Relationship between the bases of a trapezoid. Trapezoid

We encounter such a shape as a trapezoid in life quite often. For example, any bridge that is made of concrete blocks is a prime example. A more visual option is the steering of each vehicle, etc. The properties of the figure were known back in Ancient Greece, which Aristotle described in more detail in his scientific work “Elements”. And the knowledge developed thousands of years ago is still relevant today. Therefore, let's take a closer look at them.

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Basic Concepts

Figure 1. Classic trapezoid shape.

A trapezoid is essentially a quadrilateral consisting of two segments that are parallel and two other segments that are not parallel. When talking about this figure, it is always necessary to remember such concepts as: bases, height and midline. Two segments of a quadrilateral which are called bases to each other (segments AD and BC). The height is the segment perpendicular to each of the bases (EH), i.e. intersect at an angle of 90° (as shown in Fig. 1).

If we add up all the internal degree measures, then the sum of the angles of the trapezoid will be equal to 2π (360°), like that of any quadrilateral. A segment whose ends are the midpoints of the sides (IF) called the midline. The length of this segment is the sum of bases BC and AD divided by 2.

There are three types of geometric figures: straight, regular and isosceles. If at least one angle at the vertices of the base is right (for example, if ABD = 90°), then such a quadrilateral is called a right trapezoid. If the side segments are equal (AB and CD), then it is called isosceles (accordingly, the angles at the bases are equal).

How to find area

For that, to find the area of a quadrilateral ABCD use the following formula:

Figure 2. Solving the problem of finding an area

For a more clear example, let’s solve an easy problem. For example, let the upper and lower bases be 16 and 44 cm, respectively, and the sides – 17 and 25 cm. Let’s construct a perpendicular segment from vertex D so that DE II BC (as shown in Figure 2). From here we get that

Let DF be . From ΔADE (which will be isosceles), we get the following:

That is, in simple terms, we first found the height ΔADE, which is also the height of the trapezoid. From here we calculate, using the already known formula, the area of the quadrilateral ABCD, with the already known value of the height DF.

Hence, the required area ABCD is 450 cm³. That is, we can say with confidence that in order To calculate the area of a trapezoid, you only need the sum of the bases and the length of the height.

Important! When solving the problem, it is not necessary to find the value of the lengths separately; it is quite acceptable if other parameters of the figure are used, which, with appropriate proof, will be equal to the sum of the bases.

Types of trapezoids

Depending on what sides the figure has and what angles are formed at the bases, there are three types of quadrilaterals: rectangular, uneven and equilateral.

Versatile

There are two forms: acute and obtuse. ABCD is acute only if the base angles (AD) are acute and the lengths of the sides are different. If the value of one angle is greater than Pi/2 (the degree measure is more than 90°), then we get an obtuse angle.

If the sides are equal in length

Figure 3. View of an isosceles trapezoid

If the non-parallel sides are equal in length, then ABCD is called isosceles (regular). Moreover, in such a quadrilateral the degree measure of the angles at the base is the same, their angle will always be less than a right angle. It is for this reason that an isosceles line is never divided into acute-angled and obtuse-angled. A quadrilateral of this shape has its own specific differences, which include:

The segments connecting opposite vertices are equal.
Acute angles with a larger base are 45° (illustrative example in Figure 3).
If you add up the degrees of opposite angles, they add up to 180°.
You can build around any regular trapezoid.
If you add up the degree measure of opposite angles, it is equal to π.

Moreover, due to their geometric arrangement of points, there are basic properties of an isosceles trapezoid:

Angle value at base 90°

The perpendicularity of the side of the base is a capacious characteristic of the concept of “rectangular trapezoid”. There cannot be two sides with corners at the base, because otherwise it will already be a rectangle. In quadrilaterals of this type, the second side will always form an acute angle with the larger base, and an obtuse angle with the smaller one. In this case, the perpendicular side will also be the height.

The segment between the middles of the sidewalls

If we connect the midpoints of the sides, and the resulting segment is parallel to the bases and equal in length to half their sum, then the resulting straight line will be the middle line. The value of this distance is calculated by the formula:

For a more clear example, consider a problem using a center line.

Task. The midline of the trapezoid is 7 cm; it is known that one of the sides is 4 cm larger than the other (Fig. 4). Find the lengths of the bases.

Figure 4. Solving the problem of finding the lengths of the bases

Solution. Let the smaller base DC be equal to x cm, then the larger base will be equal to (x+4) cm, respectively. From here, using the formula for the midline of a trapezoid, we obtain:

It turns out that the smaller base DC is 5 cm, and the larger one is 9 cm.

Important! The concept of a midline is key in solving many geometry problems. Based on its definition, many proofs for other figures are constructed. Using the concept in practice, a more rational solution and search for the required value is possible.

Determination of height, and ways to find it

As noted earlier, the height is a segment that intersects the bases at an angle of 2Pi/4 and is the shortest distance between them. Before finding the height of the trapezoid, it is necessary to determine what input values are given. For a better understanding, let's look at the problem. Find the height of the trapezoid provided that the bases are 8 and 28 cm, the sides are 12 and 16 cm, respectively.

Figure 5. Solving the problem of finding the height of a trapezoid

Let us draw segments DF and CH at right angles to the base AD. According to the definition, each of them will be the height of a given trapezoid (Fig. 5). In this case, knowing the length of each sidewall, using the Pythagorean theorem, we will find what the height in triangles AFD and BHC is equal to.

The sum of the segments AF and HB is equal to the difference of the bases, i.e.:

Let the length AF be x cm, then the length of the segment HB= (20 – x) cm. As it was established, DF=CH, from here.

Then we get the following equation:

It turns out that the segment AF in the triangle AFD is equal to 7.2 cm, from here we calculate the height of the trapezoid DF using the same Pythagorean theorem:

Those. the height of the trapezoid ADCB will be equal to 9.6 cm. How can you be sure that calculating the height is a more mechanical process, and is based on calculating the sides and angles of triangles. But, in a number of geometry problems, only the degrees of angles can be known, in which case calculations will be made through the ratio of the sides of the internal triangles.

Important! In essence, a trapezoid is often thought of as two triangles, or as a combination of a rectangle and a triangle. To solve 90% of all problems found in school textbooks, the properties and characteristics of these figures. Most of the formulas for this GMT are derived relying on the “mechanisms” for the two types of figures indicated.

How to quickly calculate the length of the base

Before finding the base of the trapezoid, it is necessary to determine what parameters are already given and how to use them rationally. A practical approach is to extract the length of the unknown base from the midline formula. For a clearer understanding of the picture, let’s use an example task to show how this can be done. Let it be known that the middle line of the trapezoid is 7 cm, and one of the bases is 10 cm. Find the length of the second base.

Solution: Knowing that the middle line is equal to half the sum of the bases, we can say that their sum is 14 cm.

(14 cm = 7 cm × 2). From the conditions of the problem, we know that one of them is equal to 10 cm, hence the smaller side of the trapezoid will be equal to 4 cm (4 cm = 14 – 10).

Moreover, for a more comfortable solution to problems of this kind, We recommend that you thoroughly learn such formulas from the trapezoid area as:

middle line;
square;
height;
diagonals.

Knowing the essence (precisely the essence) of these calculations, you can easily find out the desired value.

Video: trapezoid and its properties

Video: features of a trapezoid

Conclusion

From the considered examples of problems, we can draw a simple conclusion that the trapezoid, in terms of calculating problems, is one of the simplest figures of geometry. To successfully solve problems, first of all, you should not decide what information is known about the object being described, in what formulas they can be applied, and decide what you need to find. By following this simple algorithm, no task using this geometric figure will be effortless.

A trapezoid is a convex quadrilateral in which one pair of opposite sides is parallel to each other and the other is not.

Based on the definition of a trapezoid and the characteristics of a parallelogram, the parallel sides of a trapezoid cannot be equal to each other. Otherwise, the other pair of sides would also become parallel and equal to each other. In this case, we would be dealing with a parallelogram.

The parallel opposite sides of a trapezoid are called reasons. That is, the trapezoid has two bases. Non-parallel opposite sides of a trapezoid are called sides.

Depending on which sides and what angles they form with the bases, different types of trapezoids are distinguished. Most often, trapezoids are divided into unequal (unilateral), isosceles (equilateral) and rectangular.

U lopsided trapezoids the sides are not equal to each other. Moreover, with a large base, both of them can form only acute angles, or one angle will be obtuse and the other acute. In the first case, the trapezoid is called acute-angled, in the second - obtuse.

U isosceles trapezoids the sides are equal to each other. Moreover, with a large base they can only form acute angles, i.e. All isosceles trapezoids are acute-angled. Therefore, they are not divided into acute-angled and obtuse-angled.

U rectangular trapezoids one side is perpendicular to the bases. The second side cannot be perpendicular to them, because in this case we would be dealing with a rectangle. In rectangular trapezoids, the non-perpendicular side always forms an acute angle with the larger base. A perpendicular side is perpendicular to both bases because the bases are parallel.

In order to feel confident and successfully solve problems in geometry lessons, it is not enough to learn the formulas. They need to be understood first. To be afraid, and even more so to hate formulas, is unproductive. This article will analyze in accessible language various ways to find the area of a trapezoid. To better understand the corresponding rules and theorems, we will pay some attention to its properties. This will help you understand how the rules work and in what cases certain formulas should be applied.

Defining a trapezoid

What kind of figure is this overall? A trapezoid is a polygon with four corners and two parallel sides. The other two sides of the trapezoid can be inclined at different angles. Its parallel sides are called bases, and for non-parallel sides the name “sides” or “hips” is used. Such figures are quite common in everyday life. The contours of the trapezoid can be seen in the silhouettes of clothing, interior items, furniture, dishes and many others. There are different types of trapezoid: scalene, equilateral and rectangular. We will examine their types and properties in more detail later in the article.

Properties of a trapezoid

Let us dwell briefly on the properties of this figure. The sum of the angles adjacent to any side is always 180°. It should be noted that all angles of a trapezoid add up to 360°. The trapezoid has the concept of a midline. If you connect the midpoints of the sides with a segment, this will be the middle line. It is designated m. The middle line has important properties: it is always parallel to the bases (we remember that the bases are also parallel to each other) and equal to their half-sum:

This definition must be learned and understood, because it is the key to solving many problems!

With a trapezoid, you can always lower the height to the base. An altitude is a perpendicular, often denoted by the symbol h, that is drawn from any point of one base to another base or its extension. The midline and height will help you find the area of the trapezoid. Such problems are the most common in the school geometry course and regularly appear among test and examination papers.

The simplest formulas for the area of a trapezoid

Let's look at the two most popular and simple formulas used to find the area of a trapezoid. It is enough to multiply the height by half the sum of the bases to easily find what you are looking for:

S = h*(a + b)/2.

In this formula, a, b denote the bases of the trapezoid, h - the height. For ease of perception, in this article, multiplication signs are marked with a symbol (*) in formulas, although in official reference books the multiplication sign is usually omitted.

Let's look at an example.

Given: a trapezoid with two bases equal to 10 and 14 cm, the height is 7 cm. What is the area of the trapezoid?

Let's look at the solution to this problem. Using this formula, you first need to find the half-sum of the bases: (10+14)/2 = 12. So, the half-sum is equal to 12 cm. Now we multiply the half-sum by the height: 12*7 = 84. What we are looking for is found. Answer: The area of the trapezoid is 84 square meters. cm.

The second well-known formula says: the area of a trapezoid is equal to the product of the midline and the height of the trapezoid. That is, it actually follows from the previous concept of the middle line: S=m*h.

Using diagonals for calculations

Another way to find the area of a trapezoid is actually not that complicated. It is connected to its diagonals. Using this formula, to find the area, you need to multiply the half-product of its diagonals (d 1 d 2) by the sine of the angle between them:

S = ½ d 1 d 2 sin a.

Let's consider a problem that shows the application of this method. Given: a trapezoid with the length of the diagonals equal to 8 and 13 cm, respectively. The angle a between the diagonals is 30°. Find the area of the trapezoid.

Solution. Using the above formula, it is easy to calculate what is required. As you know, sin 30° is 0.5. Therefore, S = 8*13*0.5=52. Answer: the area is 52 square meters. cm.

Finding the area of an isosceles trapezoid

A trapezoid can be isosceles (isosceles). Its sides are the same and the angles at the bases are equal, which is well illustrated by the figure. An isosceles trapezoid has the same properties as a regular one, plus a number of special ones. A circle can be circumscribed around an isosceles trapezoid, and a circle can be inscribed within it.

What methods are there for calculating the area of such a figure? The method below will require a lot of calculations. To use it, you need to know the values of the sine (sin) and cosine (cos) of the angle at the base of the trapezoid. To calculate them, you need either Bradis tables or an engineering calculator. Here is the formula:

S= c*sin a*(a - c*cos a),

Where With- lateral thigh, a- angle at the lower base.

An equilateral trapezoid has diagonals of equal length. The converse is also true: if a trapezoid has equal diagonals, then it is isosceles. Hence the following formula to help find the area of a trapezoid - the half product of the square of the diagonals and the sine of the angle between them: S = ½ d 2 sin a.

Finding the area of a rectangular trapezoid

A special case of a rectangular trapezoid is known. This is a trapezoid, in which one side (its thigh) adjoins the bases at a right angle. It has the properties of a regular trapezoid. In addition, it has a very interesting feature. The difference in the squares of the diagonals of such a trapezoid is equal to the difference in the squares of its bases. All previously described methods for calculating area are used for it.

We use ingenuity

There is one trick that can help if you forget specific formulas. Let's take a closer look at what a trapezoid is. If we mentally divide it into parts, we will get familiar and understandable geometric shapes: a square or rectangle and a triangle (one or two). If the height and sides of the trapezoid are known, you can use the formulas for the area of a triangle and a rectangle, and then add up all the resulting values.

Let's illustrate this with the following example. Given a rectangular trapezoid. Angle C = 45°, angles A, D are 90°. The upper base of the trapezoid is 20 cm, the height is 16 cm. You need to calculate the area of the figure.

This figure obviously consists of a rectangle (if two angles are equal to 90°) and a triangle. Since the trapezoid is rectangular, therefore, its height is equal to its side, that is, 16 cm. We have a rectangle with sides of 20 and 16 cm, respectively. Now consider a triangle whose angle is 45°. We know that one side of it is 16 cm. Since this side is also the height of the trapezoid (and we know that the height descends to the base at a right angle), therefore, the second angle of the triangle is 90°. Hence the remaining angle of the triangle is 45°. The consequence of this is that we get a right isosceles triangle with two equal sides. This means that the other side of the triangle is equal to the height, that is, 16 cm. It remains to calculate the area of the triangle and the rectangle and add the resulting values.

The area of a right triangle is equal to half the product of its legs: S = (16*16)/2 = 128. The area of a rectangle is equal to the product of its width and length: S = 20*16 = 320. We found the required: area of the trapezoid S = 128 + 320 = 448 sq. see. You can easily double-check yourself using the above formulas, the answer will be identical.

We use the Pick formula

Finally, we present another original formula that helps to find the area of a trapezoid. It is called the Pick formula. It is convenient to use when the trapezoid is drawn on checkered paper. Similar problems are often found in GIA materials. It looks like this:

S = M/2 + N - 1,

in this formula M is the number of nodes, i.e. intersections of the lines of the figure with the lines of the cell at the boundaries of the trapezoid (orange dots in the figure), N is the number of nodes inside the figure (blue dots). It is most convenient to use it when finding the area of an irregular polygon. However, the larger the arsenal of techniques used, the fewer errors and better the results.

Of course, the information provided does not exhaust the types and properties of a trapezoid, as well as methods for finding its area. This article provides an overview of its most important characteristics. When solving geometric problems, it is important to act gradually, start with easy formulas and problems, consistently consolidate your understanding, and move to another level of complexity.

Collected together, the most common formulas will help students navigate the various ways to calculate the area of a trapezoid and better prepare for tests and assignments on this topic.

In the materials of various tests and exams, they are very often found trapezoid problems, the solution of which requires knowledge of its properties.

Let's find out what interesting and useful properties a trapezoid has for solving problems.

After studying the properties of the midline of a trapezoid, one can formulate and prove property of a segment connecting the midpoints of the diagonals of a trapezoid. The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases.

MO is the middle line of triangle ABC and is equal to 1/2BC (Fig. 1).

MQ is the middle line of triangle ABD and is equal to 1/2AD.

Then OQ = MQ – MO, therefore OQ = 1/2AD – 1/2BC = 1/2(AD – BC).

When solving many problems on a trapezoid, one of the main techniques is to draw two heights in it.

Consider the following task.

Let BT be the height of an isosceles trapezoid ABCD with bases BC and AD, with BC = a, AD = b. Find the lengths of the segments AT and TD.

Solution.

Solving the problem is not difficult (Fig. 2), but it allows you to get property of the height of an isosceles trapezoid drawn from the vertex of an obtuse angle: the height of an isosceles trapezoid drawn from the vertex of an obtuse angle divides the larger base into two segments, the smaller of which is equal to half the difference of the bases, and the larger one is equal to half the sum of the bases.

When studying the properties of a trapezoid, you need to pay attention to such a property as similarity. So, for example, the diagonals of a trapezoid divide it into four triangles, and the triangles adjacent to the bases are similar, and the triangles adjacent to the sides are equal in size. This statement can be called property of triangles into which a trapezoid is divided by its diagonals. Moreover, the first part of the statement can be proven very easily through the sign of similarity of triangles at two angles. Let's prove second part of the statement.

Triangles BOC and COD have a common height (Fig. 3), if we take the segments BO and OD as their bases. Then S BOC /S COD = BO/OD = k. Therefore, S COD = 1/k · S BOC .

Similarly, triangles BOC and AOB have a common height if we take the segments CO and OA as their bases. Then S BOC /S AOB = CO/OA = k and S A O B = 1/k · S BOC .

From these two sentences it follows that S COD = S A O B.

Let's not dwell on the formulated statement, but find the relationship between the areas of the triangles into which the trapezoid is divided by its diagonals. To do this, let's solve the following problem.

Let point O be the intersection point of the diagonals of the trapezoid ABCD with the bases BC and AD. It is known that the areas of triangles BOC and AOD are equal to S 1 and S 2, respectively. Find the area of the trapezoid.

Since S COD = S A O B, then S ABC D = S 1 + S 2 + 2S COD.

From the similarity of triangles BOC and AOD it follows that BO/OD = √(S₁/S 2).

Therefore, S₁/S COD = BO/OD = √(S₁/S 2), which means S COD = √(S 1 · S 2).

Then S ABC D = S 1 + S 2 + 2√(S 1 · S 2) = (√S 1 + √S 2) 2.

Using similarity it is proved that property of a segment passing through the point of intersection of the diagonals of a trapezoid parallel to the bases.

Let's consider task:

Let point O be the intersection point of the diagonals of the trapezoid ABCD with the bases BC and AD. BC = a, AD = b. Find the length of the segment PK passing through the point of intersection of the diagonals of the trapezoid parallel to the bases. What segments is PK divided by point O (Fig. 4)?

From the similarity of triangles AOD and BOC it follows that AO/OC = AD/BC = b/a.

From the similarity of triangles AOP and ACB it follows that AO/AC = PO/BC = b/(a + b).

Hence PO = BC b / (a + b) = ab/(a + b).

Similarly, from the similarity of triangles DOK and DBC, it follows that OK = ab/(a + b).

Hence PO = OK and PK = 2ab/(a + b).

So, the proven property can be formulated as follows: a segment parallel to the bases of the trapezoid, passing through the point of intersection of the diagonals and connecting two points on the lateral sides, is divided in half by the point of intersection of the diagonals. Its length is the harmonic mean of the bases of the trapezoid.

Following four point property: in a trapezoid, the point of intersection of the diagonals, the point of intersection of the continuation of the sides, the midpoints of the bases of the trapezoid lie on the same line.

Triangles BSC and ASD are similar (Fig. 5) and in each of them the medians ST and SG divide the vertex angle S into equal parts. Therefore, points S, T and G lie on the same line.

In the same way, points T, O and G are located on the same line. This follows from the similarity of triangles BOC and AOD.

This means that all four points S, T, O and G lie on the same line.

You can also find the length of the segment dividing the trapezoid into two similar ones.

If trapezoids ALFD and LBCF are similar (Fig. 6), then a/LF = LF/b.

Hence LF = √(ab).

Thus, a segment dividing a trapezoid into two similar trapezoids has a length equal to the geometric mean of the lengths of the bases.

Let's prove property of a segment dividing a trapezoid into two equal areas.

Let the area of the trapezoid be S (Fig. 7). h 1 and h 2 are parts of the height, and x is the length of the desired segment.

Then S/2 = h 1 (a + x)/2 = h 2 (b + x)/2 and

S = (h 1 + h 2) · (a + b)/2.

Let's create a system

(h 1 (a + x) = h 2 (b + x)
(h 1 · (a + x) = (h 1 + h 2) · (a + b)/2.

Solving this system, we obtain x = √(1/2(a 2 + b 2)).

Thus, the length of the segment dividing the trapezoid into two equal ones is equal to √((a 2 + b 2)/2)(mean square of base lengths).

So, for the trapezoid ABCD with bases AD and BC (BC = a, AD = b) we proved that the segment:

1) MN, connecting the midpoints of the lateral sides of the trapezoid, is parallel to the bases and equal to their half-sum (the arithmetic mean of the numbers a and b);

2) PK passing through the point of intersection of the diagonals of the trapezoid parallel to the bases is equal to
2ab/(a + b) (harmonic mean of numbers a and b);

3) LF, which splits a trapezoid into two similar trapezoids, has a length equal to the geometric mean of the numbers a and b, √(ab);

4) EH, dividing a trapezoid into two equal ones, has length √((a 2 + b 2)/2) (the root mean square of the numbers a and b).

Sign and property of an inscribed and circumscribed trapezoid.

Property of an inscribed trapezoid: a trapezoid can be inscribed in a circle if and only if it is isosceles.

Properties of the described trapezoid. A trapezoid can be described around a circle if and only if the sum of the lengths of the bases is equal to the sum of the lengths of the sides.

Useful consequences of the fact that a circle is inscribed in a trapezoid:

1. The height of the circumscribed trapezoid is equal to two radii of the inscribed circle.

2. The side of the circumscribed trapezoid is visible from the center of the inscribed circle at a right angle.

The first is obvious. To prove the second corollary, it is necessary to establish that the angle COD is right, which is also not difficult. But knowing this corollary allows you to use a right triangle when solving problems.

Let's specify corollaries for an isosceles circumscribed trapezoid:

The height of an isosceles circumscribed trapezoid is the geometric mean of the bases of the trapezoid
h = 2r = √(ab).

The considered properties will allow you to understand the trapezoid more deeply and ensure success in solving problems using its properties.

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\[(\Large(\text(Free trapezoid)))\]

Definitions

A trapezoid is a convex quadrilateral in which two sides are parallel and the other two sides are not parallel.

The parallel sides of a trapezoid are called its bases, and the other two sides are called its lateral sides.

The height of a trapezoid is the perpendicular drawn from any point of one base to another base.

Theorems: properties of a trapezoid

1) The sum of the angles at the side is \(180^\circ\) .

2) The diagonals divide the trapezoid into four triangles, two of which are similar, and the other two are equal in size.

Proof

1) Because \(AD\parallel BC\), then the angles \(\angle BAD\) and \(\angle ABC\) are one-sided for these lines and the transversal \(AB\), therefore, \(\angle BAD +\angle ABC=180^\circ\).

2) Because \(AD\parallel BC\) and \(BD\) are a secant, then \(\angle DBC=\angle BDA\) lie crosswise.
Also \(\angle BOC=\angle AOD\) as vertical.
Therefore, at two angles \(\triangle BOC \sim \triangle AOD\).

Let's prove that \(S_(\triangle AOB)=S_(\triangle COD)\). Let \(h\) be the height of the trapezoid. Then \(S_(\triangle ABD)=\frac12\cdot h\cdot AD=S_(\triangle ACD)\). Then: \

Definition

The midline of a trapezoid is a segment connecting the midpoints of the sides.

Theorem

The midline of the trapezoid is parallel to the bases and equal to their half-sum.

Proof*

1) Let's prove parallelism.

Let us draw through the point \(M\) the straight line \(MN"\parallel AD\) (\(N"\in CD\) ). Then, according to Thales’ theorem (since \(MN"\parallel AD\parallel BC, AM=MB\)) point \(N"\) is the middle of the segment \(CD\). This means that the points \(N\) and \(N"\) will coincide.

2) Let's prove the formula.

Let's do \(BB"\perp AD, CC"\perp AD\) . Let \(BB"\cap MN=M", CC"\cap MN=N"\).

Then, by Thales' theorem, \(M"\) and \(N"\) are the midpoints of the segments \(BB"\) and \(CC"\), respectively. This means that \(MM"\) is the middle line of \(\triangle ABB"\) , \(NN"\) is the middle line of \(\triangle DCC"\) . That's why: \

Because \(MN\parallel AD\parallel BC\) and \(BB", CC"\perp AD\), then \(B"M"N"C"\) and \(BM"N"C\) are rectangles. According to Thales' theorem, from \(MN\parallel AD\) and \(AM=MB\) it follows that \(B"M"=M"B\) . Hence, \(B"M"N"C"\) and \(BM"N"C\) are equal rectangles, therefore, \(M"N"=B"C"=BC\) .

Thus:

\ \[=\dfrac12 \left(AB"+B"C"+BC+C"D\right)=\dfrac12\left(AD+BC\right)\]

Theorem: property of an arbitrary trapezoid

The midpoints of the bases, the point of intersection of the diagonals of the trapezoid and the point of intersection of the extensions of the lateral sides lie on the same straight line.

Proof*
It is recommended that you familiarize yourself with the proof after studying the topic “Similarity of triangles”.

1) Let us prove that the points \(P\), \(N\) and \(M\) lie on the same line.

Let's draw a straight line \(PN\) (\(P\) is the point of intersection of the extensions of the lateral sides, \(N\) is the middle of \(BC\)). Let it intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .

Consider \(\triangle BPN\) and \(\triangle APM\) . They are similar at two angles (\(\angle APM\) – general, \(\angle PAM=\angle PBN\) as corresponding at \(AD\parallel BC\) and \(AB\) secant). Means: \[\dfrac(BN)(AM)=\dfrac(PN)(PM)\]

Consider \(\triangle CPN\) and \(\triangle DPM\) . They are similar at two angles (\(\angle DPM\) – general, \(\angle PDM=\angle PCN\) as corresponding at \(AD\parallel BC\) and \(CD\) secant). Means: \[\dfrac(CN)(DM)=\dfrac(PN)(PM)\]

From here \(\dfrac(BN)(AM)=\dfrac(CN)(DM)\). But \(BN=NC\) therefore \(AM=DM\) .

2) Let us prove that the points \(N, O, M\) lie on the same line.

Let \(N\) be the midpoint of \(BC\) and \(O\) be the point of intersection of the diagonals. Let's draw a straight line \(NO\) , it will intersect the side \(AD\) at the point \(M\) . Let us prove that \(M\) is the midpoint of \(AD\) .

\(\triangle BNO\sim \triangle DMO\) along two angles (\(\angle OBN=\angle ODM\) lying crosswise at \(BC\parallel AD\) and \(BD\) secant; \(\angle BON=\angle DOM\) as vertical). Means: \[\dfrac(BN)(MD)=\dfrac(ON)(OM)\]

Likewise \(\triangle CON\sim \triangle AOM\). Means: \[\dfrac(CN)(MA)=\dfrac(ON)(OM)\]

From here \(\dfrac(BN)(MD)=\dfrac(CN)(MA)\). But \(BN=CN\) therefore \(AM=MD\) .

\[(\Large(\text(Isosceles trapezoid)))\]

Definitions

A trapezoid is called rectangular if one of its angles is right.

A trapezoid is called isosceles if its sides are equal.

Theorems: properties of an isosceles trapezoid

1) An isosceles trapezoid has equal base angles.

2) The diagonals of an isosceles trapezoid are equal.

3) Two triangles formed by diagonals and a base are isosceles.

Proof

1) Consider the isosceles trapezoid \(ABCD\) .

From the vertices \(B\) and \(C\), we drop the perpendiculars \(BM\) and \(CN\) to the side \(AD\), respectively. Since \(BM\perp AD\) and \(CN\perp AD\) , then \(BM\parallel CN\) ; \(AD\parallel BC\) , then \(MBCN\) is a parallelogram, therefore, \(BM = CN\) .

Consider the right triangles \(ABM\) and \(CDN\) . Since their hypotenuses are equal and the leg \(BM\) is equal to the leg \(CN\) , then these triangles are equal, therefore, \(\angle DAB = \angle CDA\) .

Because \(AB=CD, \angle A=\angle D, AD\)- general, then according to the first sign. Therefore, \(AC=BD\) .

3) Because \(\triangle ABD=\triangle ACD\), then \(\angle BDA=\angle CAD\) . Therefore, the triangle \(\triangle AOD\) is isosceles. Similarly, it is proved that \(\triangle BOC\) is isosceles.

Theorems: signs of an isosceles trapezoid

1) If a trapezoid has equal base angles, then it is isosceles.

2) If a trapezoid has equal diagonals, then it is isosceles.

Proof

Consider the trapezoid \(ABCD\) such that \(\angle A = \angle D\) .

Let's complete the trapezoid to the triangle \(AED\) as shown in the figure. Since \(\angle 1 = \angle 2\) , then the triangle \(AED\) is isosceles and \(AE = ED\) . Angles \(1\) and \(3\) are equal as corresponding angles for parallel lines \(AD\) and \(BC\) and secant \(AB\). Similarly, angles \(2\) and \(4\) are equal, but \(\angle 1 = \angle 2\), then \(\angle 3 = \angle 1 = \angle 2 = \angle 4\), therefore, the triangle \(BEC\) is also isosceles and \(BE = EC\) .

Eventually \(AB = AE - BE = DE - CE = CD\), that is, \(AB = CD\), which is what needed to be proved.

2) Let \(AC=BD\) . Because \(\triangle AOD\sim \triangle BOC\), then we denote their similarity coefficient as \(k\) . Then if \(BO=x\) , then \(OD=kx\) . Similar to \(CO=y \Rightarrow AO=ky\) .