We calculate the sum of angles and area of ​​a parallelogram: properties and characteristics. Definition of a parallelogram and its properties Proof of the properties of opposite sides and angles of a parallelogram

Lesson topic

• Properties of the diagonals of a parallelogram.

Lesson Objectives

• Get acquainted with new definitions and remember some already studied.
• State and prove the property of the diagonals of a parallelogram.
• Learn to apply the properties of shapes when solving problems.
• 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

• Test students' problem-solving skills.

Lesson Plan

1. Introduction.
2. Repetition of previously studied material.
3. Parallelogram, its properties and features.
4. Examples of tasks.
5. Self-check.

Introduction

“A major scientific discovery provides a solution to a major problem, but in the solution of any problem there is a grain of discovery.”

Property of opposite sides of a parallelogram

A parallelogram has opposite sides that are equal.

Proof.

Let ABCD be the given parallelogram. And let its diagonals intersect at point O.
Since Δ AOB = Δ COD by the first criterion of equality of triangles (∠ AOB = ∠ COD, as vertical ones, AO=OC, DO=OB, by the property of the diagonals of a parallelogram), then AB=CD. In the same way, from the equality of triangles BOC and DOA, it follows that BC = DA. The theorem has been proven.

Property of opposite angles of a parallelogram

In a parallelogram, opposite angles are equal.

Proof.

Let ABCD be the given parallelogram. And let its diagonals intersect at point O.
From what was proven in the theorem about the properties of the opposite sides of a parallelogram Δ ABC = Δ CDA on three sides (AB=CD, BC=DA from what was proven, AC – general). From the equality of triangles it follows that ∠ ABC = ∠ CDA.
It is also proved that ∠ DAB = ∠ BCD, which follows from ∠ ABD = ∠ CDB. The theorem has been proven.

Property of the diagonals of a parallelogram

The diagonals of a parallelogram intersect and are bisected at the point of intersection.

Proof.

Let ABCD be the given parallelogram. Let's draw the diagonal AC. Let's mark the middle O on it. On the continuation of the segment DO, we'll put aside the segment OB 1 equal to DO.
By the previous theorem, AB 1 CD is a parallelogram. Therefore, line AB 1 is parallel to DC. But through point A only one line parallel to DC can be drawn. This means that straight AB 1 coincides with straight AB.
It is also proved that BC 1 coincides with BC. This means that point C coincides with C 1. parallelogram ABCD coincides with parallelogram AB 1 CD. Consequently, the diagonals of the parallelogram intersect and are bisected at the point of intersection. The theorem has been proven.

In textbooks for regular schools (for example, in Pogorelovo) it is proven like this: diagonals divide a parallelogram into 4 triangles. Let's consider one pair and find out - they are equal: their bases are opposite sides, the corresponding angles adjacent to it are equal, like vertical angles with parallel lines. That is, the segments of the diagonals are equal in pairs. All.

Is that all?
It was proven above that the intersection point bisects the diagonals - if it exists. The above reasoning does not prove its very existence in any way. That is, part of the theorem “the diagonals of a parallelogram intersect” remains unproven.

The funny thing is that this part is much harder to prove. This follows, by the way, from a more general result: any convex quadrilateral will have diagonals intersecting, but any non-convex quadrilateral will not.

On the equality of triangles along a side and two adjacent angles (the second sign of equality of triangles) and others.

Thales found an important practical application to the theorem on the equality of two triangles along a side and two adjacent angles. A rangefinder was built in the harbor of Miletus to determine the distance to a ship at sea. It consisted of three driven pegs A, B and C (AB = BC) and a marked straight line SC, perpendicular to CA. When a ship appeared on the SK straight line, we found point D such that points D, .B and E were on the same straight line. As is clear from the drawing, the distance CD on the ground is the desired distance to the ship.

Questions

1. Are the diagonals of a square divided in half by the point of intersection?
   
2. Are the diagonals of a parallelogram equal?
3. Are the opposite angles of a parallelogram equal?
4. State the definition of a parallelogram?
5. How many signs of a parallelogram?
   
6. Can a rhombus be a parallelogram?
   

List of sources used

1. Kuznetsov A.V., mathematics teacher (grades 5-9), Kiev
2. “Unified State Exam 2006. Mathematics. Educational and training materials for preparing students / Rosobrnadzor, ISOP - M.: Intellect-Center, 2006"
3. Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”
4. L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina “Geometry, 7 – 9: textbook for educational institutions”

We worked on the lesson

Kuznetsov A.V.

Poturnak S.A.

Evgeniy Petrov

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A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. This definition is already sufficient, since the remaining properties of the parallelogram follow from it and are proved in the form of theorems.

The main properties of a parallelogram are:

• a parallelogram is a convex quadrilateral;
• A parallelogram has opposite sides that are equal in pairs;
• In a parallelogram, opposite angles are equal in pairs;
• The diagonals of a parallelogram are divided in half by the point of intersection.

Parallelogram - convex quadrilateral

Let us first prove the theorem that a parallelogram is a convex quadrilateral. A polygon is convex if whichever side of it is extended to a straight line, all other sides of the polygon will be on the same side of this straight line.

Let a parallelogram ABCD be given, in which AB is the opposite side for CD, and BC is the opposite side for AD. Then from the definition of a parallelogram it follows that AB || CD, BC || A.D.

Parallel segments have no common points and do not intersect. This means that CD lies on one side of AB. Since segment BC connects point B of segment AB with point C of segment CD, and segment AD connects other points AB and CD, segments BC and AD also lie on the same side of line AB where CD lies. Thus, all three sides - CD, BC, AD - lie on the same side of AB.

Similarly, it is proved that in relation to the other sides of the parallelogram, the other three sides lie on the same side.

Opposite sides and angles are equal

One of the properties of a parallelogram is that In a parallelogram, opposite sides and opposite angles are equal in pairs. For example, if a parallelogram ABCD is given, then it has AB = CD, AD = BC, ∠A = ∠C, ∠B = ∠D. This theorem is proven as follows.

A parallelogram is a quadrilateral. This means it has two diagonals. Since a parallelogram is a convex quadrilateral, any of them divides it into two triangles. In the parallelogram ABCD, consider the triangles ABC and ADC obtained by drawing the diagonal AC.

These triangles have one side in common - AC. Angle BCA is equal to angle CAD, as are vertical when BC and AD are parallel. Angles BAC and ACD are also equal to vertical angles when AB and CD are parallel. Therefore, ∆ABC = ∆ADC at two angles and the side between them.

In these triangles, side AB corresponds to side CD, and side BC corresponds to AD. Therefore, AB = CD and BC = AD.

Angle B corresponds to angle D, i.e. ∠B = ∠D. Angle A of a parallelogram is the sum of two angles - ∠BAC and ∠CAD. Angle C is equal to ∠BCA and ∠ACD. Since pairs of angles are equal to each other, then ∠A = ∠C.

Thus, it is proven that in a parallelogram opposite sides and angles are equal.

Diagonals are divided in half

Since a parallelogram is a convex quadrilateral, it has two diagonals, and they intersect. Let parallelogram ABCD be given, its diagonals AC and BD intersect at point E. Consider the triangles ABE and CDE formed by them.

These triangles have sides AB and CD equal to the opposite sides of a parallelogram. Angle ABE is equal to angle CDE as crosswise lying with parallel lines AB and CD. For the same reason, ∠BAE = ∠DCE. This means ∆ABE = ∆CDE at two angles and the side between them.

You can also notice that angles AEB and CED are vertical and therefore also equal to each other.

Since triangles ABE and CDE are equal to each other, then all their corresponding elements are equal. Side AE ​​of the first triangle corresponds to side CE of the second, which means AE = CE. Similarly BE = DE. Each pair of equal segments constitutes a diagonal of a parallelogram. Thus it is proven that The diagonals of a parallelogram are bisected by their intersection point.

A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).

Theorem 1. On the properties of the sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.

Proof. In this parallelogram ABCD we draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).

These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (crosswise angles for parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB = CD, BC = AD, ∠ B = ∠ D. The sum of angles adjacent to one side, for example angles A and D, is equal to 180° as one-sided for parallel lines. The theorem has been proven.

Comment. The equality of opposite sides of a parallelogram means that the segments of parallels cut off by parallel ones are equal.

Corollary 1. If two lines are parallel, then all points on one line are at the same distance from the other line.

Proof. Indeed, let a || b (Fig. 3).

Let us draw perpendiculars BA and CD to straight line a from some two points B and C of line b. Since AB || CD, then figure ABCD is a parallelogram, and therefore AB = CD.

The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.

According to what has been proven, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.

Example 1. The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm larger than the other. Find the sides of the parallelogram.

Solution. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram by x and the other by y. Then, by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we obtain x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.

Example 2.

Solution. Let Figure 4 meet the conditions of the problem.

Let us denote AB by x, and BC by y. According to the condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of triangle ABD is 8 cm. And since AB + AD = x + y = 5 then BD = 8 - 5 = 3. So BD = 3 cm.

Example 3. Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.

Solution. Let Figure 5 meet the conditions of the problem.

Let us denote the degree measure of angle A by x. Then the degree measure of angle D is x + 50°.

Angles BAD and ADC are one-sided interior angles with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.

Example 4. The sides of the parallelogram are 4.5 dm and 1.2 dm. A bisector is drawn from the vertex of an acute angle. What parts does it divide the larger side of the parallelogram into?

Solution. Let Figure 6 meet the conditions of the problem.

AE is the bisector of an acute angle of a parallelogram. Therefore, ∠ 1 = ∠ 2.

A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The area of ​​a parallelogram is equal to the product of its base (a) and height (h). You can also find its area through two sides and an angle and through diagonals.

Properties of a parallelogram

1. Opposite sides are identical

First of all, let's draw the diagonal \(AC\) . We get two triangles: \(ABC\) and \(ADC\).

Since \(ABCD\) is a parallelogram, the following is true:

\(AD || BC \Rightarrow \angle 1 = \angle 2\) like lying crosswise.

\(AB || CD \Rightarrow \angle3 = \angle 4\) like lying crosswise.

Therefore, (according to the second criterion: and \(AC\) is common).

And that means \(\triangle ABC = \triangle ADC\), then \(AB = CD\) and \(AD = BC\) .

2. Opposite angles are identical

According to the proof properties 1 We know that \(\angle 1 = \angle 2, \angle 3 = \angle 4\). Thus the sum of opposite angles is: \(\angle 1 + \angle 3 = \angle 2 + \angle 4\). Considering that \(\triangle ABC = \triangle ADC\) we get \(\angle A = \angle C \) , \(\angle B = \angle D \) .

3. Diagonals are divided in half by the intersection point

By property 1 we know that opposite sides are identical: \(AB = CD\) . Once again, note the crosswise lying equal angles.

Thus it is clear that \(\triangle AOB = \triangle COD\) according to the second sign of equality of triangles (two angles and the side between them). That is, \(BO = OD\) (opposite the angles \(\angle 2\) and \(\angle 1\) ) and \(AO = OC\) (opposite the angles \(\angle 3\) and \( \angle 4\) respectively).

Signs of a parallelogram

If only one feature is present in your problem, then the figure is a parallelogram and you can use all the properties of this figure.

For better memorization, note that the parallelogram sign will answer the following question - "how to find out?". That is, how to find out that a given figure is a parallelogram.

1. A parallelogram is a quadrilateral whose two sides are equal and parallel

\(AB = CD\) ; \(AB || CD \Rightarrow ABCD\)- parallelogram.

Let's take a closer look. Why \(AD || BC \) ?

\(\triangle ABC = \triangle ADC\) By property 1: \(AB = CD \) , \(\angle 1 = \angle 2 \) lying crosswise when \(AB \) and \(CD \) and the secant \(AC \) are parallel.

But if \(\triangle ABC = \triangle ADC\), then \(\angle 3 = \angle 4 \) (lie opposite \(AD || BC \) (\(\angle 3 \) and \(\angle 4 \) - those lying crosswise are also equal).

The first sign is correct.

2. A parallelogram is a quadrilateral whose opposite sides are equal

\(AB = CD \) , \(AD = BC \Rightarrow ABCD \) is a parallelogram.

Let's consider this sign. Let's draw the diagonal \(AC\) again.

By property 1\(\triangle ABC = \triangle ACD\).

It follows that: \(\angle 1 = \angle 2 \Rightarrow AD || BC \) And \(\angle 3 = \angle 4 \Rightarrow AB || CD \), that is, \(ABCD\) is a parallelogram.

The second sign is correct.

3. A parallelogram is a quadrilateral whose opposite angles are equal

\(\angle A = \angle C\) , \(\angle B = \angle D \Rightarrow ABCD\)- parallelogram.

\(2 \alpha + 2 \beta = 360^(\circ) \)(since \(\angle A = \angle C\) , \(\angle B = \angle D\) by condition).

It turns out, . But \(\alpha \) and \(\beta \) are internal one-sided at the secant \(AB \) .

And what \(\alpha + \beta = 180^(\circ) \) also says that \(AD || BC \) .

Proof

First of all, let's draw the diagonal AC. We get two triangles: ABC and ADC.

Since ABCD is a parallelogram, the following is true:

AD || BC \Rightarrow \angle 1 = \angle 2 like lying crosswise.

AB || CD\Rightarrow\angle3 =\angle 4 like lying crosswise.

Therefore, \triangle ABC = \triangle ADC (according to the second criterion: and AC is common).

And, therefore, \triangle ABC = \triangle ADC, then AB = CD and AD = BC.

Proven!

2. Opposite angles are identical.

Proof

According to the proof properties 1 We know that \angle 1 = \angle 2, \angle 3 = \angle 4. Thus the sum of opposite angles is: \angle 1 + \angle 3 = \angle 2 + \angle 4. Considering that \triangle ABC = \triangle ADC we get \angle A = \angle C , \angle B = \angle D .

Proven!

3. The diagonals are divided in half by the intersection point.

Proof

Let's draw another diagonal.

By property 1 we know that opposite sides are identical: AB = CD. Once again, note the crosswise lying equal angles.

Thus, it is clear that \triangle AOB = \triangle COD according to the second criterion for the equality of triangles (two angles and the side between them). That is, BO = OD (opposite the corners \angle 2 and \angle 1) and AO = OC (opposite the corners \angle 3 and \angle 4, respectively).

Proven!

Signs of a parallelogram

If only one feature is present in your problem, then the figure is a parallelogram and you can use all the properties of this figure.

For better memorization, note that the parallelogram sign will answer the following question - "how to find out?". That is, how to find out that a given figure is a parallelogram.

1. A parallelogram is a quadrilateral whose two sides are equal and parallel.

AB = CD ; AB || CD\Rightarrow ABCD is a parallelogram.

Proof

Let's take a closer look. Why AD || BC?

\triangle ABC = \triangle ADC by property 1: AB = CD, AC - common and \angle 1 = \angle 2 lying crosswise with parallel AB and CD and secant AC.

But if \triangle ABC = \triangle ADC , then \angle 3 = \angle 4 (lie opposite AB and CD, respectively). And therefore AD || BC (\angle 3 and \angle 4 - those lying crosswise are also equal).

The first sign is correct.

2. A parallelogram is a quadrilateral whose opposite sides are equal.

AB = CD, AD = BC \Rightarrow ABCD is a parallelogram.

Proof

Let's consider this sign. Let's draw the diagonal AC again.

By property 1\triangle ABC = \triangle ACD .

It follows that: \angle 1 = \angle 2 \Rightarrow AD || B.C. And \angle 3 = \angle 4 \Rightarrow AB || CD, that is, ABCD is a parallelogram.

The second sign is correct.

3. A parallelogram is a quadrilateral whose opposite angles are equal.

\angle A = \angle C , \angle B = \angle D \Rightarrow ABCD- parallelogram.

Proof

2 \alpha + 2 \beta = 360^(\circ)(since ABCD is a quadrilateral, and \angle A = \angle C , \angle B = \angle D by condition).

It turns out that \alpha + \beta = 180^(\circ) . But \alpha and \beta are internal one-sided at the secant AB.

And the fact that \alpha + \beta = 180^(\circ) also means that AD || B.C.

Moreover, \alpha and \beta are internal one-sided at the secant AD . And that means AB || CD.

The third sign is correct.

4. A parallelogram is a quadrilateral whose diagonals are divided in half by the point of intersection.

AO = OC ; BO = OD\Rightarrow parallelogram.

Proof

BO = OD; AO = OC , \angle 1 = \angle 2 as vertical \Rightarrow \triangle AOB = \triangle COD, \Rightarrow \angle 3 = \angle 4, and \Rightarrow AB || CD.

Similarly BO = OD; AO = OC, \angle 5 = \angle 6 \Rightarrow \triangle AOD = \triangle BOC \Rightarrow \angle 7 = \angle 8, and \Rightarrow AD || B.C.

The fourth sign is correct.