Lesson summary.
Algebra 8th grade
Teacher Sysoy 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 the skills and abilities to solve geometric problems;
3) Development of a culture of mathematical speech
Lesson plan:
1. Organizational moment
(Slide 1)
The slide shows a statement by 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 board are tasks 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 part
(Slide 3)
Definition of a parallelogram on the board. The teacher talks about a new figure and formulates a definition, making the necessary explanations using a drawing. Then, on the checkered part of the presentation, using a marker and a ruler, he shows how to draw a parallelogram (several cases are possible)
(Slide 4)
The teacher formulates the first property of a parallelogram. Invites students to tell from the drawing what is given and what needs to be proven. After this, the given task appears on the board. Students guess (maybe with the help of the teacher) that the required equalities must be proven through the equalities of triangles, which can be obtained by drawing a diagonal (a diagonal appears on the board). Next, students guess why the triangles are equal and name the sign that triangles are equal (the corresponding shape appears). They verbally communicate the facts that are necessary to make the triangles equal (as they name them, a corresponding visualization appears). Next, students formulate the property of congruent triangles, it appears as 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 suggests using the picture to tell what is given and what needs to be proven. After students correctly report what is given and what needs to be proven, the condition of the theorem appears. Students guess that the equality of the parts of the diagonals can be proven through the equality of trianglesAOB And C.O.D.. Using the previous property of a parallelogram, one guesses that the sides are equalAB And CD. Then they understand that they need to find equal angles and, using the properties of parallel lines, prove the equality of angles adjacent to equal sides. These stages are visualized on the slide. The truth of the theorem follows from the equality of the triangles - the students say it and a corresponding visualization appears on the slide.
(Slide 6)
The teacher formulates the third property of a parallelogram. Depending on the time remaining until the end of the lesson, the teacher can give the students the opportunity to prove this property on their own, or limit themselves 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 an inscribed polygon, which was repeated at the beginning of the lesson, or on the sum of the internal one-sided angles of two parallel linesAD And B.C., and a secant, for exampleAB.
4. Fixing the material
At this stage, students use previously learned theorems to solve problems. Students select ideas for solving the problem independently. Since there are many possible design options 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 recording the solution in a notebook.
(Slide 7)
The task condition appears. The teacher suggests formulating “Given” according to the condition. After the students correctly write down a short statement of the condition, “Given” appears on the board. The process for solving the problem might look like this:
Let's draw the height BH (visualized)
Triangle AHB is a right triangle. Angle A is equal to angle C and equals 30 0 (according to the property of opposite angles in a parallelogram). 2BH =AB (by the property of the leg lying opposite the 30 0 angle in a right triangle). So AB = 13 cm.
AB = CD, BC = AD (according to the property of opposite sides in a parallelogram) So AB = CD = 13 cm. Since the perimeter of the parallelogram is 50 cm, then BC = AD = (50 – 26): 2 = 12 cm.
Answer: AB = CD = 13 cm, BC = AD = 12 cm.
(Slide 8)
The task condition appears. The teacher suggests formulating “Given” according to the condition. Then “Given” appears on the screen. Using red lines, a quadrilateral is highlighted, about which you need to prove that it is a parallelogram. The process for solving the problem might look like this:
Because BK and MD are perpendicular to one line, then lines BK and MD are parallel.
Through adjacent angles it can be shown that the sum of the internal one-sided angles at straight lines BM and KD and the secant MD is equal to 180 0. Therefore, these lines are parallel.
Since the quadrilateral BMDK has opposite sides parallel in pairs, then this quadrilateral is a parallelogram.
5. End of the lesson. Behavior of the results.
(Slide 8)
Questions on the new topic appear on the slide, to which students answer.
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
From the history of the parallelogram
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 are proven 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 lines BC 
AD and secant AB, which means A+ B=
2
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
3
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: VKDP 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)
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
Pogorelov A.V. Geometry 7-9: textbook for general education. institutions - M.: Education, 2014
L.S.Atanasyan and others. Geometry. Add. Chapters for the 8th grade textbook: textbook. manual for students of schools and advanced classes. studied mathematics. – M.: Vita-press, 2003
Internet resources
Wikipedia materials
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
- Introduction.
- Repetition of previously studied material.
- Parallelogram, its properties and features.
- Examples of tasks.
- 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
- Are the diagonals of a square divided in half by the point of intersection?
- Are the diagonals of a parallelogram equal?
- Are the opposite angles of a parallelogram equal?
- State the definition of a parallelogram?
- How many signs of a parallelogram?
- Can a rhombus be a parallelogram?
List of sources used
- Kuznetsov A.V., mathematics teacher (grades 5-9), Kiev
- “Unified State Exam 2006. Mathematics. Educational and training materials for preparing students / Rosobrnadzor, ISOP - M.: Intellect-Center, 2006"
- Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”
- 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 (Fig. 233).
For an arbitrary parallelogram the following properties hold:
1. Opposite sides of a parallelogram are equal.
Proof. In the parallelogram ABCD we draw the diagonal AC. Triangles ACD and AC B are equal, as having a common side AC and two pairs of equal angles adjacent to it:
(like crosswise angles with parallel lines AD and BC). This means, and like the sides of equal triangles lying opposite equal angles, which is what needed to be proven.
2. Opposite angles of a parallelogram are equal:
3. Adjacent angles of a parallelogram, i.e., angles adjacent to one side, add up, etc.
The proof of properties 2 and 3 is immediately obtained from the properties of angles for parallel lines.
4. The diagonals of a parallelogram bisect each other at their intersection point. In other words,
Proof. Triangles AOD and BOC are congruent, since their sides AD and BC are equal (property 1) and the angles adjacent to them (like crosswise angles for parallel lines). From here it follows that the corresponding sides of these triangles are equal: AO, which is what needed to be proven.
Each of these four properties characterizes a parallelogram, or, as they say, is its characteristic property, i.e., every quadrilateral that has at least one of these properties is a parallelogram (and, therefore, has all the other three properties).
Let us carry out the proof for each property separately.
1". If the opposite sides of a quadrilateral are equal in pairs, then it is a parallelogram.
Proof. Let the quadrilateral ABCD have sides AD and BC, AB and CD respectively equal (Fig. 233). Let's draw the diagonal AC. Triangles ABC and CDA will be congruent as having three pairs of equal sides.
But then the angles BAC and DCA are equal and . The parallelism of sides BC and AD follows from the equality of angles CAD and ACB.
2. If a quadrilateral has two pairs of opposite angles equal, then it is a parallelogram.
Proof. Let . Since then both sides AD and BC are parallel (based on the parallelism of lines).
3. We leave the formulation and proof to the reader.
4. If the diagonals of a quadrilateral bisect each other at the point of intersection, then the quadrilateral is a parallelogram.
Proof. If AO = OS, BO = OD (Fig. 233), then the triangles AOD and BOC are equal, as having equal angles (vertical!) at the vertex O, enclosed between pairs of equal sides AO and CO, BO and DO. From the equality of triangles we conclude that sides AD and BC are equal. The sides AB and CD are also equal, and the quadrilateral turns out to be a parallelogram according to the characteristic property G.
Thus, in order to prove that a given quadrilateral is a parallelogram, it is enough to verify the validity of any of the four properties. The reader is invited to independently prove another characteristic property of a parallelogram.
5. If a quadrilateral has a pair of equal, parallel sides, then it is a parallelogram.
Sometimes any pair of parallel sides of a parallelogram is called its bases, then the other two are called lateral sides. A straight line segment perpendicular to two sides of a parallelogram, enclosed between them, is called the height of the parallelogram. Parallelogram in Fig. 234 has a height h drawn to the sides AD and BC, its second height is represented by the segment .
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