A parallelogram is a quadrilateral whose opposite sides are parallel, that is, they lie on parallel lines (Fig. 1).
Theorem 1. On the properties of sides and angles of a parallelogram. In a parallelogram, opposite sides are equal, opposite angles are equal, and the sum of the angles adjacent to one side of the parallelogram is 180°.
Proof. In this parallelogram ABCD, draw a diagonal AC and get two triangles ABC and ADC (Fig. 2).
These triangles are equal, since ∠ 1 = ∠ 4, ∠ 2 = ∠ 3 (cross-lying angles at parallel lines), and side AC is common. From the equality Δ ABC = Δ ADC it follows that AB = CD, BC = AD, ∠ B = ∠ D. The sum of the angles adjacent to one side, for example, angles A and D, is equal to 180 ° as one-sided with parallel lines. The theorem has been proven.
Comment. The equality of the opposite sides of a parallelogram means that the segments of the parallel ones cut off by the parallel ones are equal.
Corollary 1. If two lines are parallel, then all points of one line are at the same distance from the other line.
Proof. Indeed, let a || b (Fig. 3).
Let us draw from some two points B and C of the line b the perpendiculars BA and CD to the line a. Since AB || CD, then the figure ABCD is a parallelogram, and therefore AB = CD.
The distance between two parallel lines is the distance from an arbitrary point on one of the lines to the other line.
By what has been proved, it is equal to the length of the perpendicular drawn from some point of one of the parallel lines to the other line.
Example 1 The perimeter of the parallelogram is 122 cm. One of its sides is 25 cm longer than the other. Find the sides of the parallelogram.
Solution. By Theorem 1, opposite sides of a parallelogram are equal. Let's denote one side of the parallelogram as x, the other as y. Then by condition $$\left\(\begin(matrix) 2x + 2y = 122 \\x - y = 25 \end(matrix)\right.$$ Solving this system, we get x = 43, y = 18. Thus Thus, the sides of the parallelogram are 18, 43, 18 and 43 cm.
Example 2
Solution. Let figure 4 correspond to the condition of the problem.
Denote AB by x and BC by y. By condition, the perimeter of the parallelogram is 10 cm, i.e. 2(x + y) = 10, or x + y = 5. The perimeter of the triangle ABD is 8 cm. And since AB + AD = x + y = 5, then BD = 8 - 5 = 3 . So BD = 3 cm.
Example 3 Find the angles of the parallelogram, knowing that one of them is 50° greater than the other.
Solution. Let figure 5 correspond to the condition of the problem.
Let us denote the degree measure of angle A as x. Then the degree measure of the angle D is x + 50°.
Angles BAD and ADC are internal one-sided with parallel lines AB and DC and secant AD. Then the sum of these named angles will be 180°, i.e.
x + x + 50° = 180°, or x = 65°. Thus, ∠ A = ∠ C = 65°, a ∠ B = ∠ D = 115°.
Example 4 The sides of the parallelogram are 4.5 dm and 1.2 dm. A bisector is drawn from the vertex of an acute angle. What parts does it divide the long side of the parallelogram into?
Solution. Let figure 6 correspond to the condition of the problem.
AE is the bisector of the acute angle of the parallelogram. Therefore, ∠ 1 = ∠ 2.
As in Euclidean geometry, the point and the straight line are the main elements of the theory of planes, so the parallelogram is one of the key figures of convex quadrilaterals. From it, like threads from a ball, flow the concepts of "rectangle", "square", "rhombus" and other geometric quantities.
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Definition of a parallelogram
convex quadrilateral, consisting of segments, each pair of which is parallel, is known in geometry as a parallelogram.
What a classic parallelogram looks like is a quadrilateral ABCD. The sides are called the bases (AB, BC, CD and AD), the perpendicular drawn from any vertex to the opposite side of this vertex is called the height (BE and BF), the lines AC and BD are the diagonals.
Attention! Square, rhombus and rectangle are special cases of parallelogram.
Sides and angles: ratio features
Key properties, by and large, predetermined by the designation itself, they are proved by the theorem. These characteristics are as follows:
- Sides that are opposite are identical in pairs.
- Angles that are opposite to each other are equal in pairs.
Proof: consider ∆ABC and ∆ADC, which are obtained by dividing quadrilateral ABCD by line AC. ∠BCA=∠CAD and ∠BAC=∠ACD, since AC is common to them (vertical angles for BC||AD and AB||CD, respectively). It follows from this: ∆ABC = ∆ADC (the second criterion for the equality of triangles).
Segments AB and BC in ∆ABC correspond in pairs to lines CD and AD in ∆ADC, which means that they are identical: AB = CD, BC = AD. Thus, ∠B corresponds to ∠D and they are equal. Since ∠A=∠BAC+∠CAD, ∠C=∠BCA+∠ACD, which are also identical in pairs, then ∠A = ∠C. The property has been proven.
Characteristics of the figure's diagonals
Main feature these parallelogram lines: the point of intersection bisects them.
Proof: let m. E be the intersection point of the diagonals AC and BD of the figure ABCD. They form two commensurate triangles - ∆ABE and ∆CDE.
AB=CD since they are opposite. According to lines and secants, ∠ABE = ∠CDE and ∠BAE = ∠DCE.
According to the second sign of equality, ∆ABE = ∆CDE. This means that the elements ∆ABE and ∆CDE are: AE = CE, BE = DE and, moreover, they are commensurate parts of AC and BD. The property has been proven.
Features of adjacent corners
At adjacent sides, the sum of the angles is 180°, since they lie on the same side of the parallel lines and the secant. For quadrilateral ABCD:
∠A+∠B=∠C+∠D=∠A+∠D=∠B+∠C=180º
Bisector properties:
- , dropped to one side, are perpendicular;
- opposite vertices have parallel bisectors;
- the triangle obtained by drawing the bisector will be isosceles.
Determining the characteristic features of a parallelogram by the theorem
The features of this figure follow from its main theorem, which reads as follows: quadrilateral is considered a parallelogram in the event that its diagonals intersect, and this point divides them into equal segments.
Proof: Let lines AC and BD of quadrilateral ABCD intersect in t. E. Since ∠AED = ∠BEC, and AE+CE=AC BE+DE=BD, then ∆AED = ∆BEC (by the first sign of equality of triangles). That is, ∠EAD = ∠ECB. They are also the interior crossing angles of the secant AC for lines AD and BC. Thus, by definition of parallelism - AD || BC. A similar property of the lines BC and CD is also derived. The theorem has been proven.
Calculating the area of a figure
The area of this figure found in several ways one of the simplest: multiplying the height and the base to which it is drawn.
Proof: Draw perpendiculars BE and CF from vertices B and C. ∆ABE and ∆DCF are equal since AB = CD and BE = CF. ABCD is equal to the rectangle EBCF, since they also consist of proportionate figures: S ABE and S EBCD, as well as S DCF and S EBCD. It follows from this that the area of this geometric figure is located in the same way as a rectangle:
S ABCD = S EBCF = BE×BC=BE×AD.
For determining general formula the area of the parallelogram, denote the height as hb, and the side b. Respectively:
Other ways to find area
Area calculations through the sides of the parallelogram and the angle, which they form, is the second known method.
,
Spr-ma - area;
a and b are its sides
α - angle between segments a and b.
This method is practically based on the first, but in case it is unknown. always cuts off a right triangle whose parameters are found by trigonometric identities, i.e. . Transforming the ratio, we get . In the equation of the first method, we replace the height with this product and obtain a proof of the validity of this formula.
Through the diagonals of a parallelogram and an angle, which they create when they intersect, you can also find the area.
Proof: AC and BD intersecting form four triangles: ABE, BEC, CDE and AED. Their sum is equal to the area of this quadrilateral.
The area of each of these ∆ can be found from the expression , where a=BE, b=AE, ∠γ =∠AEB. Since , then a single value of the sine is used in the calculations. That is . Since AE+CE=AC= d 1 and BE+DE=BD= d 2 , the area formula reduces to:
.
Application in vector algebra
The features of the constituent parts of this quadrangle have found application in vector algebra, namely: the addition of two vectors. The parallelogram rule states that if given vectorsandnotare collinear, then their sum will be equal to the diagonal of this figure, the bases of which correspond to these vectors.
Proof: from an arbitrarily chosen beginning - that is. - we build vectors and . Next, we build a parallelogram OASV, where the segments OA and OB are sides. Thus, the OS lies on the vector or sum.
Formulas for calculating the parameters of a parallelogram
The identities are given under the following conditions:
- a and b, α - sides and the angle between them;
- d 1 and d 2 , γ - diagonals and at the point of their intersection;
- h a and h b - heights lowered to sides a and b;
| Parameter | Formula |
| Finding sides | |
| along the diagonals and the cosine of the angle between them | 
|
| diagonally and sideways | 
|
| through height and opposite vertex | |
| Finding the length of the diagonals | |
| on the sides and the size of the top between them | |
| along the sides and one of the diagonals | 
ConclusionThe parallelogram, as one of the key figures of geometry, is used in life, for example, in construction when calculating the area of \u200b\u200bthe site or other measurements. Therefore, knowledge about hallmarks and ways to calculate its various parameters can come in handy at any time in life. |
Definition
A parallelogram is a quadrilateral whose opposite sides are pairwise parallel.
Theorem (the first sign of a parallelogram)
If two sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.
Proof
Let the sides \(AB\) and \(CD\) of the quadrilateral \(ABCD\) be parallel and \(AB = CD\) .
Draw a diagonal \(AC\) dividing the given quadrilateral into two equal triangles: \(ABC\) and \(CDA\) . These triangles are equal in two sides and the angle between them (\(AC\) is a common side, \(AB = CD\) by condition, \(\angle 1 = \angle 2\) as the crosswise angles at the intersection of parallel lines \ (AB\) and \(CD\) secant \(AC\) ), so \(\angle 3 = \angle 4\) . But the angles \(3\) and \(4\) are crosswise lying at the intersection of the lines \(AD\) and \(BC\) of the secant \(AC\) , therefore, \(AD\parallel BC\) . Thus, in the quadrilateral \(ABCD\) the opposite sides are pairwise parallel, and hence the quadrilateral \(ABCD\) is a parallelogram.
Theorem (second feature of a parallelogram)
If the opposite sides of a quadrilateral are equal in pairs, then the quadrilateral is a parallelogram.
Proof
Draw a diagonal \(AC\) of the given quadrilateral \(ABCD\) dividing it into triangles \(ABC\) and \(CDA\) .
These triangles are equal in three sides (\(AC\) is common, \(AB = CD\) and \(BC = DA\) by assumption), so \(\angle 1 = \angle 2\) are crosswise lying at \(AB\) and \(CD\) and the secant \(AC\) . It follows that \(AB\parallel CD\) . Since \(AB = CD\) and \(AB\parallel CD\) , then by the first criterion of a parallelogram, the quadrilateral \(ABCD\) is a parallelogram.
Theorem (the third sign of a parallelogram)
If in a quadrilateral the diagonals intersect and the intersection point is bisected, then this quadrilateral is a parallelogram.
Proof
Consider a quadrilateral \(ABCD\) in which the diagonals \(AC\) and \(BD\) intersect at the point \(O\) and bisect this point.
Triangles \(AOB\) and \(COD\) are equal by the first criterion of equality of triangles (\(AO = OC\) , \(BO = OD\) by condition, \(\angle AOB = \angle COD\) as vertical corners), so \(AB = CD\) and \(\angle 1 = \angle 2\) . From the equality of the angles \(1\) and \(2\) (cross-lying at \(AB\) and \(CD\) and the secant \(AC\) ) it follows that \(AB\parallel CD\) .
So, in the quadrilateral \(ABCD\), the sides \(AB\) and \(CD\) are equal and parallel, which means that, by the first sign of a parallelogram, the quadrilateral \(ABCD\) is a parallelogram.
Parallelogram properties:
1. In a parallelogram, opposite sides are equal and opposite angles are equal.
2. The diagonals of the parallelogram are bisected by the intersection point.
Properties of the bisector of a parallelogram:
1. The bisector of a parallelogram cuts off an isosceles triangle from it.
2. Bisectors of adjacent angles of a parallelogram intersect at a right angle.
3. Bisector segments of opposite angles are equal and parallel.
Proof
1) Let \(ABCD\) be a parallelogram, \(AE\) be the bisector of the angle \(BAD\) .
The angles \(1\) and \(2\) are equal as they lie across the parallel lines \(AD\) and \(BC\) and the secant \(AE\) . The angles \(1\) and \(3\) are equal because \(AE\) is a bisector. Eventually \(\angle 3 = \angle 1 = \angle 2\), whence it follows that the triangle \(ABE\) is isosceles.
2) Let \(ABCD\) be a parallelogram, \(AN\) and \(BM\) be the bisectors of the angles \(BAD\) and \(ABC\), respectively.
Since the sum of one-sided angles at parallel lines and a secant is \(180^(\circ)\) , then \(\angle DAB + \angle ABC = 180^(\circ)\).
Since \(AN\) and \(BM\) are bisectors, then \(\angle BAN + \angle ABM = 0.5(\angle DAB + \angle ABC) = 0.5\cdot 180^\circ = 90^(\circ)\), where \(\angle AOB = 180^\circ - (\angle BAN + \angle ABM) = 90^\circ\).
3. Let \(AN\) and \(CM\) be the angle bisectors of the parallelogram \(ABCD\) .
Since opposite angles in a parallelogram are equal, \(\angle 2 = 0.5\cdot\angle BAD = 0.5\cdot\angle BCD = \angle 1\). In addition, the angles \(1\) and \(3\) are equal as if they lie across parallel lines \(AD\) and \(BC\) and the secant \(CM\) , then \(\angle 2 = \angle 3\) , which implies that \(AN\parallel CM\) . Also, \(AM\parallel CN\) , then \(ANCM\) is a parallelogram, hence \(AN = CM\) .
When solving problems on this topic, in addition to basic properties parallelogram and the corresponding formulas, you can remember and apply the following:
- The bisector of the interior angle of a parallelogram cuts off an isosceles triangle from it
- Bisectors of internal angles adjacent to one of the sides of a parallelogram are mutually perpendicular
- Bisectors coming from opposite internal angles of a parallelogram, parallel to each other or lie on one straight line
- The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides
- The area of a parallelogram is half the product of the diagonals times the sine of the angle between them.
Let's consider the tasks in the solution of which these properties are used.
Task 1.
The bisector of angle C of parallelogram ABCD intersects side AD at point M and the continuation of side AB beyond point A at point E. Find the perimeter of the parallelogram if AE \u003d 4, DM \u003d 3.
Solution.
1. Triangle CMD isosceles. (Property 1). Therefore, CD = MD = 3 cm.
2. Triangle EAM is isosceles.
Therefore, AE = AM = 4 cm.
3. AD = AM + MD = 7 cm.
4. Perimeter ABCD = 20 cm.
Answer. 20 cm
Task 2.
Diagonals are drawn in a convex quadrilateral ABCD. It is known that the areas of triangles ABD, ACD, BCD are equal. Prove that the given quadrilateral is a parallelogram.
Solution.
1. Let BE be the height of triangle ABD, CF be the height of triangle ACD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base AD, then the heights of these triangles are equal. BE = CF.
2. BE, CF are perpendicular to AD. Points B and C are located on the same side of the line AD. BE = CF. Therefore, the line BC || AD. (*)
3. Let AL be the altitude of triangle ACD, BK the altitude of triangle BCD. Since, according to the condition of the problem, the areas of the triangles are equal and they have a common base CD, then the heights of these triangles are equal. AL = BK.
4. AL and BK are perpendicular to CD. Points B and A are located on the same side of the straight line CD. AL = BK. Therefore, the line AB || CD (**)
5. Conditions (*), (**) imply that ABCD is a parallelogram.
Answer. Proven. ABCD is a parallelogram.
Task 3.
On the sides BC and CD of the parallelogram ABCD, the points M and H are marked, respectively, so that the segments BM and HD intersect at the point O;<ВМD = 95 о,
Solution.
1. In the triangle DOM<МОD = 25 о (Он смежный с <ВОD = 155 о); <ОМD = 95 о. Тогда <ОDМ = 60 о.
2. In a right triangle DHC Then<НСD = 30 о. СD: НD = 2: 1 But CD = AB. Then AB: HD = 2: 1. 3. <С = 30 о, 4. <А = <С = 30 о, <В = Answer: AB: HD = 2: 1,<А = <С = 30 о, <В = Task 4. One of the diagonals of a parallelogram of length 4√6 makes an angle of 60° with the base, and the second diagonal makes an angle of 45° with the same base. Find the second diagonal. Solution.
1. AO = 2√6. 2. Apply the sine theorem to the triangle AOD. AO/sin D = OD/sin A. 2√6/sin 45 o = OD/sin 60 o. OD = (2√6sin 60 o) / sin 45 o = (2√6 √3/2) / (√2/2) = 2√18/√2 = 6. Answer: 12.
Task 5. For a parallelogram with sides 5√2 and 7√2, the smaller angle between the diagonals is equal to the smaller angle of the parallelogram. Find the sum of the lengths of the diagonals. Solution.
Let d 1, d 2 be the diagonals of the parallelogram, and the angle between the diagonals and the smaller angle of the parallelogram be φ. 1. Let's count two different S ABCD \u003d AB AD sin A \u003d 5√2 7√2 sin f, S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin f. We obtain the equality 5√2 7√2 sin f = 1/2d 1 d 2 sin f or 2 5√2 7√2 = d 1 d 2 ; 2. Using the ratio between the sides and diagonals of the parallelogram, we write the equality (AB 2 + AD 2) 2 = AC 2 + BD 2. ((5√2) 2 + (7√2) 2) 2 = d 1 2 + d 2 2 . d 1 2 + d 2 2 = 296. 3. Let's make a system: (d 1 2 + d 2 2 = 296, Multiply the second equation of the system by 2 and add it to the first. We get (d 1 + d 2) 2 = 576. Hence Id 1 + d 2 I = 24. Since d 1, d 2 are the lengths of the diagonals of the parallelogram, then d 1 + d 2 = 24. Answer: 24.
Task 6. The sides of the parallelogram are 4 and 6. The acute angle between the diagonals is 45 o. Find the area of the parallelogram. Solution.
1. From the triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals. AB 2 \u003d AO 2 + VO 2 2 AO VO cos AOB. 4 2 \u003d (d 1 / 2) 2 + (d 2 / 2) 2 - 2 (d 1 / 2) (d 2 / 2) cos 45 o; d 1 2/4 + d 2 2/4 - 2 (d 1/2) (d 2/2)√2/2 = 16. d 1 2 + d 2 2 - d 1 d 2 √2 = 64. 2. Similarly, we write the relation for the triangle AOD. We take into account that<АОD = 135 о и cos 135 о = -cos 45 о = -√2/2. We get the equation d 1 2 + d 2 2 + d 1 d 2 √2 = 144. 3. We have a system Subtracting the first from the second equation, we get 2d 1 d 2 √2 = 80 or d 1 d 2 = 80/(2√2) = 20√2 4. S ABCD \u003d 1/2 AC BD sin AOB \u003d 1/2 d 1 d 2 sin α \u003d 1/2 20√2 √2/2 \u003d 10. Note: In this and in the previous problem, there is no need to solve the system completely, foreseeing that in this problem we need the product of diagonals to calculate the area. Answer: 10. Task 7. The area of the parallelogram is 96 and its sides are 8 and 15. Find the square of the smaller diagonal. Solution.
1. S ABCD \u003d AB AD sin VAD. Let's do a substitution in the formula. We get 96 = 8 15 sin VAD. Hence sin VAD = 4/5. 2. Find cos BAD. sin 2 VAD + cos 2 VAD = 1. (4/5) 2 + cos 2 BAD = 1. cos 2 BAD = 9/25. According to the condition of the problem, we find the length of the smaller diagonal. Diagonal BD will be smaller if angle BAD is acute. Then cos BAD = 3 / 5. 3. From the triangle ABD, using the cosine theorem, we find the square of the diagonal BD. BD 2 \u003d AB 2 + AD 2 - 2 AB BD cos BAD. ВD 2 \u003d 8 2 + 15 2 - 2 8 15 3 / 5 \u003d 145. Answer: 145.
Do you have any questions? Don't know how to solve a geometry problem? site, with full or partial copying of the material, a link to the source is required. A parallelogram is a quadrilateral whose opposite sides are parallel in pairs (Fig. 233). An arbitrary parallelogram has the following properties: 1. Opposite sides of a parallelogram are equal. Proof. Draw a diagonal AC in parallelogram ABCD. Triangles ACD and AC B are equal as having a common side AC and two pairs of equal angles adjacent to it: (as cross-lying angles with parallel lines AD and BC). Hence, and as sides of equal triangles lying opposite equal angles, which was required to be proved. 2. Opposite angles of a parallelogram are: 3. Neighboring angles of a parallelogram, that is, angles adjacent to one side, add up, etc. The proof of properties 2 and 3 immediately follows from the properties of angles at parallel lines. 4. The diagonals of a parallelogram bisect each other at the point of their intersection. In other words, Proof. Triangles AOD and BOC are equal, since their sides AD and BC are equal (property 1) and the angles adjacent to them (as cross-lying angles with parallel lines). This implies the equality of the corresponding sides of these triangles: AO which was required to be proved. Each of these four properties characterizes the parallelogram, or, as they say, is its characteristic property, i.e., any quadrangle that has at least one of these properties is a parallelogram (and, therefore, has all the other three properties). We carry out the proof for each property separately. 1". If the opposite sides of a quadrilateral are pairwise equal, 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 the sides BC and AD follows from the equality of the angles CAD and DIA. 2. If a quadrilateral has two pairs of opposite angles equal, then it is a parallelogram. Proof. Let . Since both the sides AD and BC are parallel (on the basis of parallel lines). 3. We leave the formulation and proof to the reader. 4. 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 \u003d OS, BO \u003d 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 the 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 Г. Thus, in order to prove that a given quadrilateral is a parallelogram, it suffices to verify the validity of any of the four properties. The reader is invited to independently prove one more 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, while the other two are called lateral sides. The segment of a straight line perpendicular to two sides of a parallelogram, enclosed between them, is called the height of the parallelogram. The parallelogram in fig. 234 has a height h drawn to the sides AD and BC, its second height is represented by a segment .
(
(Since in a right triangle, the leg that lies opposite an angle of 30 o is equal to half the hypotenuse).
ways of its area.
(d 1 + d 2 = 140.
(d 1 2 + d 2 2 - d 1 d 2 √2 = 64,
(d 1 2 + d 2 2 + d 1 d 2 √2 = 144.
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