Trapeze and its types. Trapeze. Full lessons - Knowledge Hypermarket. Attention exercise

A polygon is a part of a plane bounded by a closed broken line. The corners of a polygon are indicated by the points of the vertices of the polyline. Polygon corner vertices and polygon vertices are congruent points.

Definition. A parallelogram is a quadrilateral whose opposite sides are parallel.

Parallelogram Properties

1. Opposite sides are equal.
On fig. eleven AB = CD; BC = AD.

2. Opposite angles are equal (two acute and two obtuse angles).
On fig. 11∠ A = ∠C; ∠B = ∠D.

3 Diagonals (line segments connecting two opposite vertices) intersect and the intersection point is divided in half.

On fig. 11 segments AO = OC; BO = OD.

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

Parallel sides called her grounds, and the other two sides sides.

Types of trapezium

1. Trapeze, whose sides are not equal,
called versatile(Fig. 12).

2. A trapezoid whose sides are equal is called isosceles(Fig. 13).

3. A trapezoid, in which one side makes a right angle with the bases, is called rectangular(Fig. 14).

The segment connecting the midpoints of the sides of the trapezoid (Fig. 15) is called the midline of the trapezoid ( MN). The median line of the trapezoid is parallel to the bases and equal to half their sum.

A trapezoid can be called a truncated triangle (Fig. 17), therefore the names of trapeziums are similar to the names of triangles (triangles are versatile, isosceles, rectangular).

Area of ​​a parallelogram and a trapezoid

Rule. Parallelogram area is equal to the product of its side by the height drawn to this side.

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Therefore, we will call one of them big , second - small base trapezoid. Height a trapezoid can be called any segment of a perpendicular drawn from the vertices to the corresponding opposite side (for each vertex there are two opposite sides), enclosed between the taken vertex and the opposite side. But it is possible to single out a "special type" of heights.
Definition 8. The height of the base of a trapezoid is the segment of a straight line perpendicular to the bases, enclosed between the bases.
Theorem 7 . The median line of the trapezoid is parallel to the bases and equal to half their sum.
Proof. Let a trapezoid ABCD be given and middle line KM. Draw a line through points B and M. We continue the side AD through point D until it intersects with BM. Triangles BCm and MPD are equal in side and two angles (CM=MD, ∠ BCM=∠ MDP - overlapping, ∠ BMC=∠ DMP - vertical), so VM=MP or point M is the midpoint of BP. KM is the midline in the triangle ABP. According to the property of the middle line of the triangle, KM is parallel to AP and in particular AD and is equal to half of AP:

Theorem 8 . The diagonals divide the trapezoid into four parts, two of which, adjacent to the sides, are equal.
Let me remind you that figures are called equal if they have the same area. Triangles ABD and ACD are equal: they have equal heights (indicated in yellow) and a common base. These triangles are general part AOD. Their area can be expanded as follows:

Types of trapezium:
Definition 9. (Figure 1) An acute-angled trapezoid is a trapezoid in which the angles adjacent to the larger base are acute.
Definition 10. (Figure 2) An obtuse trapezoid is a trapezoid in which one of the angles adjacent to the larger base is obtuse.
Definition 11. (Figure 4) A trapezoid is called rectangular, in which one side is perpendicular to the bases.
Definition 12. (Figure 3) Isosceles (isosceles, isosceles) is a trapezoid, in which the sides are equal.

Properties of an isosceles trapezoid:
Theorem 10 . The angles adjacent to each of the bases of an isosceles trapezoid are equal.
Proof. Let us prove, for example, the equality of angles A and D with a larger base AD of an isosceles trapezoid ABCD. For this purpose, we draw a straight line through the point C parallel to the lateral side AB. It will intersect the large base at point M. The quadrilateral ABCM is a parallelogram, because by construction it has two pairs of parallel sides. Therefore, the segment CM of the secant line enclosed inside the trapezoid is equal to its lateral side: CM=AB. From here it is clear that CM=CD, the triangle CMD is isosceles, ∠CMD=∠CDM, and, therefore, ∠A=∠D. The angles adjacent to the smaller base are also equal, because are for those found internal one-sided and have a sum of two lines.
Theorem 11 . The diagonals of an isosceles trapezoid are equal.
Proof. Consider triangles ABD and ACD. It is equal on two sides and the angle between them (AB=CD, AD is common, angles A and D are equal according to Theorem 10). Therefore AC=BD.

Theorem 13 . The diagonals of an isosceles trapezoid are divided by the intersection point into correspondingly equal segments. Consider triangles ABD and ACD. It is equal on two sides and the angle between them (AB=CD, AD is common, angles A and D are equal according to Theorem 10). Therefore, ∠ ОАD=∠ ОDA, hence the angles ОВС and OSV are equal as correspondingly overlapping angles ODA and ОАD. Recall the theorem: if two angles in a triangle are equal, then it is isosceles, therefore triangles ОВС and ОAD are isosceles, which means OS=OB and ОА=OD, etc.
An isosceles trapezoid is a symmetrical figure.
Definition 13. The axis of symmetry of an isosceles trapezoid is called a straight line passing through the midpoints of its bases.
Theorem 14 . The symmetry axis of an isosceles trapezoid is perpendicular to its bases.
In Theorem 9, we proved that the line joining the midpoints of the bases of a trapezoid passes through the intersection point of the diagonals. Next (Theorem 13) we proved that the triangles AOD and BOC are isosceles. OM and OK are the medians of these triangles, respectively, by definition. Recall the property of an isosceles triangle: the median of an isosceles triangle, lowered to the base, is also the height of the triangle. Due to the perpendicularity of the bases of the parts of the straight line KM, the axis of symmetry is perpendicular to the bases.
Signs that distinguish an isosceles trapezoid among all trapeziums:
Theorem 15 . If the angles adjacent to one of the bases of a trapezoid are equal, then the trapezoid is isosceles.
Theorem 16 . If the diagonals of a trapezoid are equal, then the trapezoid is isosceles.
Theorem 17 . If the lateral sides of the trapezoid, extended to the intersection, form an isosceles triangle together with its large base, then the trapezoid is isosceles.
Theorem 18 . If a trapezoid can be inscribed in a circle, then it is isosceles.
Sign of a rectangular trapezoid:
Theorem 19 . Any quadrilateral with only two right angles at adjacent vertices is a right-angled trapezoid (obviously, two sides are parallel, because one-sided are equal. in the case when three right angles are a rectangle)
Theorem 20 . The radius of a circle inscribed in a trapezoid is equal to half the height of the base.
The proof of this theorem is to explain that the radii drawn to the bases lie at the height of the trapezoid. From the point O - the center of the circle ABCD inscribed in this trapezoid, we draw the radii to the points of contact with its bases of the trapezoid. As you know, the radius drawn to the point of contact is perpendicular to the tangent, therefore OK ^ BC and OM ^ AD. Recall the theorem: if a line is perpendicular to one of the parallel lines, then it is also perpendicular to the second. Hence, the line OK is also perpendicular to AD. Thus, two lines perpendicular to the line AD pass through the point O, which cannot be, therefore these lines coincide and make up the common perpendicular of the KM, which is equal to the sum two radii and is the diameter of the inscribed circle, so r=KM/2 or r=h/2.
Theorem 21 . The area of ​​a trapezoid is equal to the product of half the sum of the bases and the height of the bases.

Proof: Let ABCD be a given trapezoid and AB and CD be its bases. Let also AH be the height dropped from point A to line CD. Then S ABCD = S ACD + S ABC .
But S ACD = 1/2AH CD and S ABC = 1/2AH AB.
Therefore, S ABCD = 1/2AH (AB + CD).
Q.E.D.

The second formula has moved from the quadrilateral.

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

Definitions

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

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

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

Theorems: properties of a trapezoid

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

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

Proof

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

2) Because \(AD\parallel BC\) and \(BD\) is a secant, then \(\angle DBC=\angle BDA\) as lying across.
Also \(\angle BOC=\angle AOD\) as vertical.
Therefore, in two corners \(\triangle BOC \sim \triangle AOD\).

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

Definition

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

Theorem

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


Proof*

1) Let's prove the parallelism.


Draw a line \(MN"\parallel AD\) (\(N"\in CD\) ) through the point \(M\) ). Then, by the Thales theorem (because \(MN"\parallel AD\parallel BC, AM=MB\)) the point \(N"\) is the midpoint of the segment \(CD\)... Hence, the points \(N\) and \(N"\) will coincide.

2) Let's prove the formula.

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


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

Because \(MN\parallel AD\parallel BC\) and \(BB", CC"\perp AD\) , then \(B"M"N"C"\) and \(BM"N"C\) are rectangles. By the Thales theorem, \(MN\parallel AD\) and \(AM=MB\) imply that \(B"M"=M"B\) . Hence, \(B"M"N"C"\) and \(BM"N"C\) are equal rectangles, hence \(M"N"=B"C"=BC\) .

In this way:

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

Theorem: property of an arbitrary trapezoid

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


Proof*
It is recommended that you familiarize yourself with the proof after studying the topic “Similar Triangles”.

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


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

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

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

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

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


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

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

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

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

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

Definitions

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

A trapezoid is called isosceles if its sides are equal.

Theorems: properties of an isosceles trapezoid

1) An isosceles trapezoid has equal base angles.

2) The diagonals of an isosceles trapezoid are equal.

3) The two triangles formed by the diagonals and the base are isosceles.

Proof

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

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

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

2)

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

3) Because \(\triangle ABD=\triangle ACD\), then \(\angle BDA=\angle CAD\) . Therefore, the triangle \(\triangle AOD\) is isosceles. It can be proved similarly that \(\triangle BOC\) is isosceles.

Theorems: signs of an isosceles trapezoid

1) If the angles at the base of a trapezoid are equal, then it is isosceles.

2) If the diagonals of a trapezoid are equal, then it is isosceles.

Proof

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


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

Eventually \(AB = AE - BE = DE - CE = CD\), i.e. \(AB = CD\) , which was to be proved.

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


Because \(AC=BD\) , then \(x+kx=y+ky \Rightarrow x=y\) . So \(\triangle AOD\) is isosceles and \(\angle OAD=\angle ODA\) .

Thus, according to the first sign \(\triangle ABD=\triangle ACD\) (\(AC=BD, \angle OAD=\angle ODA, AD\)- general). So \(AB=CD\) , so.

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 find 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. Rectangular trapezoid- this 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. Essentially, there is no need to type in theoretical course geometry of the new properties of this figure. They can be discovered and formulated in the process of solving various tasks(better than system). 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 the 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 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 straight 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 right triangle is equal to the sum of the squares of the legs. We get: BN \u003d √ (X2-F2). Next, we use 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 those adjacent to the sides 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. Their intersection point 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 relationship between the areas of the triangles obtained, 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 line segment passing through the point where the diagonals intersect parallel to bases and connecting the two sides, is bisected by the intersection point. 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/AD. 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 the following similarity property. 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 given property allows you to use a right triangle when 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 top B to the base 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.

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