The center of a circle circumscribed about an isosceles trapezoid. Diagonals of a trapezoid

Design work "Interesting properties of a trapezoid" Completed by: 10th grade students Kudzaeva Elina Bazzaeva Diana MKOU secondary school p. N.Batako Head: Gagieva A.O. 11/20/2015

Purpose of the work: To consider the properties of the trapezoid, which are not studied in the school geometry course, but when solving the geometric problems of the Unified State Examination from the expanded part C 4, it may be necessary to know and be able to apply precisely these properties.

Properties of a trapezoid: If a trapezoid is divided by a straight line parallel to its bases, equal to a and b, into two equal-sized trapezoids. Then the segment to this straight line, enclosed between the sides, is equal to a B k

Property of a segment passing through the point of intersection of the diagonals of a trapezoid. The segment parallel to the bases, passing through the point of intersection of the diagonals is: a in c

Properties of a trapezoid: A segment of a straight line parallel to the bases of a trapezoid, enclosed inside the trapezoid, is divided by its diagonals into three parts. Then the segments adjacent to the sides are equal to each other. MP=OK R M O K

Properties of an isosceles trapezoid: If a circle can be inscribed in a trapezoid, then the radius of the circle is the average proportional to the segments into which the tangent point divides the side. O S W A D. E O

Properties of an isosceles trapezoid: If the center of the circumscribed circle lies on the base of the trapezoid, then its diagonal is perpendicular to the side O A B C D

Properties of an isosceles trapezoid: A circle can be inscribed in an isosceles trapezoid if the lateral side is equal to its midline. C V A D h

1) If the condition of the problem says that a circle is inscribed in a rectangular trapezoid, the following properties can be used: 1. The sum of the bases of the trapezoid is equal to the sum of the sides. 2. The distances from the vertex of the trapezoid to the tangent points of the inscribed circle are equal. 3. The height of a rectangular trapezoid is equal to its smaller lateral side and equal to the diameter of the inscribed circle. 4. The center of the inscribed circle is the intersection point of the bisectors of the angles of the trapezoid. 5. If the tangent point divides the lateral side into segments m and n, then the radius of the inscribed circle is equal to

Properties of a rectangular trapezoid into which a circle is inscribed: 1) A quadrilateral formed by the center of the inscribed circle, the tangent points and the vertex of the trapezoid is a square whose side is equal to the radius. (AMOE and BKOM are squares with side r). 2) If a circle is inscribed in a rectangular trapezoid, then the area of the trapezoid is equal to the product of its bases: S=AD*BC

Proof: The area of a trapezoid is equal to the product of half the sum of its bases and its height: Denote CF=m , FD=n . Since the distances from the vertices to the points of contact are equal, the height of the trapezoid is equal to two radii of the inscribed circle, and

I. The bisectors of the angles at the lateral side of the trapezoid intersect at an angle of 90º. 1)∠ABC+∠BAD=180º (as internal one-sided with AD∥BC and secant AB). 2) ∠ABK+∠KAB=(∠ABC+∠BAD):2=90º (because the bisectors bisect the angles). 3) Since the sum of the angles of a triangle is 180º, in triangle ABK we have: ∠ABK+∠KAB+∠AKB=180º, hence ∠AKB=180-90=90º. Conclusion: The bisectors of the angles at the lateral side of the trapezoid intersect at right angles. This statement is used in solving problems on a trapezoid in which a circle is inscribed.

I I Let the bisector of angle ABC intersect side AD at point S. Then triangle ABS is isosceles with base BS. Hence, its bisector AK is also a median, that is, point K is the midpoint of BS. If M and N are the midpoints of the sides of the trapezoid, then MN is the midline of the trapezoid and MN∥AD. Since M and K are the midpoints of AB and BS, MK is the midline of the triangle ABS and MK∥AS. Since only one line can be drawn through the point M and parallel to the given one, the point K lies on the midline of the trapezoid.

III. The point of intersection of the bisectors of acute angles at the base of the trapezoid belongs to another base. In this case, triangles ABK and DCK are isosceles triangles with bases AK and DK, respectively. Thus, BC=BK+KC=AB+CD. Conclusion: If the bisectors of the acute angles of a trapezoid intersect at a point belonging to the smaller base, then the smaller base is equal to the sum of the sides of the trapezoid. In an isosceles trapezoid, in this case, the smaller base is twice the size of the lateral side.

I V. The point of intersection of the bisectors of obtuse angles at the base of the trapezoid belongs to another base. In this case triangles ABF and DCF are isosceles triangles with bases BF and CF respectively. Hence AD=AF+FD=AB+CD. Conclusion: If the bisectors of the obtuse angles of a trapezoid intersect at a point belonging to the larger base, then the larger base is equal to the sum of the sides of the trapezoid. An isosceles trapezoid in this case has a larger base twice the side.

If an isosceles trapezoid with sides a, b, c, d can be inscribed and circles can be circumscribed around it, then the area of the trapezoid is

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

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 be \(BB"\cap MN=M", CC"\cap MN=N"\).

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 \(\triangle DCC"\) . So: \

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 \(M"N"=B"C"=BC\) .

Thus:

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

Theorem: property of an arbitrary trapezoid

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

Proof*
It is recommended that you familiarize yourself with the proof after studying the topic “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, \(\angle DAB = \angle CDA\) .

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.

In this article, we will try to reflect the properties of the trapezoid as fully as possible. In particular, we will talk about the 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 an isosceles and rectangular trapezoid.

An example of solving a problem using the considered properties will help you sort things out in your head and better remember the material.

Trapeze and all-all-all

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

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

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

About the various properties associated with all these elements and their combinations, we will now talk.

Properties of the diagonals of a trapezoid

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

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 XT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: XT \u003d (a - b) / 2.
Before us is the same ACME trapezoid. The diagonals intersect at point O. Let's consider the triangles AOE and IOC formed by the segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient of k triangles is expressed in terms of the ratio of the bases of the trapezoid: k = AE/KM.
The ratio of the areas of triangles AOE and IOC is described by the coefficient k 2 .
All the same trapezium, the same diagonals intersecting at point O. Only this time we will consider triangles that the diagonal segments formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal - their areas are the same.
Another property of a trapezoid includes the construction of diagonals. So, if we continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect to some point. Next, draw a straight line through the midpoints 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 join together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the midpoints of the bases of X and T intersect.
Through the point of intersection of the diagonals, we draw a segment that will connect the bases of the trapezoid (T lies on the smaller base of KM, X - on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OH = KM/AE.
And now through the point of intersection of the diagonals we 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 a segment using the formula 2ab/(a + b).

Properties of the midline of a trapezoid

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

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.
If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.

Property of the bisector of a trapezoid

Pick any angle of the trapezoid and draw a bisector. Take, for example, the angle KAE of our trapezoid ACME. Having completed the construction on your own, you can easily see 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.

Trapezoid angle properties

Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in a pair is always 180 0: α + β = 180 0 and γ + δ = 180 0 .
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 TX segment is easy to calculate based on the difference in the lengths of the bases, divided in half: TX \u003d (AE - KM) / 2.
If parallel lines are drawn through the sides of the angle of a trapezoid, they will divide the sides of the angle into proportional segments.

Properties of an isosceles (isosceles) trapezoid

In an isosceles trapezoid, the angles at any of the bases are equal.
Now build a trapezoid again to make it easier to imagine what it is about. Look carefully at the base of AE - the vertex of the opposite base of 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 midline of an isosceles trapezoid are equal.
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.
Only near an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral 180 0 is a prerequisite for this.
The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near a trapezoid, it is isosceles.
From the features of an isosceles trapezoid, the property of the height of a trapezoid follows: if its diagonals intersect at a right angle, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
Draw the line TX again 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.
This time lower to the larger base (let's call it a) the height from the opposite vertex of the trapezoid. You will get two cuts. 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's dwell on this issue in more detail. In particular, where is the center of the circle in relation to the trapezoid. Here, too, it is recommended not to be too lazy to pick up a pencil and draw what will be discussed below. So you will understand faster, and remember better.

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 emerge 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).
The diagonal and the side can also meet at an acute angle - then the center of the circle is inside the trapezoid.
The center of the circumscribed circle may be outside the trapezium, beyond its large base, if there is an obtuse angle between the diagonal of the trapezoid and the lateral side.
The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half of the central angle that corresponds to it: MAE = ½MY.
Briefly about two ways to find the radius of the circumscribed circle. Method one: look carefully at your drawing - what do you see? You will easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found through the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * sinAME. Similarly, the formula can be written for any of the sides of both triangles.
Method two: we find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R \u003d AM * ME * AE / 4 * S AME.

Properties of a trapezoid circumscribed about a circle

You can inscribe a circle in a trapezoid if one condition is met. More about it below. And together this combination of figures has a number of interesting properties.

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.
For a trapezoid ACME, circumscribed 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.
From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in that trapezoid, the sum of the bases of which is equal to the sum of the sides.
The tangent point of a circle with radius r inscribed in a trapezoid divides the lateral side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
And one more property. In order not to get confused, draw this example yourself. We have the good old ACME trapezoid, circumscribed around a circle. Diagonals are drawn in it, intersecting at the point O. The triangles AOK and EOM formed by the segments of the diagonals and the sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid is the same as the diameter of the inscribed circle.

Properties of a rectangular trapezoid

A trapezoid is called rectangular, one of the corners of which is right. And its properties stem from this circumstance.

A rectangular trapezoid has one of the sides perpendicular to the bases.
The height and side of the trapezoid adjacent to the 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.
For a rectangular trapezoid, the general properties of the trapezoid diagonals already described above are relevant.

Proofs of some properties of a trapezoid

Equality of angles at the base of an isosceles trapezoid:

You probably already guessed that here we again need the ACME trapezoid - draw an isosceles trapezoid. Draw a 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 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 trapezium ACME is isosceles:

To begin with, let's draw a straight line МХ – МХ || KE. We get a parallelogram KMHE (base - MX || KE and KM || EX).

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

MX || KE, KEA = MXE, therefore MAE = MXE.

It turned out that the triangles AKE and EMA are equal to each other, because AM \u003d KE and AE is the common side of the two triangles. And also MAE \u003d MXE. We can conclude that AK = ME, and hence it follows that the trapezoid AKME is isosceles.

Task to repeat

The bases of the trapezoid ACME are 9 cm and 21 cm, the side of the KA, equal to 8 cm, forms an angle of 150 0 with a 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. Which means they add up to 1800. Therefore, KAN = 30 0 (based on the property of the angles of the trapezoid).

Consider now the rectangular ∆ANK (I think this point is obvious to readers without further proof). From it we find the height of the trapezoid KH - in a triangle it is a leg, which lies opposite the angle of 30 0. Therefore, KN \u003d ½AB \u003d 4 cm.

The area of the trapezoid is found by the formula: S AKME \u003d (KM + AE) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.

Afterword

If you carefully and thoughtfully studied this article, were not too lazy to draw trapezoids for all the above 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 saw that the difference is huge.

Now you have a detailed summary of all the general properties of a trapezoid. As well as specific properties and features 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!

site, with full or partial copying of the material, a link to the source is required.

An example of solving a problem using the considered properties will help you sort things out in your head and better remember the material.

Trapeze and all-all-all

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

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

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

About the various properties associated with all these elements and their combinations, we will now talk.

Properties of the diagonals of a trapezoid

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

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 XT lies on the midline. And its length can be obtained by dividing the difference of the bases by two: XT \u003d (a - b) / 2.
Before us is the same ACME trapezoid. The diagonals intersect at point O. Let's consider the triangles AOE and IOC formed by the segments of the diagonals together with the bases of the trapezoid. These triangles are similar. The similarity coefficient of k triangles is expressed in terms of the ratio of the bases of the trapezoid: k = AE/KM.
The ratio of the areas of triangles AOE and IOC is described by the coefficient k 2 .
All the same trapezium, the same diagonals intersecting at point O. Only this time we will consider triangles that the diagonal segments formed together with the sides of the trapezoid. The areas of triangles AKO and EMO are equal - their areas are the same.
Another property of a trapezoid includes the construction of diagonals. So, if we continue the sides of AK and ME in the direction of the smaller base, then sooner or later they will intersect to some point. Next, draw a straight line through the midpoints 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 join together the point of intersection of the diagonals of the trapezoid O, the point at which the extensions of the sides and the midpoints of the bases of X and T intersect.
Through the point of intersection of the diagonals, we draw a segment that will connect the bases of the trapezoid (T lies on the smaller base of KM, X - on the larger AE). The intersection point of the diagonals divides this segment in the following ratio: TO/OH = KM/AE.
And now through the point of intersection of the diagonals we 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 a segment using the formula 2ab/(a + b).

Properties of the midline of a trapezoid

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

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.
If you draw any segment (height, for example) through both bases of the trapezoid, the middle line will divide it into two equal parts.

Property of the bisector of a trapezoid

Trapezoid angle properties

Whichever of the two pairs of angles adjacent to the side you choose, the sum of the angles in a pair is always 180 0: α + β = 180 0 and γ + δ = 180 0 .
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 TX segment is easy to calculate based on the difference in the lengths of the bases, divided in half: TX \u003d (AE - KM) / 2.
If parallel lines are drawn through the sides of the angle of a trapezoid, they will divide the sides of the angle into proportional segments.

Properties of an isosceles (isosceles) trapezoid

In an isosceles trapezoid, the angles at any of the bases are equal.
Now build a trapezoid again to make it easier to imagine what it is about. Look carefully at the base of AE - the vertex of the opposite base of 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 midline of an isosceles trapezoid are equal.
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.
Only near an isosceles trapezoid can a circle be described, since the sum of the opposite angles of a quadrilateral 180 0 is a prerequisite for this.
The property of an isosceles trapezoid follows from the previous paragraph - if a circle can be described near a trapezoid, it is isosceles.
From the features of an isosceles trapezoid, the property of the height of a trapezoid follows: if its diagonals intersect at a right angle, then the length of the height is equal to half the sum of the bases: h = (a + b)/2.
Draw the line TX again 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.
This time lower to the larger base (let's call it a) the height from the opposite vertex of the trapezoid. You will get two cuts. 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

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 emerge 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).
The diagonal and the side can also meet at an acute angle - then the center of the circle is inside the trapezoid.
The center of the circumscribed circle may be outside the trapezium, beyond its large base, if there is an obtuse angle between the diagonal of the trapezoid and the lateral side.
The angle formed by the diagonal and the large base of the trapezoid ACME (inscribed angle) is half of the central angle that corresponds to it: MAE = ½MY.
Briefly about two ways to find the radius of the circumscribed circle. Method one: look carefully at your drawing - what do you see? You will easily notice that the diagonal splits the trapezoid into two triangles. The radius can be found through the ratio of the side of the triangle to the sine of the opposite angle, multiplied by two. For example, R \u003d AE / 2 * sinAME. Similarly, the formula can be written for any of the sides of both triangles.
Method two: we find the radius of the circumscribed circle through the area of the triangle formed by the diagonal, side and base of the trapezoid: R \u003d AM * ME * AE / 4 * S AME.

Properties of a trapezoid circumscribed about a circle

You can inscribe a circle in a trapezoid if one condition is met. More about it below. And together this combination of figures has a number of interesting properties.

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.
For a trapezoid ACME, circumscribed 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.
From this property of the bases of a trapezoid, the converse statement follows: a circle can be inscribed in that trapezoid, the sum of the bases of which is equal to the sum of the sides.
The tangent point of a circle with radius r inscribed in a trapezoid divides the lateral side into two segments, let's call them a and b. The radius of a circle can be calculated using the formula: r = √ab.
And one more property. In order not to get confused, draw this example yourself. We have the good old ACME trapezoid, circumscribed around a circle. Diagonals are drawn in it, intersecting at the point O. The triangles AOK and EOM formed by the segments of the diagonals and the sides are rectangular.
The heights of these triangles, lowered to the hypotenuses (i.e., the sides of the trapezoid), coincide with the radii of the inscribed circle. And the height of the trapezoid is the same as the diameter of the inscribed circle.

Properties of a rectangular trapezoid

A trapezoid is called rectangular, one of the corners of which is right. And its properties stem from this circumstance.

A rectangular trapezoid has one of the sides perpendicular to the bases.
The height and side of the trapezoid adjacent to the 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.
For a rectangular trapezoid, the general properties of the trapezoid diagonals already described above are relevant.

Proofs of some properties of a trapezoid

Equality of angles at the base of an isosceles trapezoid:

You probably already guessed that here we again need the ACME trapezoid - draw an isosceles trapezoid. Draw a 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 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 trapezium ACME is isosceles:

To begin with, let's draw a straight line МХ – МХ || KE. We get a parallelogram KMHE (base - MX || KE and KM || EX).

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

MX || KE, KEA = MXE, therefore MAE = MXE.

Task to repeat

The bases of the trapezoid ACME are 9 cm and 21 cm, the side of the KA, equal to 8 cm, forms an angle of 150 0 with a 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. Which means they add up to 1800. Therefore, KAN = 30 0 (based on the property of the angles of the trapezoid).

The area of the trapezoid is found by the formula: S AKME \u003d (KM + AE) * KN / 2 \u003d (9 + 21) * 4/2 \u003d 60 cm 2.

Afterword

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The segment connecting the midpoints of the diagonals of a trapezoid is equal to half the difference of the bases
The triangles formed by the bases of the trapezoid and the segments of the diagonals up to the point of their intersection are similar
Triangles formed by segments of the diagonals of a trapezoid, the sides of which lie on the sides of the trapezoid - are equal (have the same area)
If we extend the sides of the trapezoid towards the smaller base, then they will intersect at one point with the straight line connecting the midpoints of the bases
The segment connecting the bases of the 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
A segment parallel to the bases of the trapezoid and drawn through the intersection point of the diagonals is bisected by this point, and its length is 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

Connect the midpoints of the diagonals of the trapezoid ABCD, as a result of which we will have a segment LM.
A line segment that joins the midpoints of the diagonals of a trapezoid lies on the midline of the trapezium.

This segment parallel to the bases of the trapezium.

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

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

Properties of triangles formed by the diagonals of a trapezoid

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

Since all three angles of one triangle are equal to the corresponding angles of another triangle, 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 the two corresponding elements of 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 sides of the trapezoid AB and CD. These are triangles AOB and COD. Despite the fact that the dimensions of the individual sides of these triangles can be completely different, but the areas of the triangles formed by the sides and the point of intersection of the diagonals of the trapezoid are, that is, the triangles are equal.

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

Thus, any trapezoid can be extended to a triangle. Wherein:

The 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 the trapezoid, which lies at the intersection point of the diagonals of the trapezoid (KN), then the ratio of its constituent segments from the side of the base to the intersection point 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 diagonals of the trapezoid, then it will have the following properties:

Preset distance (KM) bisects the point of intersection of the diagonals of the trapezoid
Cut 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- bases of a trapezoid

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 the 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 the diagonals of a trapezoid can be proved as a separate theorem

2 . This formula is obtained by transforming the previous formula. The square of the second diagonal is thrown over 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) is similar in meaning and expresses a similar relationship.

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

Formulas for finding the diagonals of a trapezoid in terms of height

Note. In this lesson, the solution of problems in geometry about trapezoids is given. If you have not found a solution to the 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.

Decision.
The solution of this task is absolutely identical to the previous tasks in terms of ideology.

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, as the geometric dimensions of the segments AO and OC known to us according to the condition of the problem. I.e

AO/OC=AD/BC
9 / 6 = 24 / B.C.
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.

Decision .
To find the height of a trapezoid from the vertices of the smaller base B and C, we lower two heights onto the larger base. Since the trapezoid is unequal, we denote the length AM = a, the length KD = b ( not to be confused with the symbols in the formula finding the area of a trapezoid). Since the bases of the trapezoid are parallel and we have omitted 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 right-angled, so their right angles are formed by the heights of the trapezoid. Let's denote the height of the trapezoid as h. Then by the Pythagorean theorem

H 2 + (24 - a) 2 \u003d (5√17) 2
and
h 2 + (24 - b) 2 \u003d 13 2

Consider that a \u003d 16 - b, then in the first equation
h 2 + (24 - 16 + b) 2 \u003d 425
h 2 \u003d 425 - (8 + b) 2

Substitute the value of the square of the height into the second equation, obtained by 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

Thus, KD = 12
Where
h 2 \u003d 425 - (8 + b) 2 \u003d 425 - (8 + 12) 2 \u003d 25
h = 5

Find the area of a trapezoid using its height and half the sum of the bases
, where a b - the bases of the trapezoid, h - the height of the trapezoid
S \u003d (24 + 8) * 5 / 2 \u003d 80 cm 2

Answer: the area of a trapezoid is 80 cm2.

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