All sides of a quadrangle are to a circle. The inscribed and described quadrilaterals and their properties are materials for preparing for the exam in mathematics. Criterion that a quadrilateral composed of two triangles is inscribed in some circle

Theorem 1. The sum of the opposite angles of an inscribed quadrilateral is 180°.

Let quadrilateral ABCD be inscribed in a circle with center O (Fig. 412). It is required to prove that ∠A + ∠C = 180° and ∠B + ∠D = 180°.

∠A, as inscribed in circle O, measures 1 / 2 \(\breve(BCD)\).

∠С, as inscribed in the same circle, is measured by 1 / 2 \(\breve(BAD)\).

Therefore, the sum of angles A and C is measured by half the sum of the arcs BCD and BAD; in sum, these arcs make up a circle, i.e. have 360°.

Hence ∠A + ∠C = 360°: 2 = 180°.

It is proved similarly that ∠B + ∠D = 180°. However, this can also be derived in another way. We know that the sum of the interior angles of a convex quadrilateral is 360°. The sum of the angles A and C is 180°, which means that the sum of the other two angles of the quadrilateral also remains 180°.

Theorem 2 (reverse). If the sum of two opposite angles in a quadrilateral is 180° , then a circle can be circumscribed about such a quadrilateral.

Let the sum of opposite angles of quadrilateral ABCD be 180°, namely

∠A + ∠C = 180° and ∠B + ∠D = 180° (Fig. 412).

Let us prove that a circle can be circumscribed around such a quadrilateral.

Proof. A circle can be drawn through any 3 vertices of this quadrilateral, for example, through points A, B and C. Where will point D be located?

Point D can take only one of the following three positions: be inside the circle, be outside the circle, be on the circumference of the circle.

Let us assume that the vertex is inside the circle and takes position D' (Fig. 413). Then in the quadrilateral ABCD' we will have:

∠B + ∠D' = 2 d.

Continuing the side AD' to the intersection with the circle at the point E and connecting the points E and C, we obtain the inscribed quadrilateral ABCE, in which, according to the direct theorem

∠B + ∠E = 2 d.

From these two equalities follows:

∠D' = 2 d-∠B;

∠E = 2 d-∠B;

but this cannot be, since ∠D', as external to the triangle CD'E, must be greater than angle E. Therefore, the point D cannot be inside the circle.

It is also proved that the vertex D cannot occupy the position D" outside the circle (Fig. 414).

It remains to recognize that the vertex D must lie on the circumference of the circle, i.e., coincide with the point E, which means that a circle can be circumscribed near the quadrilateral ABCD.

Consequences.

1. A circle can be circumscribed around any rectangle.

2. A circle can be circumscribed around an isosceles trapezoid.

In both cases, the sum of the opposite angles is 180°.

Theorem 3. In the circumscribed quadrilateral, the sums of opposite sides are equal. Let the quadrilateral ABCD be circumscribed about a circle (Fig. 415), that is, its sides AB, BC, CD and DA are tangent to this circle.

It is required to prove that AB + CD = AD + BC. Let's denote the points of contact with the letters M, N, K, P. Based on the properties of the tangents drawn to the circle from one point, we have:

Let us add these equalities term by term. We get:

AR + BP + DN + CN = AK + BM + DK + SM,

i.e., AB + CD = AD + BC, which was to be proved.

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Sections: Maths , Competition "Presentation for the lesson"

Presentation for the lesson

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Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

Goals.

Educational. Creating conditions for the successful assimilation of the concept of the described quadrilateral, its properties, features and mastering the skills to apply them in practice.

Developing. The development of mathematical abilities, the creation of conditions for the ability to generalize and apply the direct and reverse train of thought.

Educational. Raising a sense of beauty with the aesthetics of drawings, surprise with the unusual

decision, the formation of organization, responsibility for the results of their work.

1. Study the definition of the circumscribed quadrilateral.

2. Prove the property of the sides of the circumscribed quadrilateral.

3. Introduce the duality of the properties of the sums of opposite sides and opposite angles of the inscribed and circumscribed quadrilaterals.

4. To give experience in the practical application of the considered theorems in solving problems.

5. Conduct primary control of the level of assimilation of new material.

Equipment:

• computer, projector;
• textbook “Geometry. Grades 10-11” for general education. institutions: basic and profile. auto levels. A.V. Pogorelov.

Software: Microsoft Word, Microsoft Power Point.

Using a computer to prepare a teacher for a lesson.

Using the standard program of the Windows operating system created for the lesson:

1. Presentation.
2. Tables.
3. Blueprints.
4. Handout.

Lesson plan

Organizing time. (2 minutes.)

Checking homework. (5 minutes.)

Learning new material. (28 min.)

Independent work. (7 min.)

Homework.(1 min.)

Summary of the lesson. (2 minutes.)

During the classes

1. Organizational moment. Greetings. Message about the topic and purpose of the lesson. Write in the notebook the date and topic of the lesson.

2. Checking homework.

3. Learning new material.

Work on the concept of the circumscribed polygon.

Definition. The polygon is called described around the circle if all his side concern some circle.

Question. Which of the proposed polygons are circumscribed and which are not, and why?

<Презентация. Слайд №2>

Proof of the properties of the circumscribed quadrilateral.

<Презентация. Слайд №3>

Theorem. In the circumscribed quadrilateral, the sums of opposite sides are equal.

Students work with the textbook, write down the formulation of the theorem in a notebook.

1. Present the statement of the theorem in the form of a conditional sentence.

2. What is the condition of the theorem?

3. What is the conclusion of the theorem?

Answer. If quadrilateral circumscribed about a circle, then sums of opposite sides are equal.

Proof is being carried out, students make notes in a notebook.

<Презентация. Слайд №4>

Teacher. Note duality situations for sides and angles of circumscribed and inscribed quadrilaterals.

Consolidation of acquired knowledge.

Tasks.

The opposite sides of the circumscribed quadrilateral are 8 m and 12 m. Can you find the perimeter?

Tasks according to ready-made drawings.<Презентация. Слайд №5>

Answer. 1. 10 m 2. 20 m 3. 21 m

Proof of the attribute of the circumscribed quadrilateral.

State the inverse theorem.

Answer. If in a quadrilateral the sums of opposite sides are equal, then a circle can be inscribed in it. (Return to slide 2, Fig. 7) <Презентация. Слайд №2>

Teacher. Refine the statement of the theorem.

Theorem. If the sums of opposite sides convex quadrilaterals are equal, then a circle can be inscribed in it.

Work with the textbook. To get acquainted with the proof of the sign of the described quadrangle according to the textbook.

Application of acquired knowledge.

3. Tasks according to ready-made drawings.

1. Is it possible to inscribe a circle in a quadrilateral with opposite sides 9 m and 4 m, 10 m and 3 m?

2. Is it possible to inscribe a circle in an isosceles trapezoid with bases 1 m and 9 m, height 3 m?

<Презентация. Слайд №6>

Written work in notebooks

.

A task. Find the radius of a circle inscribed in a rhombus with diagonals 6 m and 8 m.

<Презентация. Слайд № 7>

4. Independent work.

1 option

1. Is it possible to inscribe a circle

1) into a rectangle with sides of 7 m and 10 m,

2. Opposite sides of a quadrilateral circumscribed about a circle are 7 m and 10 m.

Find the perimeter of the quadrilateral.

3. An isosceles trapezoid with bases of 4 m and 16 m is circumscribed about a circle.

1) the radius of the inscribed circle,

Option 2

1. Is it possible to inscribe a circle:

1) in a parallelogram with sides of 6 m and 13 m,

2) in a square?

2. Opposite sides of a quadrilateral circumscribed about a circle are 9 m and 11 m. Find the perimeter of the quadrilateral.

3. An isosceles trapezoid with a lateral side of 5 m is circumscribed about a circle with a radius of 2 m.

1) the base of the trapezoid,

2) the radius of the circumscribed circle.

5. Homework. P.86, No. 28, 29, 30.

6. The result of the lesson. Independent work is checked, grades are given.

<Презентация. Слайд № 8>

From Wikipedia, the free encyclopedia

• In Euclidean geometry, inscribed quadrilateral is a quadrilateral in which all vertices lie on the same circle. This circle is called circumscribed circle quadrilateral, and the vertices are said to lie on the same circle. The center of this circle and its radius are called respectively center And radius circumscribed circle. Other terms for this quadrilateral: quadrilateral lies on the same circle, the sides of the last quadrilateral are the chords of the circle. It is usually assumed that a convex quadrilateral is a convex quadrilateral. The formulas and properties given below are valid in the convex case.
• They say that if a circle can be circumscribed around a quadrilateral, then the quadrilateral is inscribed in this circle, and vice versa.

General criteria for a quadrilateral to be inscribed

• About a convex quadrilateral $\backslash pi$ radian), that is:

$\backslash angle\; A+\backslash angle\; C\; =\; \backslash angle\; B\; +\; \backslash angle\; D\; =\; 180^\backslash circ$

or in the notation of the figure:

$\backslash alpha\; +\; \backslash gamma\; =\; \backslash beta\; +\; \backslash delta\; =\; \backslash pi\; =\; 180^(\backslash circ).$

• It is possible to describe a circle around any quadrilateral, in which four perpendicular bisectors of its sides (or mediatrices of its sides, that is, perpendiculars to the sides passing through their midpoints) intersect at one point.
• It is possible to circumscribe a circle about any quadrilateral that has one external angle adjacent to given internal angle, exactly equal to another interior angle opposite given inner corner. In fact, this condition is the condition of antiparallelism of two opposite sides of the quadrilateral. On fig. the outer and adjacent inner corners of the green pentagon are shown below.

$\backslash displaystyle\; AX\backslash cdot\; XC\; =\; BX\backslash cdot\; XD.$

• intersection X may be internal or external to the circle. In the first case, we get the inscribed quadrilateral is ABCD, and in the latter case we get an inscribed quadrilateral ABDC. When crossing inside a circle, the equation says that the product of the lengths of the segments in which the point X divides one diagonal is equal to the product of the lengths of the segments in which the point X divides the other diagonal. This condition is known as the "intersecting chords theorem". In our case, the diagonals of the inscribed quadrilateral are the chords of the circle.
• Another eligibility criterion. Convex quadrilateral ABCD a circle is inscribed if and only if

$\backslash tan(\backslash frac(\backslash alpha)(2))\backslash tan(\backslash frac(\backslash gamma)(2))=\backslash tan(\backslash frac(\backslash beta)(2))\backslash tan(\backslash frac(\backslash delta)(\; 2))=1.$

Particular criteria for a quadrilateral to be inscribed

An inscribed simple (without self-intersections) quadrilateral is convex. A circle can be circumscribed about a convex quadrilateral if and only if the sum of its opposite angles is 180° ( $\backslash pi$ radian). You can describe a circle around:

• any antiparallelogram
• any rectangle (a special case of a square)
• any isosceles trapezoid
• any quadrilateral with two opposite angles right.

Properties

Formulas with diagonals

$ef=ac+bd$; $\backslash frac(e)(f)\; =\; \backslash frac(a\backslash cdot\; d+b\backslash cdot\; c)(a\backslash cdot\; b+c\backslash cdot\; d).$

In the last formula of the pair of adjacent sides of the numerator a And d, b And c rest their ends on a diagonal of length e. A similar statement holds for the denominator.

• Formulas for Diagonal Lengths(consequences ):

$e\; =\; \backslash sqrt(\backslash frac((ac+bd)(ad+bc))(ab+cd))$ And $f\; =\; \backslash sqrt(\backslash frac((ac+bd)(ab+cd))(ad+bc))$

Formulas with corners

For an inscribed quadrilateral with a sequence of sides a , b , c , d, with semiperimeter p and angle A between the parties a And d, trigonometric functions of the angle A are given by formulas

$\backslash cos\; A\; =\; \backslash frac(a^2\; +\; d^2\; -\; b^2\; -\; c^2)(2(ad\; +\; bc)),$ $\backslash sin\; A\; =\; \backslash frac(2\backslash sqrt((p-a)(p-b)(p-c)(p-d)))((ad+bc)),$ $\backslash tan\; \backslash frac(A)(2)\; =\; \backslash sqrt(\backslash frac((p-a)(p-d))((p-b)(p-c))).$

Injection θ between the diagonals is :p.26

$\backslash tan\; \backslash frac(\backslash theta)(2)\; =\; \backslash sqrt(\backslash frac((p-b)(p-d))((p-a)(p-c))).$

• If opposite sides a And c intersect at an angle φ , then it is equal to

$\backslash cos(\backslash frac(\backslash varphi)(2))=\backslash sqrt(\backslash frac((p-b)(p-d)(b+d)^2)((ab+cd)(ad+bc))),$

where p is a semi-perimeter. :p.31

Radius of a circle circumscribed about a quadrilateral

Formula of Parameshvara (Parameshvara)

If a quadrilateral with consecutive sides a , b , c , d and semiperimeter p a circle is inscribed, then its radius is Parameswar formula:p. 84

$R=\; \backslash frac(1)(4)\; \backslash sqrt(\backslash frac((ab+cd)(ad+bc)(ac+bd))((p-a)(p-b)(p-c)(p-d))).$

It was developed by the Indian mathematician Parameswar in the 15th century (c. 1380-1460)

• A convex quadrilateral (see the figure on the right) formed by four data direct Mikel, is inscribed in a circle if and only if the Miquel point M of the quadrilateral lies on the line joining two of the six points of intersection of the lines (those that are not vertices of the quadrilateral). That is, when M lies on EF.

Criterion that a quadrilateral composed of two triangles is inscribed in some circle

$f^2\; =\; \backslash frac((ac+bd)(ad+bc))((ab+cd)).$

• The last condition gives an expression for the diagonal f a quadrilateral inscribed in a circle, through the lengths of its four sides ( a, b, c, d). This formula immediately follows when multiplying and equating to each other the left and right parts of the formulas expressing the essence Ptolemy's first and second theorems(see above).

Criterion that a quadrilateral cut off by a straight line from a triangle is inscribed in some circle

• A straight line, antiparallel to the side of the triangle and intersecting it, cuts off a quadrilateral from it, around which a circle can always be circumscribed.
• Consequence. Near an antiparallelogram, in which two opposite sides are antiparallel, it is always possible to describe a circle.

Area of ​​a quadrilateral inscribed in a circle

Variants of the Brahmagupta Formula

$S=\backslash sqrt((p-a)(p-b)(p-c)(p-d)),$ where p is the semiperimeter of the quadrilateral. $S=\; \backslash frac(1)(4)\; \backslash sqrt(-\; \backslash begin(vmatrix)$

a & b & c & -d \\ b & a & -d & c \\ c & -d & a & b \\ -d & c & b & a \end(vmatrix))

Other area formulas

$S\; =\; \backslash tfrac(1)(2)(ab+cd)\backslash sin(B)$ $S\; =\; \backslash tfrac(1)(2)(ac+bd)\backslash sin(\backslash theta),$

where θ any of the angles between the diagonals. Provided that the angle A is not straight, area can also be expressed as :p.26

$S\; =\; \backslash tfrac(1)(4)(a^2-b^2-c^2+d^2)\backslash tan(A).$ $\backslash displaystyle\; S=2R^2\backslash sin(A)\backslash sin(B)\backslash sin(\backslash theta),$

where R is the radius of the circumscribed circle. As a direct consequence, we have the inequality

$S\backslash le\; 2R^2,$

where equality is possible if and only if this quadrilateral is a square.

Quadrangles of Brahmagupta

Brahmagupta Quadrangle is a quadrilateral inscribed in a circle with integer side lengths, integer diagonals, and integer area. All possible Brahmagupta quadrilaterals with sides a , b , c , d, with diagonals e , f, with area S, and the radius of the circumscribed circle R can be obtained by removing the denominators of the following expressions involving rational parameters t , u, And v :

$a=$ $b=(1+u^2)(v-t)(1+tv)$ $c=t(1+u^2)(1+v^2)$ $d=(1+v^2)(u-t)(1+tu)$ $e=u(1+t^2)(1+v^2)$ $f=v(1+t^2)(1+u^2)$ $S=uv$ $4R=(1+u^2)(1+v^2)(1+t^2).$

Examples

• Private quadrilaterals inscribed in a circle are: rectangle, square, isosceles or isosceles trapezoid, antiparallelogram.

Quadrilaterals inscribed in a circle with perpendicular diagonals (inscribed orthodiagonal quadrilaterals)

Properties of quadrilaterals inscribed in a circle with perpendicular diagonals

Radius of the circumscribed circle and area

For a quadrilateral inscribed in a circle with perpendicular diagonals, suppose that the intersection of the diagonals divides one diagonal into segments of length p 1 and p 2 , and divides the other diagonal into segments of length q 1 and q 2. Then (The first equality is Proposition 11 in Archimedes " Book of Lemmas)

$D^2=p\_1^2+p\_2^2+q\_1^2+q\_2^2=a^2+c^2=b^2+d^2,$

where D- diameter of the circle. This is true because the diagonals are perpendicular to the circle's chord. From these equations it follows that the radius of the circumscribed circle R can be written in the form

$R=\backslash tfrac(1)(2)\backslash sqrt(p\_1^2+p\_2^2+q\_1^2+q\_2^2)$

or in terms of the sides of a quadrilateral in the form

$R=\backslash tfrac(1)(2)\backslash sqrt(a^2+c^2)=\backslash tfrac(1)(2)\backslash sqrt(b^2+d^2).$

It also follows from this that

$a^2+b^2+c^2+d^2=8R^2.$

• For inscribed orthodiagonal quadrilaterals, Brahmagupta's theorem holds:

If an inscribed quadrilateral has perpendicular diagonals that intersect at a point $M$, then two pairs of antimediatris pass through the point $M$.

Comment. In this theorem, antimediatris understand the segment $F.E.$ quadrilateral in the figure on the right (by analogy with the perpendicular bisector (mediatrix) to the side of the triangle). It is perpendicular to one side and simultaneously passes through the midpoint of the opposite side of the quadrilateral.

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Notes

1. Bradley, Christopher J. (2007), The Algebra of Geometry: Cartesian, Areal and Projective Co-Ordinates, Highperception, p. 179, ISBN 1906338000, OCLC
2. . Inscribed quadrilaterals.
3. Siddons, A. W. & Hughes, R. T. (1929), Trigonometry, Cambridge University Press, p. 202, OCLC
4. Durell, C.V. & Robson, A. (2003), Courier Dover, ISBN 978-0-486-43229-8 ,
5. Alsina, Claudi & Nelsen, Roger B. (2007), "", Forum Geometricorum T. 7: 147–9 ,
6. Johnson, Roger A., Advanced Euclidean Geometry, Dover Publ., 2007 (orig. 1929).
7. Hoehn, Larry (March 2000), "Circumradius of a cyclic quadrilateral", Mathematical Gazette T. 84 (499): 69–70
8. .
9. Altshiller-Court, Nathan (2007), College Geometry: An Introduction to the Modern Geometry of the Triangle and the Circle(2nd ed.), Courier Dover, ss. 131, 137–8, ISBN 978-0-486-45805-2, OCLC
10. Honsberger, Ross (1995), . Episodes in Nineteenth and Twentieth Century Euclidean Geometry, vol. 37, New Mathematical Library, Cambridge University Press, pp. 35–39, ISBN 978-0-88385-639-0
11. Weisstein, Eric W.(English) on the Wolfram MathWorld website.
12. Bradley, Christopher (2011) ,
13. .
14. Coxeter, Harold Scott MacDonald & Greitzer, Samuel L. (1967), . Geometry Revisited, Mathematical Association of America, pp. 57, 60, ISBN 978-0-88385-619-2
15. .
16. Andreescu, Titu & Enescu, Bogdan (2004), . Mathematical Olympiad Treasures, Springer, ss. 44–46, 50, ISBN 978-0-8176-4305-8
17. .
18. Buchholz, R. H. & MacDougall, J. A. (1999), "", Bulletin of the Australian Mathematical Society T. 59(2): 263–9 , DOI 10.1017/S0004972700032883
19. .
20. Johnson, Roger A., Advanced Euclidean Geometry, Dover Publ. co., 2007
21. , from. 74.
22. .
23. .
24. .
25. Peter, Thomas (September 2003), "Maximizing the area of ​​a quadrilateral", The College Mathematics Journal T. 34 (4): 315–6
26. Prasolov, Viktor, ,
27. Alsina, Claudi & Nelsen, Roger (2009), , , Mathematical Association of America, p. 64, ISBN 978-0-88385-342-9 ,
28. Sastry, K.R.S. (2002). "" (PDF). Forum Geometricorum 2 : 167–173.
29. Posamentier, Alfred S. & Salkind, Charles T. (1970), . Challenging Problems in Geometry(2nd ed.), Courier Dover, ss. 104–5, ISBN 978-0-486-69154-1
30. .
31. .
32. .

see also

By number of sides Correct triangles Quadrilaterals see also

The article contains short ("Harvard") references to publications not listed or incorrectly described in the bibliographic section. List of non-working links: , , , , , , , , , , - Well, what, my Cossack? (Marya Dmitrievna called Natasha a Cossack) - she said, caressing Natasha with her hand, who approached her hand without fear and cheerfully. - I know that the potion is a girl, but I love it. From her huge reticule, she took out yakhon earrings with pears and, giving them to Natasha, who was beaming and blushing on her birthday, immediately turned away from her and turned to Pierre. – Eh, eh! kind! come here,” she said in a mockingly quiet and thin voice. - Come on, my dear... And she rolled up her sleeves menacingly even higher. Pierre came up, naively looking at her through his glasses. "Come, come, dear!" I told your father the truth alone, when he happened to be, and then God commands you. She paused. Everyone was silent, waiting for what was to come, and feeling that there was only a preface. - Okay, nothing to say! good boy! ... The father lies on the bed, and he amuses himself, he puts the quarter on a bear on horseback. Shame on you, dad, shame on you! Better to go to war. She turned away and offered her hand to the count, who could hardly help laughing. - Well, well, to the table, I have tea, is it time? said Marya Dmitrievna. The count went ahead with Marya Dmitrievna; then the countess, who was led by a hussar colonel, the right person with whom Nikolai was supposed to catch up with the regiment. Anna Mikhailovna is with Shinshin. Berg offered his hand to Vera. Smiling Julie Karagina went with Nikolai to the table. Behind them came other couples, stretching across the hall, and behind them all alone, children, tutors and governesses. The waiters stirred, chairs rattled, music played in the choir stalls, and the guests settled in. The sounds of the count's home music were replaced by the sounds of knives and forks, the voices of guests, the quiet footsteps of waiters. At one end of the table, the countess sat at the head. On the right is Marya Dmitrievna, on the left is Anna Mikhailovna and other guests. At the other end sat a count, on the left a hussar colonel, on the right Shinshin and other male guests. On one side of the long table, older youth: Vera next to Berg, Pierre next to Boris; on the other hand, children, tutors and governesses. From behind the crystal, bottles and vases of fruit, the count glanced at his wife and her high cap with blue ribbons and diligently poured wine to his neighbors, not forgetting himself. The Countess, also, because of the pineapples, not forgetting her duties as a hostess, threw significant glances at her husband, whose bald head and face, it seemed to her, were sharply distinguished by their redness from gray hair. There was a regular babble at the ladies' end; voices were heard louder and louder on the male, especially the hussar colonel, who ate and drank so much, blushing more and more that the count already set him as an example to other guests. Berg, with a gentle smile, spoke to Vera about the fact that love is a feeling not earthly, but heavenly. Boris called his new friend Pierre the guests who were at the table and exchanged glances with Natasha, who was sitting opposite him. Pierre spoke little, looked at new faces and ate a lot. Starting from two soups, from which he chose a la tortue, [turtle,] and kulebyaki, and up to grouse, he did not miss a single dish and not a single wine, which the butler in a bottle wrapped in a napkin mysteriously protruded from his neighbor’s shoulder, saying or “drey Madeira, or Hungarian, or Rhine wine. He substituted the first of the four crystal glasses with the count's monogram, which stood in front of each device, and drank with pleasure, looking more and more pleasantly at the guests. Natasha, who was sitting opposite him, looked at Boris, as girls of thirteen look at the boy with whom they had just kissed for the first time and with whom they are in love. This same look of hers sometimes turned to Pierre, and under the look of this funny, lively girl he wanted to laugh himself, not knowing why. Nikolai was sitting far away from Sonya, next to Julie Karagina, and again, with the same involuntary smile, he spoke something to her. Sonya smiled grandly, but apparently she was tormented by jealousy: she turned pale, then blushed, and with all her might listened to what Nikolai and Julie were saying to each other. The governess looked around uneasily, as if preparing herself for a rebuff, if anyone thought of offending the children. The German tutor tried to memorize the categories of foods, desserts and wines in order to describe everything in detail in a letter to his family in Germany, and was very offended by the fact that the butler, with a bottle wrapped in a napkin, surrounded him. The German frowned, tried to show that he did not want to receive this wine, but was offended because no one wanted to understand that he needed wine not to quench his thirst, not out of greed, but out of conscientious curiosity. At the male end of the table the conversation became more and more lively. The colonel said that the manifesto declaring war had already been published in Petersburg, and that the copy, which he himself had seen, had now been delivered by courier to the commander-in-chief. - And why is it difficult for us to fight with Bonaparte? Shinshin said. - II a deja rabattu le caquet a l "Autriche. Je crains, que cette fois ce ne soit notre tour. [He has already knocked down arrogance from Austria. I'm afraid our turn would not come now.] The colonel was a stout, tall and sanguine German, obviously a campaigner and a patriot. He was offended by Shinshin's words. “And then, we are a fat sovereign,” he said, pronouncing e instead of e and b instead of b. “Then, that the emperor knows this. He said in his manifesto that he cannot look indifferently at the dangers threatening Russia, and that the security of the empire, its dignity and the holiness of alliances,” he said, for some reason especially leaning on the word "unions", as if this was the whole essence of the matter. And with his infallible, official memory, he repeated the introductory words of the manifesto ... “and the desire, the sole and indispensable goal of the sovereign, is to establish peace in Europe on solid grounds - they decided to send part of the army now abroad and make new efforts to achieve“ this intention “. “Here’s why, we are a worthy sovereign,” he concluded, instructively drinking a glass of wine and looking back at the count for encouragement. - Connaissez vous le proverbe: [You know the proverb:] “Yerema, Yerema, if you would sit at home, sharpen your spindles,” said Shinshin, wincing and smiling. – Cela nous convient a merveille. [This is by the way for us.] Why Suvorov - and he was split, a plate couture, [on the head,] and where are our Suvorovs now? Je vous demande un peu, [I ask you] - he constantly jumped from Russian to French, he said. “We must fight until the day after the drop of blood,” said the colonel, banging on the table, “and die rrret for our emperor, and then everything will be fine.” And to argue as much as possible (he especially drew out his voice on the word “possible”), as little as possible,” he finished, again turning to the count. - So we judge the old hussars, that's all. And how do you judge, young man and young hussar? he added, turning to Nikolai, who, hearing that the matter was about the war, left his interlocutor and looked with all his eyes and listened with all his ears to the colonel. “I completely agree with you,” answered Nikolai, flushing all over, turning the plate and rearranging the glasses with such a determined and desperate look, as if at the present moment he was in great danger, “I am convinced that the Russians must die or win,” he said, himself feeling as well as others, after the word had already been said, that it was too enthusiastic and pompous for the present occasion and therefore awkward. - C "est bien beau ce que vous venez de dire, [Wonderful! what you said is wonderful]," said Julie, who was sitting next to him, sighing. Sonya trembled all over and blushed to her ears, behind her ears and to her neck and shoulders, while Nikolai spoke. Pierre listened to the colonel's speeches and nodded his head approvingly. “That's nice,” he said. “A real hussar, young man,” the colonel shouted, striking the table again. - What are you talking about there? Marya Dmitrievna's bass voice was suddenly heard across the table. What are you banging on the table for? she turned to the hussar, “who are you getting excited about? right, you think that the French are in front of you? "I'm telling the truth," said the hussar, smiling. “It’s all about the war,” the count shouted across the table. “After all, my son is coming, Marya Dmitrievna, my son is coming. - And I have four sons in the army, but I don’t grieve. Everything is the will of God: you will die lying on the stove, and God will have mercy in battle, ”the thick voice of Marya Dmitrievna sounded without any effort, from the other end of the table. - This is true. And the conversation again focused - the ladies at their end of the table, the men at theirs. “But you won’t ask,” the little brother said to Natasha, “but you won’t ask!” “I’ll ask,” Natasha answered. Her face suddenly flared up, expressing a desperate and cheerful determination. She half rose, inviting Pierre, who was sitting opposite her, to listen with a glance, and turned to her mother: - Mother! her childlike chest voice sounded all over the table. - What do you want? the countess asked frightened, but, seeing from her daughter's face that it was a prank, she waved her hand sternly, making a threatening and negative gesture with her head. The conversation hushed. - Mother! what cake will it be? - Natasha's voice sounded even more resolutely, without breaking. The Countess wanted to frown, but she couldn't. Marya Dmitrievna shook her thick finger. “Cossack,” she said threateningly. Most of the guests looked at the elders, unsure how to take this stunt. - Here I am! said the Countess. - Mother! what will the cake be? Natasha shouted already boldly and capriciously cheerfully, confident in advance that her trick would be well received. Sonya and fat Petya were hiding from laughter. “So I asked,” Natasha whispered to her little brother and Pierre, whom she looked at again. “Ice cream, but they won’t give you,” said Marya Dmitrievna. Natasha saw that there was nothing to be afraid of, and therefore she was not afraid of Marya Dmitrievna either. — Marya Dmitrievna? what an ice cream! I don't like butter. - Carrot. – No, what? Marya Dmitrievna, which one? she almost screamed. - I want to know! Marya Dmitrievna and the countess laughed, and all the guests followed. Everyone laughed not at Marya Dmitrievna's answer, but at the incomprehensible courage and dexterity of this girl, who knew how and dared to treat Marya Dmitrievna in this way. Natasha lagged behind only when she was told that there would be pineapple. Champagne was served before ice cream. Again the music began to play, the count kissed the countess, and the guests, rising, congratulated the countess, clinked glasses across the table with the count, the children, and each other. Again the waiters ran in, the chairs rattled, and in the same order, but with redder faces, the guests returned to the drawing room and the count's study. The Boston tables were moved apart, parties were made up, and the count's guests were accommodated in two living rooms, a sofa room and a library. The count, spreading his cards like a fan, could hardly resist the habit of an afternoon nap and laughed at everything. The youth, incited by the countess, gathered around the clavichord and harp. Julie was the first, at the request of everyone, to play a piece with variations on the harp and, together with other girls, began to ask Natasha and Nikolai, known for their musicality, to sing something. Natasha, who was addressed as a big one, was apparently very proud of this, but at the same time she was shy. - What are we going to sing? she asked. “The key,” answered Nikolai. - Well, let's hurry. Boris, come here, - said Natasha. - Where is Sonya? She looked around and, seeing that her friend was not in the room, ran after her. Running into Sonya's room and not finding her friend there, Natasha ran into the nursery - and Sonya was not there. Natasha realized that Sonya was in the corridor on a chest. The chest in the corridor was the place of sorrows of the female young generation of the Rostovs' house. Indeed, Sonya, in her airy pink dress, crushing it, lay face down on the dirty striped nanny's featherbed, on the chest, and, covering her face with her fingers, wept bitterly, trembling with her bare shoulders. Natasha's face, lively, all day long, suddenly changed: her eyes stopped, then her broad neck shuddered, the corners of her lips drooped. – Sonya! what are you?… What, what is the matter with you? Woo woo!… And Natasha, spreading her big mouth and becoming completely ugly, roared like a child, not knowing the reason and only because Sonya was crying. Sonya wanted to raise her head, wanted to answer, but she could not and hid even more. Natasha was crying, sitting down on a blue featherbed and hugging her friend. Gathering her strength, Sonya got up, began to wipe her tears and tell. - Nikolenka is going in a week, his ... paper ... came out ... he told me himself ... Yes, I wouldn’t cry ... (she showed the paper she held in her hand: it was poetry written by Nikolai) I wouldn’t cry, but you won’t you can... no one can understand... what kind of soul he has. And she began to cry again because his soul was so good. “It’s good for you ... I don’t envy ... I love you, and Boris too,” she said, gathering her strength a little, “he’s cute ... there are no obstacles for you. And Nikolai is my cousin... it is necessary... the metropolitan himself... and that is impossible. And then, if my mother ... (Sonya considered the countess and called her mother), she will say that I spoil Nikolai's career, I have no heart, that I am ungrateful, but right ... by God ... (she crossed herself) I love her so much too , and all of you, only Vera is one ... For what? What did I do to her? I am so grateful to you that I would be glad to sacrifice everything, but I have nothing ... Sonya could no longer speak and again hid her head in her hands and feather bed. Natasha began to calm down, but it was clear from her face that she understood the importance of her friend's grief. – Sonya! she said suddenly, as if guessing the real reason for her cousin's grief. “Right, did Vera talk to you after dinner?” Yes? - Yes, Nikolai himself wrote these poems, and I wrote off others; she found them on my table and said that she would show them to mamma, and also said that I was ungrateful, that mamma would never allow him to marry me, and he would marry Julie. You see how he is with her all day ... Natasha! For what?… And again she wept bitterly. Natasha lifted her up, hugged her and, smiling through her tears, began to comfort her. “Sonya, don’t trust her, darling, don’t. Do you remember how all three of us talked with Nikolenka in the sofa room; remember after dinner? After all, we have decided how it will be. I don’t remember how, but remember how everything was fine and everything is possible. Uncle Shinshin's brother is married to a cousin, and we are second cousins. And Boris said that it is very possible. You know, I told him everything. And he is so smart and so good,” said Natasha ... “You, Sonya, don’t cry, my dear, darling, Sonya. And she kissed her, laughing. - Faith is evil, God be with her! And everything will be fine, and she will not tell her mother; Nikolenka will tell himself, and he did not even think about Julie. And she kissed her on the head. Sonya got up, and the kitten perked up, his eyes sparkled, and he seemed ready to wave his tail, jump on his soft paws and play with the ball again, as it was proper for him. - You think? Right? By God? she said, quickly straightening her dress and hair. - Right, by God! - answered Natasha, straightening her friend under a scythe a strand of coarse hair that had fallen out. And they both laughed. - Well, let's go sing "Key". - Let's go to. - And you know, this fat Pierre, who was sitting opposite me, is so funny! Natasha suddenly said, stopping. - I have a lot of fun! And Natasha ran down the corridor. Sonya, brushing off the fluff and hiding the poems in her bosom, to the neck with protruding breast bones, with light, cheerful steps, with a flushed face, ran after Natasha along the corridor to the sofa. At the request of the guests, the young people sang the "Key" quartet, which everyone liked very much; then Nikolai sang the song he had learned again. On a pleasant night, by moonlight, Imagine being happy That there is someone else in the world Who thinks about you too! That she, with a beautiful hand, Walking along the golden harp, With its passionate harmony Calling to itself, calling you! Another day, two, and paradise will come ... But ah! your friend will not live! And he had not yet finished singing the last words, when in the hall the youth prepared for dancing and the musicians in the choirs clattered their feet and coughed. Pierre was sitting in the living room, where Shinshin, as with a visitor from abroad, started a political conversation with him that was boring for Pierre, which was joined by others. When the music started, Natasha entered the living room and, going straight up to Pierre, laughing and blushing, said: “Mom told me to ask you to dance. “I’m afraid to confuse the figures,” said Pierre, “but if you want to be my teacher ... And he gave his thick hand, lowering it low to the thin girl. While the couples were setting up and the musicians were building, Pierre sat down with his little lady. Natasha was perfectly happy; she danced with a big one who came from abroad. She sat in front of everyone and talked to him like a big one. She had a fan in her hand, which a young lady gave her to hold. And, adopting the most secular pose (God knows where and when she learned this), she, fanning herself with a fan and smiling through the fan, spoke with her gentleman. - What is it, what is it? Look, look, - said the old countess, passing through the hall and pointing to Natasha. Natasha blushed and laughed. - Well, what are you, mom? Well, what are you looking for? What is surprising here? In the middle of the third ecossaise, the chairs in the drawing-room where the count and Marya Dmitrievna were playing began to stir, and most of the honored guests and the old men, stretching after a long sitting and putting wallets and purses in their pockets, went out through the doors of the hall. Marya Dmitrievna walked in front with the count, both with merry faces. With playful politeness, as if in a ballet manner, the count extended his rounded hand to Marya Dmitrievna. He straightened up, and his face lit up with a particularly valiantly sly smile, and as soon as the last figure of the ecossaise had been danced, he clapped his hands to the musicians and shouted at the choirs, turning to the first violin: - Semyon! Do you know Danila Kupor? It was the count's favorite dance, danced by him in his youth. (Danilo Kupor was actually one Anglaise figure.) “Look at dad,” Natasha shouted to the whole hall (completely forgetting that she was dancing with a big one), bending her curly head to her knees and bursting into her sonorous laughter throughout the hall. Indeed, everyone in the hall looked with a smile of joy at the cheerful old man, who, next to his dignitary lady, Marya Dmitrievna, who was taller than he, rounded his arms, shaking them in time, straightened his shoulders, twisted his legs, slightly stamping his feet, and with a more and more blossoming smile on his round face he prepared the audience for what was to come. As soon as the cheerful, defiant sounds of Danila Kupor, similar to a merry rattler, were heard, all the doors of the hall were suddenly forced on one side by male, on the other side by female smiling faces of courtyards who came out to look at the merry gentleman. - Father is ours! Eagle! the nanny said loudly from one door. The count danced well and knew it, but his lady did not know how and did not want to dance well. Her huge body stood upright with her powerful arms hanging down (she handed the purse to the countess); only her stern but beautiful face danced. What was expressed in the whole round figure of the count, with Marya Dmitrievna was expressed only in a more and more smiling face and a twitching nose. But on the other hand, if the count, more and more dispersing, captivated the audience with the unexpectedness of deft tricks and light jumps of her soft legs, Marya Dmitrievna, with the slightest zeal in moving her shoulders or rounding her arms in turns and stomping, made no less impression on the merit, which was appreciated by everyone at her corpulence and everlasting severity. The dance became more and more lively. The counterparts could not draw attention to themselves for a minute and did not even try to do so. Everything was occupied by the count and Marya Dmitrievna. Natasha pulled the sleeves and dresses of all those present, who already did not take their eyes off the dancers, and demanded that they look at papa. During the intervals of the dance, the count took a deep breath, waved and shouted to the musicians to play faster. Quicker, faster and faster, more and more and more, the count unfolded, now on tiptoe, now on heels, rushing around Marya Dmitrievna and, finally, turning his lady to her place, made the last step, raising his soft leg upward from behind, bending his sweating head with a smiling face and roundly waving his right hand amid the roar of applause and laughter, especially Natasha. Both dancers stopped, breathing heavily and wiping themselves with cambric handkerchiefs. “This is how they danced in our time, ma chere,” said the count. - Oh yes Danila Kupor! ' said Marya Dmitrievna, letting out her breath heavily and continuously, and rolling up her sleeves. While the sixth anglaise was being danced in the hall at the Rostovs' to the sounds of tired musicians who were out of tune, and the tired waiters and cooks were preparing dinner, the sixth stroke took place with Count Bezukhim. The doctors announced that there was no hope of recovery; the patient was given a deaf confession and communion; preparations were made for the unction, and the house was full of fuss and anxiety of expectation, common at such moments. Outside the house, behind the gates, undertakers crowded, hiding from the approaching carriages, waiting for a rich order for the count's funeral. The Commander-in-Chief of Moscow, who constantly sent adjutants to learn about the position of the count, that evening he himself came to say goodbye to the famous Catherine's nobleman, Count Bezukhim. The magnificent reception room was full. Everyone stood up respectfully when the commander-in-chief, after being alone with the patient for about half an hour, came out of there, slightly answering the bows and trying as soon as possible to get past the eyes of doctors, clergy and relatives fixed on him. Prince Vasily, who had grown thinner and paler these days, saw off the commander-in-chief and quietly repeated something to him several times. After seeing off the commander-in-chief, Prince Vasily sat alone in the hall on a chair, throwing his legs high over his legs, resting his elbow on his knee and closing his eyes with his hand. After sitting like this for some time, he got up and with unusually hasty steps, looking around with frightened eyes, went through a long corridor to the back half of the house, to the elder princess. Those who were in the dimly lit room spoke in an uneven whisper among themselves and fell silent every time, and with eyes full of question and expectation looked back at the door that led to the chambers of the dying man and made a faint sound when someone left it or entered it. “The human limit,” the old man, a clergyman, said to the lady who sat down next to him and listened naively to him, “the limit is set, but you can’t pass it.” – I think it’s not too late to unction? - adding a spiritual title, the lady asked, as if she did not have any opinion on this matter.

Examples of quadrilaterals described are deltoids, which include rhombuses, which in turn include squares. Deltoids are exactly those circumscribed quadrilaterals that are also orthodiagonal. If a quadrilateral is a circumscribed and inscribed quadrilateral, it is called bicentral.

Properties

In the described quadrilateral, four bisectors intersect at the center of the circle. Conversely, a convex quadrilateral in which four bisectors intersect at one point must be circumscribed, and the intersection point of the bisectors is the center of the inscribed circle.

If opposite sides in a convex quadrilateral ABCD(which is not a trapezoid) intersect at points E And F, then they are tangent to the circle if and only if

B E + B F = D E + D F (\displaystyle \displaystyle BE+BF=DE+DF) A E − E C = A F − F C . (\displaystyle \displaystyle AE-EC=AF-FC.)

The second equality is almost the same as the equality in Urquhart's theorem. The difference is only in signs - in Urquhart's theorem, the sums, and here the differences (see the figure on the right).

Another necessary and sufficient condition is a convex quadrilateral ABCD is described if and only if the inscribed triangles ABC And ADC the circles touch each other.

Description at the corners formed by the diagonal BD with sides of a quadrilateral ABCD, belongs to Iosifescu. He proved in 1954 that a convex quadrilateral has an incircle if and only if

tan ⁡ ∠ A B D 2 ⋅ tan ⁡ ∠ B D C 2 = tan ⁡ ∠ A D B 2 ⋅ tan ⁡ ∠ D B C 2 . (\displaystyle \tan (\frac (\angle ABD)(2))\cdot \tan (\frac (\angle BDC)(2))=\tan (\frac (\angle ADB)(2))\cdot \tan (\frac (\angle DBC)(2)).) R a R c = R b R d (\displaystyle R_(a)R_(c)=R_(b)R_(d)),

where R a , R b , R c , R d are the radii of circles externally tangent to the sides a, b, c, d respectively and the continuations of adjacent sides on each side.

Some other descriptions are known for the four triangles formed by the diagonals.

Special cuts

Eight tangent segments of the circumscribed quadrilateral are the segments between the vertices and the points of contact on the sides. Each vertex has two equal tangent segments.

The points of contact form an inscribed quadrilateral.

Area

Nontrigonometric formulas

K = 1 2 p 2 q 2 − (ac − bd) 2 (\displaystyle K=(\tfrac (1)(2))(\sqrt (p^(2)q^(2)-(ac-bd) ^(2)))),

giving area in terms of diagonals p, q and parties a, b, c, d tangent quadrilateral.

The area can also be represented in terms of tangent segments (see above). If they are denoted by e, f, g, h, then the tangent quadrilateral has area

K = (e + f + g + h) (e f g + f g h + g h e + h e f) . (\displaystyle K=(\sqrt ((e+f+g+h)(efg+fgh+ghe+hef))).)

Moreover, the area of ​​a tangent quadrilateral can be expressed in terms of sides a, b, c, d and corresponding lengths of tangent segments e, f, g, h

K = a b c d − (e g − f h) 2 . (\displaystyle K=(\sqrt (abcd-(eg-fh)^(2))).)

Insofar as eg = fh if and only if it is also inscribed, we obtain that the maximum area a b c d (\displaystyle (\sqrt(abcd))) can only be achieved on quadrilaterals that are both circumscribed and inscribed at the same time.

Trigonometric formulas

K = a b c d sin ⁡ A + C 2 = a b c d sin ⁡ B + D 2 . (\displaystyle K=(\sqrt (abcd))\sin (\frac (A+C)(2))=(\sqrt (abcd))\sin (\frac (B+D)(2)).)

For a given product of sides, the area will be maximum when the quadrilateral is also an inscribed. In this case K = a b c d (\displaystyle K=(\sqrt (abcd))), since opposite angles are complementary. This can be proved in another way, using mathematical analysis.

Another formula for the area of ​​a circumscribed quadrilateral ABCD, using two opposite angles

K = (OA ⋅ OC + OB ⋅ OD) sin ⁡ A + C 2 (\displaystyle K=\left(OA\cdot OC+OB\cdot OD\right)\sin (\frac (A+C)(2) )),

where O is the center of the inscribed circle.

In fact, area can only be expressed in terms of two adjacent sides and two opposite angles.

K = a b sin ⁡ B 2 csc ⁡ D 2 sin ⁡ B + D 2 . (\displaystyle K=ab\sin (\frac (B)(2))\csc (\frac (D)(2))\sin (\frac (B+D)(2)).) K = 1 2 | (a c − b d) tan ⁡ θ | , (\displaystyle K=(\tfrac (1)(2))|(ac-bd)\tan (\theta )|,)

where θ angle (any) between the diagonals. The formula is not applicable to the case of deltoids, since in this case θ is 90° and the tangent is not defined.

inequalities

As mentioned in passing above, the area of ​​a tangent polygon with sides a, b, c, d satisfies the inequality

K ≤ a b c d (\displaystyle K\leq (\sqrt (abcd)))

and equality is achieved if and only if the quadrilateral is bicentral.

According to T. A. Ivanova (1976), the semiperimeter s circumscribed quadrilateral satisfies the inequality

s ≥ 4r (\displaystyle s\geq 4r),

where r is the radius of the inscribed circle. Inequality turns into equality if and only if the quadrilateral is a square. This means that for the area K = rs, the inequality

K ≥ 4 r 2 (\displaystyle K\geq 4r^(2))

with the transition to equality if and only if the quadrilateral is a square.

Properties of parts of a quadrilateral

Four line segments between the center of the inscribed circle and the points of contact divide the quadrilateral into four rectangular deltoid?!.

If a straight line divides the circumscribed quadrilateral into two polygons with equal areas and equal perimeters, then this line passes through the incenter.

Inscribed circle radius

Radius of the inscribed circle of an inscribed quadrilateral with sides a, b, c, d given by the formula

r = K s = K a + c = K b + d (\displaystyle r=(\frac (K)(s))=(\frac (K)(a+c))=(\frac (K)( b+d))),

where K is the area of ​​the quadrilateral, and s- semiperimeter. For circumscribed quadrangles with a given semiperimeter, the radius of the inscribed circle is maximum when the quadrilateral is also an inscribed one.

In terms of tangent segments, the radius of the inscribed circle.

r = e f g + f g h + g h e + h e f e + f + g + h . (\displaystyle \displaystyle r=(\sqrt (\frac (efg+fgh+ghe+hef)(e+f+g+h))).)

The radius of the inscribed circle can also be expressed in terms of the distance from the incenter O to the vertices of the circumscribed quadrilateral ABCD. If u = AO, v=BO, x=CO And y=DO, then

r = 2 (σ − uvx) (σ − vxy) (σ − xyu) (σ − yuv) uvxy (uv + xy) (ux + vy) (uy + vx) (\displaystyle r=2(\sqrt (\ frac ((\sigma -uvx)(\sigma -vxy)(\sigma -xyu)(\sigma -yuv))(uvxy(uv+xy)(ux+vy)(uy+vx))))),

where σ = 1 2 (u v x + v x y + x y u + y u v) (\displaystyle \sigma =(\tfrac (1)(2))(uvx+vxy+xyu+yuv)) .

Formulas for angles

If e, f, g And h tangent segments from vertices A, B, C And D respectively to the points of tangency of the circle by the quadrilateral ABCD, then the angles of the quadrilateral can be calculated by the formulas

sin ⁡ A 2 = efg + fgh + ghe + hef (e + f) (e + g) (e + h) , (\displaystyle \sin (\frac (A)(2))=(\sqrt (\frac (efg+fgh+ghe+hef)((e+f)(e+g)(e+h)))),) sin ⁡ B 2 = efg + fgh + ghe + hef (f + e) ​​(f + g) (f + h) , (\displaystyle \sin (\frac (B)(2))=(\sqrt (\frac (efg+fgh+ghe+hef)((f+e)(f+g)(f+h)))),) sin ⁡ C 2 = efg + fgh + ghe + hef (g + e) ​​(g + f) (g + h) , (\displaystyle \sin (\frac (C)(2))=(\sqrt (\frac (efg+fgh+ghe+hef)((g+e)(g+f)(g+h)))),) sin ⁡ D 2 = e f g + f g h + g h e + h e f (h + e) ​​(h + f) (h + g) . (\displaystyle \sin (\frac (D)(2))=(\sqrt (\frac (efg+fgh+ghe+hef)((h+e)(h+f)(h+g)))) .)

Angle between chords KM And LN given by formula (see figure)

sin ⁡ φ = (e + f + g + h) (e f g + f g h + g h e + h e f) (e + f) (f + g) (g + h) (h + e) ​​. (\displaystyle \sin (\varphi )=(\sqrt (\frac ((e+f+g+h)(efg+fgh+ghe+hef))((e+f)(f+g)(g+ h)(h+e)))).)

Diagonals

If e, f, g And h are segments of tangents from A, B, C And D to the points of tangency of the inscribed circle by the quadrilateral ABCD, then the lengths of the diagonals p=AC And q=BD equal

p = e + gf + h ((e + g) (f + h) + 4 fh) , (\displaystyle \displaystyle p=(\sqrt ((\frac (e+g)(f+h))(\ Big ()(e+g)(f+h)+4fh(\Big)))),) q = f + h e + g ((e + g) (f + h) + 4 e g) . (\displaystyle \displaystyle q=(\sqrt ((\frac (f+h)(e+g))(\Big ()(e+g)(f+h)+4eg(\Big)))). )

Touch point chords

If e, f, g And h are segments from the vertices to the tangent points, then the lengths of the chords to the opposite tangent points are

k = 2 (efg + fgh + ghe + hef) (e + f) (g + h) (e + g) (f + h) , (\displaystyle \displaystyle k=(\frac (2(efg+fgh+ ghe+hef))(\sqrt ((e+f)(g+h)(e+g)(f+h)))),) l = 2 (efg + fgh + ghe + hef) (e + h) (f + g) (e + g) (f + h) , (\displaystyle \displaystyle l=(\frac (2(efg+fgh+ ghe+hef))(\sqrt ((e+h)(f+g)(e+g)(f+h)))),)

where is the chord k connects sides to lengths a = e + f And c = g + h, and the chord l connects the sides with a length b = f + g And d = h + e. The square of the ratio of the chords satisfies the relation

k 2 l 2 \u003d b d a c. (\displaystyle (\frac (k^(2))(l^(2)))=(\frac (bd)(ac)).)

Two chords

Chord between sides AB And CD in the circumscribed quadrilateral ABCD longer than the chord between the sides BC And DA if and only if the median line between the sides AB And CD shorter than the median line between the sides BC And DA .

If the circumscribed quadrilateral ABCD has points of contact M on the AB And N on the CD and chord MN crosses the diagonal BD at the point P, then the ratio of the segments of the tangents B M D N (\displaystyle (\tfrac (BM)(DN))) is equal to the ratio B P D P (\displaystyle (\tfrac (BP)(DP))) diagonal segments BD.

collinear points

If M1 And M2 are the midpoints of the diagonals AC And BD respectively in the circumscribed quadrilateral ABCD O, and pairs of opposite sides intersect at points E And F And M3- middle of the segment EF, then the points M3, M1, O, And M2 lie on one straight line. The straight line connecting these points is called the Newton line of the quadrilateral.

E And F, and the extensions of the opposite sides of the quadrilateral formed by the points of contact intersect at the points T And S, then four points E, F, T And S lie on the same line

AB, BC, CD, DA at points M, K, N And L respectively, and if T M, T K, T N, T L are isotomically conjugate points of these points (i.e. AT M = BM etc.), then Nagel point defined as the intersection of lines T N T M And T K T L. Both of these lines divide the perimeter of the quadrilateral into two equal parts. More importantly, however, the Nagel point Q, "area centroid" G and the center of the inscribed circle O lie on the same straight line, and QG = 2GO. This line is called straight Nagel circumscribed quadrilateral.

In the circumscribed quadrilateral ABCD with the center of the inscribed circle O P, let be HM, H K, H N, H L are the orthocenters of the triangles AOB, BOC, COD And DOA respectively. Then the points P, HM, H K, H N And H L lie on the same line.

Competitive and perpendicular lines

Two diagonals of a quadrilateral and two chords connecting opposite points of contact (opposite vertices of an inscribed quadrilateral) are contiguous (that is, they intersect at one point). In order to show this, we can use a special case of Brianchon's theorem, which states that a hexagon, all sides of which are tangent to a conic section, has three diagonals that intersect at one point. From the described quadrilateral, it is easy to obtain a hexagon with two 180° angles by inserting two new vertices at opposite tangent points. All six sides of the resulting hexagon are tangent to the incircle, so that its diagonals intersect at one point. But two diagonals of the hexagon coincide with the diagonals of the quadrilateral, and the third diagonal passes through the opposite points of contact. Repeating the same reasoning for the other two touch points, we obtain the required result.

If the inscribed circle is tangent to the sides AB, BC, CD And DA at points M, K, N, L respectively, then straight lines MK, LN And AC competitive.

If the extensions of opposite sides of the circumscribed quadrilateral intersect at points E And F, and the diagonals intersect at a point P, then the straight line EF perpendicular to the continuation OP, where O is the center of the inscribed circle.

Inscribed circle properties

The ratio of two opposite sides of the circumscribed quadrilateral can be expressed in terms of distances from the center of the inscribed circle O to relevant parties

A B C D = O A ⋅ O B O C ⋅ O D , B C D A = O B ⋅ O C O D ⋅ O A . (\displaystyle (\frac (AB)(CD))=(\frac (OA\cdot OB)(OC\cdot OD)),\quad \quad (\frac (BC)(DA))=(\frac ( OB\cdot OC)(OD\cdot OA)).)

The product of two adjacent sides of a circumscribed quadrilateral ABCD with the center of the inscribed circle O satisfies the relation

A B ⋅ B C = O B 2 + O A ⋅ O B ⋅ O C O D . (\displaystyle AB\cdot BC=OB^(2)+(\frac (OA\cdot OB\cdot OC)(OD)).)

If O- the center of the inscribed circle of the quadrilateral ABCD, then

O A ⋅ O C + O B ⋅ O D = A B ⋅ B C ⋅ C D ⋅ D A . (\displaystyle OA\cdot OC+OB\cdot OD=(\sqrt (AB\cdot BC\cdot CD\cdot DA)).)

Center of the inscribed circle O coincides with the "centroid vertices" of the quadrilateral if and only if

O A ⋅ O C = O B ⋅ O D . (\displaystyle OA\cdot OC=OB\cdot OD.)

If M1 And M2 are the midpoints of the diagonals AC And BD respectively, then

OM 1 OM 2 = OA ⋅ OCOB ⋅ OD = e + gf + h , (\displaystyle (\frac (OM_(1))(OM_(2)))=(\frac (OA\cdot OC)(OB\cdot OD))=(\frac (e+g)(f+h)),)

where e, f, g And h- segments of tangents at the vertices A, B, C And D respectively. Combining the first equality with the last one, we get that the "centroid of the vertices" of the circumscribed quadrilateral coincides with the center of the inscribed circle if and only if the center of the inscribed circle lies midway between the midpoints of the diagonals.

1 r 1 + 1 r 3 = 1 r 2 + 1 r 4 . (\displaystyle (\frac (1)(r_(1)))+(\frac (1)(r_(3)))=(\frac (1)(r_(2)))+(\frac (1 )(r_(4))).)

This property had been proven five years earlier by Weinstein. In solving his problem, a similar property was given by Vasiliev and Senderov. If through h M , h K , h N and h L denote the heights of the same triangles (dropped from the intersection of the diagonals P), then the quadrilateral is circumscribed if and only if

1 h M + 1 h N = 1 h K + 1 h L . (\displaystyle (\frac (1)(h_(M)))+(\frac (1)(h_(N)))=(\frac (1)(h_(K)))+(\frac (1 )(h_(L))).)

Another similar property applies to the radii of excircles r M , r K , r N And r L for the same four triangles (four excircles are tangent to each of the sides of the quadrilateral and the extensions of the diagonals). A quadrilateral is circumscribed if and only if

1 r M + 1 r N = 1 r K + 1 r L . (\displaystyle (\frac (1)(r_(M)))+(\frac (1)(r_(N)))=(\frac (1)(r_(K)))+(\frac (1 )(r_(L))).)

If R M , R K , R N and R L - radii of circumcircles of triangles APB, BPC, CPD And DPA respectively, then the triangle ABCD is described if and only if

R M + R N = R K + R L . (\displaystyle R_(M)+R_(N)=R_(K)+R_(L).)

In 1996, Weinstein appears to have been the first to prove another remarkable property of circumscribed quadrilaterals, which later appeared in several magazines and websites. The property states that if a convex quadrilateral is divided into four non-overlapping triangles by its diagonals, the incircle centers of those triangles lie on the same circle if and only if the quadrilateral is a circumscribed one. In fact, the centers of the inscribed circles form an orthodiagonal inscribed four-corner. Here the inscribed circles can be replaced by excircles (tangent to the sides and continuations of the diagonals of the quadrilateral). Then a convex quadrilateral is circumscribed if and only if the centers of the excircles are the vertices of the inscribed quadrilateral.

Convex quadrilateral ABCD where the diagonals intersect at a point P, is circumscribed if and only if the four centers of the triangles' excircles APB, BPC, CPD And DPA lie on the same circle (here the excircles intersect the sides of the quadrilateral, in contrast to the analogous statement above, where the excircles lie outside the quadrilateral). If Rm, R n, Rk And Rl- radii of excircles APB, BPC, CPD And DPA respectively opposite to the vertices B And D, then another necessary and sufficient condition for the quadrilateral to be circumscribed is

1 R m + 1 R n = 1 R k + 1 R l . (\displaystyle (\frac (1)(R_(m)))+(\frac (1)(R_(n)))=(\frac (1)(R_(k)))+(\frac (1 )(R_(l))).) m △ (APB) + n △ (CPD) = k △ (BPC) + l △ (DPA) (\displaystyle (\frac (m)(\triangle (APB)))+(\frac (n)(\triangle (CPD)))=(\frac (k)(\triangle (BPC)))+(\frac (l)(\triangle (DPA))))

where m, k, n, l- side lengths AB, BC, CD And DA, and ∆( APB) - area of ​​a triangle APB.

Let us denote the segments on which the point P divides the diagonal AC how AP = p a and PC = p c. The same way P divide the diagonal BD into segments BP = p b and PD = p d. Then the quadrilateral is circumscribed if and only if one of the equalities holds:

m p c p d + n p a q b = k p a p d + l p c p b (\displaystyle mp_(c)p_(d)+np_(a)q_(b)=kp_(a)p_(d)+lp_(c)p_(b)) (pa + pb − m) (pc + pd − n) (pa + pb + m) (pc + pd + n) = (pc + pb − k) (pa + pd − l) (pc + pb + k) (pa + pd + l) (\displaystyle (\frac ((p_(a)+p_(b)-m)(p_(c)+p_(d)-n))((p_(a)+p_( b)+m)(p_(c)+p_(d)+n)))=(\frac ((p_(c)+p_(b)-k)(p_(a)+p_(d)-l ))((p_(c)+p_(b)+k)(p_(a)+p_(d)+l)))) (m + pa − pb) (n + pc − pd) (m − pa + pb) (n − pc + pd) = (k + pc − pb) (l + pa − pd) (k − pc + pb) (l − pa + pd) . (\displaystyle (\frac ((m+p_(a)-p_(b))(n+p_(c)-p_(d)))((m-p_(a)+p_(b))(n -p_(c)+p_(d))))=(\frac ((k+p_(c)-p_(b))(l+p_(a)-p_(d)))((k-p_ (c)+p_(b))(l-p_(a)+p_(d)))).)

Conditions for the circumscribed quadrilateral to be another type of quadrilateral

rhombus if and only if the opposite angles are equal.

If the inscribed circle is tangent to the sides AB, BC, CD, DA at points M, K, N, L respectively, then ABCD is also an inscribed quadrilateral if and only if

The first of these three statements means that touch quadrilateral MKNL is orthodiagonal.

A circumscribed quadrilateral is bicentric (i.e. circumscribed and inscribed at the same time) if and only if the radius of the inscribed circle is the largest among all circumscribed quadrilaterals having the same sequence of side lengths .

The described quadrilateral is a deltoid if and only if any of the following conditions is met:

• The area is half the product of the diagonals
• Diagonals are perpendicular
• Two line segments connecting opposite points of contact are of equal length
• One pair of opposite segments from the vertex to the point of contact have the same length
• Middle lines are the same length
• The products of opposite sides are equal
• The center of the inscribed circle lies on the diagonal, which is the axis of symmetry.