In a parallelogram, the base angles are equal. Parallelogram and its properties

A parallelogram is a quadrilateral whose opposite sides are pairwise parallel. Also, a parallelogram has such properties as opposite sides are equal, opposite angles are equal, the sum of all angles is 360 degrees.

You will need

• Geometry knowledge.

Instruction

1. Imagine given one of the corners of the parallelogram and equal to A. Find the values ​​of the remaining 3. By the property of the parallelogram, the opposite angles are equal. So the angle lying opposite to the given one is equal to the given one and its value is equal to A.

2. Find the remaining two corners. Because the sum of all the angles in a parallelogram is 360 degrees, and the opposite angles are equal to each other, it turns out that the angle belonging to the same side with the given one is equal to (360 - 2A) / 2. Well, either after reforming we get 180 - A. Thus, in a parallelogram, two angles are equal to A, and the other two angles are equal to 180 - A.

Note!
The value of one angle cannot exceed 180 degrees. The obtained values ​​of the angles can be easily verified. To do this, add them up and, if the sum is 360, everything is calculated correctly.

Useful advice
A rectangle and a rhombus are a special case of a parallelogram, so all the properties and methods for calculating angles apply to them as well.

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 intersection point.

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.

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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 \u003d Δ ADC it follows that AB \u003d CD, BC \u003d AD, ∠ B \u003d ∠ 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).

Draw from some two points B and C of the line b 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.

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 a 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.