(Sign of equality right triangles along the 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 triangles are congruent.
ΔABC, ΔA 1 B 1 C 1 ,
∠C=90°, ∠C 1 =90°,
BC=B 1 C 1 , AB=A 1 B 1 ,
Prove:
∆ABC= ∆A 1 B 1 C 1
Proof:
I. On the ray BC, on the other side of the point C, we set aside the segment CD, CD=CB.
Connect points A and D with a line segment.
On the ray B 1 C 1 on the other side of the point C 1 set aside the segment C 1 D 1 , C 1 D 1 =C 1 B 1 .
Let's draw a segment A 1 D 1 .
II. In triangles ACD and ACB:
From the equality of triangles follows the equality of the corresponding sides: AD=AB.
Similarly, the equality of triangles A 1 C 1 D 1 and A 1 C 1 B 1 and the equality of their sides A 1 D 1 =A 1 B 1 are proved.
III. Since AB=A 1 B 1 , then AD=A 1 D 1 .
IV. In triangles ABD and A 1 B 1 D 1:
1) AB=A 1 B 1 (by condition);
2) AD=A 1 D 1 (as proved);
3) BD=2BC=2B 1 C 1 =B 1 D 1 .
From the equality of triangles follows the equality of the corresponding angles: ∠B=∠B 1 .
In the geometry course of grade 7, they were studied, and in the last lesson they were repeated, the so-called signs of equality of triangles. Recall them:
1st sign (on 2 sides and the angle between them): if two triangles have equal sides and the angle between them, then such triangles are congruent.
2nd sign (on the side and two adjacent angles): if triangles have equal side and two angles adjacent to a given side, then such triangles are congruent. Note: using the fact that the sum of the angles of a triangle is constant and equal to , it is easy to prove that the condition of "adjacency" of the angles is not necessary, that is, the sign will be true in the following formulation: "... a side and two angles are equal, then ...".
3rd sign (on 3 sides): if all three sides of a triangle are equal, then such triangles are congruent.
Naturally, all these signs remain true for right triangles. However, right triangles have one essential feature - they always have a pair of equal right angles. Therefore, these signs are simplified for them. So, let's formulate the signs of equality of right triangles:
1st sign (on two legs): if the legs of right-angled triangles are pairwise equal, then such triangles are equal to each other (see Fig. 2).
Given:
Rice. 2. Illustration of the first sign of equality of right triangles
Prove: 
Proof: remember that in right triangles: 
. So, we can use the first sign of equality of triangles (on 2 sides and the angle between them) and get: .
Proven.
2-th sign (on the leg and angle): if the leg and acute angle of one right triangle are equal to the leg and acute angle of another right triangle, then such triangles are equal to each other (see Fig. 3).
Given:
Rice. 3. Illustration of the second sign of equality of right triangles
Prove:
Proof: we note right away that the fact that the angles adjacent to equal legs are equal is not fundamental. Indeed, the sum of acute angles of a right triangle (by property 1) is equal to . Hence, if one pair of these angles is equal, then the other is equal (since their sums are the same).
The proof of this feature comes down to using second sign of equality of triangles(at 2 corners and side). Indeed, by condition, the legs and a pair of angles adjacent to them are equal. But the second pair of angles adjacent to them consists of the angles . So, we can use the second criterion for the equality of triangles and get: .
Proven.
3rd sign (by hypotenuse and angle): if the hypotenuse and the acute angle of one right triangle are equal to the hypotenuse and the acute angle of another right triangle, then such triangles are equal to each other (see Fig. 4).
Given:
Rice. 4. Illustration of the third sign of equality of right triangles
Prove:
Proof: to prove this sign, you can immediately use the second sign of the equality of triangles- by a side and two angles (more precisely, by a consequence, which states that the angles do not have to be adjacent to the side). Indeed, by the condition: , , and from the properties of right triangles it follows that . So, we can use the second criterion for the equality of triangles, and get: .
Proven.
4th sign (by hypotenuse and leg): if the hypotenuse and leg of one right triangle are equal, respectively, to the hypotenuse and leg of another right triangle, then such triangles are equal to each other (see Fig. 5).
Given:
Rice. 5. Illustration of the fourth sign of equality of right triangles
Prove:
Proof: to prove this sign, we will use the sign of equality of triangles, which we formulated and proved in the last lesson, namely: if triangles have equal two sides and a larger angle, then such triangles are equal. Indeed, by condition we have two equal sides. In addition, by the property of right triangles: . It remains to prove that the right angle is the largest in the triangle. Let's assume that this is not the case, which means that there must be at least one more angle that is greater than . But then the sum of the angles of the triangle will already be greater. But this is impossible, which means that there cannot be such an angle in a triangle. Hence, the right angle is the largest in a right triangle. So, you can use the sign formulated above, and get: .
Proven.
We now formulate one more property, which is characteristic only for right triangles.
Property
The leg lying opposite the angle at is 2 times smaller than the hypotenuse (see Fig. 6).
Given:
Prove:AB
Proof: perform an additional construction: extend the line beyond the point by a segment equal to . Let's get a point. Since the angles and are adjacent, their sum is equal to . Since , then the angle .
So right triangles 
(according to two legs: - general, - by construction) - the first sign of the equality of right triangles.
From the equality of triangles follows the equality of all corresponding elements. Means, . Where: . In addition, (from the equality of all the same triangles). This means that the triangle is isosceles (since it has equal angles at the base), but an isosceles triangle, one of whose angles is equal, is equilateral. It follows from this, in particular, that 
, which was to be proved.
Proven.
4. Property of the leg lying opposite the angle in
It is worth noting that the converse statement is also true: if in a right triangle the hypotenuse is twice as large as one of the legs, then the acute angle opposite this leg is equal to.
We formulate another important sign of a right triangle.
Note:sign means that if some statement is true, then the triangle is a right triangle. That is, the feature allows you to identify a right triangle.
It is important not to confuse the sign withproperty - that is, if the triangle is right-angled, then it has such properties ... Often the signs and properties are mutually inverse, but not always. For example, the property of an equilateral triangle: an equilateral triangle has an angle . But this will not be a sign of an equilateral triangle, since not every triangle that has an angle , is equilateral.
You can bring more life example: property of the word "bread" - in the word "bread" 4 letters. But the presence of 4 letters is not a sign of the word "bread", since there are many words of 4 letters.
5. Sign of a right triangle (the median is equal to half the side to which it is drawn)
So, right triangle sign:
If in a triangle the median is equal to half of the side to which it is drawn, then this triangle is right-angled, and the median is drawn from the vertex of the right angle.
Note: we recall that median- a line connecting the vertex of the triangle with the middle of the opposite side (see Fig. 7).
Given:
Prove:
Proof: because the 
, then the triangles are isosceles. This means that the angles at the bases of each of these triangles are equal. That is, , . Then the sum of the angles of the triangle is So, . But: , which was required to be proved.
Proven.
In this lesson, we examined the basic properties of right triangles, studied earlier in grade 7. In particular, they remembered the signs of equality, as well as other signs and properties of right triangles.
1. In a right triangle , is the bisector, . Find the length of the leg if cm.
2. A point is marked on the hypotenuse of a right triangle so that 
. Prove that the point is equidistant from the points , and .
3. Find the acute angles of a right triangle if their ratio is 5:13.
4. The median drawn to the hypotenuse is cm.
5. In a triangle , is the bisector, . The segment is cm less than the segment . Find the bisector.
Lesson 5: Polygons
In this lesson, we will start with new topic and introduce a new concept for us "polygon". We will look at the basic concepts associated with polygons: sides, vertices, corners, convexity and non-convexity. Then we will prove the most important facts, such as the theorem on the sum of the interior angles of a polygon, the theorem on the sum of the exterior angles of a polygon. As a result, we will come close to studying special cases of polygons, which will be considered in future lessons.
1. The concept of "polygon"
In the course of geometry, we study the properties of geometric shapes and have already considered the simplest of them: triangles and circles. At the same time, we also discussed specific special cases of these figures, such as right-angled, isosceles and regular triangles. Now it's time to talk about more general and complex shapes − polygons.
With a special case polygons we are already familiar - this is a triangle (see Fig. 1).
Rice. 1. Triangle
The name itself already emphasizes that this is a figure that has three corners. Therefore, in polygon there can be many of them, i.e. more than three. For example, let's draw a pentagon (see Fig. 2), i.e. figure with five corners.
Rice. 2. Pentagon. Convex polygon
Definition. Polygon- a figure consisting of several points (more than two) and the corresponding number of segments that connect them in series. These points are called peaks polygon, and the segments parties. In this case, no two adjacent sides lie on the same straight line and no two non-adjacent sides intersect.
Definition. regular polygon is a convex polygon in which all sides and angles are equal.
Any polygon divides the plane into two regions: internal and external. The interior is also referred to as polygon.
In other words, for example, when they talk about a pentagon, they mean both its entire inner region and its border. And the inner area also includes all points that lie inside the polygon, i.e. the point also belongs to the pentagon (see Fig. 2).
Polygons are sometimes also called n-gons to emphasize that the general case of having some unknown number of corners (n pieces) is being considered.
Definition. Polygon Perimeter is the sum of the lengths of the sides of the polygon.
Now we need to get acquainted with the types of polygons. They are divided into convex And non-convex. For example, the polygon shown in Fig. 2 is convex, and in Fig. 3 non-convex.
Rice. 3. Non-convex polygon
2. Convex and non-convex polygons
Definition 1. Polygon called convex, if when drawing a straight line through any of its sides, the entire polygon lies only on one side of this line. non-convex are all the rest polygons.
It is easy to imagine that when extending any side of the pentagon in Fig. 2 it will all be on one side of this straight line, i.e. he is convex. But when drawing a straight line through the quadrilateral in Fig. 3 we already see that it divides it into two parts, i.e. he is non-convex.
But there is another definition of the convexity of a polygon.
Definition 2. Polygon called convex if, when choosing any two of its interior points and connecting them with a segment, all points of the segment are also interior points of the polygon.
A demonstration of the use of this definition can be seen in the example of constructing segments in Fig. 2 and 3.
Definition. Diagonal A polygon is any segment that connects two non-adjacent vertices.
3. Theorem on the sum of interior angles of a convex n-gon
To describe the properties of polygons, there are two most important theorems about their angles: convex polygon interior angle sum theorem And convex polygon exterior angle sum theorem. Let's consider them.
Theorem. On the sum of interior angles of a convex polygon (n-gon).
Where is the number of its angles (sides).
Proof 1. Let's depict in Fig. 4 convex n-gon.
Rice. 4. Convex n-gon
Draw all possible diagonals from the vertex. They divide the n-gon into triangles, because each of the sides of the polygon forms a triangle, except for the sides adjacent to the vertex. It is easy to see from the figure that the sum of the angles of all these triangles will just be equal to the sum of the interior angles of the n-gon. Since the sum of the angles of any triangle is , then the sum of the interior angles of an n-gon is:
Which is what needed to be proven.
Proof 2. Another proof of this theorem is also possible. Let's draw a similar n-gon in Fig. 5 and connect any of its interior points to all vertices.
.Proven.
From the proved theorem follows an interesting fact that the sum of the external angles of a convex n-gon is equal to 
on the number of its angles (sides). By the way, unlike the sum of interior angles.
Next, we will work in more detail with a special case of polygons - quadrangles. In the next lesson, we will get acquainted with such a figure as a parallelogram and discuss its properties.
Homework
1. Is there a convex polygon whose sum of angles is equal to: a) ; b) ; V) ?
2. Find the corners of the quadrangle if they are proportional to the numbers 2, 3, 10 and 21. Is this quadrangle convex or non-convex?
3. The vertices of a convex pentagon are connected through one. Find the sum of the angles at the vertices of the resulting "star".
Lesson 6: Parallelogram
This lesson is devoted to one of the types of convex quadrilaterals, namely, a parallelogram. A parallelogram is one of the private types of quadrilaterals, which includes such subspecies as a rectangle, rhombus, square - figures that each of us has been familiar with since childhood. We will look at the definition and properties of a parallelogram and also solve some examples using these properties.
Definition of a parallelogram
In the last lesson, we considered the concept of a convex polygon. Now let's study special case polygon - a quadrangle, or rather, a special case of a quadrilateral - parallelogram.
Parallelogram is a quadrilateral with opposite sides are pairwise parallel (see Fig. 1).
Rice. 1. Parallelogram
That is, if two parallel lines are given that intersect two more parallel lines, then they form a figure called a parallelogram.
From the fact that it is a parallelogram, we can draw the following conclusions: 
. The converse is also true: if , then the quadrilateral is a parallelogram.
Apart from this definition, we can give a few more equivalent ones, but we will focus on this, the classical definition of a parallelogram, and formulate the properties of this figure, using the parallelism of its opposite sides.
Recall from the material of the previous lesson that a right-angled triangle is called a triangle if it has at least one of the angles of the line (i.e., equal to 90 o).
Consider first sign triangle equality: 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:
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 stated by the first sign of the equality of triangles, that is:
A similar proof follows for right triangles:
.
The triangles are equal in the first sign.
Consider the second criterion for the equality of right triangles. If the leg and the acute angle adjacent to it 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:
A similar proof for right triangles:
The triangles are equal in the second criterion.
Consider the third criterion for the equality of right triangles: if the hypotenuse and the angle adjacent to it of one right triangle are respectively equal to the hypotenuse and the angle adjacent to another triangle, then such triangles are congruent.
Proof:
Rice. 5
Recall the second criterion for the equality of triangles:
Rice. 6
These triangles are congruent if:
Since it is known that one pair of acute angles in right triangles is equal to (∠А = ∠А 1), then the equality of the other pair of angles (∠B = ∠B 1) is proved as follows:
Since AB \u003d A 1 B 1 (by condition), ∠B \u003d ∠B 1, ∠A \u003d ∠A 1. Therefore, triangles ABC and A 1 B 1 C 1 are equal in the second sign.
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 superimpose the triangles ABC and A 1 B 1 C 1. Assume that vertices A and A 1 , as well as C and C 1 overlap, but vertex B and point B 1 do not match. This case is shown in the following figure:
Rice. 8
In this case, we can notice an isosceles triangle ABB 1 (by definition - by the condition AB = AB 1). Therefore, by property, ∠AB 1 B = ∠ABV 1 . Consider the definition of an external corner. outside corner triangle is the angle adjacent to any corner of the triangle. Its degree measure is equal to the sum of the 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 the exterior angle is greater than each of the angles that are not adjacent to it.
Thus, ∠ABB 1 is an external angle for the 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 a right triangle ABB 1) cannot be equal to the angle ∠ABB 1, because this angle is obtuse as 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 the 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 in solving some problems.
- Omsk State University ().
- Reference portal calc.ru ().
- Teacher Portal ().
1. No. 38. Butuzov V.F., Kadomtsev S.B., Prasolov V.V., edited by Sadovnichiy V.A. Geometry 7. M .: Education. 2010
2. Based on the data shown in the figure, indicate equal triangles, if any.
3. Based on the data shown in the figure, indicate equal triangles, if any. Keep in mind that AC = AF.
4. In a right triangle, the median and height are drawn to the hypotenuse. The angle between them is 20 o. Determine the size of each of the acute angles of the given right triangle.
Sections: Mathematics
Topic: “Signs of equality of right triangles”
Purpose: consolidation of knowledge (properties of right triangles), acquaintance with some signs of equality of right triangles.
During the classes:
I. Organizational moment.II. Orally.
1. Answer the questions:
- Name the elements of a right triangle.
- What are the properties of the elements of a right triangle?
- Prove that the leg of a right triangle opposite the angle of 30 0 is equal to half of the hypotenuse.
- Prove that if the leg of a right triangle is equal to half of the hypotenuse, then the angle opposite this leg is 30 0 .
- Find x. Choose your answer from the triangle. The letters of a word are in the sectors of the triangle. Pair discussion (3 min).
Picture 1.
They made up the word "sign".
III. Learning new materialStudying triangles, we say that it has certain properties and features. What signs of triangle equality do you know? We have formulated and proved the properties of right triangles, and today we will consider the signs of equality of right triangles, we will solve problems using them.
Proving the equality of triangles, how many pairs of correspondingly equal elements were found? Is it possible to prove the equality of right triangles on two legs?
Before you are two right-angled triangles ABC and A 1 B 1 C 1, their legs are equal, respectively. Prove, if possible, their equality.
No. 1. (On two legs)
Figure 2.
Given: ABC and A 1 B 1 C 1, B \u003d B 1 \u003d 90 0, AB \u003d A 1 B 1, BC \u003d B 1 C 1
Prove: ABC = A 1 B 1 C 1
How will the sign sound? (Then task #1)
No. 2. (According to the leg and the acute angle adjacent to it)
Figure 3
Given: ABC and A 1 B 1 C 1, B \u003d B 1 \u003d 90 0, BC \u003d B 1 C 1, C \u003d C 1
Prove: ABC = A 1 B 1 C 1
How will the sign sound? (Then task #2)
No. 3. (By hypotenuse and acute angle)
Figure 4
Given: ABC and A 1 B 1 C 1, B \u003d B 1 \u003d 90 0, AC \u003d A 1 C 1, A \u003d A 1
Prove: ABC = A 1 B 1 C 1
How will the sign sound? (Then task #3)
Tasks. Find equal triangles and prove their equality.
Figure 5
IV. Consolidation of what was learned in the lesson.Solve the following problem.
Figure 6
Given: ABC, A 1 B 1 C 1, DAB \u003d CBA \u003d 90 0, AD \u003d BD
Prove: CAB=DBA.
Discussion in groups of four (3 min).
Why a task from textbook No. 261 with a note.
Figure 7
Given: ABC - isosceles, AD and CE - the height of ABC
Prove: AD=CE
Proof:
V. Homework.
P.35 (three signs), No. 261 (prove that AOS is isosceles), No. 268 (a sign of equality of right triangles along the leg and opposite angle).
In the next geometry lesson, we will continue our acquaintance with the signs of equality of right triangles. I will also post marks next time based on the results for 2 lessons.
Additionally. Find equal triangles.
Three criteria for the equality of any triangles are known:
- on two sides and the angle between them;
- on two corners and the side between them;
- on three sides.
Two right triangles always have one pair of angles equal to each other - these are right angles. Therefore, the tests for triangle equality for right triangles are simplified in the sense that in order to say that the triangles are equal, one needs to know about the equality of a smaller number of elements.
The first criterion for the equality of triangles for right triangles is reduced to the equality of two legs: if the legs of one right triangle are equal to the legs of another, then these triangles are congruent. Indeed, between the legs lies a right angle, which for both triangles is 90 °.
Based on the second criterion for the equality of triangles, it is stated that if in one right triangle the leg and the indirect angle adjacent to it are equal to the leg and the indirect angle adjacent to it of another right triangle, then such triangles are equal. Indeed, after all, the legs are obtained lying between equal angles. On the one hand, acute angles are equal, and on the other - straight lines.
Since acute angles in right triangles always add up to 90°, if two right triangles have one acute angle, then the other will also be equal. For example, a is one angle, then 90° is the other angle for both triangles.
Therefore, right triangles are equal if the hypotenuse and acute angle of one is equal to the hypotenuse and acute angle of the other, since in fact we know all the acute angles of right triangles. And it turns out equality in two angles and the side between them.
Also due to the fact that if one acute angle of a right triangle is known, then the other one is also known, it follows equality of right triangles along the leg and opposite acute angle. In this case, the second sign of the equality of triangles “works”: along the side and two angles adjacent to it (one is straight, the other is calculated).
In addition to the listed signs of equality of right triangles, there is another one that does not directly follow from the three signs of equality of triangles: if right triangles have one leg and hypotenuse equal, then such triangles are congruent.
This sign of equality can be proved.
Let us attach right-angled triangles to each other with equal legs so that the right angles are on different sides of the resulting common side, and the hypotenuses are on different sides of it. These hypotenuses are equal by condition, which means we got an isosceles triangle. This means that the angles at the vertices that are separated from the common side (by which they were attached to each other) are equal. This in turn means that triangles have equal hypotenuse, leg and opposite angle. But there are signs of equality along the hypotenuse and the acute angle, along the leg and the opposite angle. This means that these right-angled triangles, in which the leg and hypotenuse are equal, are equal.
