Parallelogram concept
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: The opposite sides and angles of a parallelogram are equal, respectively.
Property 2: Diagonals drawn in a parallelogram are bisected by their point of intersection.
Signs of a parallelogram
Let's consider three characteristics 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 criss-crossed corners.
According to the $I$ criterion of equality of triangles,
since $AC$ is their common side, and $AB=CD$ by condition. Means
\[\angle DAC=\angle ACB\]
Consider the lines $AD$ and $CB$ and their secant $AC$; by the last equality across the lying angles we obtain that $AD||CB$.) Consequently, by definition $1$, this quadrilateral is a parallelogram.
The theorem has been proven.
Theorem 2
If the opposite sides of a quadrilateral are equal to each other, 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 $III$ criterion for the equality of triangles,
\[\triangle DAC=\triangle ACB\]
\[\angle DAC=\angle ACB\]
Let us consider the lines $AD$ and $CB$ and their secant $AC$; by the last equality across the lying angles we obtain that $AD||CB$. Therefore, by definition $1$, this quadrilateral is a parallelogram.
\[\angle DCA=\angle CAB\]
Let us consider the lines $AB$ and $CD$ and their secant $AC$; by the last equality across the lying angles we obtain 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 point of intersection, then this quadrilateral is a parallelogram.
Proof.
Let us be given a quadrilateral $ABCD$. Let us draw diagonals $AC$ and $BD$ in it. Let them intersect at point $O$ (Fig. 4).
Figure 4.
Since, by condition, $BO=OD,\ AO=OC$, and the angles $\angle COB=\angle DOA$ are vertical, then, by the $I$ criterion for the equality of triangles,
\[\triangle BOC=\triangle AOD\]
\[\angle DBC=\angle BDA\]
Consider the lines $BC$ and $AD$ and their secant $BD$; by the last equality across the lying angles we obtain that $BC||AD$. Also $BC=AD$. Therefore, by Theorem $1$, this quadrilateral is a parallelogram.
A parallelogram is a quadrilateral whose opposite sides are parallel, i.e. lie on parallel lines
Properties of a parallelogram: 
Theorem 22.
Opposite sides of a parallelogram are equal.
Proof. In the parallelogram ABCD we draw a diagonal AC. Triangles ACD and ACB are equal, as having a common side AC and two pairs equal angles. adjacent to it: ∠ CAB=∠ ACD, ∠ ACB=∠ DAC (as crosswise angles with parallel lines AD and BC). This means that AB = CD and BC = AD, as the corresponding sides of equal triangles, etc. From the equality of these triangles it also follows that the corresponding angles of the triangles are equal:
Theorem 23.
The opposite angles of the parallelogram are equal: ∠ 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.
Adjacent angles of a parallelogram, i.e. angles adjacent to one side add up to 180 degrees.
This is so because they are interior one-sided angles.
Theorem 25.
The diagonals of a parallelogram bisect each other at their intersection point.
Proof. Consider triangles BOC and AOD. According to the first property AD=BC ∠ OAD=∠ OCB and ∠ ODA=∠ OBC lying crosswise for parallel lines AD and BC. Therefore, triangles BOC and AOD are equal in side and adjacent angles. This means BO=OD and AO=OS, like the corresponding sides of equal triangles, etc.
Signs of a parallelogram
Theorem 26.
If the 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. Triangles ABC and ACD are equal on three sides. Then angles BAC and DCA are equal and, therefore, AB is parallel to CD. The parallelism of sides BC and AD follows from the equality of angles CAD and ACB.
Theorem 27.
If the opposite angles of a quadrilateral are equal in pairs, then it is a parallelogram.
Let ∠ A=∠ C and ∠ B=∠ D. Because ∠ A+∠ B+∠ C+∠ D=360 o, then ∠ A+∠ B=180 o and sides AD and BC are parallel (based on the parallelism of straight lines). We will also prove the parallelism of sides AB and CD and conclude that ABCD is a parallelogram by definition.
Theorem 28.
If adjacent corners of a quadrilateral, i.e. The 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 straight lines are parallel. So AB is parallel to CD and BC is parallel to AD. A quadrilateral turns out to be a parallelogram by definition.
Theorem 29.
If the diagonals of a quadrilateral bisect each other at the point of intersection, then the quadrilateral is a parallelogram.
Proof. If AO = OC, BO = OD, then triangles AOD and BOC are equal, as having equal (vertical) angles at 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 quadrilateral turns out to be a parallelogram according to criterion 1.
Theorem 30.
If a quadrilateral has a pair of equal, parallel sides, then it is a parallelogram.
Let the sides AB and CD of quadrilateral ABCD be parallel and equal. Let's draw diagonals AC and BD. From the parallelism of these lines it follows that the crosswise angles ABO = CDO and BAO = OCD are equal. Triangles ABO and CDO are equal in side and adjacent angles. Therefore AO=OS, VO=ОD, i.e. The diagonals are divided in half by the intersection point and the quadrilateral turns out to be a parallelogram according to criterion 4.
In geometry, special cases of parallelograms are considered.
When solving problems on this topic, except basic properties parallelogram and the corresponding formulas, you can remember and apply the following:
- The bisector of an interior angle of a parallelogram cuts off an isosceles triangle from it
- Bisectors of interior angles adjacent to one of the sides of a parallelogram are mutually perpendicular
- Bisectors coming from opposite interior corners of a parallelogram are parallel to each other or lie on the same 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 equal to half the product of the diagonals and the sine of the angle between them
Let us consider problems in 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 = 4, DM = 3.
Solution.
1. Triangle CMD is 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 this 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 conditions 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 relative to straight line AD. BE = CF. Therefore, straight line BC || A.D. (*)
3. Let AL be the altitude of triangle ACD, BK the altitude of triangle BCD. Since, according to the conditions 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 relative to straight line CD. AL = BK. Therefore, straight line AB || CD (**)
5. From conditions (*), (**) it follows that ABCD is a parallelogram.
Answer. Proven. ABCD is a parallelogram.
Task 3.
On sides BC and CD of the parallelogram ABCD, points M and H are marked, respectively, so that the segments BM and HD intersect at point O;<ВМD = 95 о,
Solution.
1. In 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 with a length of 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. We apply the sine theorem to triangle AOD. AO/sin D = OD/sin A. 2√6/sin 45 o = OD/sin 60 o. ОD = (2√6sin 60 о) / sin 45 о = (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 is equal to φ. 1. Let's count two different S ABCD = AB AD sin A = 5√2 7√2 sin f, S ABCD = 1/2 AC ВD sin AOB = 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 relationship 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 create a system: (d 1 2 + d 2 2 = 296, Let's 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 degrees. Find the area of the parallelogram. Solution.
1. From triangle AOB, using the cosine theorem, we write the relationship between the side of the parallelogram and the diagonals. AB 2 = AO 2 + VO 2 2 · AO · VO · cos AOB. 4 2 = (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. Let's 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 = 1/2 AC ВD sin AOB = 1/2 d 1 d 2 sin α = 1/2 20√2 √2/2 = 10. Note: In this and the previous problem there is no need to solve the system completely, anticipating 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 = AB · AD · sin ВAD. Let's make a substitution in the formula. We get 96 = 8 · 15 · sin ВAD. Hence sin ВAD = 4/5. 2. Let's find cos VAD. sin 2 VAD + cos 2 VAD = 1. (4 / 5) 2 + cos 2 VAD = 1. cos 2 VAD = 9 / 25. According to the conditions of the problem, we find the length of the smaller diagonal. The diagonal ВD will be smaller if the angle ВАD is acute. Then cos VAD = 3 / 5. 3. From the triangle ABD, using the cosine theorem, we find the square of the diagonal BD. ВD 2 = АВ 2 + АD 2 – 2 · АВ · ВD · cos ВAD. ВD 2 = 8 2 + 15 2 – 2 8 15 3 / 5 = 145. Answer: 145.
Still have questions? Don't know how to solve a geometry problem? website, when copying material in full or in part, a link to the source is required. A parallelogram is a quadrilateral whose opposite sides are parallel in pairs. The following figure shows parallelogram ABCD. It has side AB parallel to side CD and side BC parallel to side AD. As you may have guessed, a parallelogram is a convex quadrilateral. Let's consider the basic properties of a parallelogram. 1. In a parallelogram, opposite angles and opposite sides are equal. Let's prove this property - consider the parallelogram presented in the following figure. Diagonal BD divides it into two equal triangles: ABD and CBD. They are equal along the side BD and the two angles adjacent to it, since the angles lying crosswise at the secant BD of parallel lines BC and AD and AB and CD, respectively. Therefore AB = CD and 2. The diagonals of a parallelogram are divided in half by the point of intersection. Let point O be the intersection point of diagonals AC and BD of parallelogram ABCD. Then triangle AOB and triangle COD are equal to each other, along the side and two adjacent angles. (AB = CD since these are opposite sides of the parallelogram. And angle1 = angle2 and angle3 = angle4 are like crosswise angles when the lines AB and CD intersect with the secants AC and BD, respectively.) From this it follows that AO = OC and OB = OD, which and needed to be proven. All main properties are illustrated in the following three figures. “A major scientific discovery provides a solution to a major problem, but in the solution of any problem there is a grain of discovery.” A parallelogram has opposite sides that are equal. Proof. Let ABCD be the given parallelogram. And let its diagonals intersect at point O. In a parallelogram, opposite angles are equal. Proof. The diagonals of a parallelogram intersect and are bisected at the point of intersection. Proof. Let ABCD be the given parallelogram. Let's draw the diagonal AC. Let's mark the middle O on it. On the continuation of the segment DO, we'll put aside the segment OB 1 equal to DO. In textbooks for regular schools (for example, in Pogorelovo) it is proven like this: diagonals divide a parallelogram into 4 triangles. Let's consider one pair and find out - they are equal: their bases are opposite sides, the corresponding angles adjacent to it are equal, like vertical angles with parallel lines. That is, the segments of the diagonals are equal in pairs. All. Is that all? The funny thing is that this part is much harder to prove. This follows, by the way, from a more general result: any convex quadrilateral will have diagonals intersecting, but any non-convex quadrilateral will not. On the equality of triangles along a side and two adjacent angles (the second sign of equality of triangles) and others. Thales found an important practical application to the theorem on the equality of two triangles along a side and two adjacent angles. A rangefinder was built in the harbor of Miletus to determine the distance to a ship at sea. It consisted of three driven pegs A, B and C (AB = BC) and a marked straight line SC, perpendicular to CA. When a ship appeared on the SK straight line, we found point D such that points D, .B and E were on the same straight line. As is clear from the drawing, the distance CD on the ground is the desired distance to the ship. We worked on the lesson Kuznetsov A.V. Poturnak S.A. Evgeniy Petrov You can raise a question about modern education, express an idea or solve a pressing problem at Educational forum, where an educational council of fresh thought and action meets internationally. Having created blog, You will not only improve your status as a competent teacher, but also make a significant contribution to the development of the school of the future. Guild of Educational Leaders opens doors to top-ranking specialists and invites them to cooperate in creating the best schools in the world.
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(Since in a right triangle the leg that lies opposite the angle of 30° is equal to half the hypotenuse).
ways 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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Properties of a parallelogram
BC = AD. And from the equality of angles 1, 2, 3 and 4 it follows that angle A = angle1 + angle3 = angle2 + angle4 = angle C.
Lesson topic
Lesson Objectives
Lesson Objectives
Lesson Plan
Introduction
Property of opposite sides of a parallelogram
Since Δ AOB = Δ COD by the first criterion of equality of triangles (∠ AOB = ∠ COD, as vertical ones, AO=OC, DO=OB, by the property of the diagonals of a parallelogram), then AB=CD. In the same way, from the equality of triangles BOC and DOA, it follows that BC = DA. The theorem has been proven. Property of opposite angles of a parallelogram
Let ABCD be the given parallelogram. And let its diagonals intersect at point O.
From what was proven in the theorem about the properties of the opposite sides of a parallelogram Δ ABC = Δ CDA on three sides (AB=CD, BC=DA from what was proven, AC – general). From the equality of triangles it follows that ∠ ABC = ∠ CDA.
It is also proved that ∠ DAB = ∠ BCD, which follows from ∠ ABD = ∠ CDB. The theorem has been proven.Property of the diagonals of a parallelogram
By the previous theorem, AB 1 CD is a parallelogram. Therefore, line AB 1 is parallel to DC. But through point A only one line parallel to DC can be drawn. This means that straight AB 1 coincides with straight AB.
It is also proved that BC 1 coincides with BC. This means that point C coincides with C 1. parallelogram ABCD coincides with parallelogram AB 1 CD. Consequently, the diagonals of the parallelogram intersect and are bisected at the point of intersection. The theorem has been proven.
It was proven above that the intersection point bisects the diagonals - if it exists. The above reasoning does not prove its very existence in any way. That is, part of the theorem “the diagonals of a parallelogram intersect” remains unproven.
Questions
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