From ancient times to this day, the search for signs of equality of figures is considered a basic task, which is the basis of the foundations of geometry; hundreds of theorems are proven using equality tests. The ability to prove equality and similarity of figures is an important task in all areas of construction.
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Putting the skill into practice
Suppose we have a figure drawn on a piece of paper. At the same time, we have a ruler and a protractor with which we can measure the lengths of segments and the angles between them. How to transfer a figure of the same size to a second sheet of paper or double its scale.
We know that a triangle is a figure made up of three segments called sides that form the angles. Thus, there are six parameters - three sides and three angles - that define this figure.
However, having measured the size of all three sides and angles, transferring this figure to another surface will be a difficult task. In addition, it makes sense to ask the question: wouldn’t it be enough to know the parameters of two sides and one angle, or just three sides?
Having measured the length of the two sides and between them, we will then put this angle on a new piece of paper, so we can completely recreate the triangle. Let's figure out how to do this, learn how to prove the signs by which they can be considered the same, and decide what minimum number of parameters is enough to know in order to be confident that the triangles are the same.
Important! Figures are called identical if the segments forming their sides and angles are equal to each other. Similar figures are those whose sides and angles are proportional. Thus, equality is similarity with a proportionality coefficient of 1.
What are the signs of equality of triangles? Let’s give their definition:
- the first sign of equality: two triangles can be considered identical if two of their sides are equal, as well as the angle between them.
- the second sign of equality of triangles: two triangles will be the same if two angles are the same, as well as the corresponding side between them.
- third sign of equality of triangles : Triangles can be considered identical when all their sides are of equal length.
How to prove that triangles are congruent. Let us give a proof of the equality of triangles.
Evidence of 1 sign
For a long time, among the first mathematicians this sign was considered an axiom, however, as it turned out, it can be proven geometrically based on more basic axioms.
Consider two triangles - KMN and K 1 M 1 N 1 . The KM side has the same length as K 1 M 1, and KN = K 1 N 1. And the angle MKN is equal to the angles KMN and M 1 K 1 N 1.
If we consider KM and K 1 M 1, KN and K 1 N 1 as two rays that come out from the same point, then we can say that the angles between these pairs of rays are the same (this is specified by the condition of the theorem). Let us carry out a parallel transfer of rays K 1 M 1 and K 1 N 1 from point K 1 to point K. As a result of this transfer, rays K 1 M 1 and K 1 N 1 will completely coincide. Let us plot on the ray K 1 M 1 a segment of length KM, originating at point K. Since, by condition, the resulting segment will be equal to the segment K 1 M 1, then the points M and M 1 coincide. Similarly with the segments KN and K 1 N 1. Thus, by transferring K 1 M 1 N 1 so that the points K 1 and K coincide, and the two sides overlap, we obtain a complete coincidence of the figures themselves.
Important! On the Internet there are proofs of the equality of triangles by two sides and an angle using algebraic and trigonometric identities with numerical values of the sides and angles. However, historically and mathematically, this theorem was formulated long before algebra and earlier than trigonometry. To prove this feature of the theorem, it is incorrect to use anything other than the basic axioms.
Evidence 2 signs
Let us prove the second sign of equality in two angles and a side, based on the first.
Evidence 2 signs
Let's consider KMN and PRS. K is equal to P, N is equal to S. Side KN has the same length as PS. It is necessary to prove that KMN and PRS are the same.
Let us reflect the point M relative to the ray KN. Let's call the resulting point L. In this case, the length of the side KM = KL. NKL is equal to PRS. KNL is equal to RSP.
Since the sum of the angles is equal to 180 degrees, then KLN is equal to PRS, which means PRS and KLN are the same (similar) on both sides and angle, according to the first sign.
But, since KNL is equal to KMN, then KMN and PRS are two identical figures.
Evidence 3 signs
How to determine that triangles are congruent. This follows directly from the proof of the second feature.
Length KN = PS. Since K = P, N = S, KL=KM, and KN = KS, MN=ML, then:
This means that both figures are similar to each other. But since their sides are the same, they are also equal.
Many consequences follow from the signs of equality and similarity. One of them is that in order to determine whether two triangles are equal or not, it is necessary to know their properties, whether they are the same:
- all three sides;
- both sides and the angle between them;
- both angles and the side between them.
Using the triangle equality test to solve problems
Consequences of the first sign
In the course of the proof, one can come to a number of interesting and useful consequences.
- . The fact that the point of intersection of the diagonals of a parallelogram divides them into two identical parts is a consequence of the signs of equality and is quite amenable to proof. The sides of the additional triangle (with a mirror construction, as in the proofs that we performed) are the sides of the main one (the sides of the parallelogram).
- If there are two right triangles that have the same acute angles, then they are similar. If the leg of the first is equal to the leg of the second, then they are equal. This is quite easy to understand - all right triangles have a right angle. Therefore, the signs of equality are simpler for them.
- Two triangles with right angles, in which two legs have the same length, can be considered identical. This is due to the fact that the angle between the two legs is always 90 degrees. Therefore, according to the first criterion (by two sides and the angle between them), all triangles with right angles and identical legs are equal.
- If there are two right triangles, and their one leg and hypotenuse are equal, then the triangles are the same.
Let's prove this simple theorem.
There are two right triangles. One has sides a, b, c, where c is the hypotenuse; a, b - legs. The second has sides n, m, l, where l is the hypotenuse; m, n - legs.
According to the Pythagorean theorem, one of the legs is equal to:
;
.
Thus, if n = a, l = c (equality of legs and hypotenuses), respectively, the second legs will be equal. The figures, accordingly, will be equal according to the third characteristic (on three sides).
Let us note one more important consequence. If there are two equal triangle, and they are similar with a similarity coefficient k, that is, the pairwise ratios of all their sides are equal to k, then the ratio of their areas is equal to k2.
The first sign of equality of triangles. Video lesson on geometry 7th grade
Geometry 7 The first sign of equality of triangles
Conclusion
The topic we have discussed will help any student better understand basic geometric concepts and improve their skills in most interesting world mathematics.
This time I propose to organize something like an “evidence-based marathon” to solve problems that are offered to ninth-graders in the State Academic Examination in mathematics. They are connected with the proof of simple, but at the same time very useful geometric facts. The article deliberately does not provide detailed solutions to the problems, only some sketches and tips. Try to overcome this marathon distance on your own, without mistakes and in one approach.
Task 1. Prove that the bisectors of adjacent angles are perpendicular.
Angle α is designated by one arc, β by two
Proof: from the figure it is clear that α + α + β + β = 2α + 2β = 180 0 (straight angle), therefore, α + β = 90 0 . Q.E.D.
Task 2. Two segments A.C. And BD intersect at a point O, which is the middle of each of them. Prove the equality of triangles ACD And CAB.
ABCD, of course, will be a parallelogram, but this is not given in the condition
Proof: lateral triangles are equal in two sides and the angle between them ( B.O. = O.D.- by condition, A.O. = O.C.— by condition, ∠ DOC = ∠AOB- vertical), that is ∠ ACD = ∠CAB, and since they are crosswise lying at straight lines AB, CD and secant A.C., That AB parallel DC. We similarly prove the parallelism of lines B.C. And A.D. So, ABCD is a parallelogram by definition. B.C. = AD, AB = CD(in a parallelogram, opposite sides are equal), A.C.- common for triangles ACD And CAB, so they are equal on three sides. Q.E.D.
Task 3. Prove that the median drawn to the base of an isosceles triangle is the bisector of the angle opposite the base and is also perpendicular to the base.
The angles formed by the median and the base will be called “lower”, the median and sides - “upper”
Proof: the side triangles in the figure are equal on three sides, from which it follows that, firstly, the “upper” angles are equal (they proved that the bisector), secondly, the “lower” angles, in total as adjacent ones giving 180 0, and therefore equal in 90 0 each (proved perpendicularity). Q.E.D.
Task 4. Prove that the medians drawn to the lateral sides of an isosceles triangle are equal.
The triangles formed by the medians, base and lower halves of the lateral sides of the original triangle are called “lower”
Proof: the angles at the base of an isosceles triangle are equal, therefore the “lower” triangles are equal on two sides and the angle between them, which implies the equality of the drawn medians. Q.E.D.
Task 5. Prove that the bisectors drawn from the vertices of the base of an isosceles triangle are equal.
All angles marked in the figure are, of course, equal, although they are indicated by different arcs
Proof: The “bottom” triangle is isosceles, which follows from the equality of the angles at its base, the “side” triangles are equal in side (equal from the bisectors proved above) and two angles (the first are equal by condition, the second are vertical), therefore the remaining parts of the bisectors are also equal each other, which means the entire bisectors themselves are equal. Q.E.D.
Task 6. Prove that the length of the segment connecting the midpoints of two sides of a triangle is equal to half of the third side.
We will call the clean sides “bases”, the crossed out ones – “sides”
Proof: the lateral sides of the small and large triangle in the figure are related as 1: 2, in addition, they have one common angle, which means they are similar in the second attribute with a similarity coefficient of 1: 2, therefore the bases are related as 1: 2. Which is what needed to be proven .
Task 7. Prove that the diagonal of a parallelogram divides it into two equal triangles.
A parallelogram with a diagonal, there’s probably nothing more to add
Proof: Opposite sides of a parallelogram are equal, the diagonal is the common side for these triangles, so they are equal on three sides. Q.E.D.
Task 8. Prove that the median of a right triangle drawn to the hypotenuse is equal to half the hypotenuse.
In other words, the median is drawn from the vertex of the right angle
Proof: if we describe a circle around a given right triangle, then the right angle of the triangle inscribed in this circle will be described by a semicircle, so the hypotenuse will be the diameter of this circle, and the halves of the hypotenuse and the median given to us in the problem will be radii, so they are all equal. Q.E.D.
Task 9. Prove that the tangent segments drawn to a circle from one point are equal.
Additional construction: connect point C to point O (mentally)
Proof: angles B And A straight lines (the radii of the circle drawn to the swing point are perpendicular to the tangents), which means right triangles AOC And BOC equal in hypotenuse (the side we imagine is common to them O.C.) and leg (radii of the circle O.B. = O.A.), which means A.C. = C.B.. Q.E.D.
Problem 10. Prove that the diameter passing through the midpoint of a chord of a circle is perpendicular to it.
The line connecting two points in the figure is the median of the triangle we will consider
Proof: V isosceles triangle, formed by the points of intersection of the chord with the circle and the center of this circle, the depicted median will be the height, which means the diameter containing this height is perpendicular to the chord. Q.E.D.
Problem 11. Prove that if two circles have a common chord, then the line passing through the center of these circles is perpendicular to this chord.
Mentally connect together all the points marked in the figure, let’s call the point of intersection of the horizontal and vertical H
Proof: triangles O 1 A.O. 2 and O 1 B.O. 2 are equal on three sides, therefore, ∠ HO 2 A = ∠HO 2 B, then triangles HAO 2 and HBO 2 are equal on both sides and the angle between them, which means ∠ AHO 2 = ∠BHO 2, and in total two equal angles can give 180 0 only if each of them is equal to 90 0. Q.E.D.
Problem 12. Prove that if a circle can be inscribed in a quadrilateral, then the sums of the lengths of its opposite sides are equal.
Circumscribed quadrilateral. Let's call it ABCD. Let M, E, X and L be tangent points
Proof: We use the theorem on tangent segments (Problem 9). VC = VR, SR = CH, DX = D.L. And AT = AK. Let's sum up the sides AB And CD: AB + CD= (A.M.+ M.B.) + (DX+ XC) = AL+ BE+ D.L.+ C.E.= (AL+ LD) + (BE+ E.C.) = AD+ B.C. Q.E.D.
Problem 13. Prove that if a circle can be circumscribed around a quadrilateral, then the sums of its opposite angles are equal.
Circumcircle
Proof: According to the inscribed angle theorem, the sum of the opposite angles of this quadrilateral is equal to 180 0, since together they rest on a complete circle, the degree measure of which is 360 0. Q.E.D.
Problem 14. Prove that if a circle can be circumscribed around a trapezoid, then the trapezoid is isosceles.
Proof: the sum of the opposite angles of a quadrilateral inscribed in a circle is equal to α + β = 180 0 (see problem 13), the sum of the angles at the lateral side of the trapezoid is also equal to α + γ = 180 0 (these angles are one-sided with parallel bases and a secant side), from comparing these formulas we find that β = γ , that is, the angles at the base of such a trapezoid are equal, and it is truly isosceles. Q.E.D.
Problem 15. Squared ABCD points TO And E- midpoints of the sides AB And AD respectively. Prove that KD perpendicular C.E..
Two angles are called adjacent if they have one side in common, and the other sides of these angles are complementary rays. In Figure 20, angles AOB and BOC are adjacent.
The sum of adjacent angles is 180°
Theorem 1. The sum of adjacent angles is 180°.
Proof. Beam OB (see Fig. 1) passes between the sides of the unfolded angle. That's why ∠ AOB + ∠ BOS = 180°.
From Theorem 1 it follows that if two angles are equal, then their adjacent angles are equal.
Vertical angles are equal
Two angles are called vertical if the sides of one angle are complementary rays of the sides of the other. The angles AOB and COD, BOD and AOC, formed at the intersection of two straight lines, are vertical (Fig. 2).
Theorem 2. Vertical angles are equal.
Proof. Let's consider the vertical angles AOB and COD (see Fig. 2). Angle BOD is adjacent to each of angles AOB and COD. By Theorem 1 ∠ AOB + ∠ BOD = 180°, ∠ COD + ∠ BOD = 180°.
From this we conclude that ∠ AOB = ∠ COD.
Corollary 1. An angle adjacent to a right angle is a right angle.
Consider two intersecting straight lines AC and BD (Fig. 3). They form four corners. If one of them is straight (angle 1 in Fig. 3), then the remaining angles are also right (angles 1 and 2, 1 and 4 are adjacent, angles 1 and 3 are vertical). In this case, they say that these lines intersect at right angles and are called perpendicular (or mutually perpendicular). The perpendicularity of lines AC and BD is denoted as follows: AC ⊥ BD.
A perpendicular bisector to a segment is a line perpendicular to this segment and passing through its midpoint.
AN - perpendicular to a line
Consider a straight line a and a point A not lying on it (Fig. 4). Let's connect point A with a segment to point H with straight line a. The segment AN is called a perpendicular drawn from point A to line a if lines AN and a are perpendicular. Point H is called the base of the perpendicular.
Drawing square
The following theorem is true.
Theorem 3. From any point not lying on a line, it is possible to draw a perpendicular to this line, and, moreover, only one.
To draw a perpendicular from a point to a straight line in a drawing, use a drawing square (Fig. 5).
Comment. The formulation of the theorem usually consists of two parts. One part talks about what is given. This part is called the condition of the theorem. The other part talks about what needs to be proven. This part is called the conclusion of the theorem. For example, the condition of Theorem 2 is that the angles are vertical; conclusion - these angles are equal.
Any theorem can be expressed in detail in words so that its condition begins with the word “if” and its conclusion with the word “then”. For example, Theorem 2 can be stated in detail as follows: “If two angles are vertical, then they are equal.”
Example 1. One of the adjacent angles is 44°. What is the other equal to?
Solution.
Let us denote the degree measure of another angle by x, then according to Theorem 1.
44° + x = 180°.
Solving the resulting equation, we find that x = 136°. Therefore, the other angle is 136°.
Example 2. Let the angle COD in Figure 21 be 45°. What are the angles AOB and AOC?
Solution.
Angles COD and AOB are vertical, therefore, by Theorem 1.2 they are equal, i.e. ∠ AOB = 45°. Angle AOC is adjacent to angle COD, which means according to Theorem 1.
∠ AOC = 180° - ∠ COD = 180° - 45° = 135°.
Example 3. Find adjacent angles if one of them is 3 times larger than the other.
Solution.
Let us denote the degree measure of the smaller angle by x. Then the degree measure of the larger angle will be 3x. Since the sum of adjacent angles is equal to 180° (Theorem 1), then x + 3x = 180°, whence x = 45°.
This means that adjacent angles are 45° and 135°.
Example 4. The sum of two vertical angles is 100°. Find the size of each of the four angles.
Solution.
Let Figure 2 meet the conditions of the problem. The vertical angles COD to AOB are equal (Theorem 2), which means that their degree measures are also equal. Therefore, ∠ COD = ∠ AOB = 50° (their sum according to the condition is 100°). Angle BOD (also angle AOC) is adjacent to angle COD, and therefore, by Theorem 1
∠ BOD = ∠ AOC = 180° - 50° = 130°.
Geometry as a separate subject begins for schoolchildren in the 7th grade. Until that time they touch geometric problems fairly light in shape and basically what can be seen with illustrative examples: the area of the room, land plot, length and height of walls in rooms, flat objects, etc. At the beginning of studying geometry itself, the first difficulties appear, such as, for example, the concept of a straight line, since it is not possible to touch this straight line with your hands. As for triangles, this is the simplest type of polygon, containing only three angles and three sides.
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The theme of triangles is one of the main ones important and big topics school curriculum in geometry 7-9 grades. Having mastered it well, it is possible to decide very complex tasks. In this case, you can initially consider a completely different geometric figure, and then divide it for convenience into suitable triangular parts.
To work on the proof of equality ∆ ABC And ∆A1B1C1 You need to thoroughly understand the signs of equality of figures and be able to use them. Before studying the signs, you need to learn determine equality sides and angles of the simplest polygons.
To prove that the angles of triangles are equal, the following options will help:
- ∠ α = ∠ β based on the construction of the figures.
- Given in the task conditions.
- With two parallel lines and the presence of a secant, both internal cross-lying and corresponding ones can be formed ∠ α = ∠ β.
- By adding (subtracting) to (from) ∠ α = ∠ β equal angles.
- Vertical ∠ α and ∠ β are always similar
- General ∠ α, simultaneously belonging to ∆MNK And ∆MNH .
- The bisector divides ∠ α into two equal ones.
- Adjacent to ∠ 90°- angle equal to the original one.
- Adjacent equal angles are equal.
- The height forms two adjacent ∠ 90° .
- In isosceles ∆MNK at the base ∠ α = ∠ β.
- Equal ∆MNK And ∆SDH corresponding ∠ α = ∠ β.
- Previously proven equality ∆MNK And ∆SDH .
This is interesting: How to find the perimeter of a triangle.
3 signs that triangles are equal
Proof of equality ∆ ABC And ∆A1B1C1 very convenient to produce, based on basic signs the identity of these simplest polygons. There are three such signs. They are very important in solving many geometric problems. Each one is worth considering.
The characteristics listed above are theorems and are proven by the method of superimposing one figure on another, connecting the vertices of the corresponding angles and the beginning of the rays. Proofs for the equality of triangles in grade 7 are described in a very accessible form, but are difficult for schoolchildren to study in practice, since they contain a large number of elements designated by capital Latin letters. This is not entirely familiar to many students when they start studying the subject. Teenagers get confused about the names of sides, rays, and angles.
A little later, another important topic “Similarity of triangles” appears. The very definition of “similarity” in geometry means similarity of shape with different sizes. For example, you can take two squares, the first with a side of 4 cm, and the second 10 cm. These types of quadrangles will be similar and, at the same time, have a difference, since the second will be larger, with each side increased by the same number of times.
In considering the topic of similarity, 3 signs are also given:
- The first one is about two respectively equal angles two triangular figures in question.
- The second is about the angle and the sides that form it ∆MNK, which are equal to the corresponding elements ∆SDH .
- The third one indicates the proportionality of all corresponding sides of the two desired figures.
How can you prove that the triangles are similar? It is enough to use one of the above signs and correctly describe the entire process of proving the task. Theme of similarity ∆MNK And ∆SDH is easier to perceive by schoolchildren based on the fact that by the time of studying it, students are already fluent in using the notation of elements in geometric constructions, do not get confused by a huge number of names and know how to read drawings.
Concluding the passage of the extensive topic of triangular geometric shapes, students should already know perfectly how to prove equality ∆MNK = ∆SDH on two sides, set the two triangles to be equal or not. Considering that a polygon with exactly three angles is one of the most important geometric figures, you should take the material seriously, paying special attention to even the smallest facts of the theory.
