Property of adjacent angles of a parallelogram. parallelogram and its properties

A parallelogram is a quadrilateral whose opposite sides are pairwise parallel. The following figure shows the parallelogram ABCD. It has side AB parallel to side CD and side BC parallel to side AD.

As you may have guessed, a parallelogram is convex quadrilateral. Consider the basic properties of a parallelogram.

Parallelogram properties

1. In a parallelogram, opposite angles and opposite sides are equal. Let's prove this property - consider the parallelogram shown in the following figure.

Diagonal BD divides it into two equal triangles: ABD and CBD. They are equal in side BD and two angles adjacent to it, since the angles lying at the secant of BD are parallel lines BC and AD and AB and CD, respectively. Therefore, AB = CD and
BC=AD. And from the equality of angles 1, 2,3 and 4 it follows that angle A = angle1 + angle3 = angle2 + angle4 = angle C.

2. The diagonals of the parallelogram are bisected by the intersection point. Let the point O be the point of intersection of the diagonals AC and BD of the parallelogram ABCD.

Then the triangle AOB and the triangle COD are equal to each other, along the side and two angles adjacent to it. (AB=CD since they are opposite sides of the parallelogram. And angle1 = angle2 and angle3 = angle4 as cross-lying angles at the intersection of lines AB and CD by secants AC and BD, respectively.) It follows that AO = OC and OB = OD, which and needed to be proven.

All main properties are illustrated in the following three figures.

The concept of a parallelogram

Definition 1

Parallelogram is a quadrilateral in which opposite sides are parallel to each other (Fig. 1).

Picture 1.

The parallelogram has two basic properties. Let's consider them without proof.

Property 1: Opposite sides and angles of a parallelogram are equal, respectively, to each other.

Property 2: Diagonals drawn in a parallelogram are bisected by their intersection point.

Parallelogram features

Consider three features of a parallelogram and present them in the form of theorems.

Theorem 1

If two sides of a quadrilateral are equal to each other and also parallel, then this quadrilateral will be a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. In which $AB||CD$ and $AB=CD$ Let us draw a diagonal $AC$ in it (Fig. 2).

Figure 2.

Consider parallel lines $AB$ and $CD$ and their secant $AC$. Then

\[\angle CAB=\angle DCA\]

like crosswise corners.

According to the $I$ criterion for the equality of triangles,

since $AC$ is their common side, and $AB=CD$ by condition. Means

\[\angle DAC=\angle ACB\]

Consider the lines $AD$ and $CB$ and their secant $AC$; by the last equality of the cross-lying angles, we obtain that $AD||CB$.) Therefore, by the definition of $1$, this quadrilateral is a parallelogram.

The theorem has been proven.

Theorem 2

If opposite sides of a quadrilateral are equal, then it is a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. In which $AD=BC$ and $AB=CD$. Let us draw a diagonal $AC$ in it (Fig. 3).

Figure 3

Since $AD=BC$, $AB=CD$, and $AC$ is a common side, then by the $III$ triangle equality test,

\[\triangle DAC=\triangle ACB\]

\[\angle DAC=\angle ACB\]

Consider the lines $AD$ and $CB$ and their secant $AC$, by the last equality of the cross-lying angles we get that $AD||CB$. Therefore, by the definition of $1$, this quadrilateral is a parallelogram.

\[\angle DCA=\angle CAB\]

Consider the lines $AB$ and $CD$ and their secant $AC$, by the last equality of the cross-lying angles we get that $AB||CD$. Therefore, by Definition 1, this quadrilateral is a parallelogram.

The theorem has been proven.

Theorem 3

If the diagonals drawn in a quadrilateral are divided into two equal parts by their intersection point, then this quadrilateral is a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. Let us draw the diagonals $AC$ and $BD$ in it. Let them intersect at the point $O$ (Fig. 4).

Figure 4

Since, by the condition $BO=OD,\ AO=OC$, and the angles $\angle COB=\angle DOA$ are vertical, then, by the $I$ triangle equality test,

\[\triangle BOC=\triangle AOD\]

\[\angle DBC=\angle BDA\]

Consider the lines $BC$ and $AD$ and their secant $BD$, by the last equality of the cross-lying angles we get that $BC||AD$. Also $BC=AD$. Therefore, by Theorem $1$, this quadrilateral is a parallelogram.

Lesson outline.

Algebra Grade 8

Teacher Sysoi A.K.

School 1828

Lesson topic: "Parallelogram and its properties"

Lesson type: combined

Lesson Objectives:

1) Ensure the assimilation of a new concept - a parallelogram and its properties

2) Continue developing skills and abilities to solve geometric problems;

3) Development of culture mathematical speech

Lesson plan:

1. Organizing time

(Slide 1)

The slide shows the statement of Lewis Carroll. Students are informed about the purpose of the lesson. The readiness of students for the lesson is checked.

2. Updating knowledge

(Slide 2)

On the task board for oral work. The teacher invites students to think about these problems and raise their hands to those who understand how to solve the problem. After solving two problems, a student is called to the board to prove the theorem on the sum of angles, who independently makes additional constructions on the drawing and proves the theorem orally.

Students use the formula for the sum of the angles of a polygon:

3. Main body

(Slide 3)

On the board is the definition of a parallelogram. The teacher talks about a new figure and formulates a definition, making the necessary explanations using the drawing. Then, on the checkered part of the presentation, using a marker and a ruler, shows how to draw a parallelogram (several cases are possible)

(Slide 4)

The teacher formulates the first property of a parallelogram. Invites students to say, according to the picture, what is given and what needs to be proved. After that, the given task appears on the board. Students guess (perhaps with the help of a teacher) that the desired equalities must be proved through the equalities of triangles, which can be obtained by drawing a diagonal (a diagonal appears on the board). Next, the students guess why the triangles are equal and call the sign of the equality of triangles (appears corresponding form). Orally communicate the facts that are necessary for the equality of triangles (as they name them, the corresponding visualization appears). Next, the students formulate the property equal triangles, it appears in the form of point 3 of the proof and then independently complete the proof of the theorem orally.

(Slide 5)

The teacher formulates the second property of a parallelogram. A drawing of a parallelogram appears on the board. The teacher offers to say from the picture what is given, what needs to be proved. After the students correctly report what is given and what needs to be proved, the condition of the theorem appears. Students guess that the equality of parts of the diagonals can be proved through the equality of trianglesAOB And COD. Using the previous property of a parallelogram, guess about the equality of the sidesAB And CD. Then they understand that it is necessary to find equal angles and, using the properties of parallel lines, they prove the equality of adjacent to equal parties corners. These stages are visualized on the slide. The truth of the theorem follows from the equality of triangles - the students pronounce the corresponding visualization on the slide.

(Slide 6)

The teacher formulates the third property of a parallelogram. Depending on the time that remains until the end of the lesson, the teacher can give the students the opportunity to prove this property on their own, or limit it to its formulation, and leave the proof itself to the students as homework. The proof can be based on the sum of the angles of the inscribed polygon, which was repeated at the beginning of the lesson, or on the sum of the interior one-sided angles for two parallel linesAD And BC, and a secant, for exampleAB.

4. Fixing the material

At this stage, students, using previously studied theorems, solve problems. Ideas for solving the problem are selected by students on their own. Because options there are a lot of designs and they all depend on how the students will look for a solution to the problem, there is no visualization of the solution to the problems, and the students independently draw up each stage of the solution on a separate board with the solution written in a notebook.

(Slide 7)

The task condition appears. The teacher suggests formulating “Given” according to the condition. After the students correctly compose short note conditions on the board appears "Given". The problem solving process might look like this:

Draw height BH (rendered)

Triangle AHB is a right triangle. Angle A equal to the angle C and is equal to 30 0 (by the property of opposite angles in a parallelogram). 2BH \u003d AB (by the property of the leg lying opposite the angle at 30 0 in right triangle). So AB = 13 cm.

AB \u003d CD, BC \u003d AD (by property opposite sides in a parallelogram) So AB \u003d CD \u003d 13cm. Since the perimeter of the parallelogram is 50 cm, then BC \u003d AD \u003d (50 - 26): 2 \u003d 12 cm.

Answer: AB=CD=13cm, BC=AD=12cm.

(Slide 8)

The task condition appears. The teacher suggests formulating “Given” according to the condition. Then “Dano” appears on the screen. With the help of red lines, a quadrilateral is selected, about which you need to prove that it is a parallelogram. The problem solving process might look like this:

Because BK and MD are perpendicular to the same line, then lines BK and MD are parallel.

Through adjacent corners it can be shown that the sum of internal one-sided angles at lines BM and KD and secant MD is equal to 180 0 . Therefore, these lines are parallel.

Since the opposite sides of the quadrilateral BMDK are pairwise parallel, this quadrilateral is a parallelogram.

5. End of the lesson. outcome behavior.

(Slide 8)

Questions on the slide new topic to which students respond.

Municipal budgetary educational institution

Savinskaya average comprehensive school

Research

Parallelogram and its new properties

Done by: 8B grade student

MBOU Savinskaya secondary school

Kuznetsova Svetlana, 14 years old

Leader: math teacher

Tulchevskaya N.A.

Savino

Ivanovo region, Russia

2016

I. Introduction ______________________________________________ page 3

II. From the history of the parallelogram ___________________________________page 4

III Additional properties of a parallelogram ______________________page 4

IV. Proof of properties _____________________________________ page 5

V. Solving problems using additional properties __________page 8

VI. Application of the properties of a parallelogram in life ___________________page 11

VII. Conclusion _________________________________________________page 12

VIII. Literature _________________________________________________ page 13

Introduction

"Among equal minds

at similarity of other conditions

superior to those who know geometry"

(Blaise Pascal).

While studying the topic “Parallelogram” in geometry lessons, we considered two properties of a parallelogram and three features, but when we started solving problems, it turned out that this was not enough.

I had a question, does the parallelogram have any other properties, and how they will help in solving problems.

And I decided to study additional properties of a parallelogram and show how they can be applied to solve problems.

Subject of study : parallelogram

Object of study : parallelogram properties
Goal of the work:

formulation and proof of additional properties of a parallelogram that are not studied at school;

application of these properties to solve problems.

Tasks:

To study the history of the parallelogram and the history of the development of its properties;

Find additional literature on the research question;

Study additional properties of a parallelogram and prove them;

Show the application of these properties to solve problems;

Consider the application of the properties of a parallelogram in life.
Research methods:

Work with educational and scientific - popular literature, Internet resources;

The study of theoretical material;

Selection of a range of tasks that can be solved using additional properties of a parallelogram;

Observation, comparison, analysis, analogy.

Study duration : 3 months: January-March 2016

In a geometry textbook, we read the following definition of a parallelogram: A parallelogram is a quadrilateral whose opposite sides are parallel in pairs.

The word "parallelogram" is translated as " parallel lines"(from Greek words Parallelos - parallel and gramme - line), this term was introduced by Euclid. In his book Elements, Euclid proved following properties parallelogram: opposite sides and angles of a parallelogram are equal, and a diagonal bisects it. Euclid does not mention the point of intersection of the parallelogram. Only towards the end of the Middle Ages was developed full theory Parallelograms And only in the 17th century, parallelogram theorems appeared in textbooks, which are proved using Euclid's theorem on the properties of a parallelogram.

III Additional properties of a parallelogram

In the textbook on geometry, only 2 properties of a parallelogram are given:

Opposite angles and sides are equal

The diagonals of a parallelogram intersect and the point of intersection is bisected

In various sources on geometry, the following additional properties can be found:

The sum of adjacent angles of a parallelogram is 180 0

The bisector of the angle of a parallelogram cuts off from it isosceles triangle;

Bisectors of opposite angles of a parallelogram lie on parallel lines;

Bisectors of adjacent angles of a parallelogram intersect at right angles;

The bisectors of all angles of a parallelogram form a rectangle when they intersect;

The distances from opposite corners of a parallelogram to one and the same diagonal are equal.

If you connect opposite vertices in a parallelogram with the midpoints of opposite sides, you get another parallelogram.

The sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its adjacent sides.

If we draw heights from two opposite angles in a parallelogram, we get a rectangle.

IV Proof of the properties of a parallelogram

The sum of adjacent angles of a parallelogram is 180 0

Given:

ABCD is a parallelogram

Prove:

A+
B=

Proof:

A and
B - internal one-sided corners with parallel straight lines BC AD and secant AB, so
A+
B=

Given: ABCD - parallelogram,

AK -bisector
A.

Prove: AVK - isosceles

Proof:

1)
1=
3 (cross-lying with BC AD and secant AK ),

2)
2=
3 because AK is a bisector,

means 1=
2.

3) ABK is isosceles because 2 angles of a triangle are equal

. The angle bisector of a parallelogram cuts off an isosceles triangle from it

Given: ABCD is a parallelogram

AK is the bisector of A,

СР is the bisector of C.

Prove: AK ║ SR

Proof:

1) 1=2 since AK-bisector

2) 4=5 because SR - bisector

3) 3=1 (cross-lying angles at

BC ║ AD and AK-secant),

4) A \u003d C (by the property of a parallelogram), which means 2 \u003d 3 \u003d 4 \u003d 5.

4) From paragraphs 3 and 4 it follows that 1 = 4, and these angles are corresponding with straight lines AK and SR and a secant BC,

hence, AK ║ SR (on the basis of parallel lines)

. Bisectors of opposite angles of a parallelogram lie on parallel lines

Bisectors of adjacent angles of a parallelogram intersect at right angles

Given: ABCD - parallelogram,

AC bisector A,

DP-bisector D

Prove: DP AK.

Proof:

1) 1=2, because AK - bisector

Let 1=2=x, then A=2x,

2) 3=4, because D P - bisector

Let 3=4=y, then D=2y

3) A + D \u003d 180 0, because the sum of adjacent angles of a parallelogram is 180

2) Consider A OD

1+3=90 0 then
<5=90 0 (сумма углов треугольников равна 180 0)

5. The bisectors of all angles of a parallelogram form a rectangle when they intersect

Given: ABCD - parallelogram, AK-bisector A,

DP-bisector D,

CM is the bisector of C,

BF -bisector of B .

Prove: KRNS -rectangle

Proof:

Based on the previous property 8=7=6=5=90 0 ,

means KRNS is a rectangle.

The distances from opposite corners of a parallelogram to one and the same diagonal are equal.

Given: ABCD-parallelogram, AC-diagonal.

VC AU, D.P. AC

Prove: BK=DP

Proof: 1) DCP \u003d KAB, as internal crosswise lying at AB ║ CD and secant AC.

2) AKB= CDP (along the side and two corners adjacent to it AB=CD CD P=AB K).

And in equal triangles, the corresponding sides are equal, so DP \u003d BK.

If you connect opposite vertices in a parallelogram with the midpoints of opposite sides, you get another parallelogram.

Given: ABCD parallelogram.

Prove: VKDP is a parallelogram.

Proof:

1) BP=KD (AD=BC, points K and P

bisect these sides)

2) BP ║ KD (lie on AD BC)

If the opposite sides of a quadrilateral are equal and parallel, then this quadrilateral is a parallelogram.

If we draw heights from two opposite angles in a parallelogram, we get a rectangle.

The sum of the squares of the diagonals of a parallelogram is equal to twice the sum of the squares of its adjacent sides.

Given: ABCD is a parallelogram. BD and AC are diagonals.

Prove: AC 2 + BD 2 =2(AB 2 + AD 2 )

Proof: 1)ASK: AC ²=
+

2)B RD : BD 2 = B R 2 + PD 2 (according to the Pythagorean theorem)

3) AC ²+ BD ²=SC²+A K²+B Р²+РD ²

4) SK = BP = H(height )

5) AC 2 +VD 2 = H 2 + A TO 2 + H 2 +PD 2

6) Let D K=A P=x, Then C TOD : H 2 = CD 2 - X 2 according to the Pythagorean theorem )

7) AC²+BD ² = CD 2 - x²+ AK 1 ²+ CD 2 -X 2 +PD 2 ,

AC²+VD ²=2CD 2 -2x 2 + A TO 2 +PD 2

8) A TO=AD+ X, RD=AD- X,

AC²+VD ² =2CD 2 -2x 2 +(AD +x) 2 +(AD -X) 2 ,

AC²+ IND²=2 WITHD²-2 X²+AD 2 +2AD X+ X 2 + AD 2 -2AD X+ X 2 ,
AC²+ IND²=2CD 2 +2AD 2 =2(CD 2 + AD 2 ).

V . Solving problems using these properties

The intersection point of the bisectors of two angles of a parallelogram adjacent to one side belongs to the opposite side. The shorter side of the parallelogram is 5 . Find his big side.

Given: ABCD is a parallelogram,

AK - bisector
A,

D K - bisector
D, AB=5

Find: sun

solution

Solution

Because AK - bisector
A, then ABC is isosceles.

Because D K - bisector
D , then DCK - isosceles

DC \u003d C K \u003d 5

Then, VS=VK+SK=5+5 = 10

Answer: 10

2. Find the perimeter of the parallelogram if the bisector of one of its angles divides the side of the parallelogram into segments of 7 cm and 14 cm.

1 case

Given:
A,

VK=14 cm, KS=7 cm

Find: R parallelogram

Solution

BC=VK+KS=14+7=21 (cm)

Because AK - bisector
A, then ABC is isosceles.

AB=BK=14cm

Then P \u003d 2 (14 + 21) \u003d 70 (cm)

happening

Given: ABCD is a parallelogram,

D K - bisector
D,

VK=14 cm, KS=7 cm

Find: R parallelogram

Solution

BC=VK+KS=14+7=21 (cm)

Because D K - bisector
D , then DCK - isosceles

DC \u003d C K \u003d 7

Then, P \u003d 2 (21 + 7) \u003d 56 (cm)

Answer: 70cm or 56cm

3. The sides of the parallelogram are 10 cm and 3 cm. The bisectors of two angles adjacent to the larger side divide the opposite side into three segments. Find these segments.

1 case: bisectors intersect outside the parallelogram

Given: ABCD - parallelogram, AK - bisector
A,

D K - bisector
D , AB=3 cm, BC=10 cm

Find: BM, MN, NC

Solution

Because AM - bisector
And, then AVM is isosceles.

Because DN - bisector
D , then DCN - isosceles

DC=CN=3

Then, MN \u003d 10 - (BM + NC) \u003d 10 - (3 + 3) \u003d 4 cm

2 case: bisectors intersect inside a parallelogram

Because AN - bisector
A, then ABN is isosceles.

AB=BN = 3 D

And the sliding grille - move to the required distance in the doorway

Parallelogram mechanism- a four-link mechanism, the links of which form a parallelogram. It is used to implement the translational movement of hinged mechanisms.

Parallelogram with fixed link- one link is motionless, the opposite one makes a rocking motion, remaining parallel to the motionless one. Two parallelograms connected one behind the other give the final link two degrees of freedom, leaving it parallel to the fixed one.

Examples: bus windshield wipers, forklifts, tripods, hangers, car hangers.

Parallelogram with fixed hinge- the property of a parallelogram is used to maintain a constant ratio of distances between three points. Example: drawing pantograph - a device for scaling drawings.

Rhombus- all links are of the same length, the approach (contraction) of a pair of opposite hinges leads to the expansion of the other two hinges. All links work in compression.

Examples are a car diamond jack, a tram pantograph.

scissor or X-shaped mechanism, also known as Nuremberg scissors- a variant of a rhombus - two links connected in the middle by a hinge. The advantages of the mechanism are compactness and simplicity, the disadvantage is the presence of two sliding pairs. Two (or more) such mechanisms, connected in series, form a rhombus (s) in the middle. It is used in lifts, children's toys.

VII Conclusion

Who has been involved in mathematics since childhood,

he develops attention, trains his brain,

own will, cultivates perseverance

and perseverance in achieving the goal

A. Markushevich

In the course of work, I proved additional properties of a parallelogram.

I was convinced that by applying these properties, you can solve problems faster.

I showed how these properties are applied on examples of solving specific problems.

I learned a lot about the parallelogram that is not in our geometry textbook

I was convinced that knowledge of geometry is very important in life by examples of applying the properties of a parallelogram.

The purpose of my research work has been accomplished.

The importance of mathematical knowledge is evidenced by the fact that a prize was established for the one who publishes a book about a person who has lived all his life without the help of mathematics. No one has received this award so far.

VIII Literature

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 its height (h). You can also find its area through two sides and an angle and through the diagonals.

Parallelogram properties

1. Opposite sides are identical.

First of all, draw the diagonal $AC $ . Two triangles are obtained: $ABC $ and $ADC $ .

Since $ABCD $ is a parallelogram, the following is true:

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

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

Therefore, (on the second basis: and $AC$ is common).

And, therefore, $\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 $. So the sum of the opposite angles is: $\angle 1 + \angle 3 = \angle 2 + \angle 4 $. Given that $\triangle ABC = \triangle ADC $ we get $\angle A = \angle C $ , $\angle B = \angle D $ .

3. The diagonals are bisected by the intersection point.

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

Thus, it is seen that $\triangle AOB = \triangle COD $ according to the second criterion for the equality of triangles (two angles and a 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).

Parallelogram features

If only one sign 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 sign of a parallelogram 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 consider in more detail. Why $AD || BC $ ?

$\triangle ABC = \triangle ADC $ By property 1: $AB = CD $ , $\angle 1 = \angle 2 $ as crosswise with parallel $AB $ and $CD $ and secant $AC $ .

But if $\triangle ABC = \triangle ADC $, then $\angle 3 = \angle 4 $ (they lie opposite $AD || BC $ ($\angle 3 $ and $\angle 4 $ - lying opposite 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 feature. 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) $(because $\angle A = \angle C $ , $\angle B = \angle D $ by definition).

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