles is simply impossible, except in cases where the figure is mirror reflection first.

right triangles

In the structure of such triangles there are always vertices with an angle of 90°. Therefore, the following statements are true:

triangles with a right angle are equal if the legs of one are identical to the legs of the second;
figures are equal if their hypotenuses and one of the legs are equal;
such triangles are congruent if their legs and sharp corner are identical.

This sign refers to To prove the theorem, the figures are applied to each other, as a result of which the triangles are folded with legs so that two straight lines come out with sides CA and CA 1.

Practical use

In most cases, in practice, the first sign of the equality of triangles is used. In fact, such a seemingly simple topic of grade 7 in geometry and planimetry is also used to calculate the length, for example, of a telephone cable without measuring the terrain along which it will pass. Using this theorem, it is easy to make the necessary calculations to determine the length of an island in the middle of a river without swimming across it. Either strengthen the fence by placing the bar in the span so that it divides it into two equal triangles, or calculate complex elements work in carpentry, or when calculating the roof truss system during construction.

The first sign of equality of triangles is widely used in real "adult" life. Although in school years It is this topic that seems boring and completely unnecessary for many.

The third criterion for the equality of triangles on three sides is formulated as a theorem.

Theorem : If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are congruent.

Proof. consider ΔABC and ΔA 1 B 1 C 1 in which AB=A 1 B 1 , AC=A 1 C 1 , BC=B 1 C 1 . Let us prove that ΔABC=ΔA 1 B 1 C 1

Let ABC and A 1 B 1 C 1 be triangles with AB=A 1 B 1 , AC=A 1 C 1 , BC=B 1 C 1 . We impose ∆ABC on ∆A 1 B 1 C 1 so that the vertex A is aligned with A 1 , and the vertices B and B 1 , and the vertices C and C 1 are aligned different sides from straight line A 1 B 1 . Three cases are possible: 1) the beam C 1 C passes inside the angle A 1 C 1 B 1 (Fig. a)); 2) beam C 1 C coincides with one of the sides of this angle (Fig. b)); beam C 1 C passes outside the angle A 1 C 1 B 1 (Fig. c)). Let's consider the first case. Since, according to the condition of the theorem, the sides AC and A 1 C 1, BC and B 1 C 1 are equal, then the triangles A 1 C 1 C and B 1 C 1 C are isosceles. According to the theorem on the property of angles of an isosceles triangle, l = l2, l3 = l4, therefore lA 1 CB 1 = = lA 1 C 1 B 1 . So, AC=A 1 C 1 , BC=B 1 C 1 , РС = РС 1 . Therefore, triangles ABC and A 1 B 1 C 1 are equal according to the first criterion for the equality of triangles.

Board writing:

Given:ΔABC, ΔA 1 B 1 C 1 , AB=A 1 B 1 , AC=A 1 C 1 , BC=B 1 C 1

Prove:∆ABC=∆A 1 B 1 C 1

Proof. Let's impose ∆ABC on ∆A 1 B 1 C 1 so that A →A 1 , and B → B 1 , and C and C 1 are on opposite sides of the line A 1 B 1 . Let's consider a case. beam C 1 C passes inside RA 1 C 1 B 1 (Fig. a)).

AC \u003d A 1 C 1, BC \u003d B 1 C 1 ═> ΔA 1 C 1 C and ΔB 1 C 1 C - equal. ═> Ðl \u003d Ð2, Ð3 \u003d Ð4 (according to the property of angles equilateral Δ), ═> ÐA 1 CB 1 \u003d ÐA 1 C 1 B 1 ═> AC \u003d A 1 C 1, BC \u003d B 1 C 1, ÐС = РС 1 ═>

ΔABC=ΔA 1 B 1 C 1 according to the first sign of equality of triangles.

2.Rhombus. Definition, properties, signs.

A rhombus is a kind of quadrilateral.

Definition: A rhombus is a parallelogram in which all sides are equal.

The figure shows a parallelogram ABCD with AB=BC=CD=DA. By definition, this parallelogram is a rhombus. AC and BD are the diagonals of the rhombus. Since a rhombus is a parallelogram, all the properties and signs of a parallelogram are valid for it.

Properties:

1) In a rhombus, opposite angles are equal (ÐA=ÐC, ÐB=ÐD)

2) The diagonals of the rhombus are bisected by the intersection point. (BO=OD, AO=OC)

3) The diagonals of a rhombus are mutually perpendicular and its angles are divided in half. (AC DВ, ‌‌РАBO=РОВС, РАДО=РОDC, ‌‌ÐBСО=РDСО, РDАО=РВАО) ( special property)

4) The sum of the angles adjacent to one side is 180 0 (ÐA+ÐB= ÐC+ÐD=ÐB+ÐC=ÐA+ÐD=180 0)

signs rhombus:

1) If the diagonals of a parallelogram are mutually perpendicular, then this parallelogram is a rhombus

2) If the diagonal of a parallelogram bisects its angles, then the parallelogram is a rhombus.

3) If all sides of a parallelogram are equal, then it is a rhombus.

Writing on the board.

Properties:

1) ÐA=ÐC, ÐB=ÐD 2) BO=OD, AO=OC

3) AC DВ, ‌‌РАBO=РОВС, RADO=РОDC, ‌‌ÐBСО=РDСО, РDАО=РВАО

4) ÐA+ÐB= ÐC+ÐD=ÐB+ÐC=ÐA+ÐD=180 0

The inverse statements are signs rhombus:

1 ) If ABCD is parallel, and AC DB, then - ABCD is a rhombus.

2) If ABCD is parallel, and AC and DB are bisectors, then ABCD is a rhombus.

3) If ABCD is parallel, and AC \u003d DB and BC \u003d AD, then - ABCD is a rhombus.

Task.

Two triangles are said to be congruent if they can be overlapped. Figure 1 shows equal triangles ABC and A 1 B 1 C 1. Each of these triangles can be superimposed on another so that they are completely compatible, that is, their vertices and sides are paired together. It is clear that in this case the angles of these triangles will be combined in pairs.

Thus, if two triangles are equal, then the elements (i.e., sides and angles) of one triangle are respectively equal to the elements of the other triangle. Note that in equal triangles against respectively equal sides(i.e. overlapping when superimposed) lie equal angles and back: opposite respectively equal angles lie equal sides.

So, for example, in equal triangles ABC and A 1 B 1 C 1, shown in Figure 1, equal angles C and C 1 lie against respectively equal sides AB and A 1 B 1. The equality of triangles ABC and A 1 B 1 C 1 will be denoted as follows: Δ ABC = Δ A 1 B 1 C 1. It turns out that the equality of two triangles can be established by comparing some of their elements.

Theorem 1. The first sign 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 equal (Fig. 2).

Proof. Consider triangles ABC and A 1 B 1 C 1, in which AB \u003d A 1 B 1, AC \u003d A 1 C 1 ∠ A \u003d ∠ A 1 (see Fig. 2). Let us prove that Δ ABC = Δ A 1 B 1 C 1 .

Since ∠ A \u003d ∠ 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 A 1, and the sides AB and AC overlap, respectively, on the rays A 1 B 1 and A 1 C one . 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, triangles ABC and A 1 B 1 C 1 are completely compatible, which means they are equal.

Theorem 2 is proved similarly by the superposition method.

Theorem 2. The second sign of the equality of triangles. If 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, then such triangles are equal (Fig. 34).

Comment. Based on Theorem 2, Theorem 3 is established.

Theorem 3. The sum of any two interior angles of a triangle is less than 180°.

Theorem 4 follows from the last theorem.

Theorem 4. The external angle of a triangle is greater than any inner corner, not adjacent to it.

Theorem 5. The third sign of the equality of triangles. If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are equal ().

Example 1 In triangles ABC and DEF (Fig. 4)

∠ A = ∠ E, AB = 20 cm, AC = 18 cm, DE = 18 cm, EF = 20 cm. Compare triangles ABC and DEF. What is the angle in triangle DEF equal to the angle AT?

Decision. These triangles are equal in the first sign. Angle F of triangle DEF is equal to angle B triangle ABC, since these angles lie opposite respectively equal sides DE and AC.

Example 2 Segments AB and CD (Fig. 5) intersect at point O, which is the midpoint of each of them. What is segment BD equal to if segment AC is 6 m?

Decision. Triangles AOC and BOD are equal (by the first criterion): ∠ AOC = ∠ BOD (vertical), AO = OB, CO = OD (by condition).
From the equality of these triangles follows the equality of their sides, i.e. AC = BD. But since, according to the condition, AC = 6 m, then BD = 6 m.

The second sign of equality of triangles

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.

MN=PR N=R M=P

As in the proof of the first sign, you need to make sure that this is enough for the triangles to be equal, can they be completely combined?

1. Since MN = PR, then these segments are combined if their end points are combined.

2. Since N = R and M = P , then the rays \(MK\) and \(NK\) overlap the rays \(PT\) and \(RT\), respectively.

3. If the rays coincide, then their points of intersection \(K\) and \(T\) coincide.

4. All the vertices of the triangles are combined, that is, Δ MNK and Δ PRT are completely compatible, which means they are equal.

The third sign of equality of triangles

If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are congruent.

MN = PR KN = TR MK = PT

Again, let's try to combine the triangles Δ MNK and Δ PRT by overlapping and make sure that the correspondingly equal sides guarantee the equality of the corresponding angles of these triangles and they completely coincide.

Let's combine, for example, identical segments \(MK\) and\(PT\). Let us assume that the points \(N\) and \(R\) do not coincide in this case.

Let \(O\) be the midpoint of the segment \(NR\). According to this information MN = PR , KN = TR . Triangles \(MNR\) and \(KNR\) are isosceles with common ground\(NR\).

Therefore, their medians \(MO\) and \(KO\) are heights, so they are perpendicular to \(NR\). The lines \(MO\) and \(KO\) do not coincide, because the points \(M\), \(K\), \(O\) do not lie on the same line. But through the point \(O\) of the line \(NR\) it is possible to draw only one line perpendicular to it. We have come to a contradiction.

It is proved that the vertices \(N\) and \(R\) must also coincide.

The third sign allows us to call the triangle a very strong, stable figure, sometimes they say that triangle - rigid figure . If the lengths of the sides do not change, then the angles do not change either. For example, a quadrilateral does not have this property. Therefore, various supports and fortifications are made triangular.

But a kind of stability, stability and perfection of the number \ (3 \) people have been evaluating and highlighting for a long time.

Fairy tales talk about it.

There we meet "Three Bears", "Three Winds", "Three Little Pigs", "Three Comrades", "Three Brothers", "Three Lucky Men", "Three Craftsmen", "Three Princes", "Three Friends", "Three hero", etc.

There are given “three attempts”, “three advice”, “three instructions”, “three meetings”, “three wishes” are fulfilled, you need to endure “three days”, “three nights”, “three years”, go through “three states ”,“ three underground kingdoms ”, endure“ three trials ”, swim through the“ three seas ”.