Inscribed and circumscribed quadrilaterals and their properties - materials for preparing for the Unified State Exam in mathematics. A criterion that a quadrilateral cut by a straight line from a triangle is inscribed in a certain circle

A circle is said to be inscribed in a quadrilateral if all sides of the quadrilateral are tangent to the circle.

The center of this circle is the intersection point of the bisectors of the corners of the quadrilateral. In this case, the radii drawn to the tangent points are perpendicular to the sides of the quadrilateral

A circle is called circumscribed about a quadrilateral if it passes through all its vertices.

The center of this circle is the point of intersection of the perpendicular bisectors to the sides of the quadrilateral

Not every quadrilateral can be inscribed with a circle, and not every quadrilateral can be inscribed with a circle.

PROPERTIES OF INSCRIBED AND CIRCULAR Quadrilaterals

THEOREM In a convex inscribed quadrilateral, the sums of opposite angles are equal to each other and equal to 180°.

THEOREM Conversely: if in a quadrilateral the sums of opposite angles are equal, then a circle can be described around the quadrilateral. Its center is the point of intersection of the perpendicular bisectors to the sides.

THEOREM If a circle is inscribed in a quadrilateral, then the sums of its opposite sides are equal.

THEOREM Conversely: if in a quadrilateral the sums of opposite sides are equal, then a circle can be inscribed in it. Its center is the point of intersection of the bisectors.

Corollaries: of all parallelograms, only around a rectangle (in particular, around a square) can a circle be described.

Of all the parallelograms, only a rhombus (in particular a square) can inscribe a circle (the center is the point of intersection of the diagonals, the radius is equal to half height).

If a circle can be described around a trapezoid, then it is isosceles. A circle can be described around any isosceles trapezoid.

If a circle is inscribed in a trapezoid, then its radius is equal to half the height.

Tasks with solutions

1. Find the diagonal of a rectangle inscribed in a circle whose radius is 5.

The center of a circle circumscribed about a rectangle is the point of intersection of its diagonals. Therefore, the diagonal AC equals 2 R. That is AC=10
Answer: 10.

2. A circle is described around a trapezoid, the bases of which are 6 cm and 8 cm, and the height is 7 cm. Find the area of this circle.

Let DC=6, AB=8. Since a circle is circumscribed around a trapezoid, it is isosceles.

Let's draw two heights DM and CN.Since the trapezoid is isosceles, then AM=NB=

Then AN=6+1=7

From a triangle ANS using the Pythagorean theorem we find AC.

From a triangle CВN using the Pythagorean theorem we find Sun.

The circumscribed circle of a trapezoid is also the circumscribed circle of a triangle. DIA

Let's find the area this triangle in two ways using the formulas

Where h- height and - base of triangle

Where R is the radius of the circumscribed circle.

From these expressions we obtain the equation. Where

The area of the circle will be equal to

3. Angles and quadrilaterals are related as . Find the angle if a circle can be described around a given quadrilateral. Give your answer in degrees

It follows from the condition that .Since a circle can be described around a quadrilateral, then

We get the equation . Then . The sum of all the angles of a quadrilateral is 360º. Then

. where do we get that

4.The sides of a trapezoid circumscribed about a circle are 3 and 5. Find the midline of the trapezoid.

Then midline equal to

5. Perimeter rectangular trapezoid circumscribed about the circle is 22, its major side is 7. Find the radius of the circle.

In a trapezoid, the radius of the inscribed circle is equal to half the height. Let's draw the height of the SC.

Then .

Since a circle is inscribed in a trapezoid, the sums of the lengths opposite sides are equal. Then

Then the perimeter

We get the equation

6. The bases of an isosceles trapezoid are 8 and 6. The radius of the circumscribed circle is 5. Find the height of the trapezoid.

Let O be the center of the circle circumscribed about the trapezoid. Then .

Let's draw the height KH through point O

Then , where KO and OH are heights and at the same time medians isosceles triangles DOC and AOB. Then

According to the Pythagorean theorem.

Inscribed quadrilateral - a quadrilateral whose vertices all lie on the same circle.
Obviously, this circle will be called described around the quadrangle.

Described a quadrilateral is such that all its sides touch one circle. In this case the circle inscribed into a quadrangle.

The figure shows inscribed and circumscribed quadrilaterals and their properties.

Let's see how these properties are used in solving Unified State Examination problems.

1. Two angles of a quadrilateral inscribed in a circle are 82° and 58°. Find the largest remaining angle. Give your answer in degrees.

The sum of the opposite angles of an inscribed quadrilateral is 180°. Let angle A be 82°. Then there is an angle of 98 degrees opposite it. If angle B is 58°, then angle D is 180° - 58° = 122°.

Answer: 122.

2. The three sides of a quadrilateral circumscribed around a circle are in the ratio (in sequential order) as 1:2:3. Find the longest side of this quadrilateral if it is known that its perimeter is 32.

Let side AB be x, AD be 2x, and DC be 3x. According to the property of the described quadrilateral, the sums of opposite sides are equal, and therefore
x + 3x = BC + 2x.
It turns out that BC is equal to 2x. Then the perimeter of the quadrilateral is 8x. We get that x = 4, and big side equals 12.

3. A trapezoid is described around a circle, the perimeter of which is 40. Find its midline.

We remember that the midline of a trapezoid is equal to half the sum of the bases. Let the bases of the trapezoid be equal to a and c, and sides- b and d. According to the property of the described quadrilateral,
a + c = b + d, which means the perimeter is 2(a + c).
We get that a + c = 20, and the middle line is 10.

Let us once again repeat the properties of an inscribed and circumscribed quadrilateral.

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

A quadrilateral can be circumscribed around a circle if and only if the sums of the lengths of its opposite sides are equal.

"Circle" We have seen that a circle can be circumscribed around any triangle. That is, for every triangle there is a circle such that all three vertices of the triangle “sit” on it. Like this:

Question: can the same be said about a quadrilateral? Is it true that there will always be a circle on which all four vertices of the quadrilateral will “sit”?

It turns out that this is NOT TRUE! A quadrilateral can NOT ALWAYS be inscribed in a circle. There is a very important condition:

In our picture:

Look, the angles and lie opposite each other, which means they are opposite. What then about the angles and? They seem to be opposites too? Is it possible to take angles and instead of angles and?

Of course you can! The main thing is that the quadrilateral has some two opposite angles, the sum of which will be. The remaining two angles will then also add up by themselves. Don't believe me? Let's make sure. Look:

Let it be. Do you remember what the sum of all four angles of any quadrilateral is? Certainly, . That is - always! . But, → .

Magic right there!

So remember this very firmly:

If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal to

and vice versa:

If a quadrilateral has two opposite angles whose sum is equal, then the quadrilateral is cyclic.

We will not prove all this here (if you are interested, look into the next levels of theory). But let's see where this leads wonderful fact that the sum of the opposite angles of an inscribed quadrilateral is equal.

For example, the question comes to mind: is it possible to describe a circle around a parallelogram? Let's try the “poke method” first.

Somehow it doesn't work out.

Now let's apply the knowledge:

Let's assume that we somehow managed to fit a circle onto a parallelogram. Then there must certainly be: , that is.

Now let's remember the properties of a parallelogram:

Every parallelogram has equal opposite angles.

It turned out that

What about the angles and? Well, the same thing of course.

Inscribed → →

Parallelogram→ →

Amazing, right?

It turns out that if a parallelogram is inscribed in a circle, then all its angles are equal, that is, it is a rectangle!

And at the same time - the center of the circle coincides with the intersection point of the diagonals of this rectangle. This is included as a bonus, so to speak.

Well, that means we found out that a parallelogram inscribed in a circle is rectangle.

Now let's talk about the trapezoid. What happens if a trapezoid is inscribed in a circle? But it turns out there will be isosceles trapezoid . Why?

Let the trapezoid be inscribed in a circle. Then again, but due to the parallelism of the lines and.

This means we have: → → isosceles trapezoid.

Even easier than with a rectangle, right? But you need to firmly remember - it will come in handy:

Let's list the most important ones again main statements tangent to a quadrilateral inscribed in a circle:

A quadrilateral is inscribed in a circle if and only if the sum of its two opposite angles is equal to
A parallelogram inscribed in a circle - certainly rectangle and the center of the circle coincides with the intersection point of the diagonals
A trapezoid inscribed in a circle is equilateral.

Inscribed quadrilateral. Intermediate level

It is known that for every triangle there is a circumscribed circle (we proved this in the topic “The Circumscribed Circle”). What can be said about the quadrilateral? It turns out that NOT EVERY quadrilateral can be inscribed in a circle, and there is such a theorem:

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

In our drawing -

Let's try to understand why this is so? In other words, we will now prove this theorem. But before you prove it, you need to understand how the statement itself works. Did you notice the words “then and only then” in the statement? Such words mean that harmful mathematicians have crammed two statements into one.

Let's decipher:

“Then” means: If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal.
“Only then” means: If a quadrilateral has two opposite angles whose sum is equal, then such a quadrilateral can be inscribed in a circle.

Just like Alice: “I think what I say” and “I say what I think.”

Now let’s figure out why both 1 and 2 are true?

First 1.

Let a quadrilateral be inscribed in a circle. Let's mark its center and draw radii and. What will happen? Do you remember that an inscribed angle is half the size of the corresponding central angle? If you remember, we’ll apply it now, and if not, take a look at the topic "Circle. Inscribed angle".

Inscribed

But look: .

We get that if - is inscribed, then

Well, it’s clear that it also adds up. (we also need to consider).

Now “vice versa”, that is, 2.

Let it turn out that in a quadrilateral the sum of some two opposite angles is equal. Let's say let

We don't know yet whether we can describe a circle around it. But we know for sure that we are guaranteed to be able to describe a circle around a triangle. So let's do it.

If a point does not “sit” on the circle, then it inevitably ends up either outside or inside.

Let's consider both cases.

Let the point be outside first. Then the segment intersects the circle at some point. Let's connect and. The result is an inscribed (!) quadrilateral.

We already know about it that the sum of its opposite angles is equal, that is, and according to our condition.

It turns out that it should be so that.

But this cannot possibly be because - external corner for and means .

What about inside? Let's do similar things. Let the point be inside.

Then the continuation of the segment intersects the circle at a point. Again - an inscribed quadrilateral, and according to the condition it must be satisfied, but - an external angle for and means, that is, again it cannot be that.

That is, a point cannot be either outside or inside the circle - that means it is on the circle!

The whole theorem has been proven!

Now let's see what good consequences this theorem gives.

Corollary 1

A parallelogram inscribed in a circle can only be a rectangle.

Let's understand why this is so. Let a parallelogram be inscribed in a circle. Then it should be done.

But from the properties of a parallelogram we know that.

And the same, naturally, regarding the angles and.

So it turns out to be a rectangle - all the corners are along.

But, in addition, there is an additional pleasant fact: the center of the circle circumscribed about the rectangle coincides with the point of intersection of the diagonals.

Let's understand why. I hope you remember very well that the angle subtended by the diameter is a straight line.

Diameter,

Diameter

which means it is the center. That's it.

Corollary 2

A trapezoid inscribed in a circle is isosceles.

Let the trapezoid be inscribed in a circle. Then.

And the same.

Have we discussed everything? Not really. In fact, there is another, “secret” way to recognize an inscribed quadrilateral. We will formulate this method not very strictly (but understandably), and we will prove it only in last level theories.

If in a quadrilateral one can observe such a picture as here in the figure (here the angles “looking” at the side of the points and are equal), then such a quadrilateral is inscribed.

This is a very important drawing - in problems it is often easier to find equal angles, than the sum of angles and.

Despite the complete lack of rigor in our formulation, it is correct, and moreover, it is always accepted by the Unified State Exam examiners. You should write something like this:

“- inscribed” - and everything will be fine!

Don't forget this one important sign- remember the picture, and perhaps it will catch your eye in time when solving the problem.

Inscribed quadrilateral. Brief description and basic formulas

If a quadrilateral is inscribed in a circle, then the sum of any two of its opposite angles is equal to

and vice versa:

If a quadrilateral has two opposite angles whose sum is equal, then the quadrilateral is cyclic.

A quadrilateral is inscribed in a circle if and only if the sum of its two opposite angles is equal.

Parallelogram inscribed in a circle- certainly a rectangle, and the center of the circle coincides with the intersection point of the diagonals.

A trapezoid inscribed in a circle is isosceles.

INSCRIBED AND CIRCULAR POLYGONS,

§ 106. PROPERTIES OF INSCRIBED AND DESCRIBED QUADRIAGONS.

Theorem 1. The sum of the opposite angles of a cyclic quadrilateral is 180°.

Let a 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 BCD.
/ C, as inscribed in the same circle, measures 1/2 BAD.

Consequently, the sum of angles A and C is measured by the half-sum of arcs BCD and BAD; in sum, these arcs make up a circle, i.e. they have 360°.
From here / A+ / C = 360°: 2 = 180°.

Similarly, it is proved that / B + / D = 180°. However, this can be deduced in another way. We know that the sum of interior angles convex quadrilateral equal to 360°. The sum of angles A and C is equal to 180°, which means that the sum of the other two angles of the quadrilateral also remains 180°.

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

Let the sum of the opposite angles of the quadrilateral ABCD be equal to 180°, namely
/ A+ / C = 180° and / B + / D = 180° (drawing 412).

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

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

Point D can occupy only one of next three positions: to be inside the circle, to be outside the circle, to be on the circumference of the circle.

Let's 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 side AD" to the intersection with the circle at point E and connecting points E and C, we obtain the cyclic quadrilateral ABCE, in which, by the direct theorem

/ B+ / E = 2 d.

From these two equalities it follows:

/ D" = 2 d - / B;
/ E=2 d - / B;

/ D" = / E,

but this cannot be, because / D", as external relative to the triangle CD"E, must be more angle E. Therefore, point D cannot be inside the circle.

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

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

Consequences. 1. A circle can be described around any rectangle.

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

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

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

It is required to prove that AB + CD = AD + BC. Let us denote the points of tangency by the letters M, N, K, P. Based on the properties of tangents drawn to a circle from one point (§ 75), we have:

AR = AK;
VR = VM;
DN = DK;
CN = CM.

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

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

i.e. AB + CD = AD + BC, which is what needed to be proven.

Exercises.

1. In an inscribed quadrilateral, two opposite angles are in the ratio 3:5,
and the other two are in the ratio 4:5. Determine the magnitude of these angles.

2. In the described quadrilateral, the sum of two opposite sides is 45 cm. The remaining two sides are in the ratio 0.2: 0.3. Find the length of these sides.

Task 6: in an isosceles trapezoid the bases are 21 and 9 centimeters, the height is 8 centimeters. Find the radius of the circumcircle.

1. Let's carry out perpendicular bisectors to the bases H and K, then the center of the circle O lies on the straight line NK.

2. AO=OB=R. Point O divides the segment NK into two parts: let HO = x, then OK = 8 - x.

3. AO 2 = AK 2 + KO 2; OB 2 = VN 2 + NO 2;

since OA 2 = OB 2, we get:

AK 2 + KO 2 = VN 2 + NO 2

90 + 64 - 16x = 0

OB 2 = HV 2 + NO 2

Answer: OB = 10.625

Problems with a circle inscribed in a quadrilateral

Task 7: A circle of radius R is inscribed in a rhombus. Find the area of the rhombus if its large diagonal 4 times greater than radius inscribed circle.

Given: rhombus, inscribed circle radius - R, BD r 4 times

1. Let OE = R, BD = 4OE = 4R

Problem 8: Find the area of an isosceles trapezoid circumscribed about a circle with radius 4 if it is known that the lateral side of the trapezoid is 10.

Given: ABCD - isosceles trapezoid, r = 4, AB = 10

1. AB = CD = 10 by condition

2. AB + CD = AD + BC by the incircle property

3. AD + BC = 10 + 10 = 20

4. FE = 2r = 2 4 = 8

Problem 9: inside regular triangle with side a there are three equal circles, each of which touches two sides of the triangle and two other circles. Find the area of the part of the triangle located outside these circles.

1. Let AB = BC = AC = a.

2. Let us denote O 1 E = O 1 K = ED = r, then AD = AE + ED = AE + r = .

3. AO 1 is the bisector of angle A, therefore, ? O 1 AE = 30? and in the rectangular?AO 1 E we have AO 1 = 2O 1 E = 2r and AE ===. Then AE + r = == , whence.

Problem 10: the entire arc of a circle of radius R is divided into 4 large and 4 small parts, which alternate one after another. Most twice as long as the small one. Determine the area of an octagon whose vertices are the dividing points of the circular arc.

1. Let?AOB = 2x, ?BOC = x, then by condition 8x + 4x = 360°, x = 30°, 2x = 60°, ?AOB = 60°, ?BOC = 30°

Problem 11: The sides of the triangle are 12 m, 16 m and 20 m. Find the height drawn from the vertex of the larger angle.

1. 202 = 122 + 162

400 = 400 is correct, therefore ? ABC - rectangular (according to the theorem, converse of the theorem Pythagoras)

Answer: VN = 9.6

Problem 12: V right triangle a square is inscribed with it common angle. Find the area of the square if the sides of the triangle are 10 m and 15 m.

Given: ? ABC - rectangular, AC = 15, CB = 10

1. ? ADE ~ ? ACB (? A - common, ? ADE = ? ACB = 90°)

2. Let DE = DC = X, then AD = 15 - X