Determine the true form of the parallelogram. Parallelogram and its properties

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. The 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 criterion 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, \(\alpha + \beta = 180^(\circ) \). But \(\alpha \) and \(\beta \) are internal one-sided at the secant \(AB \) .

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.

Important notes!
1. If you see gobbledygook instead of formulas, clear your cache. How to do this in your browser is written here:
2. Before you start reading the article, pay attention to our navigator for the most useful resources for

1. Parallelogram

Compound word "parallelogram"? And behind it lies a very simple figure.

Well, that is, we took two parallel lines:

Crossed by two more:

And inside there is a parallelogram!

What properties does a parallelogram have?

Properties of a parallelogram.

That is, what can you use if the problem is given a parallelogram?

The following theorem answers this question:

Let's draw everything in detail.

What does it mean first point of the theorem? And the fact is that if you HAVE a parallelogram, then you will certainly

The second point means that if there IS a parallelogram, then, again, certainly:

Well, finally, the third point means that if you HAVE a parallelogram, then be sure to:

Do you see what a wealth of choice there is? What to use in the problem? Try to focus on the question of the problem, or just try everything one by one - some “key” will do.

Now let’s ask ourselves another question: how can we recognize a parallelogram “by sight”? What must happen to a quadrilateral for us to have the right to give it the “title” of a parallelogram?

Several signs of a parallelogram answer this question.

Signs of a parallelogram.

Attention! Begin.

Parallelogram.

Please note: if you found at least one sign in your problem, then you definitely have a parallelogram, and you can use all the properties of a parallelogram.

2. Rectangle

I think that it will not be news to you at all that

First question: is a rectangle a parallelogram?

Of course it is! After all, he has - remember, our sign 3?

And from here, of course, it follows that in a rectangle, like in any parallelogram, the diagonals are divided in half by the point of intersection.

But the rectangle also has one distinctive property.

Rectangle property

Why is this property distinctive? Because no other parallelogram has equal diagonals. Let's formulate it more clearly.

Please note: in order to become a rectangle, a quadrilateral must first become a parallelogram, and then demonstrate the equality of the diagonals.

3. Diamond

And again the question: is a rhombus a parallelogram or not?

With full right - a parallelogram, because it has and (remember our feature 2).

And again, since a rhombus is a parallelogram, then it must have all the properties of a parallelogram. This means that in a rhombus, opposite angles are equal, opposite sides are parallel, and the diagonals bisect at the point of intersection.

Properties of a rhombus

Look at the picture:

As in the case of a rectangle, these properties are distinctive, that is, for each of these properties we can conclude that this is not just a parallelogram, but a rhombus.

Signs of a diamond

And again, pay attention: there must be not just a quadrilateral whose diagonals are perpendicular, but a parallelogram. Make sure:

No, of course, although its diagonals are perpendicular, and the diagonal is the bisector of the angles and. But... diagonals are not divided in half by the point of intersection, therefore - NOT a parallelogram, and therefore NOT a rhombus.

That is, a square is a rectangle and a rhombus at the same time. Let's see what happens.

Is it clear why? - rhombus is the bisector of angle A, which is equal to. This means it divides (and also) into two angles along.

Well, it's quite clear: the diagonals of a rectangle are equal; The diagonals of a rhombus are perpendicular, and in general, a parallelogram of diagonals is divided in half by the point of intersection.

AVERAGE LEVEL

Properties of quadrilaterals. Parallelogram

Properties of a parallelogram

Attention! Words " properties of a parallelogram"mean that if in your task There is parallelogram, then all of the following can be used.

Theorem on the properties of a parallelogram.

In any parallelogram:

Let's understand why this is all true, in other words WE'LL PROVE theorem.

So why is 1) true?

If it is a parallelogram, then:

lying like criss-cross
lying like crosses.

This means (according to criterion II: and - general.)

Well, that’s it, that’s it! - proved.

But by the way! We also proved 2)!

Why? But (look at the picture), that is, precisely because.

Only 3 left).

To do this, you still have to draw a second diagonal.

And now we see that - according to the II characteristic (angles and the side “between” them).

Properties proven! Let's move on to the signs.

Signs of a parallelogram

Recall that the parallelogram sign answers the question “how do you know?” that a figure is a parallelogram.

In icons it's like this:

Why? It would be nice to understand why - that's enough. But look:

Well, we figured out why sign 1 is true.

Well, it's even easier! Let's draw a diagonal again.

Which means:

AND It's also easy. But...different!

Means, . Wow! But also - internal one-sided with a secant!

Therefore the fact that means that.

And if you look from the other side, then - internal one-sided with a secant! And therefore.

Do you see how great it is?!

And again simple:

Exactly the same, and.

Pay attention: if you found at least one sign of a parallelogram in your problem, then you have exactly parallelogram and you can use everyone properties of a parallelogram.

For complete clarity, look at the diagram:

Properties of quadrilaterals. Rectangle.

Rectangle properties:

Point 1) is quite obvious - after all, sign 3 () is simply fulfilled

And point 2) - very important. So, let's prove that

This means on two sides (and - general).

Well, since the triangles are equal, then their hypotenuses are also equal.

Proved that!

And imagine, equality of diagonals is a distinctive property of a rectangle among all parallelograms. That is, this statement is true^

Let's understand why?

This means (meaning the angles of a parallelogram). But let us remember once again that it is a parallelogram, and therefore.

Means, . Well, of course, it follows that each of them! After all, they have to give in total!

So they proved that if parallelogram suddenly (!) the diagonals turn out to be equal, then this exactly a rectangle.

But! Pay attention! This is about parallelograms! Not just anyone a quadrilateral with equal diagonals is a rectangle, and only parallelogram!

Properties of quadrilaterals. Rhombus

And again the question: is a rhombus a parallelogram or not?

With full right - a parallelogram, because it has (Remember our feature 2).

And again, since a rhombus is a parallelogram, it must have all the properties of a parallelogram. This means that in a rhombus, opposite angles are equal, opposite sides are parallel, and the diagonals bisect at the point of intersection.

But there are also special properties. Let's formulate it.

Properties of a rhombus

Why? Well, since a rhombus is a parallelogram, then its diagonals are divided in half.

Why? Yes, that's why!

In other words, the diagonals turned out to be bisectors of the corners of the rhombus.

As in the case of a rectangle, these properties are distinctive, each of them is also a sign of a rhombus.

Signs of a diamond.

Why is this? And look,

That means both These triangles are isosceles.

To be a rhombus, a quadrilateral must first “become” a parallelogram, and then exhibit feature 1 or feature 2.

Properties of quadrilaterals. Square

That is, a square is a rectangle and a rhombus at the same time. Let's see what happens.

Is it clear why? A square - a rhombus - is the bisector of an angle that is equal to. This means it divides (and also) into two angles along.

Why? Well, let's just apply the Pythagorean theorem to...

SUMMARY AND BASIC FORMULAS

Properties of a parallelogram:

Opposite sides are equal: , .
Opposite angles are equal: , .
The angles on one side add up to: , .
The diagonals are divided in half by the point of intersection: .

Rectangle properties:

The diagonals of the rectangle are equal: .
A rectangle is a parallelogram (for a rectangle all the properties of a parallelogram are fulfilled).

Properties of a rhombus:

The diagonals of a rhombus are perpendicular: .
The diagonals of a rhombus are the bisectors of its angles: ; ; ; .
A rhombus is a parallelogram (for a rhombus all the properties of a parallelogram are fulfilled).

Properties of a square:

A square is a rhombus and a rectangle at the same time, therefore, for a square all the properties of a rectangle and a rhombus are fulfilled. And:

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

You can use our tasks (optional) and we, of course, recommend them.

In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.

How? There are two options:

Unlock all hidden tasks in this article -
Unlock access to all hidden tasks in all 99 articles of the textbook - Buy a textbook - 499 RUR

Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.

Access to all hidden tasks is provided for the ENTIRE life of the site.

In conclusion...

If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

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;
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.

Municipal budgetary educational institution

Savinskaya secondary school

Research

Parallelogram and its new properties

Completed by: 8B grade student

MBOU Savinskaya Secondary School

Kuznetsova Svetlana, 14 years old

Head: mathematics teacher

Tulchevskaya N.A.

p. 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 equality of other conditions

he who knows geometry is superior"

(Blaise Pascal).

While studying the topic “Parallelogram” in geometry lessons, we looked at 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 a parallelogram have other properties, and how will they 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 : properties of a parallelogram
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:

Study the history of the appearance of the parallelogram and the history of the development of its properties;

Find additional literature on the issue under study;

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:

Working with educational and popular science literature, Internet resources;

Study of theoretical material;

Identification of a range of problems that can be solved using additional properties of a parallelogram;

Observation, comparison, analysis, analogy.

Duration of the study : 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 the Greek words Parallelos - parallel and gramme - line), this term was introduced by Euclid. In his book Elements, Euclid proved the following properties of a parallelogram: opposite sides and angles of a parallelogram are equal, and the diagonal bisects it. Euclid does not mention the point of intersection of a parallelogram. Only towards the end of the Middle Ages was a complete theory of parallelograms developed. And only in the 17th century did theorems about parallelograms appear in textbooks, which were proved using Euclid’s theorem on the properties of a parallelogram.

III Additional properties of a parallelogram

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

Opposite angles and sides are equal

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

In various sources on geometry you can find the following additional properties:

The sum of adjacent angles of a parallelogram is 180 0

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

The bisectors of opposite angles of a parallelogram lie on parallel lines;

The bisectors of adjacent angles of a parallelogram intersect at right angles;

When the bisectors of all angles of a parallelogram intersect, they form a rectangle;

The distances from opposite corners of a parallelogram to 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 you draw altitudes from two opposite angles in a parallelogram, you get a rectangle.

IV Proof of the properties of a parallelogram

The sum of adjacent angles of a parallelogram is 180 0

Given:

ABCD – parallelogram

Prove:

A+
B=

Proof:

A and
B – internal one-sided angles with parallel straight lines BC AD and secant AB, which means
A+
B=

Given: ABCD - parallelogram,

AK bisector
A.

Prove: AVK – isosceles

Proof:

1)
1=
3 (crosswise lying at BC AD and secant AK ),

2)
2=
3 because AK is a bisector,

means 1=
2.

3) ABC - isosceles because 2 angles of a triangle are equal

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

Given: ABCD is a parallelogram,

AK – bisector A,

CP - bisector C.

Prove: AK ║ SR

Proof:

1) 1=2 because AK is a bisector

2) 4=5 because CP – bisector

3) 3=1 (crosswise lying angles at

BC ║ AD and AK-secant),

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

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

this means AK ║ CP (based on the parallelism of 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,

AK-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 Р – bisector

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

3) A + D =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 when intersecting form a rectangle

Given: ABCD - parallelogram, AK-bisector A,

DP-bisector D,

CM bisector C,

BF - bisector 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 the same diagonal are equal.

Given: ABCD-parallelogram, AC-diagonal.

VC AC, D.P. A.C.

Prove: BC=DP

Proof: 1) DCP = KAB, as internal crosses lying with AB ║ CD and secant AC.

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

And in equal triangles the corresponding sides are equal, which means DP=BK.

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

Given: ABCD parallelogram.

Prove: VKDR is a parallelogram.

Proof:

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

divide these sides in half)

2) BP ║ KD (lie on AD BC)

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

If you draw altitudes from two opposite angles in a parallelogram, you 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 +ВD 2 =2(AB 2 + AD 2 )

Proof: 1)ASK: A.C. ²=
+

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

3) A.C. ²+ BD ²=SK²+A K²+B Р²+РD ²

4) SC = BP = N(height )

5) AC 2 +BD 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+BD ²=2СD 2 -2x 2 + A TO 2 +PD 2

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

AC²+BD 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 point of intersection of the bisectors of two angles of a parallelogram adjacent to one side belongs to the opposite side. The shortest side of a parallelogram is 5 . Find its big side.

Given: ABCD is a parallelogram,

AK – bisector
A,

D K – bisector
D , AB=5

Find: Sun

decision

Solution

Because AK - bisector
And then ABC is isosceles.

Because D K – bisector
D, then DCK - isosceles

DC =C K= 5

Then, BC=VC+SC=5+5 = 10

Answer: 10

2. Find the perimeter of a 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: P parallelogram

Solution

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

Because AK – bisector
And then ABC is isosceles.

AB=BK= 14 cm

Then P=2 (14+21) =70 (cm)

happening

Given: ABCD is a parallelogram,

D K – bisector
D

VK=14 cm, KS=7 cm

Find: P parallelogram

Solution

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

Because D K – bisector
D, then DCK - isosceles

DC =C K= 7

Then, P = 2 (21+7) = 56 (cm)

Answer: 70cm or 56cm

3. The sides of a 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: VM, MN, NC

Solution

Because AM - bisector
And then AVM is isosceles.

Because DN – bisector
D, then DCN - isosceles

DC=CN=3

Then, MN = 10 – (BM +NC) = 10 – (3+3)=4 cm

Case 2: bisectors intersect inside a parallelogram

Because AN - bisector
And, then ABN is isosceles.

AB=BN = 3 D

And the sliding grille should be moved to the required distance in the doorway

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

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

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

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

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

Examples - automobile diamond-shaped jack, tram pantograph.

Scissor or X-shaped mechanism, also known as Nuremberg scissors- rhombus version - 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 diamond(s) in the middle. Used in lifts and children's toys.

VII Conclusion

Who has been studying mathematics since childhood?

he develops attention, trains his brain,

own will, cultivates perseverance

and perseverance in achieving goals

A. Markushevich

During the work, I proved additional properties of the parallelogram.

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

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

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

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

The purpose of my research work has been completed.

The importance of mathematical knowledge is evidenced by the fact that a prize was established for the person who publishes a book about a person who lived his entire life without the help of mathematics. Not a single person has received this award yet.

VIII Literature