One of the signs of a parallelogram is that if two sides of a quadrilateral are equal and parallel, then such a quadrilateral is a parallelogram. That is, if a quadrilateral has two sides equal and parallel, then the other two sides also turn out to be equal and parallel to each other, since this fact is the definition and property of a parallelogram.
Thus, a parallelogram can only be defined by two sides that are equal and parallel to each other.
This characteristic of a parallelogram can be formulated as a theorem and proven. In this case, we are given a quadrilateral whose two sides are equal and parallel to each other. It is required to prove that such a quadrilateral is a parallelogram (that is, its other two sides are equal and parallel to each other).
Let the given quadrilateral be ABCD and its sides AB || CD and AB = CD.
By condition, we are given a quadrilateral. Nothing is said about whether it is convex or not (although only convex quadrilaterals can be parallelograms). However, even in a non-convex quadrilateral there is always one diagonal that divides it into two triangles. If this is a diagonal AC, then we get two triangles ABC and ADC. If this is the diagonal BD, then there will be ∆ABD and ∆BCD.
Let's say we get triangles ABC and ADC. They have one side in common (diagonal AC), side AB of one triangle is equal to side CD of the other (by condition), angle BAC is equal to angle ACD (as lying crosswise between the transversal and parallel lines). This means ∆ABC = ∆ADC on two sides and the angle between them.
From the equality of triangles it follows that their other sides and angles are respectively equal. But side BC of triangle ABC corresponds to side AD of triangle ADC, which means BC = AD. Angle B corresponds to angle D, which means ∠B = ∠D. These angles can be equal to each other if BC || AD (since AB || CD, these lines can be combined by parallel translation, then ∠B will become cross-lying ∠D, and their equality can only be with BC || AD).
By definition, a parallelogram is a quadrilateral whose opposite sides are equal and parallel to each other.
Thus, it was proven that if a quadrilateral ABCD has sides AB and CD equal and parallel and the diagonal AC divides it into two triangles, then its other pair of sides turns out to be equal to each other and parallel.
If quadrilateral ABCD were divided into two triangles by another diagonal (BD), then triangles ABD and BCD would be considered. Their equality would be proved similarly to the previous one. It would turn out that BC = AD and ∠A = ∠C, which would imply that BC || A.D.
Sign-ki pa-ral-le-lo-gram-ma
1. Definition and basic properties of a parallelogram
Let's start by recalling the definition of para-ral-le-lo-gram.
Definition. Parallelogram- what-you-rekh-gon-nick, which has every two pro-ti-false sides that are parallel (see Fig. . 1).
Rice. 1. Pa-ral-le-lo-gram
Let's remember basic properties of pa-ral-le-lo-gram-ma:
In order to be able to use all these properties, you need to be sure that the fi-gu-ra, about someone -roy we are talking about, - par-ral-le-lo-gram. To do this, it is necessary to know such facts as signs of pa-ral-le-lo-gram-ma. We are looking at the first two of them this year.
2. The first sign of a parallelogram
Theorem. The first sign of pa-ral-le-lo-gram-ma. If in a four-coal the two opposite sides are equal and parallel, then this four-coal nickname - parallelogram. 
.
Rice. 2. The first sign of pa-ral-le-lo-gram-ma
Proof. Let's put the dia-go-nal in the four-reh-coal-ni-ka (see Fig. 2), she split it into two tri-coal-ni-ka. Let's write down what we know about these triangles:
according to the first sign of the equality of triangles.
From the equality of the indicated triangles it follows that, by the sign of parallelism of straight lines when crossing ch-nii their s-ku-shchi. We have that:
Do-ka-za-but.
3. Second sign of a parallelogram
Theorem. The second sign is pa-ral-le-lo-gram-ma. If in a four-corner every two pro-ti-false sides are equal, then this four-corner is parallelogram. 
.
Rice. 3. The second sign of pa-ral-le-lo-gram-ma
Proof. We put the dia-go-nal into the four-corner (see Fig. 3), she splits it into two triangles. Let's write down what we know about these triangles, based on the theory's form:
according to the third sign of the equality of triangles.
From the equality of triangles it follows that, by the sign of parallel lines, when intersecting them s-ku-shchey. Let's eat:
par-ral-le-lo-gram by definition. Q.E.D.
Do-ka-za-but.
4. An example of using the first parallelogram feature
Let's look at an example of the use of signs of pa-ral-le-lo-gram.
Example 1. In the bulge there are no coals Find: a) the corners of the coals; b) hundred-ro-well.
Solution. Illustration Fig. 4.
pa-ral-le-lo-gram according to the first sign of pa-ral-le-lo-gram-ma.
A. 
by the property of a par-ral-le-lo-gram about pro-ti-false angles, by the property of a par-ral-le-lo-gram about the sum of angles, when lying to one side.
B. 
by the nature of equality of pro-false sides.
re-tiy sign pa-ral-le-lo-gram-ma
5. Review: Definition and Properties of a Parallelogram
Let's remember that parallelogram- this is a four-square-corner, which has pro-ti-false sides in pairs. That is, if - par-ral-le-lo-gram, then 
(see Fig. 1).
The parallel-le-lo-gram has a number of properties: the opposite angles are equal (), the opposite angles -we are equal ( 
). In addition, the dia-go-na-li pa-ral-le-lo-gram-ma at the point of re-se-che-niya is divided according to the sum of the angles, at-le- pressing to any side pa-ral-le-lo-gram-ma, equal, etc.
But in order to take advantage of all these properties, it is necessary to be absolutely sure that the ri-va-e-my th-you-rekh-coal-nick - pa-ral-le-lo-gram. For this purpose, there are signs of par-ral-le-lo-gram: that is, those facts from which one can draw a single-valued conclusion , that what-you-rekh-coal-nick is a par-ral-le-lo-gram-mom. In the previous lesson, we already looked at two signs. Now we're looking at the third time.
6. The third sign of a parallelogram and its proof
If in a four-coal there is a dia-go-on at the point of re-se-che-niya they do-by-lams, then the given four-you Roh-coal-nick is a pa-ral-le-lo-gram-mom.
Given:
What-you-re-coal-nick; ; .
Prove:
Parallelogram.
Proof:
In order to prove this fact, it is necessary to show the parallelism of the parties to the par-le-lo-gram. And the parallelism of straight lines most often appears through the equality of internal cross-lying angles at these right angles. Thus, here is the next method to obtain the third sign of par-ral -le-lo-gram-ma: through the equality of triangles 
.
Let's see how these triangles are equal. Indeed, from the condition it follows: . In addition, since the angles are vertical, they are equal. That is:
(first sign of equalitytri-coal-ni-cov- along two sides and the corner between them).
From the equality of triangles: (since the internal crosswise angles at these straight lines and separators are equal). In addition, from the equality of triangles it follows that . This means that we understand that in four-coal two hundred are equal and parallel. According to the first sign, pa-ral-le-lo-gram-ma: - pa-ral-le-lo-gram.
Do-ka-za-but.
7. Example of a problem on the third sign of a parallelogram and generalization
Let's look at the example of using the third sign of pa-ral-le-lo-gram.
Example 1
Given:
- parallelogram; . - se-re-di-na, - se-re-di-na, - se-re-di-na, - se-re-di-na (see Fig. 2).
Prove:- pa-ral-le-lo-gram.
Proof:
This means that in the four-coal-no-dia-go-on-whether at the point of re-se-che-niya they do-by-lam. By the third sign of pa-ral-le-lo-gram, it follows from this that - pa-ral-le-lo-gram.
Do-ka-za-but.
If you analyze the third sign of pa-ral-le-lo-gram, then you can notice that this sign is with-vet- has the property of a par-ral-le-lo-gram. That is, the fact that the dia-go-na-li de-la-xia is not just a property of the par-le-lo-gram, and its distinctive, kha-rak-te-ri-sti-che-property, by which it can be distinguished from the set what-you-rekh-coal-ni-cov.
SOURCE
http://interneturok.ru/ru/school/geometry/8-klass/chyotyrehugolniki/priznaki-parallelogramma
http://interneturok.ru/ru/school/geometry/8-klass/chyotyrehugolniki/tretiy-priznak-parallelogramma
http://www.uchportfolio.ru/users_content/675f9820626f5bc0afb47b57890b466e/images/46TThxQ8j4Y.jpg
http://cs10002.vk.me/u31195134/116260458/x_56d40dd3.jpg
http://wwww.tepka.ru/geometriya/16.1.gif
This is a quadrilateral whose opposite sides are parallel in pairs.
Property 1. Any diagonal of a parallelogram divides it into two equal triangles.
Proof . According to the II characteristic (crosswise angles and common side).
The theorem is proven.
Property 2. In a parallelogram opposing sides are equal, opposite angles are equal.
Proof .
Likewise,
The theorem is proven.
Property 3. In a parallelogram, the diagonals are bisected by the point of intersection.
Proof .
The theorem is proven.
Property 4. The angle bisector of a parallelogram, crossing the opposite side, divides it into an isosceles triangle and a trapezoid. (Ch. words - vertex - two isosceles? -ka).
Proof .
The theorem is proven.
Property 5. In a parallelogram, a line segment with ends on opposite sides passing through the point of intersection of the diagonals is bisected by this point.
Proof .
The theorem is proven.
Property 6. The angle between the altitudes dropped from the vertex of an obtuse angle of a parallelogram is equal to an acute angle of a parallelogram.
Proof .
The theorem is proven.
Property 7. The sum of the angles of a parallelogram adjacent to one side is 180°.
Proof .
The theorem is proven.
Constructing the bisector of an angle. Properties of the angle bisector of a triangle.
1) Construct an arbitrary ray DE.
2) On a given ray, construct an arbitrary circle with a center at the vertex and the same
with the center at the beginning of the constructed ray.
3) F and G - points of intersection of the circle with the sides of a given angle, H - point of intersection of the circle with the constructed ray
Construct a circle with center at point H and radius equal to FG.
5) I is the point of intersection of the circles of the constructed beam.
6) Draw a straight line through the vertex and I.
IDH is the required angle.
)
Property 1. The bisector of an angle of a triangle divides the opposite side in proportion to the adjacent sides.
Proof . Let x, y be segments of side c. Let's continue the beam BC. On ray BC we plot from C a segment CK equal to AC.
