Trapezoid-like definition. Remember and apply the properties of a trapezoid

In various materials tests and exams are very common 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 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 (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 described 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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A trapezoid is a special case of a quadrilateral in which one pair of sides is parallel. The term "trapezoid" comes from Greek wordτράπεζα, 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 the materials of exams and various tests, you can often 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. 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 (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 during the 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 the 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 the 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). Acute angle 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 midline are equal;

The center of the circle is the point at which ;

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

The quadrilateral that is formed by the points of tangency, 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: it is necessary 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-sized 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/2N = D1*D2*sinβ/2N.

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

Tra-pe-tion

1. Trapezoid and its types

Definition

Tra-pe-tion- this is a four-corner, which has two hundred parallel lines, but the other two do not.

In Fig. 1. The image is made in a free-form manner. - these are the other sides (those that are not parallel). - basics (parallel aspects).

Rice. 1. Tra-pe-tion

If we compare the trape-tion with the par-ral-le-lo-gram, then the par-le-lo-gram has two pairs of parallel sides. That is, the parallel-le-lo-gram is not a special case of tra-pe-tion, since in the definition of tra-pe-tion it is clearly -for-but that the two sides of the tra-pe-tions are not parallel.

You de-lim some types of traps (special cases):

2. The midline of the trapezoid and its properties

Definition

Midline of the trap- from a cut that connects the three sides.

In Fig. 2. image on a trapezoid with a middle line.

Rice. 2. Midline of the trap

Properties of the middle line of the trap:

1. The middle line of the tra-pe-tion pa-ral-lel-na os-no-va-ni-yam tra-pe-tion.

Proof:

Let se-re-di-na bo-ko-voy hundred-ro-ny tra-pe-tions - point. Let's pass through this point a straight line, a parallel os-no-va-ni-yam. This straight line crosses the second side of the line at point .

According to the structure: . According to the theory of Fa-le-sa, this follows: . It means, - se-re-di-a hundred-ro-ny. That means it's the middle line.

Do-ka-za-but.

2. The middle line of the tra-pe-tion is equal to the sum of the main tra-pe-tion: .

Proof:

We draw the middle line of the trapezium and one of the dia-go-na-leys: for example, (see Fig. 3).

According to the theory of Fa-le-sa, the parallel straight lines from the sides of the corner are pro-por-tsi-o-nal from the cut ki. Since the cuttings are equal: . This means that from the re-zok there is an average tri-corner, and from the re-zok there is an average tri-corner no way.

Means, .

Note: this follows from the property of the middle line of the triangle: the middle line of the triangle is par-ral-on-axis but-va-niyu and equal to his lo-vina. The first part of this property is analogous to the first property of the middle line of travel. tions, and the second part can be shown (for example, for the middle line of a triangle), passing through a straight line point, pa- ral-lel-nuyu. From the theory of Fa-le-sa it will follow that this straight line will be the middle line, and the image will be you-rekh-coal-nick - pa-ral-le-lo-gram-mom (two pairs of pair-but-par-ral-le-l-nyh sides). From here it’s no longer difficult to get my property.

Let's eat: .

Do-ka-za-but.

Let us now take a closer look at the main types of trappings and their properties.

3. Signs of an isosceles trapezoid

Let us remember that an equal-poor-ren-trap-tion is a trap-pe-tsion in which both sides are equal. Let's look at the properties of bo-ko-voy tra-pe-tions.

1. The angles at the base of the equal-to-be-ren-noy tra-pe-tion are equal.

Proof:

This is a completely standard, complete construction, which is very often used when solving problems -personal tasks on the trap: we will carry out a direct parallel-but-on-the-side side (see Fig. 4).

Parallelogram.

From here it follows that: . This means that the triangle is equal. This means that the angles at its base are equal, that is: (the last two angles are equal, as corresponding to parallel lines mykh).

Do-ka-za-but.

2. Dia-go-on-whether equal-bed-ren-noy tra-pe-tions are equal.

Proof:

To achieve this property, we use the previous one. Indeed, consider the triangle: and (see Fig. 5.).

(based on the first sign of the equality of triangles: two sides and the angle between them).

From this equality it immediately follows that: .

Do-ka-za-but.

It turns out that, as in the case of the par-ral-le-lo-gram, the equal-bed-ren-tra-pe-tion has the same properties -but-from time-to-time they appear and recognize. Let's formulate and figure out these signs.

Signs of equal-bad-ren-tra-pe-tion

1. Given: - tra-pe-tion; .

Prove:

Proof:

Before-ka-za-tel-stvo is given ab-so-lute-but ana-logic-but before-ka-za-tel-stvu with-from-vet-st-stv- y-y properties. Let's move in the trap in a straight line parallel to the side (see Fig. 6).

(corresponding angles with parallel lines). From-where-yes, using the condition-vi-e, po-lu-cha-e: - equally-poor-ren-ny

(the angles at the axis are equal). Mean-cheat: (in par-ral-le-lo-gram-ma the pro-ti-vo-false hundred-ro-ns are equal).

Do-ka-za-but.

2. Given: - tra-pe-tion; .

Prove: .

Proof:

You have completed one more standard, complete construction when solving problems with tra-pe-tsi: let's do it through top-shi-well straight par-ral-lel-but dia-go-na-li (see Fig. 7).

Par-ral-le-lo-gram (two pairs of par-but par-ral-lele-nyh sides).

(corresponding angles for parallel lines). In addition, - equally-poor-ren-ny (- by condition; - by property of par-le-lo-gram). Which means: .

Do-ka-za-but.

4. Examples of problems

Let's look at several examples of solving problems with traps.

Example 1.

Given: - tra-pe-tion; .

Solution:

The sum of the angles at the side of the trap is equal - the property of internal one-sided angles at parallel lines. From this fact we can obtain two equalities:

Example 2.

Given: - tra-pe-tion; . .

Solution:

Let's talk about you. I'm eating a four-square-corner, in which the pro-ti-false sides are in pairs, but par-ral-lel- us, and two angles are equal in . It means, - par-ral-le-lo-gram, or more precisely, rectangular.

It follows from this that . Where: .

Consider a right-angled triangle. In it, one of the acute angles, by condition, is equal to . This means that the second one is equal to , that is: . It takes advantage of the property of the ka-te-ta, lying opposite the corner: it is half the size of the gi-po-te-nu-zy.

In this lesson, we looked at the trap and its properties, studied the types of trap, and also decided on several -measures of certain tasks.

SOURCE

http://interneturok.ru/ru/school/geometry/8-klass/chyotyrehugolniki/trapetsiya

http://img3.proshkolu.ru/content/media/pic/std/1000000/983000/982960-b6b4e8f6a4e7b336.jpg

http://static.wixstatic.com/media/13679f_7ac2889143594b059462e77b25eda7c6.jpg

http://delaem-uroki.narod.ru/img/102/792/KZqhOMb.gif

Trapezoid. Trapezoid midline task.

http://cs323223.vk.me/v323223595/5e51/Gi2qlTPgLVo.jpg

http://dok.opredelim.com/pars_docs/refs/47/46420/img2.jpg