The video lesson “The third criterion for the equality of triangles” contains a proof of the theorem, which is a test for 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 equality of triangles already known to students.
The proof of this theorem is complex, therefore, in order to improve the quality of teaching and develop the ability to prove geometric statements, it is advisable to use this visual aid, which will help concentrate 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 learning.
At the beginning of the lesson, the title of the topic is demonstrated and a 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 down by students in a notebook. Next, we consider the proof of this theorem.
To prove the theorem, triangles ΔАВС and ΔА 1 В 1 С 1 are constructed. From the conditions of the theorem it follows that the sides are equal in pairs, that is, AB = A 1 B 1, BC = B 1 C 1 and AC = A 1 C 1. At the beginning of the proof, we demonstrate the imposition of the triangle ΔABC on ΔA 1 B 1 C 1 so that the vertices A and A 1, as well as B and B 1 of these triangles are aligned. In this case, the vertices 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:
- Ray C 1 C lies inside the angle ∠A 1 C 1 B 1.
- Ray C 1 C coincides with one of the sides of the angle ∠A 1 C 1 B 1.
- Ray C 1 C lies outside the angle ∠A 1 C 1 B 1.
Each case must be considered separately, since the evidence cannot be the same for all given cases. In the first case, two triangles formed as a result of construction are considered. Since, by condition, in these triangles the sides AC = A 1 C 1, and BC = B 1 C 1, then the resulting triangles ΔB 1 C 1 C and ΔA 1 C 1 are isosceles. Using the studied property of isosceles triangles, we can state that angles ∠1 and ∠2 are equal to each other, and also ∠3 and ∠4 are equal. Since these angles are equal, then 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 proven this fact, we can re-examine the triangles ΔABC and ΔA 1 B 1 C 1, in which the sides BC = B 1 C 1 and AC = A 1 C 1 according to the conditions of the theorem, and it is proven that the angles between them are ∠C and ∠C 1 are also equal. Accordingly, these triangles will be equal according to the first sign of equality of triangles, which is already known to students.
In the second case, when triangles were superimposed, points C and C 1 lay on one straight line passing through point B (B 1). The sum of two triangles ΔАВС and ΔА 1 В 1 С 1 results in a triangle ΔСАС 1, in which the two sides AC = А 1 С 1 are equal according to the conditions of the theorem. Accordingly, this triangle is isosceles. In an isosceles triangle, equal sides have equal angles, so we can say 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, similar to the first two, uses the first sign of equality of triangles. The geometric figure constructed by superimposing 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, angles ∠С and ∠С 1 are also equal. Since, according to the conditions of the theorem, the sides AC = A 1 C 1 are equal, then the angles at them in the isosceles triangle ΔАСС 1 are also equal. Taking into account the fact that the angles ∠C and ∠C 1 are equal, and the angles ∠DCA and ∠DC 1 A are equal to each other, then the angles ∠ACB and ∠AC 1 B are also equal. Considering this fact, to prove the equality of triangles ΔABC and ΔA 1 B 1 C 1, you can use the first sign of equality of triangles, since the two sides of these triangles are equal according to the conditions, and the equality of the angles between them is proven in the course of reasoning.
At the end of the video lesson, an important application of the third sign of equality of triangles is demonstrated - the rigidity of a given geometric figure. An example explains what this statement means. An example of a flexible design is two slats connected by a nail. These slats can be moved apart and moved at any angle. If we attach another one to the slats, connected at the ends to the existing slats, then we get a rigid structure in which it is impossible to change the angle between the slats. Obtaining a triangle with these sides and other angles is impossible. This corollary of the theorem has important practical significance. The screen depicts engineering structures in which this property of triangles is used.
The video lesson “The third criterion for the equality of triangles” makes it easier for the teacher to present new material on this topic in a geometry lesson. Also, the video lesson can be successfully used for distance learning in mathematics and will help students understand the complexities of proof on their own.
>>Geometry: The third sign of equality of triangles. Complete lessons
LESSON TOPIC: The third sign of equality of triangles.
Lesson objectives:
- Educational – repetition, generalization and testing of knowledge on the topic: “Signs of equality of triangles”; development of basic skills.
- Developmental – to develop students’ attention, perseverance, perseverance, logical thinking, mathematical speech.
- Educational - through the lesson, cultivate an attentive attitude towards each other, instill the ability to listen to comrades, mutual assistance, and independence.
Lesson objectives:
- Develop skills in constructing triangles using a scale ruler, protractor and drawing triangle.
- Test students' problem-solving skills.
Lesson plan:
- From the history of mathematics.
- Signs of equality of triangles.
- Updating basic knowledge.
- Right triangles.
From the history of mathematics.
The right triangle occupies a place of honor in Babylonian geometry, and mention of it is often found in the Ahmes papyrus.
The term hypotenuse comes from the Greek hypoteinsa, meaning stretching under something, contracting. The word originates from the image of ancient Egyptian harps, on which the strings were stretched over the ends of two mutually perpendicular stands.
The term leg comes from the Greek word “kathetos”, which meant plumb line, perpendicular. In the Middle Ages, the word leg meant the height of a right triangle, while its other sides were called the hypotenuse, respectively the base. In the 17th century, the word cathet began to be used in the modern sense and became widespread starting in the 18th century.
Euclid uses the expressions:
“sides concluding a right angle” - for legs;
“the side subtending a right angle” - for the hypotenuse.
First, we need to refresh our memory of the previous signs of equality of triangles. And so let's start with the first one.
1st sign of equality of triangles.
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 whether 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, the rays \(MK\) and \(NK\) will overlap the rays \(PT\) and \(RT\), respectively.
3. If the rays coincide, then their intersection points \(K\) and \(T\) coincide.
4. All the vertices of the triangles are aligned, that is, Δ MNK and Δ PRT are completely aligned, 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
Let's again try to combine the triangles Δ MNK and Δ PRT by overlapping and make sure that the corresponding equal sides guarantee that the corresponding angles of these triangles are equal and they will completely coincide.
Let us combine, for example, identical segments \(MK\) and \(PT\). Let us assume that the points \(N\) and \(R\) do not coincide.
Let \(O\) be the midpoint of the segment \(NR\). According to this information, MN = PR, KN = TR. Triangles \(MNR\) and \(KNR\) are isosceles with a common base \(NR\).
Therefore, their medians \(MO\) and \(KO\) are heights, which means they are perpendicular to \(NR\). The lines \(MO\) and \(KO\) do not coincide, since the points \(M\), \(K\), \(O\) do not lie on the same line. But through the point \(O\) of the line \(NR\) only one line perpendicular to it can be drawn. We have arrived at a contradiction.
It has been proven that the vertices \(N\) and \(R\) must 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 people have been evaluating and highlighting the peculiar stability, stability and perfection of the number \(3\) for a long time.
Fairy tales talk about this.
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 “three attempts”, “three advice”, “three instructions”, “three meetings” are given, “three wishes” are fulfilled, one must endure “three days”, “three nights”, “three years”, go through “three states” ", "three underground kingdoms", withstand "three tests", sail through the "three seas".
Two triangles are said to be congruent if they can be brought together by overlapping. Figure 1 shows equal triangles ABC and A 1 B 1 C 1. Each of these triangles can be superimposed on the other so that they are completely compatible, that is, their vertices and sides are compatible in pairs. It is clear that the angles of these triangles will also match in pairs.
Thus, if two triangles are congruent, 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 correspondingly equal sides(i.e., overlapping when superimposed) equal angles lie and back: Equal sides lie opposite respectively equal angles.
So, for example, in equal triangles ABC and A 1 B 1 C 1, shown in Figure 1, opposite equal sides AB and A 1 B 1, respectively, lie equal angles C and C 1. We will denote the equality of triangles ABC and A 1 B 1 C 1 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 congruent (Fig. 2).
Proof. Consider triangles ABC and A 1 B 1 C 1, in which AB = A 1 B 1, AC = A 1 C 1 ∠ A = ∠ A 1 (see Fig. 2). Let us prove that Δ ABC = Δ A 1 B 1 C 1 .
Since ∠ A = ∠ A 1, then triangle ABC can be superimposed on triangle A 1 B 1 C 1 so that vertex A is aligned with vertex A 1, and sides AB and AC are respectively superimposed on rays A 1 B 1 and A 1 C 1. Since AB = A 1 B 1, AC = A 1 C 1, then side AB will align with side A 1 B 1 and side AC will align with side A 1 C 1; in particular, points B and B 1, C and C 1 will coincide. Consequently, sides BC and B 1 C 1 will align. So, triangles ABC and A 1 B 1 C 1 are completely compatible, which means they are equal.
Theorem 2 is proved in a similar way using the superposition method.
Theorem 2. The second sign of equality of triangles. If a side and two adjacent angles of one triangle are respectively equal to the side and two adjacent angles of another triangle, then such triangles are congruent (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 exterior angle of a triangle is greater than any interior angle not adjacent to it.
Theorem 5. 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 ().
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 according to the first sign. Angle F of triangle DEF is equal to angle B of 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 middle of each of them. What is the length of segment BD if segment AC is 6 m?
Solution.
Triangles AOC and BOD are equal (according to the first criterion): ∠ AOC = ∠ BOD (vertical), AO = OB, CO = OD (by condition).
From the equality of these triangles it follows that their sides are equal, i.e. AC = BD. But since according to the condition AC = 6 m, then BD = 6 m.
