Define a parallelogram and draw. Property of the diagonals of a parallelogram

1. Definition of a parallelogram.

If we intersect a pair of parallel lines with another pair of parallel lines, we get a quadrilateral whose opposite sides are pairwise parallel.

In the quadrilaterals ABDC and EFNM (Fig. 224) BD || AC and AB || CD;

EF || MN and EM || F.N.

A quadrilateral whose opposite sides are pairwise parallel is called a parallelogram.

2. Properties of a parallelogram.

Theorem. The diagonal of a parallelogram divides it into two equal triangles.

Let there be a parallelogram ABDC (Fig. 225) in which AB || CD and AC || BD.

It is required to prove that the diagonal divides it into two equal triangles.

Let's draw a diagonal CB in the parallelogram ABDC. Let us prove that \(\Delta\)CAB = \(\Delta\)СDВ.

The NE side is common to these triangles; ∠ABC = ∠BCD, as internal cross lying angles with parallel AB and CD and secant CB; ∠ACB = ∠CBD, same as internal cross lying angles with parallel AC and BD and secant CB.

Hence \(\Delta\)CAB = \(\Delta\)СDВ.

In the same way, one can prove that the diagonal AD divides the parallelogram into two equal triangles ACD and ABD.

Consequences:

1 . Opposite angles of a parallelogram are equal.

∠A = ∠D, this follows from the equality of triangles CAB and CDB.

Similarly, ∠C = ∠B.

2. Opposite sides of a parallelogram are equal.

AB \u003d CD and AC \u003d BD, since these are sides of equal triangles and lie opposite equal angles.

Theorem 2. The diagonals of a parallelogram are bisected at the point of their intersection.

Let BC and AD be the diagonals of the parallelogram ABDC (Fig. 226). Let us prove that AO = OD and CO = OB.

To do this, let's compare some pair of opposite triangles, for example \(\Delta\)AOB and \(\Delta\)COD.

In these triangles AB = CD, as opposite sides of a parallelogram;

∠1 = ∠2, as interior angles crosswise lying at parallel AB and CD and secant AD;

∠3 = ∠4 for the same reason, since AB || CD and CB are their secant.

It follows that \(\Delta\)AOB = \(\Delta\)COD. And in equal triangles opposite equal angles lie equal sides. Therefore, AO = OD and CO = OB.

Theorem 3. The sum of the angles adjacent to one side of the parallelogram is equal to 180°.

Draw a diagonal AC in parallelogram ABCD and get two triangles ABC and ADC.

The triangles are congruent because ∠1 = ∠4, ∠2 = ∠3 (cross-lying angles at parallel lines), and side AC is common.
The equality \(\Delta\)ABC = \(\Delta\)ADC implies that AB = CD, BC = AD, ∠B = ∠D.

The sum of the angles adjacent to one side, for example, angles A and D, is equal to 180 ° as one-sided with parallel lines.

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Municipal budgetary educational institution

Savinskaya secondary 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
Objective:

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 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:

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 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 a diagonal bisects it. Euclid does not mention the point of intersection of the parallelogram. Only by the end of the Middle Ages, a complete theory of parallelograms was developed. 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 angle bisector of a parallelogram cuts off an isosceles triangle from it;

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
BUT.

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 be 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 FROMD²-2 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
BUT,

D K - bisector
D, AB=5

To 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:
BUT,

VK=14 cm, KS=7 cm

To 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

To 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
BUT,

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

To 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 an award 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

Proof

Let's draw the diagonal AC first. 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, \triangle ABC = \triangle ADC (by the second feature: 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. So the sum of the 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 bisected by the intersection point.

Proof

Let's draw another diagonal.

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

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

Proven!

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 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 consider in more detail. Why AD || BC?

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

But if \triangle ABC = \triangle ADC , then \angle 3 = \angle 4 (they lie opposite AB and CD respectively). And therefore AD || BC (\angle 3 and \angle 4 - lying across 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 feature. 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.

Proof

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

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

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

At the same time, \alpha and \beta are internal one-sided with a secant AD . And that means AB || CD.

The third sign is correct.

4. A parallelogram is a quadrilateral whose diagonals are bisected by the intersection point.

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 || BC.

The fourth sign is correct.

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