The theorem of equal triangles. The third sign of equality of triangles

The video lesson "The third sign of the equality of triangles" contains the proof of the theorem, which is a sign of the equality of two triangles on three sides. This theorem is an important part of geometry. It is often used to solve practical problems. Its proof is based on the signs of the equality of triangles already known to students.

The proof of this theorem is complex, therefore, in order to improve the quality of education, to form the ability to prove geometric statements, it is advisable to use this visual aid, which will help to focus students' attention on the material being studied. Also, with the help of animation, visual demonstration of constructions and proofs, it makes it possible to improve the quality of education.

At the beginning of the lesson, the title of the topic is demonstrated and the theorem is formulated that triangles are equal if all sides of one triangle are pairwise equal to all sides of the second triangle. The text of the theorem is shown on the screen and can be written by students in a notebook. Next, we consider the proof of this theorem.

To prove the theorem, triangles ΔABC and ΔA 1 B 1 C 1 are constructed. From the conditions of the theorem it follows that the sides are pairwise equal, that is, AB \u003d A 1 B 1, BC \u003d B 1 C 1 and AC \u003d A 1 C 1. At the beginning of the proof, the imposition of the triangle ΔАВС on ΔА 1 В 1 С 1 is demonstrated so that the vertices A and A 1 , as well as B and B 1 of these triangles are aligned. In this case, the peaks C and C 1 should be located on opposite sides of the superimposed sides AB and A 1 B 1. With this construction, several options for the arrangement of triangle elements are possible:

1. Beam C 1 C lies inside the angle ∠A 1 C 1 B 1 .
2. Beam C 1 C coincides with one of the sides of the angle ∠A 1 C 1 B 1.
3. Ray C 1 C lies outside the angle ∠A 1 C 1 B 1.

Each case must be considered separately, since the proof cannot be the same for all given cases. In the first case, two triangles formed as a result of construction are considered. Since, according to the condition, in these triangles the sides are AC \u003d A 1 C 1, and BC \u003d B 1 C 1, then the resulting triangles ΔB 1 C 1 C and ΔA 1 C 1 are equilateral. Using the studied property of isosceles triangles, we can assert that the angles ∠1 and ∠2 are equal to each other, and also ∠3 and ∠4 are equal. Since these angles are equal, the sum of ∠1 and ∠3, as well as ∠2 and ∠4 will also give equal angles. Therefore, the angles ∠С and ∠С 1 are equal. Having proved this fact, we can reconsider the triangles ΔABC and ΔA 1 B 1 C 1, in which the sides BC \u003d B 1 C 1 and AC \u003d A 1 C 1 according to the condition of the theorem, and it is proved that the angles between them ∠C and ∠C 1 are also equal. Accordingly, these triangles will be equal according to the first criterion for the equality of triangles, which is already known to the students.

In the second case, when the triangles are superimposed, the points C and C 1 lie on one straight line passing through the point B (B 1). In the sum of two triangles ΔABC and ΔA 1 B 1 C 1, a triangle ΔCAC 1 is obtained, in which the two sides AC \u003d A 1 C 1, according to the condition of the theorem, are equal. Accordingly, this triangle is isosceles. In an isosceles triangle with equal sides, there are equal angles, so it can be argued that the angles ∠С=∠С 1. It also follows from the conditions of the theorem that the sides BC and B 1 C 1 are equal to each other, therefore, ΔABC and ΔA 1 B 1 C 1, taking into account the stated facts, are equal to each other according to the first sign of equality of triangles.

The proof in the third case, similarly to the first two, uses the first criterion for the equality of triangles. A geometric figure constructed by imposing triangles, when connected by a segment of vertices C and C 1, is transformed into a triangle ΔB 1 C 1 C. This triangle is isosceles, since its sides B 1 C 1 and B 1 C are equal by condition. And with equal sides in an isosceles triangle, the angles ∠С and ∠С 1 are also equal. Since, according to the condition of the theorem, the sides AC \u003d A 1 C 1 are equal, then the angles at them in the isosceles triangle ΔACS 1 are also equal. Taking into account the fact that the angles ∠С and ∠С 1 are equal, and the angles ∠DCAand ∠DC 1 A are equal to each other, then the angles ∠ACB and ∠AC 1 B are also equal. Given this fact, to prove the equality of triangles ΔABC and ΔA 1 B 1 C 1, you can use the first sign of the equality of triangles, since the two sides of these triangles are equal in terms of conditions, and the equality of the angles between them is proved in the course of reasoning.

At the end of the video tutorial, an important application of the third criterion for the equality of triangles is demonstrated - the rigidity of a given geometric figure. An example explains what this statement means. As an example of a flexible design, two battens connected with a nail are given. These slats can be moved apart and shifted at any angle. If we attach another one to the rails, connected by ends to the existing rails, then we will get a rigid structure in which it is impossible to change the angle between the rails. Getting a triangle with given sides and other angles is not possible. This corollary of the theorem is of great practical importance. The screen displays engineering structures in which this property of triangles is used.

The video lesson "The third sign of the equality of triangles" makes it easier for the teacher to present new material in a geometry lesson on this topic. Also, the video lesson can be successfully used for distance learning in mathematics, it will help students to understand the complexities of the proof on their own.

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 correspondingly 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. An external angle of a triangle is greater than any internal angle 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 angle in triangle DEF is equal to angle B?

Solution. These triangles are equal in the first sign. Angle F of triangle DEF is equal to angle B of triangle ABC, since these angles lie opposite the corresponding 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?

Solution. 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.

Theorem

Proof

Consider triangles ABC and A 1 B 1 C 1, in which AB \u003d A 1 B 1, ∠A \u003d ∠A 1, ∠B \u003d ∠B 1 (Fig. 68). Let us prove that Δ ABC = Δ A 1 B 1 C 1 .

Rice. 68

We impose triangle ABC on triangle A 1 B 1 C 1 so that vertex A is aligned with vertex A 1, side AB is equal to its side AjBj, and vertices C and C 1 are on the same side of the line A 1 B 1.

Since ∠A \u003d ∠A 1 and ∠B \u003d ∠B 1, then the AC side will overlap the beam A 1 C 1, and the BC side will overlap the beam B 1 C 1. Therefore, the vertex C - the common point of the sides AC and BC - will lie both on the ray A 1 C 1 and on the ray B 1 C 1 and, therefore, will be aligned with the common point of these rays - the vertex C 1. This means that the sides AC and A 1 C 1, BC and B 1 C 1 will be combined.

So, the triangles ABC and A 1 B 1 C 1 are completely compatible, therefore they are equal. The theorem has been proven.

The third sign of equality of triangles

Theorem

Proof

Consider triangles ABC and A 1 B 1 C 1, in which AB \u003d A 1 B 1, BC \u003d B 1 C 1, CA \u003d C 1 A 1 (Fig. 69).

Rice. 69

Let us prove that Δ ABC = Δ A 1 B 1 C 1 . We apply triangle ABC to triangle A 1 B 1 C 1 so that vertex A is aligned with vertex A 1, vertex B is aligned with vertex B 1, and vertices C and C 1 are on opposite sides of the straight line A 1 B 1 (Fig. 70 ).

Rice. 70

Three cases are possible: the beam C 1 C passes inside the angle A 1 C 1 B 1 (Fig. 70, a); the beam C 1 C coincides with one of the sides of this angle (Fig. 70, b); the beam C 1 C passes outside the angle A 1 C 1 B 1 (Fig. 70, c). Consider the first case (consider the rest of the cases yourself).

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 (see Fig. 70, a). By the theorem on the property of angles of an isosceles triangle ∠1 = ∠2, ∠3 = ∠4, therefore ∠A 1 CB 1 = ∠A 1 C 1 B 1 . So, AC \u003d A 1 C 1, BC \u003d B 1 C 1, ∠C \u003d ∠C 1.

Therefore, triangles ABC and A 1 B 1 C 1 are equal according to the first sign of equality of triangles. The theorem has been proven.

From the third criterion for the equality of triangles, it follows that triangle - rigid figure. Let's explain what this means.

Imagine two slats, in which two ends are fastened with a nail (Fig. 71, a). This design is not rigid: by shifting or pushing the free ends of the rails, we can change the angle between them. Now let's take another rail and fasten its ends with the free ends of the first two rails (Fig. 71, b).

Rice. 71

The resulting construction - a triangle - will already be rigid. No two sides can be moved or moved apart in it, that is, not a single corner can be changed. Indeed, if this succeeded, then we would get a new triangle, not equal to the original one. But this is impossible, since the new triangle must be equal to the original one according to the third criterion for the equality of triangles.

This property - the rigidity of a triangle - is widely used in practice. So, in order to fix the post in a vertical position, a support is placed on it (Fig. 72, a); the same principle is used when installing the bracket (Fig. 72, b).

Rice. 72

Tasks

121. Segments AB and CD intersect at the middle O of segment AB, ∠OAD = ∠OBC.

a) Prove that ∆CBO = ∆DAO;
b) find BC and CO if CD = 26 cm, AD = 15 cm.

122. In figure 53 (see p. 31) ∠1 = ∠2, ∠3 = ∠4.

a) Prove that Δ ABC = Δ CDA;
b) find AB and BC if AO = 19 cm, CD = 11 cm.

123. Point D is taken on the bisector of angle A, and points B and C are taken on the sides of this angle such that ∠ADB = ∠ADC. Prove that BD = CD.

124. According to Figure 73, prove that OP = OT, ∠P = ∠T.

Rice. 73

125. In figure 74 ∠DAC = ∠DBC, AO = BO. Prove that ∠C = ∠D and AC = BD.

Rice. 74

126. In Figure 74, ∠DAB = ∠CBA, ∠CAB = ∠DBA, AC = 13 cm. Find BD.

127. In triangles ABC and A 1 B 1 C 1 AB \u003d A 1 B 1, BC \u003d B 1 C 1, ∠B - ∠B 1. Points D and D 1 are marked on sides AB and A 1 B 1 so that ∠ACO = ∠A 1 C 1 D 1 . Prove that ∆BCD = ∆B 1 C 1 D 1 .

128. Prove that in equal triangles the bisectors drawn to respectively equal sides are equal.

129. Segments AC and BD intersect in the middle O of segment AC, ∠BCO = ∠DAO. Prove that ∆BOA = ∆DOC.

130. In triangles ABC and A 1 B 1 C 1, segments CO and C 1 O 1 are medians, BC \u003d B 1 C 1, ∠B - ∠B 1 and ∠C \u003d ∠C 1. Prove that:

a) Δ ACO \u003d Δ A 1 C 1 O 1;
b) Δ BCO \u003d Δ B 1 C 1 O.

131. In triangles DEF and MNP, EF - NP, DF = MP and ∠F = ∠P. The bisectors of angles E and D intersect at point O, and the bisectors of angles M and N intersect at point K. Prove that ∠DOE = ∠MKN.

132. A line perpendicular to the bisector of angle A intersects the sides of the angle at points M and N. Prove that triangle AMN is isosceles.

133. Prove that if the bisector of a triangle is its height, then the triangle is isosceles.

134. Prove that isosceles triangles are congruent if the base and adjacent angle of one triangle are respectively equal to the base and adjacent angle of another triangle.

135. Prove that if the side of one equilateral triangle is equal to the side of another equilateral triangle, then the triangles are congruent.

136. In figure 52 (see p. 31) AB-AC, BD = DC and ∠BAC = 50°. Find ∠CAD.

137. In figure 53 (see p. 31) BC = AD, AB = CD. Prove that ∠B = ∠D.

138. In figure 75 AB = CD and BD = AC. Prove that: a) ∠CAD = ∠ADB; b) ∠BAC = ∠CDB.

Rice. 75

139. In Figure 76, AB = CD, AD = BC, BE is the bisector of angle ABC, and DF is the bisector of angle ADC. Prove that:

a) ∠ABE = ∠ADF;
b) Δ ABE = Δ CDF.

Rice. 76

140. In triangles ABC and A 1 B 1 C 1, the medians BM and B 1 M 1 are equal, AB \u003d A 1 B 1 AC \u003d A 1 C 1. Prove that Δ ABC = Δ A 1 B 1 C 1 .

141. In triangles ABC and A 1 B 1 C 1, segments AD and A 1 D 1 are bisectors, AB \u003d A 1 B 1, BD \u003d B 1 D 1 and AD \u003d A 1 D 1. Prove that Δ ABC = Δ A 1 B 1 C 1 .

142. Isosceles triangles ADC and BCD have a common base DC. Line AB intersects segment CD at point O. Prove that: a) ∠ADB = ∠ACB; b) DO = OC.

Answers to tasks

121. b) BC = 15 cm, CO = 13 cm.

122. b) AB = 11 cm, BC = 19 cm.

142. Instruction. Consider two cases. Point B lies: a) on the ray AO; b) on the continuation of the ray AO.

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 . The triangles \(MNR\) and \(KNR\) are isosceles with a common base \(NR\).

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 ”.

1) on two sides and the angle between them

Proof:

Let triangles ABC and A 1 B 1 C 1 have angle A equal to angle A 1, AB equal to A 1 B 1, AC equal to A 1 C 1. We prove that the triangles are congruent.

Overlay triangle ABC (or symmetrical to it) onto triangle A 1 B 1 C 1 so that angle A coincides with angle A 1 . Since AB \u003d A 1 B 1, and AC \u003d A 1 C 1, then B will coincide with B 1, and C will coincide with C 1. Hence, the triangle A 1 B 1 C 1 coincides with the triangle ABC, and therefore is equal to the triangle ABC.

The theorem has been proven.

2) along the side and adjacent corners

Proof:

Let ABC and A 1 B 1 C 1 be two triangles in which AB is equal to A 1 B 1, angle A is equal to angle A 1, and angle B is equal to angle B 1. Let's prove that they are equal.

Overlay triangle ABC (or symmetrical to it) on triangle A 1 B 1 C 1 so that AB coincides with A 1 B 1. Since ∠BAC \u003d ∠B 1 A 1 C 1 and ∠ABC \u003d ∠A 1 B 1 C 1, then the beam AC will coincide with A 1 C 1 , and BC will coincide with B 1 C 1 . It follows that the vertex C coincides with C 1. Hence, the triangle A 1 B 1 C 1 coincides with the triangle ABC, and therefore is equal to the triangle ABC.

The theorem has been proven.

3) on three sides

Proof :

Consider triangles ABC and A l B l C 1, in which AB \u003d A 1 B 1, BC \u003d B l C 1 CA \u003d C 1 A 1. Let's prove that ΔABS \u003d ΔA 1 B 1 C 1.

Apply triangle ABC (or symmetrical to it) to the triangle A 1 B 1 C 1 so that the vertex A is aligned with the vertex A 1, the vertex B is aligned with the vertex B 1, and the vertices C and C 1 are on opposite sides of the line A 1 B 1. Consider 3 cases:

1) Beam C 1 C passes inside the angle A 1 C 1 B 1. 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 ∠1 = ∠2, ∠3 = ∠4, therefore ∠ACB=∠A 1 C 1 B 1 .

2) Beam C 1 C coincides with one of the sides of this angle. A lies on CC 1 . AC=A 1 C 1 , BC=B 1 C 1 , C 1 BC - isosceles , ∠ACB=∠A 1 C 1 B 1 .

3) Beam C 1 C passes outside the angle A 1 C 1 B 1 . AC=A 1 C 1 , BC=B 1 C 1 , so ∠1 = ∠2, ∠1+∠3 = ∠2+∠4, ∠ACB=∠A 1 C 1 B 1 .

So, AC=A 1 C 1 , BC=B 1 C 1 , ∠C=∠C 1 . Therefore, triangles ABC and A 1 B 1 C 1 are equal in
the first criterion for the equality of triangles.

The theorem has been proven.

2. Division of a segment into n equal parts.

Draw a ray through A, put n equal segments on it. Through B and A n draw a straight line and parallel to it through points A 1 - A n -1. We mark their points of intersection with AB. We get n segments that are equal according to the Thales theorem.

Thales' theorem. If on one of the two straight lines several equal segments are sequentially laid aside and parallel lines are drawn through their ends, intersecting the second straight line, then they will cut off segments equal to each other on the second straight line.

Proof. AB=CD

1. Draw straight lines through points A and C parallel to the other side of the angle. We get two parallelograms AB 2 B 1 A 1 and CD 2 D 1 C 1 . According to the parallelogram property: AB 2 = A 1 B 1 and CD 2 = C 1 D 1.

2. ΔABB 2 \u003d ΔCDD 2 ABB 2 CDD 2 BAB 2 DCD 2 and are equal based on the second criterion for the equality of triangles:
AB = CD according to the condition of the theorem,
as corresponding, formed at the intersection of parallel BB 1 and DD 1 straight line BD.

3. Similarly, each of the angles and turns out to be equal to the angle with the vertex at the intersection point of the secants. AB 2 = CD 2 as corresponding elements in equal triangles.

4. A 1 B 1 = AB 2 = CD 2 = C 1 D 1