Proof of the circumscribed circle theorem of a quadrilateral. Properties of inscribed and circumscribed quadrilaterals

Examples of described quadrilaterals 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 cyclic quadrilateral, it is called bicentral.

Properties

In a circumscribed 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 point of intersection of the bisectors is the center of the incircle.

If opposite sides in a convex quadrilateral ABCD(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 only difference is in the signs - in Urquhart's theorem of the sum, 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 inscribed in triangles ABC And ADC the circles touch each other.

Description of the angles formed by the diagonal BD with the 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 four triangles formed by diagonals.

Special sections

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

The tangent points form an inscribed quadrilateral.

Square

Non-trigonometric formulas

K = 1 2 p 2 q 2 − (a c − b d) 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 we denote them 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 the 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))).)

Because the 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 cyclic. In this case K = a b c d (\displaystyle K=(\sqrt (abcd))) because opposite angles are complementary. This can be proven in another way, using mathematical analysis.

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

K = (O A ⋅ O C + O B ⋅ O D) 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 be expressed in terms of just 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 diagonals. The formula is not applicable to the case of deltoids, since in this case θ equals 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), semiperimeter s of the described quadrilateral satisfies the inequality

s ≥ 4 r (\displaystyle s\geq 4r),

Where r- 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 holds

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

If a straight line divides a 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 a circumscribed quadrilateral with sides a, b, c, d is 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- semi-perimeter. For circumscribed quadrilaterals with a given semi-perimeter, 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 an 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, That

r = 2 (σ − u v x) (σ − v x y) (σ − x y u) (σ − y u v) u v x y (u v + x y) (u x + v y) (u y + v x) (\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 using the formulas

sin ⁡ A 2 = e f g + f g h + g h e + h e f (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 = e f g + f g h + g h e + h e f (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 = e f g + f g h + g h e + h e f (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 K.M. And LN is given by the 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 tangent segments 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 + g f + h ((e + g) (f + h) + 4 f h) , (\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)))). )

Chords of tangent points

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

k = 2 (e f g + f g h + g h e + h e f) (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 (e f g + f g h + g h e + h e f) (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 sides with length b = f + g And d = h + e. The square of the chord ratio satisfies the relation

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

Two chords

Chord between sides AB And CD in a circumscribed quadrilateral ABCD longer than the chord between the sides B.C. And D.A. if and only if the midline between the sides AB And CD shorter than the midline between the sides B.C. And D.A. .

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

Collinear points

If M 1 And M 2 are the midpoints of the diagonals A.C. And BD respectively in the circumscribed quadrilateral ABCD O, and pairs of opposite sides intersect at points E And F And M 3- the middle of the segment E.F., then the points M 3, M 1, O, And M 2 lie on the same straight line. The straight line connecting these points is called the Newton straight line of the quadrilateral.

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

AB, B.C., CD, D.A. at points M, K, N And L accordingly, and if T M, T K, T N, T L are isotomically conjugate points of these points (that is, AT M = B.M. 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. However, more importantly, the Nagel point Q, "area centroid" G and the center of the inscribed circle O lie on the same straight line, and at the same time QG = 2GO. This line is called Nagel's straight line circumscribed quadrilateral.

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

Competitive and perpendicular lines

Two diagonals of a quadrilateral and two chords connecting opposite points of tangency (opposite vertices of an inscribed quadrilateral) are competitive (i.e., intersect at one point). To show this, we can use a special case of Brianchon's theorem, which states that a hexagon whose all sides 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 points of contact. All six sides of the resulting hexagon are tangent to the incircle, so that its diagonals intersect at one point. But two diagonals of a hexagon coincide with the diagonals of a quadrilateral, and the third diagonal passes through opposite points of tangency. Repeating the same reasoning for the other two points of contact, we obtain the required result.

If the incircle touches the sides AB, B.C., CD And D.A. at points M, K, N, L accordingly, then straight MK, LN And A.C. competitive.

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

Properties of the Incircle

The relationship between two opposite sides of a circumscribed quadrilateral can be expressed in terms of distances from the center of the incircle O to the 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)).)

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- center of the inscribed circle of a quadrilateral ABCD, That

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 inscribed circle O coincides with the “centroid of the vertices” of the quadrilateral if and only if

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

If M 1 And M 2 are the midpoints of the diagonals A.C. And BD accordingly, then

O M 1 O M 2 = O A ⋅ O C O B ⋅ O D = e + g f + 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- tangent segments at vertices A, B, C And D respectively. Combining the first equality with the last, we obtain 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 in the middle 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 was 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 (omitted 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 touch 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 - circumradii of triangles APB, BPC, CPD And DPA accordingly, 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 journals and websites. The property states that if a convex quadrilateral is divided into four non-overlapping triangles by its diagonals, the incenter centers of these triangles lie on the same circle if and only if the quadrilateral is circumscribed. In fact, the centers of the inscribed circles form an ordiagonal inscribed quadrangle. Here inscribed circles can be replaced by excircles (tangential sides and extensions 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, in which the diagonals intersect at the point P, is circumscribed if and only if the four centers of the excircles of the triangles APB, BPC, CPD And DPA lie on the same circle (here the excircles intersect the sides of the quadrilateral, in contrast to the similar statement above, where the excircles lie outside the quadrilateral). If R m, Rn, Rk And R l- radii of excircles APB, BPC, CPD And DPA respectively, opposite to the vertices B And D, then another necessary and sufficient condition for the fact that the quadrilateral is circumscribed will be

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 △ (A P B) + n △ (C P D) = k △ (B P C) + l △ (D P A) (\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, B.C., CD And D.A., and ∆( APB) - area of ​​a triangle APB.

Let us denote the segments on which the point P divides the diagonal A.C. How AP = p a and PC = p c. The same way P divide the diagonal BD into segments B.P. = p b and P.D. = p d. Then the quadrilateral is described 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)) (p a + p b − m) (p c + p d − n) (p a + p b + m) (p c + p d + n) = (p c + p b − k) (p a + p d − l) (p c + p b + k) (p a + p d + 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 + p a − p b) (n + p c − p d) (m − p a + p b) (n − p c + p d) = (k + p c − p b) (l + p a − p d) (k − p c + p b) (l − p a + p d) . (\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 a described quadrilateral to be another type of quadrilateral

rhombus if and only if opposite angles are equal.

If the incircle touches the sides AB, B.C., CD, D.A. at points M, K, N, L accordingly, then ABCD is also an inscribed quadrilateral if and only if

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

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

A circumscribed quadrilateral is a deltoid if and only if any of the following conditions are true:

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

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Sections: Mathematics , Competition "Presentation for the lesson"

Presentation for the lesson

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Goals.

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

Developmental. Development of mathematical abilities, creation of conditions for the ability to generalize and apply forward and backward train of thought.

Educational. Cultivating a sense of beauty through the aesthetics of drawings, surprise at the unusual

decision, formation of organization, responsibility for the results of one’s work.

1. Study the definition of a 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 inscribed and circumscribed quadrilaterals.

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

5. Conduct initial monitoring of the level of assimilation of new material.

Equipment:

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

Software: Microsoft Word, Microsoft Power Point.

Using a computer when preparing a teacher for a lesson.

Using a standard Windows operating system program, the following were 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.)

Lesson summary. (2 minutes.)

During the classes

1. Organizational moment. Greetings. State the topic and purpose of the lesson. Record the date and topic of the lesson in your notebook.

2. Checking homework.

3. Studying new material.

Work on the concept of a circumscribed polygon.

Definition. The polygon is called described about a circle, if All his sides concern some circle.

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

<Презентация. Слайд №2>

Proof of the properties of the circumscribed quadrilateral.

<Презентация. Слайд №3>

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

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

1. Present the formulation 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 a quadrilateral is circumscribed about a circle, That the sums of the opposite sides are equal.

The proof is carried out, students make notes in their notebooks.

<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 described quadrilateral are 8 m and 12 m. Is it possible to find the perimeter?

Tasks based on finished drawings.<Презентация. Слайд №5>

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

Proof of the characteristic of a circumscribed quadrilateral.

State the converse 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. Clarify the formulation of the theorem.

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

Working with the textbook. Get acquainted with the proof of the test for a circumscribed quadrilateral using the textbook.

Application of acquired knowledge.

3. Tasks based on finished 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 into an isosceles trapezoid with bases of 1 m and 9 m, and a height of 3 m?

<Презентация. Слайд №6>

Written work in notebooks

.

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 7 m and 10 m,

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

Find the perimeter of the quadrilateral.

3. An equilateral trapezoid with bases 4 m and 16 m is described around a circle.

1) radius of the inscribed circle,

Option 2

1. Is it possible to inscribe a circle:

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

2) squared?

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

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

1) the base of the trapezoid,

2) radius of the circumscribed circle.

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

6. Lesson summary. Independent work is checked and grades are given.

<Презентация. Слайд № 8>

The circumscribed circle of a quadrilateral. ? ? A circle can be described around a quadrilateral if the sum of the opposite angles is 180°: ? + ? =? + ? If a quadrilateral is inscribed in a circle, then the sum of the opposite angles is 180°. ? ? a. d. d1. PTOLOMY'S THEOREM The sum of the products of opposite sides is equal to the product of the diagonals: ac + bd = d1 d2. d2. b. c. b. Area of ​​a quadrilateral. a. c. d. Where p is the semi-perimeter of the quadrilateral.

Slide 9 from the presentation "Radius of inscribed and circumscribed circle". The size of the archive with the presentation is 716 KB.

Geometry 9th grade

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“Types and properties of triangles” - Area of ​​a triangle. Problems in coordinates. Final repetition of geometry. Properties. Regular triangle. Triangle. Check yourself. Center of the circumscribed circle. The relative position of the triangle and segments. Isosceles triangle. Right triangle. Bisector.

““Triangles” 9th grade” - Isosceles. Triangles. Sum of angles of a triangle. Rectangular. Bisector. Equilateral. Middle line. Perpendicular bisector. Median. Triangles. An obtuse triangle is a triangle in which one of the angles is obtuse. The relationship between the sides and angles of a triangle. Triangle inequality. External corner. Height.

“Circumference and Circle” - Find the circumference of a circle. Area of ​​a circle. Calculate. Find the radius of the circle. Complete the statement. Circle. Circular sector. Calculate the length of the equator. Circumference. Independent work. Circle. A game. Find the area of ​​the shaded figure. Draw a circle with center K and radius 2 cm.

“Questions on polyhedra” - What geometric figure will be obtained on the cut of the cylinder. Rectangle. Obtaining some of Archimedes' solids. V = abc. Cylinder height. Cube, parallelepiped, pyramid. Some geometric bodies. Find the volume of the aquarium shown in the figure. What objects are cylindrical in shape? Cone. Why did you classify a cube, parallelepiped, and pyramid as polyhedra? A ball and a globe are spheres. Sphere, cylinder, cone, truncated cone.

A quadrilateral is inscribed in a circle (problems). We continue to consider the tasks included in the Unified State Examination in mathematics. In this article we will solve several problems using the properties of an inscribed angle. The theory has already been outlined in detail. In this article, solving problems essentially came down to applying the property of an inscribed angle immediately, that is, these were tasks in almost one action. Here you need to think a little; the course of the decision is not always immediately obvious.

Applies: the theorem on the sum of the angles of a triangle, properties of an inscribed angle, property of a quadrilateral inscribed in a circle. More about the latter.

*This property has already been presented, but in a different interpretation. So:

Properties:

An inscribed quadrilateral is a quadrilateral whose vertices all lie on the same circle.

A quadrilateral can be inscribed in a circle if and only if the sum of its opposite angles is equal to 180 degrees.

That is, if we are such a quadrilateral, then the sum of its opposite angles is equal to 180 degrees.

Let's consider the tasks:

27870. In a circle with a center O A.C. And BD- diameters. Central angle AOD equals 110 0. Find the inscribed angle ACB. Give your answer in degrees.

Triangle BOS isosceles, because OS=OB(these are radii). It is known that the sum of the angles of a triangle is 180 degrees. Consider ∠BOC and ∠AOD:

Hence

The angles at the base of an isosceles triangle are equal, that is

Another way:

Angle AOB is the central angle for the inscribed angle ACB.By the property of an angle inscribed in a circle

The sum of adjacent angles is 180 0, which means

Thus

Answer: 35

27871. Angle A of a quadrilateral ABCD inscribed in a circle is equal to 58 0. Find angle C of this quadrilateral. Give your answer in degrees.

Here it is enough to recall the property of such a quadrilateral. It is known that the sum of its opposite angles is equal to 180 degrees, which means angle C will be equal to

Second way:

Let's build OB and OD.

By the property of the inscribed angle, the degree magnitude of the arc BCD is equal to

2∙58 0 = 116 0

Therefore, the degree magnitude of the arc BAD will be equal to

360 0 – 116 0 = 244 0

According to the property of an inscribed angle, angle C will be two times smaller, that is, 122 0.

Answer: 122

27872. Sides of a quadrilateral ABCD AB, B.C., CD And AD subtend the arcs of the circumscribed circle, the degree values ​​of which are respectively 95 0, 49 0, 71 0, 145 0. Find the angle B this quadrilateral. Give your answer in degrees.

Let's construct the radii AO, OD, OC:

The degree value of arc AD is equal to 145 0, the degree value of arc CD is equal to 71 0, which means that the degree value of arc ADC is equal to 145 0 + 71 0 = 216 0.

According to the property of the inscribed angle, angle B will be half the central angle corresponding to the arc ADC, that is

Answer: 108

27874. Quadrangle ABCD inscribed in a circle. Corner ABC equals 105 0, angle CAD equals 35 0 . Find the angle ABD. Give your answer in degrees.

This task can be challenging. It is immediately impossible to clearly see the progress of the decision. Let's remember what is known about an inscribed quadrilateral: the sum of its opposite angles is equal to 180 degrees. Let's find

At the moment, we have found the angle that can be immediately determined by a known property. If it is possible to find any value, do it, it will come in handy. We act on the principle “we find what can be found based on given values.”

Inscribed angles ABD and ACD are based on the same arc, this means that they are equal, that is

Answer: 70

27875. Quadrangle ABCD inscribed in a circle. Corner ABD equals 75 0, angle CAD equals 35 0 . Find the angle ABC. Give your answer in degrees.

It is known that inscribed angles based on the same arc and those lying on the same side of it are equal. Hence

In triangle ACD there are two known angles, we can find the third: