Properties of the midline of a trapezoid. Remembering and applying the properties of a trapezoid

The segment of the straight line connecting the midpoints of the sides of the trapezoid is called the midline of the trapezoid. How to find the middle line of the trapezoid and how it relates to other elements of this figure, we will describe below.

Midline theorem

Let's draw a trapezoid in which AD is the larger base, BC is the smaller base, EF is the middle line. Let's extend the base AD beyond point D. Draw the line BF and continue it until it intersects with the continuation of the base AD at point O. Consider the triangles ∆BCF and ∆DFO. Angles ∟BCF = ∟DFO as vertical. CF = DF, ∟BCF = ∟FDO, because VS // AO. Therefore, triangles ∆BCF = ∆DFO. Hence the sides BF = FO.

Now consider ∆ABO and ∆EBF. ∟ABO is common to both triangles. BE/AB = ½ by convention, BF/BO = ½ because ∆BCF = ∆DFO. Therefore, triangles ABO and EFB are similar. Hence the ratio of the sides EF / AO = ½, as well as the ratio of the other sides.

We find EF = ½ AO. The drawing shows that AO = AD + DO. DO = BC as sides equal triangles, so AO = AD + BC. Hence EF = ½ AO = ½ (AD + BC). Those. the length of the midline of a trapezoid is half the sum of the bases.

Is the midline of a trapezoid always equal to half the sum of the bases?

Suppose there is such special case when EF ≠ ½ (AD + BC). Then BC ≠ DO, hence ∆BCF ≠ ∆DCF. But this is impossible, since they have two equal angles and sides between them. Therefore, the theorem is true under all conditions.

The problem of the middle line

Suppose, in our trapezoid ABCD AD // BC, ∟A=90°, ∟С = 135°, AB = 2 cm, diagonal AC is perpendicular to the side. Find the midline of the trapezoid EF.

If ∟A = 90°, then ∟B = 90°, so ∆ABC is rectangular.

∟BCA = ∟BCD - ∟ACD. ∟ACD = 90° by convention, therefore ∟BCA = ∟BCD - ∟ACD = 135° - 90° = 45°.

If in a right triangle ∆ABS one angle is 45°, then the legs in it are equal: AB = BC = 2 cm.

Hypotenuse AC \u003d √ (AB² + BC²) \u003d √8 cm.

Consider ∆ACD. ∟ACD = 90° by convention. ∟CAD = ∟BCA = 45° as the angles formed by the secant of the parallel bases of the trapezoid. Therefore, the legs AC = CD = √8.

Hypotenuse AD = √(AC² + CD²) = √(8 + 8) = √16 = 4 cm.

The median line of the trapezoid EF = ½(AD + BC) = ½(2 + 4) = 3 cm.

A trapezoid is a special case of a quadrilateral in which one pair of sides is parallel. The term "trapeze" comes from Greek wordτράπεζα, meaning "table", "table". In this article we will consider the types of trapezium and its properties. In addition, we will figure out how to calculate the individual elements of this example, the diagonal of an isosceles trapezoid, the midline, area, etc. The material is presented in the style of elementary popular geometry, that is, in an easily accessible form.

General information

First, let's understand what a quadrilateral is. This figure is a special case of a polygon containing four sides and four vertices. Two vertices of a quadrilateral that are not adjacent are called opposite. The same can be said about two non-adjacent sides. The main types of quadrilaterals are parallelogram, rectangle, rhombus, square, trapezoid and deltoid.

So, back to the trapeze. As we have already said, this figure has two sides that are parallel. They are called bases. The other two (non-parallel) are the sides. In exam materials and various control works very often you can meet tasks related to trapezoids, the solution of which often requires the student to have knowledge that is not provided for by the program. The school geometry course introduces students to the properties of angles and diagonals, as well as the midline of an isosceles trapezoid. But after all, in addition to this, the mentioned geometric figure has other features. But more on them later...

Types of trapezoid

There are many types of this figure. However, most often it is customary to consider two of them - isosceles and rectangular.

1. A rectangular trapezoid is a figure in which one of the sides is perpendicular to the bases. It has two angles that are always ninety degrees.

2. An isosceles trapezoid is a geometric figure whose sides are equal to each other. This means that the angles at the bases are also pairwise equal.

The main principles of the methodology for studying the properties of a trapezoid

The main principle is the use of the so-called task approach. In fact, there is no need to introduce new properties of this figure into the theoretical course of geometry. They can be discovered and formulated in the process of solving various problems (better than systemic ones). At the same time, it is very important that the teacher knows what tasks need to be set for students at one time or another of the educational process. Moreover, each property of a trapezoid can be represented as key task in the task system.

The second principle is the so-called spiral organization of the study of the "remarkable" properties of the trapezium. This implies a return in the learning process to individual features of a given geometric figure. Thus, it is easier for students to memorize them. For example, the property of four points. It can be proved both in the study of similarity and subsequently with the help of vectors. And the equal area of ​​triangles adjacent to the sides of the figure can be proved by applying not only the properties of triangles with equal heights drawn to the sides that lie on the same line, but also using the formula S= 1/2(ab*sinα). In addition, you can work out on an inscribed trapezoid or a right triangle on a circumscribed trapezoid, etc.

The use of "out-of-program" features of a geometric figure in the content of a school course is a task technology for teaching them. The constant appeal to the studied properties when passing through other topics allows students to gain a deeper knowledge of the trapezoid and ensures the success of solving the tasks. So, let's start studying this wonderful figure.

Elements and properties of an isosceles trapezoid

As we have already noted, the sides of this geometric figure are equal. It is also known as the right trapezoid. Why is it so remarkable and why did it get such a name? The features of this figure include the fact that not only the sides and corners at the bases are equal, but also the diagonals. Also, the sum of the angles of an isosceles trapezoid is 360 degrees. But that's not all! Of all known trapezoids, only around an isosceles one can a circle be described. This is due to the fact that the sum of the opposite angles of this figure is 180 degrees, and only under this condition can a circle be described around the quadrilateral. The next property of the geometric figure under consideration is that the distance from the base vertex to the projection of the opposite vertex onto the straight line that contains this base will be equal to the midline.

Now let's figure out how to find the angles of an isosceles trapezoid. Consider a solution to this problem, provided that the dimensions of the sides of the figure are known.

Solution

Usually, a quadrilateral is usually denoted by the letters A, B, C, D, where BS and AD are the bases. In an isosceles trapezoid, the sides are equal. We will assume that their size is X, and the sizes of the bases are Y and Z (smaller and larger, respectively). To carry out the calculation, it is necessary to draw a height H from angle B. The result is a right-angled triangle ABN, where AB is the hypotenuse, and BN and AN are the legs. We calculate the size of the leg AN: we subtract the smaller one from the larger base, and divide the result by 2. We write it in the form of a formula: (Z-Y) / 2 \u003d F. Now, to calculate the acute angle of the triangle, we use the cos function. We get the following record: cos(β) = Х/F. Now we calculate the angle: β=arcos (Х/F). Further, knowing one angle, we can determine the second, for this we perform an elementary arithmetic operation: 180 - β. All angles are defined.

There is also a second solution to this problem. At the beginning, we lower the height H from the corner B. We calculate the value of the BN leg. We know that the square of the hypotenuse of a right triangle is equal to the sum squares of legs. We get: BN \u003d √ (X2-F2). Next, we use the trigonometric function tg. As a result, we have: β = arctg (BN / F). Sharp corner found. Next, we determine in the same way as the first method.

Property of the diagonals of an isosceles trapezoid

Let's write down four rules first. If the diagonals in an isosceles trapezoid are perpendicular, then:

The height of the figure will be equal to the sum of the bases divided by two;

Its height and median line are equal;

The center of the circle is the point where the ;

If the lateral side is divided by the point of contact into segments H and M, then it is equal to square root products of these segments;

The quadrilateral, which was formed by the tangent points, the vertex of the trapezoid and the center of the inscribed circle, is a square whose side is equal to the radius;

The area of ​​a figure is equal to the product of the bases and the product of half the sum of the bases and its height.

Similar trapeziums

This topic is very convenient for studying the properties of this one. For example, the diagonals divide the trapezoid into four triangles, and those adjacent to the bases are similar, and to the sides they are equal. This statement can be called a property of the triangles into which the trapezoid is divided by its diagonals. The first part of this assertion is proved through the criterion of similarity in two angles. To prove the second part, it is better to use the method given below.

Proof of the theorem

We accept that the figure ABSD (AD and BS - the bases of the trapezoid) is divided by the diagonals VD and AC. The point of their intersection is O. We get four triangles: AOS - at the lower base, BOS - at the upper base, ABO and SOD at the sides. Triangles SOD and BOS have a common height if the segments BO and OD are their bases. We get that the difference between their areas (P) is equal to the difference between these segments: PBOS / PSOD = BO / OD = K. Therefore, PSOD = PBOS / K. Similarly, the BOS and AOB triangles have a common height. We take the segments CO and OA as their bases. We get PBOS / PAOB \u003d CO / OA \u003d K and PAOB \u003d PBOS / K. It follows from this that PSOD = PAOB.

To consolidate the material, students are advised to find a connection between the areas of the resulting triangles, into which the trapezoid is divided by its diagonals, by solving the following problem. It is known that the areas of triangles BOS and AOD are equal, it is necessary to find the area of ​​the trapezoid. Since PSOD \u003d PAOB, it means that PABSD \u003d PBOS + PAOD + 2 * PSOD. From the similarity of the triangles BOS and AOD it follows that BO / OD = √ (PBOS / PAOD). Therefore, PBOS/PSOD = BO/OD = √(PBOS/PAOD). We get PSOD = √ (PBOS * PAOD). Then PABSD = PBOS+PAOD+2*√(PBOS*PAOD) = (√PBOS+√PAOD)2.

similarity properties

Continuing to develop this topic, we can prove other interesting features trapezium. So, using similarity, you can prove the property of a segment that passes through a point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, we solve the following problem: it is necessary to find the length of the segment RK, which passes through the point O. From the similarity of triangles AOD and BOS, it follows that AO/OS=AD/BS. From the similarity of triangles AOP and ASB, it follows that AO / AS \u003d RO / BS \u003d AD / (BS + AD). From here we get that RO \u003d BS * AD / (BS + AD). Similarly, from the similarity of the triangles DOK and DBS, it follows that OK \u003d BS * AD / (BS + AD). From here we get that RO=OK and RK=2*BS*AD/(BS+AD). The segment passing through the point of intersection of the diagonals, parallel to the bases and connecting the two sides, is divided by the point of intersection in half. Its length is the harmonic mean of the bases of the figure.

Consider the following property of a trapezoid, which is called the property of four points. The intersection points of the diagonals (O), the intersections of the continuation of the sides (E), as well as the midpoints of the bases (T and W) always lie on the same line. This is easily proved by the similarity method. The resulting triangles BES and AED are similar, and in each of them the medians ET and EZH divide the angle at the vertex E into equal parts. Therefore, the points E, T and W lie on the same straight line. In the same way, the points T, O, and G are located on the same straight line. All this follows from the similarity of the triangles BOS and AOD. From this we conclude that all four points - E, T, O and W - will lie on one straight line.

Using similar trapezoids, students can be asked to find the length of the segment (LF) that divides the figure into two similar ones. This segment should be parallel to the bases. Since the resulting trapezoids ALFD and LBSF are similar, then BS/LF=LF/BP. It follows that LF=√(BS*BP). We get that the segment that divides the trapezoid into two similar ones has a length equal to the geometric mean of the lengths of the bases of the figure.

Consider next property similarities. It is based on a segment that divides the trapezoid into two equal-sized figures. We accept that the trapezoid ABSD is divided by the segment EN into two similar ones. From the vertex B, the height is omitted, which is divided by the segment EH into two parts - B1 and B2. We get: PABSD / 2 \u003d (BS + EH) * B1 / 2 \u003d (AD + EH) * B2 / 2 and PABSD \u003d (BS + HELL) * (B1 + B2) / 2. Next, we compose a system whose first equation is (BS + EH) * B1 \u003d (AD + EH) * B2 and the second (BS + EH) * B1 \u003d (BS + HELL) * (B1 + B2) / 2. It follows that B2/ B1 = (BS+EN)/(AD+EN) and BS+EN = ((BS+AD)/2)*(1+B2/ B1). We get that the length of the segment dividing the trapezoid into two equal ones is equal to the mean square of the lengths of the bases: √ ((BS2 + AD2) / 2).

Similarity inferences

Thus, we have proven that:

1. The segment connecting the midpoints of the sides of the trapezoid is parallel to AD and BS and is equal to the arithmetic mean of BS and AD (the length of the base of the trapezoid).

2. The line passing through the point O of the intersection of the diagonals parallel to AD and BS will be equal to the harmonic mean of the numbers AD and BS (2 * BS * AD / (BS + AD)).

3. The segment that divides the trapezoid into similar ones has the length of the geometric mean of the bases BS and AD.

4. An element that divides a figure into two equal ones has the length of the mean square numbers AD and BS.

To consolidate the material and understand the connection between the considered segments, the student needs to build them for a specific trapezoid. He can easily display the midline and the segment that passes through the point O - the intersection of the diagonals of the figure - parallel to the bases. But where will be the third and fourth? This answer will lead the student to the discovery of the desired relationship between the averages.

A line segment that joins the midpoints of the diagonals of a trapezoid

Consider the following property of this figure. We accept that the segment MH is parallel to the bases and bisects the diagonals. Let's call the intersection points W and W. This segment will be equal to the half-difference of the bases. Let's analyze this in more detail. MSH - the middle line of the triangle ABS, it is equal to BS / 2. MS - the middle line of the triangle ABD, it is equal to AD / 2. Then we get that ShShch = MShch-MSh, therefore, Sshch = AD / 2-BS / 2 = (AD + VS) / 2.

Center of gravity

Let's look at how this element is determined for a given geometric figure. To do this, it is necessary to extend the bases in opposite directions. What does it mean? It is necessary to add the lower base to the upper base - to any of the sides, for example, to the right. And the bottom is extended by the length of the top to the left. Next, we connect them with a diagonal. The point of intersection of this segment with the middle line of the figure is the center of gravity of the trapezoid.

Inscribed and circumscribed trapezoids

Let's list the features of such figures:

1. A trapezoid can only be inscribed in a circle if it is isosceles.

2. A trapezoid can be described around a circle, provided that the sum of the lengths of their bases is equal to the sum of the lengths of the sides.

Consequences of the inscribed circle:

1. The height of the described trapezoid is always equal to two radii.

2. The lateral side of the described trapezoid is observed from the center of the circle at a right angle.

The first corollary is obvious, and to prove the second one it is required to establish that the SOD angle is right, which, in fact, will also not be difficult. But knowledge of this property will allow us to use a right-angled triangle in solving problems.

Now we specify these consequences for an isosceles trapezoid, which is inscribed in a circle. We get that the height is the geometric mean of the bases of the figure: H=2R=√(BS*AD). Practicing the main technique for solving problems for trapezoids (the principle of drawing two heights), the student must solve the following task. We accept that BT is the height of the isosceles figure ABSD. It is necessary to find segments AT and TD. Using the formula described above, this will not be difficult to do.

Now let's figure out how to determine the radius of a circle using the area of ​​the circumscribed trapezoid. We lower the height from vertex B to the base of AD. Since the circle is inscribed in a trapezoid, then BS + AD \u003d 2AB or AB \u003d (BS + AD) / 2. From the triangle ABN we find sinα = BN / AB = 2 * BN / (BS + AD). PABSD \u003d (BS + AD) * BN / 2, BN \u003d 2R. We get PABSD \u003d (BS + HELL) * R, it follows that R \u003d PABSD / (BS + HELL).

All formulas of the midline of a trapezoid

Now it's time to move on to the last element of this geometric figure. Let's figure out what the middle line of the trapezoid (M) is equal to:

1. Through the bases: M \u003d (A + B) / 2.

2. Through height, base and angles:

M \u003d A-H * (ctgα + ctgβ) / 2;

M \u003d B + H * (ctgα + ctgβ) / 2.

3. Through height, diagonals and the angle between them. For example, D1 and D2 are the diagonals of a trapezoid; α, β - angles between them:

M = D1*D2*sinα/2H = D1*D2*sinβ/2H.

4. Through the area and height: M = P / N.

The area of ​​the trapezoid. Greetings! In this publication, we will consider this formula. Why is it the way it is and how can you understand it? If there is an understanding, then you do not need to learn it. If you just want to see this formula and what is urgent, then you can immediately scroll down the page))

Now in detail and in order.

A trapezoid is a quadrilateral, two sides of this quadrilateral are parallel, the other two are not. Those that are not parallel are the bases of the trapezium. The other two are called sides.

If the sides are equal, then the trapezoid is called isosceles. If one of the sides is perpendicular to the bases, then such a trapezoid is called rectangular.

In the classical form, the trapezoid is depicted as follows - the larger base is at the bottom, respectively, the smaller one is at the top. But no one forbids depicting it and vice versa. Here are the sketches:

The next important concept.

The median line of a trapezoid is a segment that connects the midpoints of the sides. The median line is parallel to the bases of the trapezoid and is equal to their half-sum.

Now let's delve deeper. Why exactly?

Consider a trapezoid with bases a and b and with the middle line l, and perform some additional constructions: draw straight lines through the bases, and perpendiculars through the ends of the midline until they intersect with the bases:

*Letter designations of vertices and other points are not entered intentionally to avoid unnecessary designations.

Look, triangles 1 and 2 are equal according to the second sign of equality of triangles, triangles 3 and 4 are the same. From the equality of triangles follows the equality of the elements, namely the legs (they are indicated respectively in blue and red).

Now attention! If we mentally “cut off” the blue and red segments from the lower base, then we will have a segment (this is the side of the rectangle) equal to the midline. Further, if we “glue” the cut off blue and red segments to the upper base of the trapezoid, then we will also get a segment (this is also the side of the rectangle) equal to the midline of the trapezoid.

Got it? It turns out that the sum of the bases will be equal to the two medians of the trapezoid:

See another explanation

Let's do the following - build a straight line passing through the lower base of the trapezoid and a straight line that will pass through points A and B:

We get triangles 1 and 2, they are equal in side and adjacent angles (the second sign of equality of triangles). This means that the resulting segment (in the sketch it is marked in blue) is equal to the upper base of the trapezoid.

Now consider a triangle:

*The median line of this trapezoid and the median line of the triangle coincide.

It is known that the triangle is equal to half of the base parallel to it, that is:

Okay, got it. Now about the area of ​​the trapezoid.

Trapezium area formula:

They say: the area of ​​a trapezoid is equal to the product of half the sum of its bases and height.

That is, it turns out that it is equal to the product of the midline and height:

You probably already noticed that this is obvious. Geometrically, this can be expressed as follows: if we mentally cut off triangles 2 and 4 from the trapezoid and put them on triangles 1 and 3, respectively:

Then we get a rectangle in area equal to the area of ​​​​our trapezoid. The area of ​​this rectangle will be equal to the product of the midline and height, that is, we can write:

But the point here is not in writing, of course, but in understanding.

Download (view) the material of the article in *pdf format

That's all. Good luck to you!

Sincerely, Alexander.

In this article, another selection of tasks with a trapezoid has been made for you. Conditions are somehow connected with its middle line. Job types are taken from the open bank typical tasks. If you wish, you can refresh your theoretical knowledge. The blog has already covered tasks whose conditions are associated with, as well as. Briefly about the middle line:

The middle line of the trapezoid connects the midpoints of the sides. It is parallel to the bases and equal to their half-sum.

Before solving problems, let's consider a theoretical example.

Given a trapezoid ABCD. Diagonal AC intersecting with the midline forms a point K, diagonal BD a point L. Prove that the segment KL is equal to half the difference of the bases.

Let's first note the fact that the midline of a trapezoid bisects any segment whose ends lie on its bases. This conclusion suggests itself. Imagine a segment connecting two points of the bases, it will split this trapezoid into two others. It turns out that a segment parallel to the bases of the trapezoid and passing through the middle of the side on the other side will pass through its middle.

It is also based on the Thales theorem:

If on one of the two straight lines we put off successively several equal segments and through their ends draw parallel lines intersecting the second line, then they will cut off equal segments on the second line.

That is, in this case, K is the middle of AC and L is the middle of BD. Hence EK is the midline of triangle ABC, LF is the midline of triangle DCB. According to the property of the midline of a triangle:

We can now express the segment KL in terms of bases:

Proven!

This example is not just given. In tasks for independent solution, there is just such a task. Only it does not say that the segment connecting the midpoints of the diagonals lies on the midline. Consider the tasks:

27819. Find the midline of a trapezoid if its bases are 30 and 16.

We calculate by the formula:

27820. The midline of the trapezoid is 28 and the smaller base is 18. Find the larger base of the trapezoid.

Let's express the larger base:

Thus:

27836. A perpendicular dropped from the apex of an obtuse angle to the greater base of an isosceles trapezoid divides it into parts having lengths 10 and 4. Find the midline of this trapezoid.

In order to find the middle line, you need to know the bases. The base AB is easy to find: 10+4=14. Find DC.

Let's construct the second perpendicular DF:

Segments AF, FE and EB will be equal to 4, 6 and 4 respectively. Why?

In an isosceles trapezoid, the perpendiculars dropped to the larger base divide it into three segments. Two of them, which are the legs of the cut off right triangles are equal to each other. The third segment is equal to the smaller base, since when constructing the indicated heights, a rectangle is formed, and in the rectangle, the opposite sides are equal. In this task:

Thus DC=6. We calculate:

27839. The bases of the trapezoid are in ratio 2:3, and the midline is 5. Find the smaller base.

Let's introduce the coefficient of proportionality x. Then AB=3x, DC=2x. We can write:

Therefore, the smaller base is 2∙2=4.

27840. The perimeter of an isosceles trapezoid is 80, its midline is equal to the lateral side. Find the side of the trapezoid.

Based on the condition, we can write:

If we denote the middle line through x, we get:

The second equation can already be written as:

27841. The midline of the trapezoid is 7, and one of its bases is 4 more than the other. Find the larger base of the trapezoid.

Let's denote the smaller base (DC) as x, then the larger one (AB) will be equal to x + 4. We can record

We got that the smaller base is early than five, which means that the larger one is equal to 9.

27842. The midline of the trapezoid is 12. One of the diagonals divides it into two segments, the difference of which is 2. Find the larger base of the trapezoid.

We can easily find the larger base of the trapezoid if we calculate the segment EO. It is the middle line in triangle ADB, and AB=2∙EO.

What do we have? It is said that the middle line is equal to 12 and the difference between the segments EO and OF is equal to 2. We can write down two equations and solve the system:

It is clear that in this case it is possible to select a pair of numbers without calculations, these are 5 and 7. But, nevertheless, we will solve the system:

So EO=12–5=7. Thus, the larger base is equal to AB=2∙EO=14.

27844. In an isosceles trapezoid, the diagonals are perpendicular. The height of the trapezoid is 12. Find its midline.

Immediately, we note that the height drawn through the intersection point of the diagonals in an isosceles trapezoid lies on the axis of symmetry and divides the trapezoid into two equal rectangular trapezoids, that is, the bases of this height are divided in half.

It would seem that in order to calculate the average line, we must find the grounds. Here a small dead end arises ... How, knowing the height, in this case, calculate the bases? And no how! Many such trapezoids with a fixed height and diagonals intersecting at an angle of 90 degrees can be built. How to be?

Look at the formula for the midline of a trapezoid. After all, we do not need to know the bases themselves, it is enough to know their sum (or half-sum). This we can do.

Since the diagonals intersect at a right angle, isosceles right triangles are formed with height EF:

It follows from the above that FO=DF=FC, and OE=AE=EB. Now let's write down what the height expressed through the segments DF and AE is equal to:

So the middle line is 12.

* In general, this is a problem, as you understand, for an oral account. But I'm sure the detailed explanation provided is necessary. And so ... If you look at the figure (provided that the angle between the diagonals is observed during construction), the equality FO=DF=FC, and OE=AE=EB immediately catches your eye.

As part of the prototypes, there are also types of tasks with trapezoids. It was built on a sheet in a cell and it is required to find the middle line, the side of the cell is usually 1, but there may be another value.

27848. Find the midline of the trapezoid ABCD if the sides of the square cells are 1.

It's simple, we calculate the bases by cells and use the formula: (2 + 4) / 2 = 3

If the bases are built at an angle to the cell grid, then there are two ways. For example!

The concept of the midline of the trapezoid

First, let's remember what figure is called a trapezoid.

Definition 1

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

In this case, parallel sides are called the bases of the trapezoid, and not parallel - the sides of the trapezoid.

Definition 2

The midline of a trapezoid is a line segment that connects the midpoints of the sides of the trapezoid.

Trapezium midline theorem

We now introduce the theorem on the midline of a trapezoid and prove it by the vector method.

Theorem 1

The median line of the trapezoid is parallel to the bases and equal to their half sum.

Proof.

Let us be given a trapezoid $ABCD$ with bases $AD\ and\ BC$. And let $MN$ be the midline of this trapezoid (Fig. 1).

Figure 1. The middle line of the trapezoid

Let us prove that $MN||AD\ and\ MN=\frac(AD+BC)(2)$.

Consider the vector $\overrightarrow(MN)$. Next, we use the polygon rule for vector addition. On the one hand, we get that

On the other side

Adding the last two equalities, we get

Since $M$ and $N$ are the midpoints of the sides of the trapezoid, we have

We get:

Hence

From the same equality (since $\overrightarrow(BC)$ and $\overrightarrow(AD)$ are codirectional and, therefore, collinear), we get that $MN||AD$.

The theorem has been proven.

Examples of tasks on the concept of the midline of a trapezoid

Example 1

The sides of the trapezoid are $15\cm$ and $17\cm$ respectively. The perimeter of the trapezoid is $52\cm$. Find the length of the midline of the trapezoid.

Solution.

Denote the midline of the trapezoid by $n$.

The sum of the sides is

Therefore, since the perimeter is $52\ cm$, the sum of the bases is

Hence, by Theorem 1, we obtain

Answer:$10\cm$.

Example 2

The ends of the circle's diameter are $9$ cm and $5$ cm respectively from its tangent. Find the diameter of this circle.

Solution.

Let us be given a circle with center $O$ and diameter $AB$. Draw the tangent $l$ and construct the distances $AD=9\ cm$ and $BC=5\ cm$. Let's draw the radius $OH$ (Fig. 2).

Figure 2.

Since $AD$ and $BC$ are the distances to the tangent, then $AD\bot l$ and $BC\bot l$ and since $OH$ is the radius, then $OH\bot l$, hence $OH |\left|AD\right||BC$. From all this we get that $ABCD$ is a trapezoid, and $OH$ is its midline. By Theorem 1, we get