Let us recall from the material in the previous lesson that a triangle is called a right triangle if at least one of its angles is a right angle (i.e. equal to 90°).
Let's consider first sign Equality of triangles: if two legs of one right triangle are respectively equal to two legs of another right triangle, then such triangles are congruent.
Let's illustrate this case:
Rice. 1. Equal right triangles
Proof:
Let us recall the first equality of arbitrary triangles.
Rice. 2
If two sides and the angle between them of one triangle and the corresponding two sides and the angle between them of the second triangle are equal, then these triangles are congruent. This is indicated by the first sign of equality of triangles, that is:
A similar proof follows for right triangles:
.
Triangles are equal according to the first criterion.
Let's consider the second sign of equality of right triangles. If the leg and the adjacent acute angle of one right triangle are respectively equal to the leg and the adjacent acute angle of another right triangle, then such triangles are congruent.
Rice. 3
Proof:
Rice. 4
Let's use the second criterion for the equality of triangles:
Similar proof for right triangles:
Triangles are equal according to the second criterion.
Let's consider the third criterion for the equality of right triangles: if the hypotenuse and the adjacent angle of one right triangle are respectively equal to the hypotenuse and the adjacent angle of another triangle, then such triangles are congruent.
Proof:
Rice. 5
Let us recall the second criterion for the equality of triangles:
Rice. 6
These triangles are equal if:
Since it is known that one pair of acute angles in right triangles is equal to (∠A = ∠A 1), then the equality of the other pair of angles (∠B = ∠B 1) is proven as follows:
Since AB = A 1 B 1 (by condition), ∠B = ∠B 1, ∠A = ∠A 1. Therefore, triangles ABC and A 1 B 1 C 1 are equal according to the second criterion.
Consider the following criterion for the equality of triangles:
If the leg and hypotenuse of one triangle are respectively equal to the leg and hypotenuse of another triangle, such right triangles are congruent.
Rice. 7
Proof:
Let's combine triangles ABC and A 1 B 1 C 1 by overlapping. Let us assume that vertices A and A 1, as well as C and C 1 are superimposed, but vertex B and point B 1 do not coincide. This is exactly the case shown in the following figure:
Rice. 8
IN in this case we can notice the isosceles triangle ABB 1 (by definition - by the condition AB = AB 1). Therefore, according to the property, ∠AB 1 B = ∠ABV 1. Let's look at the definition of an external angle. External corner of a triangle is the angle adjacent to any angle of the triangle. Its degree measure is equal to the sum of two angles of a triangle that are not adjacent to it. The figure shows this ratio:
Rice. 9
Angle 5 is outer corner triangle and is equal to ∠5 = ∠1 + ∠2. It follows that an external angle is greater than each of the angles non-adjacent to it.
Thus, ∠ABB 1 is the external angle for triangle ABC and equal to the sum∠ABV 1 = ∠CAB + ∠ACV = ∠ABC = ∠CAB + 90 o. Thus, ∠AB 1 V (which is acute angle in a right triangle ABC 1) cannot be equal to the angle ∠ABB 1, because this angle is obtuse according to what has been proven.
This means that our assumption regarding the location of points B and B 1 turned out to be incorrect, therefore these points coincide. This means that triangles ABC and A 1 B 1 C 1 are superimposed. Therefore they are equal (by definition).
Thus, these features are not introduced in vain, because they can be used to solve some problems.
1. No. 38. Butuzov V.F., Kadomtsev S.B., Prasolov V.V., edited by Sadovnichy V.A. Geometry 7. M.: Education. 2010
2. Based on the data shown in the figure, indicate equal triangles, if they are.
3. Based on the data indicated in the figure, indicate equal triangles, if any. Keep in mind that AC = AF.
4. In a right triangle, the median and altitude are drawn to the hypotenuse. The angle between them is 20°. Determine the size of each of the acute angles of this right triangle.
Let's consider three points that do not lie on the same line, and three segments connecting these points (Fig. 1).
A triangle is a part of the plane bounded by these segments, the segments are called the sides of the triangle, and the ends of the segments (three points that do not lie on the same straight line) are the vertices of the triangle.
Table 1 lists all possible types triangles depending on the size of their angles .
Table 1 - Types of triangles depending on the size of the angles
| Drawing | Triangle type | Definition |
 | Acute triangle | A triangle with all angles are sharp , called acute-angled |
 | Right triangle | A triangle with one of the angles is right , called rectangular |
 | Obtuse triangle | A triangle with one of the angles is obtuse , called obtuse |
| Acute triangle |
Definition: A triangle with all angles are sharp , called acute-angled |
| Right triangle |
Definition: A triangle with one of the angles is right , called rectangular |
| Obtuse triangle |
Definition: A triangle with one of the angles is obtuse , called obtuse |
Depending on the lengths of the sides there are two important types triangles.
Table 2 - Isosceles and equilateral triangles
| Drawing | Triangle type | Definition |
 | Isosceles triangle | sides, and the third side is called the base of an isosceles triangle |
 | Equilateral (correct) triangle | A triangle in which all three sides are equal is called an equilateral or regular triangle. |
| Isosceles triangle |
Definition: A triangle whose two sides are equal is called an isosceles triangle. In this case two equal sides called sides, and the third side is called the base of an isosceles triangle |
| Equilateral (right) triangle |
Definition: A triangle in which all three sides are equal is called an equilateral or regular triangle. |
Triangles are said to be equal if they can be combined by overlay .
Table 3 shows signs of equality of triangles.
Table 3 – Signs of equality of triangles
| Drawing | Feature name | Attribute wording |
 | By two sides and the angle between them | |
 | Test for equivalence of triangles By side and two adjacent angles | |
 | Test for equivalence of triangles By three parties |
| Test for equivalence of triangles on two sides and the angle between them |
|
| Attribute wording. If two sides of one triangle and the angle between them are respectively equal to two sides of another triangle and the angle between them, then such triangles are congruent |
| Test for equivalence of triangles along a side and two adjacent corners |
|
| Attribute wording. 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 |
| Test for equivalence of triangles on three sides |
|
| Attribute wording. If three sides of one triangle are respectively equal to three sides of another triangle, then such triangles are congruent |
The following names are commonly used for the sides of right triangles.
The hypotenuse is the side of a right triangle that lies opposite right angle(Fig. 2), the other two sides are called legs.
Table 4 – Signs of equality of right triangles
| Drawing | Feature name | Attribute wording |
 | By two sides | If two legs of one right triangle are respectively equal to two legs of another right triangle, then such right triangles are congruent |
 | Equality test for right triangles By leg and adjacent acute angle | If the leg and the adjacent acute angle of one right triangle are respectively equal to the leg and the adjacent acute angle of another right triangle, then such right triangles are congruent |
 | Equality test for right triangles By leg and opposite acute angle | If the leg and the opposite acute angle of one right triangle are respectively equal to the leg and the opposite acute angle of another right triangle, then such right triangles are congruent |
 | Equality test for right triangles By hypotenuse and acute angle | If the hypotenuse and acute angle of one right triangle are respectively equal to the hypotenuse and acute angle of another right triangle, then such right triangles are congruent |
 | Equality test for right triangles By leg and hypotenuse | If the leg and hypotenuse of one right triangle are respectively equal to the leg and hypotenuse of another right triangle, then such right triangles are congruent |
| Sign of equality of right triangles on two sides |
To establish the equality of right triangles, it is enough to know that two elements of one triangle are respectively equal to two elements of another triangle (excluding the right angle). This, of course, does not apply to the equality of two angles of one triangle to two angles of another triangle.
Since in a right triangle the angle between two legs is straight, and any two right angles are equal, then from the first sign of equality of triangles it follows:
If the legs of one right triangle are correspondingly equal to the legs of another, then such triangles are congruent (Fig. 5).
If the leg and the adjacent acute angle of one right triangle are respectively equal to the leg and the adjacent angle of another triangle, then such triangles are congruent (Fig. 6).
Let's consider two more signs of equality of right triangles.
THEOREM . If the hypotenuse and acute angle of one right triangle are equal to the hypotenuse and acute angle of another triangle, then such triangles are congruent (Fig. 7).
PROOF. From property 1є § it follows that in such triangles the other two acute angles are also equal, therefore the triangles are equal according to the second sign of equality of triangles, that is, along the side (hypotenuse) and two adjacent angles.
Q.E.D.
THEOREM . If the hypotenuse and leg of one right triangle are respectively equal to the hypotenuse and leg of another triangle, then such triangles are congruent.
PROOF. Consider triangles ABC and A 1 B 1 C 1, whose angles C and C 1 are right angles, AB = A 1 B 1, BC = B 1 C 1 (Fig. 8).
Because< C = < C 1 , то треугольник ABC можно наложить на треугольник A 1 B 1 C 1 так, что вершина C совместится с вершиной C 1 , а стороны CA и CB наложатся соответственно на лучи C 1 A 1 и C 1 B 1 , поскольку CB = C 1 B 1 , то вершина B совместится с вершиной B 1 . Но тогда вершины A и A 1 также совместятся. В самом деле, если предположить, что точка A совместится с некоторой другой точкой A 2 луча C 1 A 1 , то получим равнобедренный треугольник A 1 B 1 A 2 , в котором углы при основании A 1 A 2 не равны (на рисунке < A 2 - острый, а < A 1 - тупой как смежный с острым углом B 1 A 1 C 1). Но это невозможно, поэтому вершины A и A 1 совместятся. Следовательно, полностью совместятся треугольники ABC A 1 B 1 C 1 , то есть они равны.
Q.E.D.
Its significance lies in the fact that most of the theorems of geometry can be deduced from it or with its help. One of the theorems makes it possible to verify that if a perpendicular and inclined lines are drawn to it from a point outside a line, then: a) the inclined lines are equal if their projections are equal; b) the one that is inclined is larger, which has a larger projection.
The Pythagorean theorem was the first statement that related the lengths of the sides of triangles. Then we learned how to find the lengths of the sides and angles of acute and obtuse triangles. The whole science of trigonometry arose (“trigon” means “triangle” in Greek). This science has found application in land surveying. But even earlier, with its help, they learned to measure imaginary triangles in the sky, the vertices of which were stars. Now trigonometry is even used to measure distances between spacecraft.
Using the properties of the areas of polygons, we will now establish a remarkable relationship between the hypotenuse and the legs of a right triangle. The theorem we will prove is called the Pythagorean theorem, which is the most important theorem in geometry.
If we are given a triangle,
And with a right angle at that,
That is the square of the hypotenuse
We can always easily find:
We square the legs,
We find the sum of powers
And in such a simple way
We will come to the result.
THEOREM. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs.
PROOF. Consider a right triangle with legs a, b and c (Fig. 9 a).
Let us prove that c 2 = a 2 + b 2 . Let's build the triangle to a square with side a+b, as shown in the figure (Fig. 9 b).
The area of such a square with side a + b is equal to (a + b) 2. On the other hand, this square is made up of four equal right triangles with area ab and a square with side c, so
Thus, (a + b) 2 =2ab + c 2, whence c 2 = a 2 + b 2.
Q.E.D.
COROLLARY 1 . In a right triangle, any of the legs is less than the hypotenuse.
PROOF. According to the Pythagorean theorem AB 2 = AC 2 + BC 2 . Since BC 2 >0, then AC 2<АВ, То есть АС<АВ.
CONCLUSION 2. For any acute angle b cosb<1.
EVIDENCE. By definition of cosine cosб = . But in Corollary 1 it was proven that AC<АВ, This means the fraction is less than 1.
Right triangles whose sides are expressed as integers are called Pythagorean triangles.
It can be proven that the legs a, b and hypotenuse c of such triangles are expressed by the formulas a=2kmn; b=k(m 2 -n 2); c=k(m 2 +n 2), where k, m and n are natural numbers such that m>n. Triangles with sides whose lengths are 3, 4, 5 are called Egyptian triangles, because they were known to the ancient Egyptians.
Converse to the Pythagorean theorem.
If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is right-angled (a sign of a right triangle).
PROOF.
Let in triangle ABC AB 2 = AC 2 + BC 2. Let us prove that angle C is a right angle. Consider a right triangle A 1 B 1 C 1 with right angle C 1, in which A 1 C 1 = AC and B 1 C 1 = BC. By the Pythagorean theorem A 1 B 1 2 =A 1 C 1 2 +B 1 C 1 2, and therefore A 1 B 1 2 = AC 2 +BC 2. But AC 2 + BC 2 = AB 2 according to the theorem. Therefore, A 1 B 1 2 = AB 2, whence A 1 B 1 = AB. Triangles ABC and A 1 B 1 C 1 are equal on three sides, therefore< C = < C 1 , то есть треугольник ABC прямоугольный с прямым углом C.
Q.E.D.
Let us recall from the material in the previous lesson that a triangle is called a right triangle if at least one of its angles is a right angle (i.e. equal to 90°).
Let's consider first sign Equality of triangles: if two legs of one right triangle are respectively equal to two legs of another right triangle, then such triangles are congruent.
Let's illustrate this case:
Rice. 1. Equal right triangles
Proof:
Let us recall the first equality of arbitrary triangles.
Rice. 2
If two sides and the angle between them of one triangle and the corresponding two sides and the angle between them of the second triangle are equal, then these triangles are congruent. This is indicated by the first sign of equality of triangles, that is:
A similar proof follows for right triangles:
.
Triangles are equal according to the first criterion.
Let's consider the second sign of equality of right triangles. If the leg and the adjacent acute angle of one right triangle are respectively equal to the leg and the adjacent acute angle of another right triangle, then such triangles are congruent.
Rice. 3
Proof:
Rice. 4
Let's use the second criterion for the equality of triangles:
Similar proof for right triangles:
Triangles are equal according to the second criterion.
Let's consider the third criterion for the equality of right triangles: if the hypotenuse and the adjacent angle of one right triangle are respectively equal to the hypotenuse and the adjacent angle of another triangle, then such triangles are congruent.
Proof:
Rice. 5
Let us recall the second criterion for the equality of triangles:
Rice. 6
These triangles are equal if:
Since it is known that one pair of acute angles in right triangles is equal to (∠A = ∠A 1), then the equality of the other pair of angles (∠B = ∠B 1) is proven as follows:
Since AB = A 1 B 1 (by condition), ∠B = ∠B 1, ∠A = ∠A 1. Therefore, triangles ABC and A 1 B 1 C 1 are equal according to the second criterion.
Consider the following criterion for the equality of triangles:
If the leg and hypotenuse of one triangle are respectively equal to the leg and hypotenuse of another triangle, such right triangles are congruent.
Rice. 7
Proof:
Let's combine triangles ABC and A 1 B 1 C 1 by overlapping. Let us assume that vertices A and A 1, as well as C and C 1 are superimposed, but vertex B and point B 1 do not coincide. This is exactly the case shown in the following figure:
Rice. 8
In this case, we can notice the isosceles triangle ABV 1 (by definition - by the condition AB = AB 1). Therefore, according to the property, ∠AB 1 B = ∠ABV 1. Let's look at the definition of an external angle. External corner of a triangle is the angle adjacent to any angle of the triangle. Its degree measure is equal to the sum of two angles of a triangle that are not adjacent to it. The figure shows this ratio:
Rice. 9
Angle 5 is the external angle of the triangle and is equal to ∠5 = ∠1 + ∠2. It follows that an external angle is greater than each of the angles non-adjacent to it.
Thus, ∠ABB 1 is the external angle for triangle ABC and is equal to the sum ∠ABB 1 = ∠CAB + ∠ACB = ∠ABC = ∠CAB + 90 o. Thus, ∠AB 1 B (which is an acute angle in the right triangle ABC 1) cannot be equal to the angle ∠ABB 1, because this angle is obtuse according to what has been proven.
This means that our assumption regarding the location of points B and B 1 turned out to be incorrect, therefore these points coincide. This means that triangles ABC and A 1 B 1 C 1 are superimposed. Therefore they are equal (by definition).
Thus, these features are not introduced in vain, because they can be used to solve some problems.
1. No. 38. Butuzov V.F., Kadomtsev S.B., Prasolov V.V., edited by Sadovnichy V.A. Geometry 7. M.: Education. 2010
2. Based on the data indicated in the figure, indicate equal triangles, if any.
3. Based on the data indicated in the figure, indicate equal triangles, if any. Keep in mind that AC = AF.
4. In a right triangle, the median and altitude are drawn to the hypotenuse. The angle between them is 20°. Determine the size of each of the acute angles of this right triangle.
1. The first two signs of equality of right triangles.
For two triangles to be equal, it is enough that three elements of one triangle are equal to the corresponding elements of the other triangle, and these elements must certainly include at least one side.
Since all right angles are equal to each other, right triangles already have one equal element, namely one right angle.
It follows that right triangles are congruent:
if the legs of one triangle are respectively equal to the legs of another triangle (Fig. 153);
if the leg and the adjacent acute angle of one triangle are respectively equal to the leg and the adjacent acute angle of the other triangle (Fig. 154).
Let us now prove two theorems that establish two more criteria for the equality of right triangles.
To prove this theorem, let us construct two rectangular angles ABC and A'B'C', in which the angles A and A' are equal, the hypotenuses AB and A'B' are also equal, and the angles C and C' are right (Fig. 157) .
Let’s superimpose triangle A’B’C’ onto triangle ABC so that vertex A’ coincides with vertex A, hypotenuse A’B’ coincides with equal hypotenuse AB. Then, due to the equality of angles A and A’, the side A’C’ will go along the side AC; leg B’C’ will coincide with leg BC: both of them are perpendiculars drawn to one straight line AC from one point B. This means that vertices C and C’ will coincide.
Triangle ABC coincides with triangle A'B'C'.
Therefore, \(\Delta\)ABC = \(\Delta\)A'B'C'.
This theorem gives the 3rd criterion for the equality of right triangles (by the hypotenuse and the acute angle).
Theorem 2. If the hypotenuse and leg of one triangle are respectively equal to the hypotenuse and leg of another triangle, then such right triangles are congruent.
To prove this, let's construct two right triangles ABC and A'B'C', in which angles C and C' are right angles, legs AC and A'C' are equal, hypotenuses AB and A'B' are also equal (Fig. 158) .
Let's draw a straight line MN and mark point C on it, from this point we draw a perpendicular SC to straight line MN. Then we will superimpose the right angle of the triangle ABC onto the right angle KSM so that their vertices are aligned and the leg AC goes along the ray SC, then the leg BC goes along the ray CM. The right angle of the triangle A'B'C' will be superimposed on the right angle KCN so that their vertices are aligned and the leg A'C' goes along the ray SK, then the leg C'B' goes along the ray CN. The vertices A and A' will coincide due to the equality of the legs AC and A'C'.
Triangles ABC and A'B'C' will together form an isosceles triangle BAB', in which AC will be the altitude and bisector, and therefore the axis of symmetry of triangle BAB'. It follows from this that \(\Delta\)ABC = \(\Delta\)A’B’C’.
This theorem gives the 4th criterion for the equality of right triangles (by hypotenuse and leg).
So, all the signs of equality of right triangles:
1. If two legs of one right triangle are respectively equal to two legs of another right triangle, then such right triangles are equal2. If the leg and the adjacent acute angle of one right triangle are respectively equal to the leg and the adjacent acute angle of another right triangle, then such right triangles are congruent
3. If the leg and the opposite acute angle of one right triangle are respectively equal to the leg and the opposite acute angle of another right triangle, then such right triangles are congruent
4. If the hypotenuse and acute angle of one right triangle are respectively equal to the hypotenuse and acute angle of another right triangle, then such right triangles are congruent
5. If the leg and hypotenuse of one right triangle are respectively equal to the leg and hypotenuse of another right triangle, then such right triangles are congruent
