A parallelogram is a quadrilateral whose opposite sides are parallel, i.e. lie on parallel lines
Parallelogram properties: 
Theorem 22.
Opposite sides of a parallelogram are equal.
Proof. Draw a diagonal AC in a parallelogram ABCD. Triangles ACD and ACB are congruent as having a common side AC and two pairs of equal angles. adjacent to it: ∠ CAB=∠ ACD, ∠ ASV=∠ DAC (as cross-lying angles with parallel lines AD and BC). Hence, AB=CD and BC=AD as corresponding sides of equal triangles, etc. The equality of these triangles also implies the equality of the corresponding angles of the triangles:
Theorem 23.
The opposite angles of a parallelogram are: ∠ A=∠ C and ∠ B=∠ D.
The equality of the first pair comes from the equality of triangles ABD and CBD, and the second - ABC and ACD.
Theorem 24.
Neighboring corners of a parallelogram, i.e. angles adjacent to one side add up to 180 degrees.
This is so because they are interior one-sided corners.
Theorem 25.
The diagonals of a parallelogram bisect each other at the point of their intersection.
Proof. Consider triangles BOC and AOD. According to the first property, AD=BC ∠ ОАD=∠ OSV and ∠ ОDA=∠ ОВС as lying across with parallel lines AD and BC. Therefore, triangles BOC and AOD are equal in side and angles adjacent to it. Hence, BO=OD and AO=OC, as the corresponding sides of equal triangles, etc.
Parallelogram features
Theorem 26.
If 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. 2). Let's draw the diagonal AC. Triangle ABC and ACD have three equal sides. Then the angles BAC and DCA are equal and therefore AB is parallel to CD. The parallelism of the sides BC and AD follows from the equality of the angles CAD and DIA.
Theorem 27.
If the opposite angles of a quadrilateral are equal in pairs, then it is a parallelogram.
Let ∠ A=∠ C and ∠ B=∠ D. ∠ A+∠ B+∠ C+∠ D=360 o, then ∠ A+∠ B=180 o and sides AD and BC are parallel (on the basis of parallel lines). We also prove the parallelism of the sides AB and CD and conclude that ABCD is a parallelogram by definition.
Theorem 28.
If the adjacent corners of the quadrilateral, i.e. angles adjacent to one side add up to 180 degrees, then it is a parallelogram.
If the interior one-sided angles add up to 180 degrees, then the lines are parallel. This means AB is a pair of CD and BC is a pair of AD. A quadrilateral turns out to be a parallelogram by definition.
Theorem 29.
If the diagonals of a quadrilateral are mutually divided at the point of intersection in half, then the quadrilateral is a parallelogram.
Proof. If AO=OC, BO=OD, then the triangles AOD and BOC are equal, as having equal angles (vertical) at the vertex O, enclosed between pairs of equal sides. From the equality of triangles we conclude that AD and BC are equal. The sides AB and CD are also equal, and the quadrangle turns out to be a parallelogram according to feature 1.
Theorem 30.
If a quadrilateral has a pair of equal, parallel sides, then it is a parallelogram.
Let sides AB and CD be parallel and equal in quadrilateral ABCD. Draw the diagonals AC and BD. From the parallelism of these lines follows the equality of the cross-lying angles ABO=CDO and BAO=OCD. Triangles ABO and CDO are equal in side and adjacent angles. Therefore, AO=OC, BO=OD, i.e. the diagonals of the intersection point are divided in half and the quadrilateral turns out to be a parallelogram according to feature 4.
In geometry, special cases of a parallelogram are considered.
Task 1. One of the angles of the parallelogram is 65°. Find the remaining angles of the parallelogram.
∠C = ∠A = 65° as opposite angles of the parallelogram.
∠A + ∠B = 180° as angles adjacent to one side of the parallelogram.
∠B = 180° - ∠A = 180° - 65° = 115°.
∠D = ∠B = 115° as opposite angles of the parallelogram.
Answer: ∠A = ∠C = 65°; ∠B = ∠D = 115°.
Task 2. The sum of two angles of a parallelogram is 220°. Find the angles of the parallelogram.
Since the parallelogram has 2 equal acute angles and 2 equal obtuse angles, we are given the sum of two obtuse angles, i.e. ∠B +∠D = 220°. Then ∠В =∠D = 220° :
2 = 110°.
∠A + ∠B = 180° as angles adjacent to one side of the parallelogram, so ∠A = 180° - ∠B = 180° - 110° = 70°. Then ∠C =∠A = 70°.
Answer: ∠A = ∠C = 70°; ∠B = ∠D = 110°.
Task 3. One of the angles of the parallelogram is 3 times the other. Find the angles of the parallelogram.
Let ∠A =x. Then ∠B = 3x. Knowing that the sum of the angles of a parallelogram adjacent to one of its sides is equal to 180 °, we compose an equation.
x = 180 : 4;
We get: ∠A \u003d x \u003d 45 °, and ∠ B \u003d 3x \u003d 3 ∙ 45 ° \u003d 135 °.
Opposite angles of a parallelogram are equal, so
∠A = ∠C = 45°; ∠B = ∠D = 135°.
Answer: ∠A = ∠C = 45°; ∠B = ∠D = 135°.
Task 4. Prove that if two sides of a quadrilateral are parallel and equal, then this quadrilateral is a parallelogram.
Proof.
Draw the diagonal BD and consider Δ ADB and Δ CBD.
AD = BC by condition. The BD side is common. ∠1 = ∠2 as internal cross-lying under parallel (by assumption) lines AD and BC and secant BD. Therefore, Δ ADB = Δ CBD on two sides and the angle between them (the 1st criterion for the equality of triangles). In congruent triangles, the corresponding angles are equal, so ∠3 = ∠4. And these angles are internal crosswise lying at lines AB and CD and secant BD. This implies the parallelism of lines AB and CD. Thus, in the given quadrilateral ABCD, the opposite sides are pairwise parallel, therefore, by definition, ABCD is a parallelogram, which was to be proved.
Task 5. The two sides of a parallelogram are related as 2 : 5, and the perimeter is 3.5 m. Find the sides of the parallelogram.
∙ (AB+AD).
Let's denote one part by x. then AB = 2x, AD = 5x meters. Knowing that the perimeter of the parallelogram is 3.5 m, we write the equation:
2 ∙ (2x + 5x) = 3.5;
2 ∙ 7x=3.5;
x=3.5 : 14;
One part is 0.25 m. Then AB = 2 ∙ 0.25 = 0.5 m; AD=5 ∙ 0.25 = 1.25 m.
Examination.
Parallelogram perimeter P ABCD = 2 ∙ (AB+AD) = 2 ∙ (0,25 + 1,25) = 2 ∙ 1.75 = 3.5 (m).
Since the opposite sides of the parallelogram are equal, then CD = AB = 0.25 m; BC = AD = 1.25 m.
Answer: CD = AB = 0.25 m; BC = AD = 1.25 m.
The video course "Get an A" includes all the topics necessary for the successful passing of the exam in mathematics by 60-65 points. Completely all tasks 1-13 of the Profile USE in mathematics. Also suitable for passing the Basic USE in mathematics. If you want to pass the exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!
Preparation course for the exam for grades 10-11, as well as for teachers. Everything you need to solve part 1 of the exam in mathematics (the first 12 problems) and problem 13 (trigonometry). And this is more than 70 points on the Unified State Examination, and neither a hundred-point student nor a humanist can do without them.
All the necessary theory. Quick solutions, traps and secrets of the exam. All relevant tasks of part 1 from the Bank of FIPI tasks have been analyzed. The course fully complies with the requirements of the USE-2018.
The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.
Hundreds of exam tasks. Text problems and probability theory. Simple and easy to remember problem solving algorithms. Geometry. Theory, reference material, analysis of all types of USE tasks. Stereometry. Cunning tricks for solving, useful cheat sheets, development of spatial imagination. Trigonometry from scratch - to task 13. Understanding instead of cramming. Visual explanation of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. Base for solving complex problems of the 2nd part of the exam.
A parallelogram is a quadrilateral whose opposite sides are pairwise parallel. This definition is already sufficient, since the remaining properties of a 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 equal in pairs;
- a parallelogram has opposite angles that are equal in pairs;
- the diagonals of a parallelogram are bisected by the point of intersection.
Parallelogram - a convex quadrilateral
Let us first prove the theorem that a parallelogram is a convex quadrilateral. A polygon is convex when whatever 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 it follows from the definition of a parallelogram that AB || CD, BC || AD.
Parallel segments do not have common points, they 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 with respect 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. For example, if a parallelogram ABCD is given, then it has AB = CD, AD = BC, ∠A = ∠C, ∠B = ∠D. This theorem is proved as follows.
A parallelogram is a quadrilateral. So it has two diagonals. Since a parallelogram is a convex quadrilateral, any of them divides it into two triangles. Consider the triangles ABC and ADC in the parallelogram ABCD obtained by drawing the diagonal AC.
These triangles have one side in common - AC. The angle BCA is equal to the angle CAD, as are the verticals with parallel BC and AD. Angles BAC and ACD are also equal, as are the vertical angles when AB and CD are parallel. Therefore, ∆ABC = ∆ADC over 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. The angle C equals consists of ∠BCA and ∠ACD. Since the pairs of angles are equal to each other, then ∠A = ∠C.
Thus, it is proved that in a parallelogram opposite sides and angles are equal.
Diagonals cut in half
Since a parallelogram is a convex quadrilateral, it has two two diagonals, and they intersect. Let a parallelogram ABCD be given, its diagonals AC and BD intersect at a point E. Consider the triangles ABE and CDE formed by them.
These triangles have sides AB and CD equal as opposite sides of a parallelogram. The angle ABE is equal to the angle CDE as they lie across parallel lines AB and CD. For the same reason, ∠BAE = ∠DCE. Hence, ∆ABE = ∆CDE over two angles and the side between them.
You can also notice that the angles AEB and CED are vertical, and therefore also equal to each other.
Since triangles ABE and CDE are equal to each other, so are all their corresponding elements. Side AE of the first triangle corresponds to side CE of the second, so AE = CE. Similarly, BE = DE. Each pair of equal segments makes up the diagonal of the parallelogram. Thus, it is proved that the diagonals of a parallelogram are bisected by the point of intersection.
Middle level
Parallelogram, rectangle, rhombus, square (2019)
1. Parallelogram
Compound word "parallelogram"? And behind it is a very simple figure.
Well, that is, we took two parallel lines:
Crossed by two more:
And inside - a parallelogram!
What are the properties of a parallelogram?
Parallelogram properties.
That is, what can be used if a parallelogram is given in the problem?
This question is answered by the following theorem:
Let's draw everything in detail.
What does first point of the theorem? And the fact that if you HAVE a parallelogram, then by all means
The second paragraph means that if there is a parallelogram, then, again, by all means:
Well, and finally, the third point means that if you HAVE a parallelogram, then be sure:
See what a wealth of choice? What to use in the task? Try to focus on the question of the task, or just try everything in turn - some kind of “key” will do.
And now let's ask ourselves another question: how to recognize a parallelogram "in the face"? What must happen to a quadrilateral in order for us to have the right to give it the “title” of a parallelogram?
This question is answered by several signs of a parallelogram.
Features of a parallelogram.
Attention! Begin.
Parallelogram.
Pay attention: if you have found at least one sign in your problem, then you have exactly a parallelogram, and you can use all the properties of a parallelogram.
2. Rectangle
I don't think it will be news to you at all.
The first question is: 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 for a rectangle, like for any parallelogram, and, and the diagonals are divided by the intersection point in half.
But there is a rectangle and one distinctive property.
Rectangle Property
Why is this property distinctive? Because no other parallelogram has equal diagonals. Let's formulate it more clearly.
Pay attention: in order to become a rectangle, a quadrilateral must first become a parallelogram, and then present the equality of the diagonals.
3. Diamond
And again the question is: is a rhombus a parallelogram or not?
With full right - a parallelogram, because it has and (remember our sign 2).
And again, since a rhombus is a parallelogram, then it must have all the properties of a parallelogram. This means that a rhombus has opposite angles equal, opposite sides are parallel, and the diagonals are bisected by the point of intersection.
Rhombus Properties
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 we have not just a parallelogram, but a rhombus.
Signs of a rhombus
And pay attention again: there should be not just a quadrangle with perpendicular diagonals, but a parallelogram. Make sure:
No, of course not, although its diagonals and are perpendicular, and the diagonal is the bisector of angles u. But ... the diagonals do not divide, the intersection point in half, 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 comes out of this.
Is it clear why? - rhombus - the bisector of angle A, which is equal to. So it divides (and also) into two angles along.
Well, it's quite clear: the rectangle's diagonals are equal; rhombus diagonals are perpendicular, and in general - parallelogram diagonals are divided by the point of intersection in half.
MIDDLE LEVEL
Properties of quadrilaterals. Parallelogram
Parallelogram Properties
Attention! The words " parallelogram properties» means that if you have a task there is parallelogram, then all of the following can be used.
Theorem on the properties of a parallelogram.
In any parallelogram:
Let's see why this is true, in other words WE WILL PROVE theorem.
So why is 1) true?
Since it is a parallelogram, then:
- like lying crosswise
- as lying across.
Hence, (on the II basis: and - general.)
Well, once, then - that's it! - proved.
But by the way! We also proved 2)!
Why? But after all (look at the picture), that is, namely, because.
Only 3 left).
To do this, you still have to draw a second diagonal.
And now we see that - according to the II sign (the angle and the side "between" them).
Properties proven! Let's move on to the signs.
Parallelogram features
Recall that the sign of a parallelogram answers the question "how to find out?" That the 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, that's even easier! Let's draw a diagonal again.
Which means:
And is also easy. But… different!
Means, . Wow! But also - internal one-sided at a secant!
Therefore the fact that means that.
And if you look from the other side, then they are internal one-sided at a secant! And therefore.
See how great it is?!
And again simply:
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
So, on two legs (and - general).
Well, since the triangles are equal, then their hypotenuses are also equal.
Proved that!
And imagine, the equality of the diagonals is a distinctive property of a rectangle among all parallelograms. That is, the following statement is true
Let's see why?
So, (meaning the angles of the parallelogram). But once again, remember that - a parallelogram, and therefore.
Means, . And, of course, it follows from this that each of them After all, in the amount they should give!
Here we have proved that if parallelogram suddenly (!) will be equal diagonals, then this exactly a rectangle.
But! Pay attention! This is about parallelograms! Not any a quadrilateral with equal diagonals is a rectangle, and only parallelogram!
Properties of quadrilaterals. Rhombus
And again the question is: is a rhombus a parallelogram or not?
With full right - a parallelogram, because it has and (Remember our sign 2).
And again, since a rhombus is a parallelogram, it must have all the properties of a parallelogram. This means that a rhombus has opposite angles equal, opposite sides are parallel, and the diagonals are bisected by the point of intersection.
But there are also special properties. We formulate.
Rhombus Properties
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 and turned out to be the 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.
Rhombus signs.
Why is that? And look
Hence, and both these triangles are isosceles.
To be a rhombus, a quadrilateral must first "become" a parallelogram, and then already demonstrate 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 comes out of this.
Is it clear why? Square - rhombus - the bisector of the angle, which is equal to. So it divides (and also) into two angles along.
Well, it's quite clear: the rectangle's diagonals are equal; rhombus diagonals are perpendicular, and in general - parallelogram diagonals are divided by the point of intersection in half.
Why? Well, just apply the Pythagorean Theorem to.
SUMMARY AND BASIC FORMULA
Parallelogram properties:
- Opposite sides are equal: , .
- Opposite angles are: , .
- The angles at one side add up to: , .
- The diagonals are divided by the intersection point in half: .
Rectangle properties:
- The diagonals of a rectangle are: .
- Rectangle is a parallelogram (all properties of a parallelogram are fulfilled for a rectangle).
Rhombus properties:
- The diagonals of the rhombus are perpendicular: .
- The diagonals of a rhombus are the bisectors of its angles: ; ; ; .
- A rhombus is a parallelogram (all properties of a parallelogram are fulfilled for a rhombus).
Square properties:
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. As well as.
