In order to determine whether this figure parallelogram there are a number of features. Let's look at the three main features of a parallelogram.
1 parallelogram sign
If two sides of a quadrilateral are equal and parallel, then this quadrilateral will be a parallelogram.
Proof:
Consider the quadrilateral ABCD. Let the sides AB and CD be parallel. And let AB=CD. Let's draw the diagonal BD in it. It will divide this quadrilateral into two equal triangles: ABD and CBD.
These triangles are equal to each other on two sides and the angle between them (BD - common side, AB = CD by condition, angle1 = angle2 as crosswise angles with the transversal BD of parallel lines AB and CD.), and therefore angle3 = angle4.
And these angles will lie crosswise when the lines BC and AD intersect with the secant BD. It follows from this that BC and AD are parallel to each other. We have that in the quadrilateral ABCD opposite sides are pairwise parallel, and therefore the quadrilateral ABCD is a parallelogram.
Parallelogram sign 2
If in a quadrilateral the opposite sides are equal in pairs, then this quadrilateral will be a parallelogram.
Proof:
Consider the quadrilateral ABCD. Let's draw the diagonal BD in it. It will divide this quadrilateral into two equal triangles: ABD and CBD.
These two triangles will be equal to each other on three sides (BD is the common side, AB = CD and BC = AD by condition). From this we can conclude that angle1 = angle2. It follows that AB is parallel to CD. And since AB = CD and AB is parallel to CD, then according to the first criterion of a parallelogram, the quadrilateral ABCD will be a parallelogram.
3 parallelogram sign
If the diagonals of a quadrilateral intersect and are bisected by the point of intersection, then this quadrilateral will be a parallelogram.
Consider the quadrilateral ABCD. Let us draw two diagonals AC and BD in it, which will intersect at point O and are bisected by this point.
Triangles AOB and COD will be equal to each other, according to the first sign of equality of triangles. (AO = OC, BO = OD by condition, angle AOB = angle COD as vertical angles.) Therefore, AB = CD and angle 1 = angle 2. From the equality of angles 1 and 2, we have that AB is parallel to CD. Then we have that in the quadrilateral ABCD the sides AB are equal to CD and parallel, and according to the first criterion of a parallelogram, the quadrilateral ABCD will be a parallelogram.
I. If two opposite sides of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
Task 1. From vertices B and D of parallelogram ABCD, in which AB≠ BC and angle A is acute, perpendiculars BK and DM are drawn to straight line AC. Prove that the quadrilateral BMDK is a parallelogram.
Proof.
Since VC and DM are perpendicular to the same straight line AC, then VC II DM. In addition, VK and DM are heights drawn in equal trianglesΔ ABC and Δ CDA from vertices equal angles∠B and ∠D to the same side AC, therefore BC = DM. We have: two sides BC and DM of the quadrilateral BMDK are parallel and equal, which means that BMDK is a parallelogram, which is what needed to be proven.
II. If opposite sides of a quadrilateral are equal in pairs, then this quadrilateral is a parallelogram.
Task 2. On sides AB, BC, CD and DA of quadrilateral ABCD the points M, N, P and Q are marked respectively so that AM=CP, BN=DQ, BM=DP, NC=QA. Prove that ABCD and MNPQ are parallelograms.
Proof.
1. According to the condition, in quadrilateral ABCD the opposite sides consist of equal segments, therefore equal, i.e. AD=BC, AB=CD. Therefore, ABCD is a parallelogram.
2. Consider Δ MBN and Δ PDQ. BM=DP and BN=DQ by condition. ∠B =∠D as opposite angles parallelogram ABCD. This means that Δ MBN = Δ PDQ on two sides and the angle between them (1st sign of equality of triangles). And in equal triangles opposite equal angles lie equal sides. Hence MN=PQ. We have proven that the opposite sides MN and PQ of the quadrilateral MNPQ are equal. Similarly, from the equality of triangles Δ MAQ and Δ PCN it follows that the sides MQ and PN are equal, which are opposite sides of the quadrilateral MNPQ. We have: opposite sides of the quadrilateral MNPQ are equal in pairs. Therefore, the quadrilateral MNPQ is a parallelogram. The problem is solved.
III. If the diagonals of a quadrilateral intersect and are bisected by the point of intersection, then the quadrilateral is a parallelogram.
Task 3. The diagonals of the parallelogram ABCD intersect at point O. Prove that the quadrilateral MNPQ, whose vertices are the midpoints of the segments OA, OB, OC and OD, is a parallelogram.
Proof.
According to the property of the diagonals of a parallelogram ABCD, its diagonals AC and BD are divided in half by the intersection point, i.e. OA=OS and OB=OD. The diagonals of the quadrilateral MNPQ also intersect at point O, which will be the midpoint of each of them. Indeed, since the vertices of the quadrilateral MNPQ are by condition the midpoints of the segments OA, OC, OB and OD, then BN=ON=OQ=DQ and AM=OM=OP=CP. Consequently, the diagonals MP and NQ of the quadrilateral MNPQ are bisected at the intersection point, therefore, the quadrilateral MNPQ is a parallelogram, which is what needed to be proven.
Back forward
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The purpose of the lesson: consider the characteristics of a parallelogram and consolidate the acquired knowledge in the process of solving problems.
Tasks:
- educational: developing the ability to apply parallelogram features to solve problems;
- developing: development logical thinking, attention, skills independent work, self-esteem skills;
- educational: nurturing interest in the subject, the ability to work in a team, a culture of communication.
Lesson type: learning new material, primary consolidation.
Equipment: interactive board, projector, task cards, presentation.
During the classes
1. Organizational moment.
U: Good afternoon, guys! Today in class we will again talk about parallelograms. We have to complete many tasks, prove theorems and learn how to apply them when solving problems. The motto of our lesson will be the words of Le Carbusier: “Everything around is geometry.”
2. Updating students' knowledge.
Theoretical survey
Give some students individual assignments on cards on the topic properties of a parallelogram(everyone chooses tasks independently on the presentation slide via a hyperlink, pointing the mouse pointer at the figure, but not at the number), listen individually to each respondent.
With the rest - prove additional properties of a parallelogram. (First discuss the proof orally, then check it with the interactive whiteboard).
1°. The bisector of the angle of a parallelogram cuts off an isosceles triangle from it.
2°. The bisectors of adjacent angles of a parallelogram are perpendicular, and the bisectors opposite angles are parallel or lie on the same straight line.
After preparation, listen to evidence of additional properties of a parallelogram.
ABCD -parallelogram,
AE is the bisector of angle BAD.
Prove: ABE is isosceles.
Proof:
Since ABCD is a parallelogram, therefore BC || AD, then angle EAD = angle BEA lying crosswise with parallel lines BC and AD and secant AE. AE is the bisector of angle BAD, which means angle BAE = angle EAD, therefore angle BAE = angle BEA.
In ABE, angle BAE = angle BEA, which means ABE is isosceles with base AE.
Suggestive questions:
Formulate the sign of an isosceles triangle.
Which angles in BAE can be equal? Why?
ABCD -parallelogram,
BE is the bisector of angle CBA,
AE is the bisector of angle BAD.
Suggestive questions:
When will lines AE and CK be parallel?
Are angles BEA and<3? Почему?
In what case will AE and CK coincide?
Preparing to study new material
Frontal work with the class (orally).
- What do the words “properties” and “character” mean? Give examples.
- What is the converse theorem?
- Is the opposite of this statement always true? Give examples.
3. Explanation of new material.
U.: Each object has its own properties and characteristics. Please tell me how properties differ from signs.
Let's try to understand this issue using a simple example. The given object is autumn. Name its properties: Its characteristics:
- What statements are the property and attribute of an object in relation to each other? (answer: inverse)
- What properties have we already studied in the geometry course? State them. (name a few)
Is it possible to construct a true converse statement for any property? (different answers).
Let's check this on the following properties:
Conclude: Is it possible to construct a true converse statement for any property? (no, not for anyone)
Now, let's return to our quadrilateral, remember its properties and formulate their converse statements, i.e.:.. (answer - characteristics of a parallelogram). So, the topic of today's lesson is: “Signs of a parallelogram.”
So, name the properties of a parallelogram.
Formulate statements that are inverse to the properties. (students formulate signs, the teacher corrects them and formulates them again)
Let us prove these signs. The first sign is in detail, the second is brief, the third is on your own at home.
4. Consolidation of the studied material.
Work in workbooks: solve problem No. 11 on the interactive whiteboard, call a less prepared student to the board.
Solution to problem No. 379 (write the solution on the interactive whiteboard). From vertices B and D of the parallelogram ABCD, in which AB BC and A are acute, perpendiculars BC and DM are drawn to the straight line AC. Prove that the quadrilateral BMDK is a parallelogram.
I. If two opposite sides of a quadrilateral are parallel and equal, then the quadrilateral is a parallelogram.
Task 1. From vertices B and D of parallelogram ABCD, in which AB≠ BC and angle A is acute, perpendiculars BK and DM are drawn to straight line AC. Prove that the quadrilateral BMDK is a parallelogram.
Proof.
Since VC and DM are perpendicular to the same straight line AC, then VC II DM. In addition, BC and DM are the heights drawn in equal triangles Δ ABC and Δ CDA from the vertices of equal angles ∠B and ∠D to the same side AC, therefore, BC = DM. We have: two sides BC and DM of the quadrilateral BMDK are parallel and equal, which means that BMDK is a parallelogram, which is what needed to be proven.
II. If opposite sides of a quadrilateral are equal in pairs, then this quadrilateral is a parallelogram.
Task 2. On sides AB, BC, CD and DA of quadrilateral ABCD the points M, N, P and Q are marked respectively so that AM=CP, BN=DQ, BM=DP, NC=QA. Prove that ABCD and MNPQ are parallelograms.
Proof.
1. By condition, in the quadrilateral ABCD, the opposite sides consist of equal segments, therefore they are equal, i.e. AD=BC, AB=CD. Therefore, ABCD is a parallelogram.
2. Consider Δ MBN and Δ PDQ. BM=DP and BN=DQ by condition. ∠B =∠D as opposite angles of parallelogram ABCD. This means that Δ MBN = Δ PDQ on two sides and the angle between them (1st sign of equality of triangles). And in equal triangles, equal sides lie opposite equal angles. Hence MN=PQ. We have proven that the opposite sides MN and PQ of the quadrilateral MNPQ are equal. Similarly, from the equality of triangles Δ MAQ and Δ PCN it follows that the sides MQ and PN are equal, which are opposite sides of the quadrilateral MNPQ. We have: opposite sides of the quadrilateral MNPQ are equal in pairs. Therefore, the quadrilateral MNPQ is a parallelogram. The problem is solved.
III. If the diagonals of a quadrilateral intersect and are bisected by the point of intersection, then the quadrilateral is a parallelogram.
Task 3. The diagonals of the parallelogram ABCD intersect at point O. Prove that the quadrilateral MNPQ, whose vertices are the midpoints of the segments OA, OB, OC and OD, is a parallelogram.
Proof.
According to the property of the diagonals of a parallelogram ABCD, its diagonals AC and BD are divided in half by the intersection point, i.e. OA=OS and OB=OD. The diagonals of the quadrilateral MNPQ also intersect at point O, which will be the midpoint of each of them. Indeed, since the vertices of the quadrilateral MNPQ are by condition the midpoints of the segments OA, OC, OB and OD, then BN=ON=OQ=DQ and AM=OM=OP=CP. Consequently, the diagonals MP and NQ of the quadrilateral MNPQ are bisected at the intersection point, therefore, the quadrilateral MNPQ is a parallelogram, which is what needed to be proven.