What is the sum of one-sided angles in a parallelogram. Parallelogram and its properties

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.

The concept of a parallelogram

Definition 1

Parallelogram is a quadrilateral in which opposite sides are parallel to each other (Fig. 1).

Picture 1.

A parallelogram has two main properties. Let's consider them without proof.

Property 1: Opposite sides and angles of a parallelogram are equal, respectively, to each other.

Property 2: Diagonals drawn in a parallelogram are bisected by their intersection point.

Parallelogram features

Consider three features of a parallelogram and present them in the form of theorems.

Theorem 1

If two sides of a quadrilateral are equal to each other and also parallel, then this quadrilateral will be a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. In which $AB||CD$ and $AB=CD$ Let us draw a diagonal $AC$ in it (Fig. 2).

Figure 2.

Consider parallel lines $AB$ and $CD$ and their secant $AC$. Then

\[\angle CAB=\angle DCA\]

like crosswise corners.

According to the $I$ criterion for the equality of triangles,

since $AC$ is their common side, and $AB=CD$ by assumption. Means

\[\angle DAC=\angle ACB\]

Consider the lines $AD$ and $CB$ and their secant $AC$; by the last equality of the cross-lying angles, we obtain that $AD||CB$.) Therefore, by the definition of $1$, this quadrilateral is a parallelogram.

The theorem has been proven.

Theorem 2

If opposite sides of a quadrilateral are equal, then it is a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. In which $AD=BC$ and $AB=CD$. Let us draw a diagonal $AC$ in it (Fig. 3).

Figure 3

Since $AD=BC$, $AB=CD$, and $AC$ is a common side, then by the $III$ triangle equality test,

\[\triangle DAC=\triangle ACB\]

\[\angle DAC=\angle ACB\]

Consider the lines $AD$ and $CB$ and their secant $AC$, by the last equality of the cross-lying angles we get that $AD||CB$. Therefore, by the definition of $1$, this quadrilateral is a parallelogram.

\[\angle DCA=\angle CAB\]

Consider the lines $AB$ and $CD$ and their secant $AC$, by the last equality of the cross-lying angles we get that $AB||CD$. Therefore, by Definition 1, this quadrilateral is a parallelogram.

The theorem has been proven.

Theorem 3

If the diagonals drawn in a quadrilateral are divided into two equal parts by their intersection point, then this quadrilateral is a parallelogram.

Proof.

Let us be given a quadrilateral $ABCD$. Let us draw the diagonals $AC$ and $BD$ in it. Let them intersect at the point $O$ (Fig. 4).

Figure 4

Since, by the condition $BO=OD,\ AO=OC$, and the angles $\angle COB=\angle DOA$ are vertical, then, by the $I$ triangle equality test,

\[\triangle BOC=\triangle AOD\]

\[\angle DBC=\angle BDA\]

Consider the lines $BC$ and $AD$ and their secant $BD$, by the last equality of the cross-lying angles we get that $BC||AD$. Also $BC=AD$. Therefore, by Theorem $1$, this quadrilateral is a parallelogram.

Definition

Parallelogram is called a quadrilateral whose opposite sides are pairwise parallel.

Figure 1 shows a parallelogram $A B C D, A B\|C D, B C\| A D$.

Parallelogram Properties

1. In a parallelogram, opposite sides are equal: $A B=C D, B C=A D$ (Fig. 1).
2. In a parallelogram, opposite angles are equal to $\angle A=\angle C, \angle B=\angle D$ (Fig. 1).
3. The diagonals of the parallelogram at the point of intersection are halved $A O=O C, B O=O D$ (Fig. 1).
4. The diagonal of a parallelogram divides it into two equal triangles.

The sum of the angles of a parallelogram adjacent to one side is $180^(\circ)$:

$$\angle A+\angle B=180^(\circ), \angle B+\angle C=180^(\circ)$$

$$\angle C+\angle D=180^(\circ), \angle D+\angle A=180^(\circ)$$

The diagonals and sides of a parallelogram are related by the following relationship:

$$d_(1)^(2)+d_(2)^(2)=2 a^(2)+2 b^(2)$$

5. In a parallelogram, the angle between the heights is equal to its acute angle: $\angle K B H=\angle A$.
6. Bisectors of angles adjacent to one side of a parallelogram are mutually perpendicular.
7. Bisectors of two opposite angles of a parallelogram are parallel.

Parallelogram features

Quadrilateral $ABCD$ is a parallelogram if

1. $A B=C D$ and $A B \| C D$
2. $A B=C D$ and $B C=A D$
3. $A O=O C$ and $B O=O D$
4. $\angle A=\angle C$ and $\angle B=\angle D$

The area of ​​a parallelogram can be calculated using one of the following formulas:

$S=a \cdot h_(a), \quad S=b \cdot h_(b)$

$S=a \cdot b \cdot \sin \alpha, \quad S=\frac(1)(2) d_(1) \cdot d_(2) \cdot \sin \phi$

Examples of problem solving

Example

Exercise. The sum of two angles of a parallelogram is $140^(\circ)$. Find the largest angle of the parallelogram.

Solution. In a parallelogram, opposite angles are equal. Let's denote the larger angle of the parallelogram $\alpha$, and the smaller angle $\beta$. The sum of the angles $\alpha$ and $\beta$ is $180^(\circ)$, so the given sum of $140^(\circ)$ is the sum of two opposite angles, then $140^(\circ) : 2=70 ^(\circ)$. Thus the smaller angle is $\beta=70^(\circ)$. We find the larger angle $\alpha$ from the relation:

$\alpha+\beta=180^(\circ) \Rightarrow \alpha=180^(\circ)-\beta \Rightarrow$

$\Rightarrow \alpha=180^(\circ)-70^(\circ) \Rightarrow \alpha=110^(\circ)$

Answer.$\alpha=110^(\circ)$

Example

Exercise. The sides of the parallelogram are 18 cm and 15 cm, and the height drawn to the smaller side is 6 cm. Find another height of the parallelogram.

Solution. Let's make a drawing (Fig. 2)

By condition, $a=15$ cm, $b=18$ cm, $h_(a)=6$ cm. For a parallelogram, the following formulas are valid for finding the area:

$$S=a \cdot h_(a), \quad S=b \cdot h_(b)$$

Equate the right parts of these equalities, and express, from the resulting equality, $h_(b) $:

$$a \cdot h_(a)=b \cdot h_(b) \Rightarrow h_(b)=\frac(a \cdot h_(a))(b)$$

Substituting the initial data of the problem, we finally get:

$h_(b)=\frac(15 \cdot 6)(18) \Rightarrow h_(b)=5$ (cm)

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.

A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The area of ​​a parallelogram is equal to the product of its base (a) and its height (h). You can also find its area through two sides and an angle and through the diagonals.

Parallelogram Properties

1. Opposite sides are identical

First of all, draw the diagonal \(AC \) . 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, (on the second basis: and \(AC\) is common).

And, therefore, \(\triangle ABC = \triangle ADC \), then \(AB = CD \) and \(AD = BC \) .

2. Opposite angles are identical

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: \(\angle 1 + \angle 3 = \angle 2 + \angle 4 \). Given that \(\triangle ABC = \triangle ADC \) we get \(\angle A = \angle C \) , \(\angle B = \angle D \) .

3. Diagonals are bisected by the intersection point

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

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

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 sign of a parallelogram 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

\(AB = CD \) ; \(AB || CD \Rightarrow ABCD \)- parallelogram.

Let's consider in more detail. Why \(AD || BC \) ?

\(\triangle ABC = \triangle ADC \) on property 1: \(AB = CD \) , \(\angle 1 = \angle 2 \) as crosswise with parallel \(AB \) and \(CD \) and secant \(AC \) .

But if \(\triangle ABC = \triangle ADC \), then \(\angle 3 = \angle 4 \) (they lie opposite \(AD || BC \) (\(\angle 3 \) and \(\angle 4 \) - lying opposite 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.

Let's consider this feature. 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.

\(2 \alpha + 2 \beta = 360^(\circ) \)(because \(\angle A = \angle C \) , \(\angle B = \angle D \) by definition).

It turns out, . But \(\alpha \) and \(\beta \) are internal one-sided at the secant \(AB \) .

And what \(\alpha + \beta = 180^(\circ) \) says also that \(AD || BC \) .