About a trapezoid, you can describe a circle if. Circumscribed circle and trapezoid. Equal area triangles of a trapezoid

Consider several directions for solving problems in which a trapezoid is inscribed in a circle.

When can a trapezoid be inscribed in a circle? A quadrilateral can be inscribed in a circle if and only if the sum of its opposite angles is 180º. Hence it follows that only an isosceles trapezoid can be inscribed in a circle.

The radius of a circle circumscribed about a trapezoid can be found as the radius of a circle circumscribed about one of the two triangles into which the trapezoid is divided by its diagonal.

Where is the center of the circle circumscribed about the trapezoid? It depends on the angle between the diagonal of the trapezoid and its side.

If the diagonal of a trapezoid is perpendicular to its lateral side, then the center of the circle circumscribed about the trapezoid lies in the middle of its larger base. The radius of the circle described near the trapezoid in this case is equal to half of its larger base:

If the diagonal of a trapezoid forms an acute angle with the lateral side, then the center of the circle circumscribed about the trapezoid lies inside the trapezoid.

If the diagonal of a trapezoid forms an obtuse angle with the lateral side, then the center of the circle circumscribed about the trapezoid lies outside the trapezoid, behind the large base.

The radius of a circle circumscribed about a trapezoid can be found from the corollary of the sine theorem. From triangle ACD

From triangle ABC

Another option to find the radius of the circumscribed circle is −

The sines of angle D and angle CAD can be found, for example, from right triangles CFD and ACF:

When solving problems for a trapezoid inscribed in a circle, you can also use the fact that the inscribed angle is equal to half of the corresponding central angle. For example,

By the way, you can use COD and CAD angles to find the area of a trapezoid. According to the formula for finding the area of a quadrilateral through its diagonals

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 trapezoid, 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 right angles, 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 trapezoid, 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!

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Trapeze is a quadrilateral with two parallel sides, which are the bases, and two non-parallel sides, which are the sides.

There are also names such as isosceles or isosceles.

It is a trapezoid with right angles on the lateral side.

Trapeze elements

a, b bases of a trapezoid(a parallel to b ),

m, n — sides trapeze,

d 1 , d 2 — diagonals trapeze,

h- height trapezoid (a segment connecting the bases and at the same time perpendicular to them),

MN- middle line(a segment connecting the midpoints of the sides).

Trapezium area

Through half the sum of the bases a, b and the height h : S = \frac(a + b)(2)\cdot h
Through the midline MN and height h : S = MN\cdot h
Through the diagonals d 1 , d 2 and the angle (\sin \varphi ) between them: S = \frac(d_(1) d_(2) \sin \varphi)(2)

Trapezoid Properties

Median line of the trapezoid

middle line is parallel to the bases, equal to their half-sum, and divides each segment with ends located on straight lines that contain the bases (for example, the height of the figure) in half:

MN || a, MN || b, MN = \frac(a + b)(2)

The sum of the angles of a trapezoid

The sum of the angles of a trapezoid, adjacent to each side, is equal to 180^(\circ) :

\alpha + \beta = 180^(\circ)

\gamma + \delta =180^(\circ)

Equal area triangles of a trapezoid

Equal-sized, that is, having equal areas, are the segments of the diagonals and the triangles AOB and DOC formed by the sides.

Similarity of formed trapezoid triangles

similar triangles are AOD and COB, which are formed by their bases and diagonal segments.

\triangle AOD \sim \triangle COB

similarity coefficient k is found by the formula:

k = \frac(AD)(BC)

Moreover, the ratio of the areas of these triangles is equal to k^(2) .

The ratio of the lengths of segments and bases

Each segment connecting the bases and passing through the point of intersection of the diagonals of the trapezoid is divided by this point in relation to:

\frac(OX)(OY) = \frac(BC)(AD)

This will also be true for the height with the diagonals themselves.

If a circle is inscribed in a trapezoid, several paths appear in the problem along which one can reason.

1. A circle can be inscribed in a quadrilateral if and only if the sums of the lengths of its opposite sides are equal. Hence it follows that If a circle is inscribed in a trapezoid, then the sum of its bases is equal to the sum of the sides.

AB+CD=AD+BC

2. Segments of tangents drawn from one point are equal. Hence it follows that

3. The height of a trapezoid is equal to the length of the diameter of the inscribed circle or two of its radii.

MK is the height of the trapezoid, MK=2r, where r is the radius of the circle inscribed in the trapezoid.

4. The center of the inscribed circle is the intersection point of the bisectors of the angles of the trapezoid.

Let's consider the basic problem.

Find the radius of the circle inscribed in the trapezoid if the point of contact divides the lateral side into segments of length m and n (CF=m, FD=n).

1) ∠ADC+∠BCD=180º (as the sum of internal one-sided angles with parallel lines AD and BC and secant CD);

2) since the point O is the point of intersection of the bisectors of the angles of the trapezoid, then ∠ODF+∠OCF=1/2∙(∠ADC+∠BCD)=90º;

3) since the sum of the angles of a triangle is 180º, then in the triangle COD ∠COD=90º;

4) thus, the triangle COD is right-angled, and OF is the height drawn to the hypotenuse, CF and FD are the projections of the leg OC and OD on the hypotenuse. Since the height drawn to the hypotenuse is between the projections of the legs on the hypotenuse,

Hence, the radius of the circle inscribed in the trapezoid is expressed in terms of the lengths of the segments in which the lateral side is divided by the point of contact, as

And since the height of the trapezoid is equal to its diameter, then the height of the trapezoid can be expressed in terms of the lengths of these segments.

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

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

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 \(CO=y \Rightarrow AO=ky\) .