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Diagonals of a trapezoid. Rectangular and isosceles trapezoid: properties and characteristics A trapezoid is isosceles if its sides are parallel

  1. A segment connecting the midpoints of the diagonals of a trapezoid equal to half base differences
  2. Triangles formed by the bases of a trapezoid and the segments of the diagonals up to their point of intersection are similar
  3. Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the lateral sides of the trapezoid - are equal in size (have the same area)
  4. If you extend sides trapezoids towards the smaller base, then they intersect at one point with the straight line connecting the midpoints of the bases
  5. A segment connecting the bases of a trapezoid and passing through the point of intersection of the diagonals of the trapezoid is divided by this point in a proportion equal to the ratio of the lengths of the bases of the trapezoid
  6. A segment parallel to the bases of the trapezoid and drawn through the point of intersection of the diagonals is divided in half by this point, and its length is equal to 2ab/(a + b), where a and b are the bases of the trapezoid

Properties of a segment connecting the midpoints of the diagonals of a trapezoid

Let's connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A segment connecting the midpoints of the diagonals of a trapezoid lies on the midline of the trapezoid.

This segment parallel to the bases of the trapezoid.

The length of the segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of its bases.

LM = (AD - BC)/2
or
LM = (a-b)/2

Properties of triangles formed by the diagonals of a trapezoid


Triangles that are formed by the bases of a trapezoid and the point of intersection of the diagonals of the trapezoid - are similar.
Triangles BOC and AOD are similar. Since angles BOC and AOD are vertical, they are equal.
Angles OCB and OAD are internal angles lying crosswise with parallel lines AD and BC (the bases of the trapezoid are parallel to each other) and a secant line AC, therefore they are equal.
Angles OBC and ODA are equal for the same reason (internal crosswise).

Since all three angles of one triangle are equal to the corresponding angles of another triangle, then these triangles are similar.

What follows from this?

To solve problems in geometry, the similarity of triangles is used as follows. If we know the lengths of two corresponding elements similar triangles, then we find the similarity coefficient (we divide one by the other). From where the lengths of all other elements are related to each other by exactly the same value.

Properties of triangles lying on the lateral side and diagonals of a trapezoid


Consider two triangles lying on the lateral sides of the trapezoid AB and CD. These are triangles AOB and COD. Despite the fact that the sizes of individual sides of these triangles may be completely different, but the areas of the triangles formed by the lateral sides and the point of intersection of the diagonals of the trapezoid are equal, that is, the triangles are equal in size.


If we extend the sides of the trapezoid towards the smaller base, then the point of intersection of the sides will be coincide with a straight line that passes through the middle of the bases.

Thus, any trapezoid can be expanded into a triangle. Wherein:

  • Triangles formed by the bases of a trapezoid with a common vertex at the point of intersection of the extended sides are similar
  • The straight line connecting the midpoints of the bases of the trapezoid is, at the same time, the median of the constructed triangle

Properties of a segment connecting the bases of a trapezoid


If you draw a segment whose ends lie on the bases of a trapezoid, which lies at the point of intersection of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the point of intersection of the diagonals (KO/ON) will be equal to the ratio of the bases of the trapezoid(BC/AD).

KO/ON = BC/AD

This property follows from the similarity of the corresponding triangles (see above).

Properties of a segment parallel to the bases of a trapezoid


If we draw a segment parallel to the bases of the trapezoid and passing through the point of intersection of the trapezoid’s diagonals, then it will have the following properties:

  • Specified distance (KM) bisected by the intersection point of the trapezoid's diagonals
  • Section length passing through the point of intersection of the diagonals of the trapezoid and parallel to the bases is equal to KM = 2ab/(a + b)

Formulas for finding the diagonals of a trapezoid


a, b- trapezoid bases

c, d- sides of the trapezoid

d1 d2- diagonals of a trapezoid

α β - angles with a larger base of the trapezoid

Formulas for finding the diagonals of a trapezoid through the bases, sides and angles at the base

The first group of formulas (1-3) reflects one of the main properties of trapezoid diagonals:

1. The sum of the squares of the diagonals of a trapezoid is equal to the sum of the squares of the sides plus twice the product of its bases. This property of trapezoid diagonals can be proven as a separate theorem

2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown through the equal sign, after which the square root is extracted from the left and right sides of the expression.

3 . This formula for finding the length of the diagonal of a trapezoid is similar to the previous one, with the difference that another diagonal is left on the left side of the expression

The next group of formulas (4-5) are similar in meaning and express a similar relationship.

The group of formulas (6-7) allows you to find the diagonal of a trapezoid if the larger base of the trapezoid, one side side and the angle at the base are known.

Formulas for finding the diagonals of a trapezoid through height

Note. This lesson provides solutions to geometry problems about trapezoids. If you have not found a solution to a geometry problem of the type you are interested in, ask a question on the forum.

Task.
The diagonals of the trapezoid ABCD (AD | | BC) intersect at point O. Find the length of the base BC of the trapezoid if the base AD = 24 cm, length AO = 9 cm, length OS = 6 cm.

Solution.
The solution to this problem is ideologically absolutely identical to the previous problems.

Triangles AOD and BOC are similar in three angles - AOD and BOC are vertical, and the remaining angles are pairwise equal, since they are formed by the intersection of one line and two parallel lines.

Since the triangles are similar, all their geometric dimensions are related to each other, just like the geometric dimensions of the segments AO and OC known to us according to the conditions of the problem. That is

AO/OC = AD/BC
9 / 6 = 24 / BC
BC = 24 * 6 / 9 = 16

Answer: 16 cm

Task .
In the trapezoid ABCD it is known that AD=24, BC=8, AC=13, BD=5√17. Find the area of ​​the trapezoid.

Solution .
To find the height of a trapezoid from the vertices of the smaller base B and C, we lower two heights to the larger base. Since the trapezoid is unequal, we denote the length AM = a, length KD = b ( not to be confused with the notation in the formula finding the area of ​​a trapezoid). Since the bases of the trapezoid are parallel, and we dropped two heights perpendicular to the larger base, then MBCK is a rectangle.

Means
AD = AM+BC+KD
a + 8 + b = 24
a = 16 - b

Triangles DBM and ACK are rectangular, so their right angles are formed by the altitudes of the trapezoid. Let us denote the height of the trapezoid by h. Then, by the Pythagorean theorem

H 2 + (24 - a) 2 = (5√17) 2
And
h 2 + (24 - b) 2 = 13 2

Let's take into account that a = 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 = 425
h 2 = 425 - (8 + b) 2

Let's substitute the value of the square of the height into the second equation obtained using the Pythagorean Theorem. We get:
425 - (8 + b) 2 + (24 - b) 2 = 169
-(64 + 16b + b) 2 + (24 - b) 2 = -256
-64 - 16b - b 2 + 576 - 48b + b 2 = -256
-64b = -768
b = 12

So KD = 12
Where
h 2 = 425 - (8 + b) 2 = 425 - (8 + 12) 2 = 25
h = 5

Find the area of ​​the trapezoid through its height and half the sum of the bases
, where a b - the base of the trapezoid, h - the height of the trapezoid
S = (24 + 8) * 5 / 2 = 80 cm 2

Answer: the area of ​​the trapezoid is 80 cm2.

A trapezoid is a special case of a quadrilateral in which one pair of sides is parallel. The term "trapezoid" comes from the Greek word τράπεζα, meaning "table", "table". In this article we will look at the types of trapezoid and its properties. In addition, we will figure out how to calculate individual elements of this For example, the diagonal of an isosceles trapezoid, the center line, area, etc. The material is presented in the style of elementary popular geometry, i.e. in an easily accessible form.

General information

First, let's figure out 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 for two non-adjacent sides. The main types of quadrilaterals are parallelogram, rectangle, rhombus, square, trapezoid and deltoid.

So let's get back to trapezoids. As we have already said, this figure has two parallel sides. They are called bases. The other two (non-parallel) are the lateral sides. In exam materials and various tests very often you can find problems related to trapezoids, the solution of which often requires the student to have knowledge not provided for in 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, in addition to this, the mentioned geometric figure has other features. But more about them a little 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. Her two angles are always equal to 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 equal in pairs.

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

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

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

The use of “extra-program” features of a geometric figure in the content school course- this is a task-based technology for teaching them. Constantly referring to the properties being studied while going through other topics allows students to gain a deeper knowledge of the trapezoid and ensures the success of solving assigned problems. So, let's start studying this wonderful figure.

Elements and properties of an isosceles trapezoid

As we have already noted, this geometric figure has equal sides. It is also known as the correct trapezoid. Why is it so remarkable and why did it get such a name? The peculiarity of this figure is that not only the sides and angles at the bases are equal, but also the diagonals. In addition, the sum of the angles of an isosceles trapezoid is 360 degrees. But that's not all! Of all the known trapezoids, only an isosceles one can be described as a circle. This is due to the fact that the sum of the opposite angles of this figure is equal to 180 degrees, and only under this condition can one describe a circle around a quadrilateral. The next property of the geometric figure under consideration is that the distance from the vertex of the base 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. Let us consider a solution to this problem, provided that the dimensions of the sides of the figure are known.

Solution

Typically, 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 equal to X, and the sizes of the bases are equal to Y and Z (smaller and larger, respectively). To carry out the calculation, it is necessary to draw the height H from angle B. The result is a right 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 = F. Now, to calculate the acute angle of the triangle, we use the cos function. We get the following entry: cos(β) = X/F. Now we calculate the angle: β=arcos (X/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 a second solution to this problem. First, we lower it from the corner to height H. We calculate the value of the leg BN. We know that the square of the hypotenuse right triangle equal to the sum of the squares of the legs. We get: BN = √(X2-F2). Next we use trigonometric function tg. As a result, we have: β = arctan (BN/F). An acute angle has been found. Next, we define it similarly to the first method.

Property of diagonals of an isosceles trapezoid

First, let's write down four rules. 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 middle line equal;

The center of the circle is the point at which ;

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

The quadrilateral that is 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 trapezoids

This topic is very convenient for studying the properties of this For example, the diagonals divide a trapezoid into four triangles, and those adjacent to the bases are similar, and those adjacent to the sides are equal in size. This statement can be called a property of the triangles into which the trapezoid is divided by its diagonals. The first part of this statement is proven through the sign of similarity at 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 are the bases of the trapezoid) is divided by 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 find 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, triangles BOS and AOB have a common height. We take the segments CO and OA as their bases. We get PBOS/PAOB = CO/OA = K and PAOB = PBOS/K. It follows from this that PSOD = PAOB.

To consolidate the material, students are recommended to find the 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 triangles BOS and AOD have equal areas; it is necessary to find the area of ​​the trapezoid. Since PSOD = PAOB, it means PABSD = PBOS+PAOD+2*PSOD. From the similarity of 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.

Properties of similarity

Continuing to develop this topic, one can prove other interesting features trapezoid. Thus, using similarity, one can prove the property of a segment that passes through the point formed by the intersection of the diagonals of this geometric figure, parallel to the bases. To do this, let's solve the following problem: we need to find the length of the segment RK that passes through 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/AC=RO/BS=AD/(BS+AD). From here we get that RO=BS*BP/(BS+BP). Similarly, from the similarity of triangles DOC and DBS, it follows that OK = BS*AD/(BS+AD). From here we get that RO=OK and RK=2*BS*AD/(BS+AD). A segment passing through the point of intersection of the diagonals, parallel to the bases and connecting two lateral sides, is divided in half by the point of intersection. Its length is the harmonic mean of the figure's bases.

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

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

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

Similarity findings

Thus, we have proven that:

1. The segment connecting the midpoints of the lateral sides of a 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 dividing the trapezoid into similar ones has the length of the geometric mean of the bases BS and AD.

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

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

A segment connecting the midpoints of the diagonals of a trapezoid

Consider the following property of this figure. We assume that the segment MH is parallel to the bases and bisects the diagonals. Let's call the intersection points Ш and Ш. This segment will be equal to half the difference of the bases. Let's look at this in more detail. MS is the middle line of the ABS triangle, it is equal to BS/2. MSH is the middle line of triangle ABD, it is equal to AD/2. Then we get that ShShch = MSh-MSh, therefore, ShShch = 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? You need to add the lower base to the upper base - in any direction, for example, to the right. And we extend the lower one by the length of the upper one to the left. Next, we connect them diagonally. The point of intersection of this segment with the midline 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 be inscribed in a circle only 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.

Corollaries of the incircle:

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

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

The first corollary is obvious, but to prove the second it is necessary to establish that the angle SOD is right, which, in fact, is also not difficult. But knowledge of this property will allow you to use a right triangle when solving problems.

Now let us specify these consequences for an isosceles trapezoid inscribed in a circle. We find that the height is the geometric mean of the bases of the figure: H=2R=√(BS*AD). While practicing the basic technique for solving problems for trapezoids (the principle of drawing two heights), the student must solve the following task. We assume that BT is the height of the isosceles figure ABSD. It is necessary to find the 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 AD. Since the circle is inscribed in a trapezoid, then BS+AD = 2AB or AB = (BS+AD)/2. From triangle ABN we find sinα = BN/AB = 2*BN/(BS+AD). PABSD = (BS+BP)*BN/2, BN=2R. We get PABSD = (BS+BP)*R, it follows that R = PABSD/(BS+BP).

All formulas for 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 = (A+B)/2.

2. Through height, base and corners:

M = A-H*(ctgα+ctgβ)/2;

M = B+N*(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/2Н = D1*D2*sinβ/2Н.

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

In this article we will try to reflect the properties of a trapezoid as fully as possible. In particular, we will talk about general signs and properties of a trapezoid, as well as about the properties of an inscribed trapezoid and about a circle inscribed in a trapezoid. We will also touch on the properties of the isosceles and rectangular trapezoid.

An example of solving a problem using the properties discussed will help you sort it into places in your head and better remember the material.

Trapeze and all-all-all

To begin with, let us briefly recall what a trapezoid is and what other concepts are associated with it.

So, a trapezoid is a quadrilateral figure, two of whose sides are parallel to each other (these are the bases). And the two are not parallel - these are the sides.

In a trapezoid, the height can be lowered - perpendicular to the bases. The center line and diagonals are drawn. It is also possible to draw a bisector from any angle of the trapezoid.

We will now talk about the various properties associated with all these elements and their combinations.

Properties of trapezoid diagonals

To make it clearer, while you are reading, sketch out the trapezoid ACME on a piece of paper and draw diagonals in it.

  1. If you find the midpoints of each of the diagonals (let's call these points X and T) and connect them, you get a segment. One of the properties of the diagonals of a trapezoid is that the segment HT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: ХТ = (a – b)/2.
  2. Before us is the same trapezoid ACME. The diagonals intersect at point O. Let's look at the triangles AOE and MOK, formed by segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient k of triangles is expressed through the ratio of the bases of the trapezoid: k = AE/KM.
    The ratio of the areas of triangles AOE and MOK is described by the coefficient k 2 .
  3. The same trapezoid, the same diagonals intersecting at point O. Only this time we will consider the triangles that the segments of the diagonals formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal in size - their areas are the same.
  4. Another property of a trapezoid involves the construction of diagonals. So, if you continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect at a certain point. Next, draw a straight line through the middle of the bases of the trapezoid. It intersects the bases at points X and T.
    If we now extend the line XT, then it will connect together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the middle of the bases X and T intersect.
  5. Through the point of intersection of the diagonals we will draw a segment that will connect the bases of the trapezoid (T lies on the smaller base KM, X on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OX = KM/AE.
  6. Now, through the point of intersection of the diagonals, we will draw a segment parallel to the bases of the trapezoid (a and b). The intersection point will divide it into two equal parts. You can find the length of the segment using the formula 2ab/(a + b).

Properties of the midline of a trapezoid

Draw the middle line in the trapezoid parallel to its bases.

  1. The length of the midline of a trapezoid can be calculated by adding the lengths of the bases and dividing them in half: m = (a + b)/2.
  2. If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.

Trapezoid Bisector Property

Select any corner of the trapezoid and draw a bisector. Let's take, for example, the angle KAE of our trapezoid ACME. Having completed the construction yourself, you can easily verify that the bisector cuts off from the base (or its continuation on a straight line outside the figure itself) a segment of the same length as the side.

Properties of trapezoid angles

  1. Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in the pair is always 180 0: α + β = 180 0 and γ + δ = 180 0.
  2. Let's connect the midpoints of the bases of the trapezoid with a segment TX. Now let's look at the angles at the bases of the trapezoid. If the sum of the angles for any of them is 90 0, the length of the segment TX can be easily calculated based on the difference in the lengths of the bases, divided in half: TX = (AE – KM)/2.
  3. If parallel lines are drawn through the sides of a trapezoid angle, they will divide the sides of the angle into proportional segments.

Properties of an isosceles (equilateral) trapezoid

  1. In an isosceles trapezoid, the angles at any base are equal.
  2. Now build a trapezoid again to make it easier to imagine what we're talking about. Look carefully at the base AE - the vertex of the opposite base M is projected to a certain point on the line that contains AE. The distance from vertex A to the projection point of vertex M and the middle line of the isosceles trapezoid are equal.
  3. A few words about the property of the diagonals of an isosceles trapezoid - their lengths are equal. And also the angles of inclination of these diagonals to the base of the trapezoid are the same.
  4. Only around an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral is 180 0 - a prerequisite for this.
  5. The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near the trapezoid, it is isosceles.
  6. From the features of an isosceles trapezoid follows the property of the height of a trapezoid: if its diagonals intersect at right angles, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
  7. Again, draw the segment TX through the midpoints of the bases of the trapezoid - in an isosceles trapezoid it is perpendicular to the bases. And at the same time TX is the axis of symmetry of an isosceles trapezoid.
  8. This time, lower the height from the opposite vertex of the trapezoid onto the larger base (let's call it a). You will get two segments. The length of one can be found if the lengths of the bases are added and divided in half: (a + b)/2. We get the second one when we subtract the smaller one from the larger base and divide the resulting difference by two: (a – b)/2.

Properties of a trapezoid inscribed in a circle

Since we are already talking about a trapezoid inscribed in a circle, let us dwell on this issue in more detail. In particular, on where the center of the circle is in relation to the trapezoid. Here, too, it is recommended that you take the time to pick up a pencil and draw what will be discussed below. This way you will understand faster and remember better.

  1. The location of the center of the circle is determined by the angle of inclination of the diagonal of the trapezoid to its side. For example, a diagonal may extend from the top of a trapezoid at right angles to the side. In this case, the larger base intersects the center of the circumscribed circle exactly in the middle (R = ½AE).
  2. The diagonal and the side can also meet under acute angle– then the center of the circle is inside the trapezoid.
  3. The center of the circumscribed circle may be outside the trapezoid, beyond its larger base, if there is an obtuse angle between the diagonal of the trapezoid and the side.
  4. The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half the central angle that corresponds to it: MAE = ½MOE.
  5. Briefly about two ways to find the radius of a circumscribed circle. Method one: look carefully at your drawing - what do you see? You can easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found by the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R = AE/2*sinAME. In a similar way, the formula can be written for any of the sides of both triangles.
  6. Method two: find the radius of the circumscribed circle through the area of ​​the triangle formed by the diagonal, side and base of the trapezoid: R = AM*ME*AE/4*S AME.

Properties of a trapezoid circumscribed about a circle

You can fit a circle into a trapezoid if one condition is met. Read more about it below. And together this combination of figures has a number of interesting properties.

  1. If a circle is inscribed in a trapezoid, the length of its midline can be easily found by adding the lengths of the sides and dividing the resulting sum in half: m = (c + d)/2.
  2. For the trapezoid ACME, described about a circle, the sum of the lengths of the bases is equal to the sum of the lengths of the sides: AK + ME = KM + AE.
  3. From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in a trapezoid whose sum of bases is equal to the sum of its sides.
  4. The tangent point of a circle with radius r inscribed in a trapezoid divides the side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
  5. And one more property. To avoid confusion, draw this example yourself too. We have the good old trapezoid ACME, described around a circle. It contains diagonals that intersect at point O. The triangles AOK and EOM formed by the segments of the diagonals and the lateral sides are rectangular.
    The heights of these triangles, lowered to the hypotenuses (i.e., the lateral sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid coincides with the diameter of the inscribed circle.

Properties of a rectangular trapezoid

A trapezoid is called rectangular if one of its angles is right. And its properties stem from this circumstance.

  1. A rectangular trapezoid has one of its sides perpendicular to its base.
  2. Height and lateral side of the trapezoid adjacent to right angle, are equal. This allows you to calculate the area of ​​a rectangular trapezoid ( general formula S = (a + b) * h/2) not only through the height, but also through the side adjacent to the right angle.
  3. For a rectangular trapezoid, the general properties of the diagonals of a trapezoid already described above are relevant.

Evidence of some properties of the trapezoid

Equality of angles at the base of an isosceles trapezoid:

  • You probably already guessed that here we will need the AKME trapezoid again - draw an isosceles trapezoid. Draw a straight line MT from vertex M, parallel to the side of AK (MT || AK).

The resulting quadrilateral AKMT is a parallelogram (AK || MT, KM || AT). Since ME = KA = MT, ∆ MTE is isosceles and MET = MTE.

AK || MT, therefore MTE = KAE, MET = MTE = KAE.

Where does AKM = 180 0 - MET = 180 0 - KAE = KME.

Q.E.D.

Now, based on the property of an isosceles trapezoid (equality of diagonals), we prove that trapezoid ACME is isosceles:

  • To begin with, let’s draw a straight line MX – MX || KE. We obtain a parallelogram KMHE (base – MX || KE and KM || EX).

∆AMX is isosceles, since AM = KE = MX, and MAX = MEA.

MH || KE, KEA = MHE, therefore MAE = MHE.

It turned out that triangles AKE and EMA are equal to each other, because AM = KE and AE – common side two triangles. And also MAE = MXE. We can conclude that AK = ME, and from this it follows that the trapezoid AKME is isosceles.

Review task

The bases of the trapezoid ACME are 9 cm and 21 cm, the side side KA, equal to 8 cm, forms an angle of 150 0 with the smaller base. You need to find the area of ​​the trapezoid.

Solution: From vertex K we lower the height to the larger base of the trapezoid. And let's start looking at the angles of the trapezoid.

Angles AEM and KAN are one-sided. This means that in total they give 180 0. Therefore, KAN = 30 0 (based on the property of trapezoidal angles).

Let us now consider the rectangular ∆ANC (I believe this point is obvious to readers without additional evidence). From it we will find the height of the trapezoid KH - in a triangle it is a leg that lies opposite the angle of 30 0. Therefore, KH = ½AB = 4 cm.

We find the area of ​​the trapezoid using the formula: S ACME = (KM + AE) * KN/2 = (9 + 21) * 4/2 = 60 cm 2.

Afterword

If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the given properties with a pencil in your hands and analyze them in practice, you should have mastered the material well.

Of course, there is a lot of information here, varied and sometimes even confusing: it is not so difficult to confuse the properties of the described trapezoid with the properties of the inscribed one. But you yourself have seen that the difference is huge.

Now you have a detailed summary of all general properties trapezoids. As well as specific properties and characteristics of isosceles and rectangular trapezoids. It is very convenient to use to prepare for tests and exams. Try it yourself and share the link with your friends!

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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 – middle line triangle ABC and equal to 1/2ВС (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 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 the 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.

Deciding this system, we get 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 (average arithmetic 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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